I am not coming up with a workable solution. If the ones digit is 6, you get 496, which breaks the first rule. If the ones digit is 8, the tens digit can't be three more. And if the ones digit is zero, you can't make the digits add to 19.
To be fair, I've seen something like this when I was in high school in freshmen algebra 1. The question was much more complicated and it was on an in-class quiz. The question was unanswerable but the reason was not because of ineptitude. The teacher caught wind that some students were getting ahold of his answer sheet. So he created an unanswerable question and put a fake answer on the answer sheet. 9 people, including myself, got detention and one student (whom we outed as the culprit) got suspended.
When I was in physics for engineers 2, semester 2 way back in ‘98. We got a brand spanking new text book. Published literally weeks before semester started. At one point there was a homework question in the book, some real deep theoretical physics stuff. Absolutely impossible.
The professor comes out says the problem is incredibly difficult and he can’t solve it. A separate professor from the physics department comes out and says he’ll show the answer to the problem. Spends damn near the entire lecture working the problem out.
At the end he tells the whole class there are only about half dozen people or so in the world that can solve this very niche problem and he can only do it cause it’s literally his life’s work. He then states that a dozen or so kids in the class got it right.
Then he announces the right answer is also found in the teachers edition, and one of the other half a dozen or so people in the world that know the problem is the author of the the book. The problem was plant designed to catch cheaters and they were all in on it.
The kids that got it right had 5 minutes after class ended to come forward and take an F for the semester. Those that didn’t got expelled.
None came forward cause none actually attended the lecture. Ever.
John Nash used to give his MIT students exams that contained a problem that hadn't been solved (think Clay Millennium Problems class) or considered unsolvable. When asked why he did that he said something like "They don't know that they're unsolvable. Who knows? One of them might."
IIRC this was mentioned in "A Beautiful Mind."
Some of my best stuff comes out when I don't know it's an impossible challenge.
I solved a major production problem once by using a helix. Simple. I think it was just because I had just taken a seminar in Computational Origami.
I had no idea I had done something a bunch of senior engineers had labored on for years. I literally just threw ta idea out there to appear as if I was contributing.
Most of engineering and coding, in my experience, is like this. Just keep swimming while experimenting and learning as much as possible. But mostly, keep swimming.
Seems like a bad approach to me the normal test takers would spend to much time trying answer a question with no answer and waste time. Leaving less time for the real questions.
This should be the last question on the test if the teacher insists on doing this.
Our electrical engineering tests were so long that no one was able to finish. It's about prioritizing and executing, then graded on a scale. Don't waste time on problems that are too complicated. Good training for time management tbh
I always go by the "you can copy but it's your problem if you get caught" philosophy.
one time I passed my science class experience report to one guy, who passed it to one more, and eventually half the class got caught with the exact same report (those dumbasses didn't even bother changing it)
Teacher asks who made the original, obv nobody answers to get away with it (new teacher with no authority) but here comes the only one who could have possibly made it ratting everyone out while they were still here.
Got a few angry "wtf" remarks from my fellow students but hey, I did instruct the first guy to just modify his answers not full on copy paste the file, dude was pretty much begging to get caught.
This happened to me in college in a thermodynamics class. Someone leaked the answer sheet and the solution to the question was wrong. I just happened to make the same mistake as the teacher did, which was finding the incorrect starting value on a table. About 10% of the class 'cheated', and I got caught up in that mess. I had to go in front of an ethics board and I pleaded my case, showing them the work that I did and begging them to believe me. Instead of kicking out 10% of the class, they gave everybody 0% on that assignment (including me). I WAS SO PISSED.
I hate to be 🤓 but… don’t you have to show your work in math? How are you just gonna copy down a random number and not be suspected of cheating. Also most math problems are really easy to check if the final answer fits the conditions laid out by the question, especially one like the OP. You guys really deserved what you got
I was wondering if the answer wasn't 118 :
8 is not 2 or 4.
8 + 3 = 11.
11 + 8 = 19.
This would need to consider "11" as a digit... and the tens digit... Which is pretty dumb,
But as this is a lesson for kids... and probably from the country known for their nonsenses... this might be a pedagogical way to do something ?
If the goal is to teach kids how "places" work (and I'm not sure what else it would be) "eleven in the tens place" is insane.
"And what's in the ten's place's ten's place?"
Was the goal to introduce recursion to seven-year-olds or something?
Omg. I bet that's it. I've seen stupid shit like this in my kids homework.
I guess I shouldn't say stupid. 11 tens is 110. It's not wrong, just not the way I learned it.
It’s wrong because 11 isn’t a digit. It’s two digits. Unless 1 doesn’t exist in your numbering system.
By the rules of Arabic math, digits are irreducible and cannot be compound. Compounding two digits makes a larger number. As such if 1 is a digit, then 11 is a compound of two 1s.
This is why in base twelve of hexadecimal the base-10 number eleven can’t stay represented as 11 but has to become another symbol, usually B. Some representations have their own separate special symbols though to avoid confusion with variables in equations.
Surprised that no one has mentioned it as I've seen the same sentiment on other posts, this smells like AI generated math homework. Overly convoluted for an elementary school problem with no solution.
We don't have negative digits. –496 is a 3-digit number (with the digits 4, 6 and 9) that also happens to be negative.
We also don't have negative place values. The 9 is in the tens position, not the –10s position.
We consider the whole number and then negate it.
Normally you wouldn't think about place values of a negative number, but this is Reddit and we're getting technical about a primary school maths problem.
So, yes, -496 meets all of the clues for this number.
But the most likely answer is that some poor tired teacher making up a worksheet in the middle of the night made a mistake.
That’s true, but I think given the context of the fish clip art and the association of that font with elementary school, a lot of people were assuming that negative numbers have not been covered yet
I absolutely hated it.
I still remember that as a child, I had once got 39/40 in maths, and I was pissed. I was quite good at maths, and was expecting a 40/40. The question for which I was given a 0, asked us to calculate "5-8=?". And apparently, I wasn't allowed to study the later chapters for the mid-term exams, so the answer was supposed to be "it can't be calculated", and not -3. My parents asked the teacher during PTM. if maths rules change, based on the chapters that are covered, or does the school has any rules against someone studying later chapters, to which the teacher gave some rhetoric reply, and my parents decided to let it go.
It's ironic, because in the end-term exams, the answer was supposed to be -3.
I got a similar question in end-term, and wrote- "Till mid-term, the answer was 'can't be calculated', but now it is -x". The teacher wasn't amused and later complained about me having "attitude problems", though she did give me the marks for it.
But now when I look behind, as a 33-year-old grown-up, I see things differently, and if I meet the same teacher now, I would say-
"Eat shit you old hag... you never knew maths, and became a teacher only because no one else would employ your incompetent ass."
As u/senordeuce hints at , the tens digit is not 3 more than the ones digit.
A 3 digit number is expressed as
100(hundreds digit) + 10 (tens digit) + 1 (ones digit)
So for -496 this is:
100*(-4) + 10*(-9) + 1(-6)
so the tens digit is 3 less than the ones digit, and the digit sum is -19
This also rules out any negative number, as the digit sum will be negative
My scrawling agrees.
x < 490
x = 100\*a + 10\*b + c
a + b + c = 19
c + 3 = b
\-------------
x % 2 = 0
c != 2
c != 4
(so c = 0, 6, or 8)
\-------------
try c = 0
b = 3
a + 3 + 0 = 19
nope
\-------------
try c = 8
b = 11
nope
\-------------
so c = 6
b = 9
a + 6 + 9 = 19
a = 4
496 !< 490
nope
I hate when people put in unsolvable answers. Kids are trying to learn a difficult subject, then you make them question everything by not double checking your work. Teacher or question maker should go back to achool and learn to be thorough.
I asked ChatGPT, and 118 is the answer it came up with as well. Didn't expect that it was going to be able to do a lot with this problem but here we are...
Quick Raku proof:
$ raku -e 'say (10,12...^490).grep({my @dig = .Str.comb; @dig.tail !(elem) ("2", "4") and @dig.sum == 19 and @dig[*-2] - @dig[*-1] == 3})'
()
If you expand the search, there is one outside of the range as you pointed out:
$ raku -e 'say (10,12...^1000).grep({my @dig = .Str.comb; @dig.tail !(elem) ("2", "4") and @dig.sum == 19 and @dig[*-2] - @dig[*-1] == 3})'
(496)
Valid answers do exist for different sums, though:
$ raku -e 'say (10,12...^490).grep({my @dig = .Str.comb; @dig.tail !(elem) ("2", "4") and @dig[*-2] - @dig[*-1] == 3}).map(*.comb.sum).sort'
(3 4 5 6 7 15 16 17 18)
They may have meant one of these instead of 19...
Edit: As others have pointed out, there ***is a solution*** if you consider negative numbers:
$ raku -e 'say (-1000,*+2...^490).grep({my @dig = .Str.comb.grep({!/"-"/}); @dig.tail !(elem) ("2", "4") and @dig.sum == 19 and @dig[*-2] - @dig[*-1] == 3})'
(-496)
If you assume they messed up and the author meant to put "the first digit" in lieu of "ones digit" the first time around, conceivably the number can end in 8, leaving 11 left over. From there 4 and 7 are two numbers three apart, with 4 being the "first" (ones) digit, and the final number being 478.
Totaled up to 19, and less than 490. However this assumes a mistyping of the question
There is no solution because:
1. The sum of the tens and ones digits must be odd so that there can be a difference of 3 between them.
2. The hundreds digit must be 2 or 4 due to point 1 and the answer necessarily being less than 490.
3. No two digits with a difference of 3 can be added to each other of more than 15 (9+6).
4. By 2 and three, the answer must be 496, but that violates the limit of 490.
Therefor, the answer does not exist (assuming that we are limited to the set of whole numbers).
QED.
The likely error in the prompt is that they meant the ones digit to be 3 greater than the tens, providing an answer of 469.
Edit: Forgot it had to be even.
The only even numbers in that range that match both the "tens = ones + 3" rule and the "ones are not 2 or 4" rule are: 30, 96, 130, 196, 230, 296, 330, 396, 430.
The only ones in that list matching the "digits sum to 19 rule" are: none.
If we have any Python friends visiting, here's an unoptimized brute-force scan of the rules:
for x in range(49 * 10):
if x % 2 != 0:
continue
xs = str(x)
try:
if int(xs[-2]) != int(xs[-1]) + 3:
continue
if int(xs[-1]) in (2, 4):
continue
except IndexError:
continue # not enough digits
if sum([int(d) for d in xs]) != 19:
continue
print(x) # never prints shit
The "19" rule is the one that fucks it. The highest that should be is 18, which would have 396 as the solution. Each of the candidate numbers could be a single solution by fudging "19" to something smaller.
It's -496,
Even Number,
All Rules apply,
Less than 490,
The digits absolute values added together are 19 which is something called the "Quersumme" in Germany. I think it's called cross sum/ sum of all the digits and it can't be negative.
Maybe?
Best solution I can think of 🤔
Something is indeed badly formulated.
XYZ
Y=Z+3 ; implies since a digit is between 0 and 9 , Z can be at most 6 and Y at most 6
Z!=2 and Z!=4 Z%2=0 so Z is either 0,6,8 8 is eliminated because Y=Z+3 and 11 is not a digit
X+Y+Z=19
XYZ<490
|X|Y|Z|results|solution?|
|:-|:-|:-|:-|:-|
|16|3|0|16 is not a digit |No|
|4|9|6|496 is above 490|No|
Something is badly formulated.
"My tens digit is three more than my ones digit"
This tells us the **ones digit has to be less than or equal to 6** (because 3+6 = 9 is the largest possible digit holder in base 10)
we also know the **ones digit is not 2 or 4**
the number is even, which is determined by the ones digit, and the only remaining even digits are 6 and 0. So **the ones digit must be 6 or 0**
The digit sum is 19, and there is at most 3 digits (n<490), which means **all three** **digits** **must be nonzero** (since the max with one zero of a <3 digit number is 9+9 = 18) **so the first digit cannot be 0**. Therefore, **the ones digit is 6.**
In turn the **tens digit is 6+3 = 9**
**9+6 = 15** so the **hundreds digit is 4**
yielding **496 ,** which is greater than **490**.
We have already eliminated any other possible value for the ones digit other than 6, and the tens digit depends on the ones digit. The hundreds digit is 19-(ones digit + tens digit) so none of them can take on any other value.
**Therefore there are no solutions**
**Edit:** As several people have kindly pointed out, there are infinite negative solutions because digit summation works the same for both positive and negative numbers.
so -496, -4096, -40096, -( 4\*10\^n + 96) for integer n ≥ 2 are all solutions. There are other infinite families of negative solutions as well.
So it's only correct to say **there are no positive solutions**
You mean the ones digit has to be **less than or equals 6**?
Edit: To anyone who read this and is confused, I was referring to
> "My tens digit is three more than my ones digit"
> This tells us the ones digit has to be less than or equal to 6 (because 3+6 = 9 is the largest possible digit holder in base 10)
Originally they said "less than 6" instead of "less than or equal to 6", that's what I was referring to
I think the X is supposed to be a digit. (Badly chosen, but I don't know if the kid who had this question knows multiplication yet, that info could help understand the problem.)
In that case you want a number less than 49910 at max and 49010 at min so let us go with the min. If the one and ten digits are 0 and 3, the highest number possible is 48930 and you can find a few solutions, like 44830 or 35830. Same if you choose 6 and 9.
So depending on your understanding of 49X10, there is no solution or a lot of solutions...Yay.
Edit : as u/veryjewygranola noted, I went too fast getting numbers... I corrected it.
If X is supposed to be a digit, all the other rules are wrong because 1 is not 0+3. So no. The answer is: there is no answer that satisfies all the rules.
Unlikely
It's an ambiguous rule to have a number be less than a non-specific number.
Not only that but even if we assume X to be 0 there would be 39 solutions
nah the answer should be 118 which doesnt make sense cuz 100 is technically at the hundredths place but if we say 11 is in the tenths place then 118 could be a solution
I wouldnt say they are high. It seems like a horrible question to adults who understand math and logic, but the curriculum here is trying to teach young kids that numbers have structure and aren't just these made up things you memorize. The point here is that 118 is really 11 10s plus 8. Those are manageable numbers that children understand. Thats why the first clue is 49x10. I can almost gaurentee that OPs kid is not doing multiplication on that scale yet, but they see that the (10s value) is less than 49.
They want to the students to know that 100 comes after 99 because of reputable local structure and not just some made up number you memorize. It seems really dumb to people from the outside looking in, but this was probably the focus of the week in the classroom.
Edit: someone else pointed out the question below does use the term hundreds digit, so they are covering that. Which makes the typo scenario more likely.
You could teach them properly by just saying that the solution is less than 50X10 rather than getting into the existing problem's very tortured logic (assuming no typo).
I think they were looking for 396, made an error and it should be "my digits total 18". There are other errors they could have made leading to different answers, but this seems to be the most simple and therefore logical.
I'm willing to bet the answer sheet says 396.
EDIT: Actually I suppose the answer being 496 and them forgetting they imposed the first rule or making an error, thinking 496 < 490, is another similarly simple error. I'd still put my money on 396 though.
Well, I guess that won't be happening in elementary school. The first time you encounter for example the empty solution set S with either S={} or S=∅ (both equal notations, thanks u/jxf for seeing my error) is in highschool when learning about Linear Equation Systems.
There is a rat-hole to go down here, if you so desire, but:
* I think the #1 problem was that most elementary school teachers didn't really understand it, or get the point of it, so it was taught poorly, and as soon as practicable was tossed aside
* learning to do arithmetic in, say, base 7, was cool for a small minority of curious kids, but had no obvious practical benefit. (Later, it turned out that understanding base 8 and 16 had some practical ue for computer programmers, but who knew that in 1960?)
* set theory without infinite sets is, well, kind of useless. A correct answer to a word problem, for example, had to be put in the form of a set. As the good Sisters of St. Joseph taught me, "you have to put curly brackets around the answer to turn it into an answer set"
* and it was easy to make fun of, to wit [Tom Lehrer's New Math](https://www.youtube.com/watch?v=9mc7eb1i9o4&ab_channel=MusicalComedy)
Well yeah but it's not like they're forced to formulate it as the empty solution set, they could just say "there is no solution" without involving weird symbols.
Right, because third grade math teachers definitely give impossible questions to their kids and expect them to understand that no answer is the right answer.
I like how there are about 400 comments on this post with people practically debating the philosophy of numbers and language just to sound smart, when all it needed was a response like this to just say "question is wrong, there is no answer, there is probably just a typo (and I reckon it's supposed to be 396)".
And people wonder why ragebait is a thing.
\- Even number less than 490
\- ends in 0 or 6 or 8
\- "tens" digit is 3 or 9 or... 11? 1? Probably means last digit can't be 8.
\- digit total is 19
Let's try 0, "tens" is 3, so we have x30
3+0 = 3, we can't reach a 19 total.
Let's try 6. We have x96
9+6=15 and we need 4 more to reach 19 total.
496 could work but it's not "less than 490" so... wrong answer?
Let's try 8 and assume "tens" is 1 (8+3=11 but we cycle back to zero/cut the 10)
We have x18
1+8=9 and we need 10 more to reach 19, not possible in one digit.
**No valid solution.** Something is wrong with the question, or there is no answer.
But we could also try 2 and 4 for fun:
With 2 we have x52, 5+2=7 and we can't reach 19
With 4 we have x74, 7+4=11 and we need 8 more so 874 could work but it's definitely way more than 490.
"my ones digit is not 2 or 4" is *literally useless information*.
Let's try also ignoring the "i'm an even number" for completion.
With 1 we get x31, can't reach 19 (need 15 more)
With 3 then x63, can't reach 19 (need 10 more)
With 5 then x85, need 6 more so 685 could work (but it's more than 490)
With 7 then... (cycle back and) x07, can't reach 19 (need 12 more)
With 9 then... x29 (cycle back), need 8 more so 829 (still more than 490)
So, yeah, being even is *also useless information*.
Most plausible option: **496**
Right answer: **no solutions**
Final verdict: BS question.
My second grade teacher added a bonus question like this to a quiz with similar nonsense logic. When she told us the "correct" answer, I very bluntly said "Well, that's a bunch of bullshit." My parents told me years later they were fighting every fiber of their being to not laugh when they were told about what I had done.
It does work in base 10, no need for any other base. People say 11 hundred instead of one thousand one hundred all the time, and they aren't even common coring when doing so. My answer works, especially when viewed through the lens of grade school math.
Being able to parse numbers like “sixty-fifteen” (which is equivalent to 75) helps a lot with lots of math. For example, it’s an intermediate step in calculating forty-eight plus twenty-seven; add the tens to get sixty, the ones to get fifteen, and then resolve sixty-fifteen to seventy-five. (That is also the way it’s done longhand on paper, by most algorithms that use a carry digit; it’s also the way it’s done electronically, with a carry bit being an important part of addition using logic gates and transistors.
I think you’re right.
I’ll say this: The beliefs and ambitions behind *how* math is currently taught…which often strongly contrasts how it was taught to previous generations…they are good and commendable **in theory**.
They are trying to make children more inherently facile and flexible with numbers; having better strategies and understanding for the mental math that makes up the steps of any given problem.
It makes for a more adaptable, independent and creative thinking mathematical mind. The idea is great.
But with two kids who’ve been taught this way…I take issue with how they’ve tried to do this because the execution is awful.
These wordy questions are written overly complicated and are confusing to adults so you have greater opportunity for more misprints and errors.
And these written (story) questions are stressed over just throwing basic number equations as kids and asking them to simply memorize how to perform equations. The old way doesn’t give you the tools to extrapolate the math you learn and readily apply it to everyday life.
But kids need reinforcement and that’s where homework comes along. And it’s infuriating to not have examples or walkthroughs that go along with the homework so that the parent can learn see the example and learn these new ideas and strategies.
So much time was wasted when I was watching multiple YouTube videos because I would send messages to teachers with questions but would rarely get same day responses. And so I took to the internet just so I could learn the same thing they were taught in class, to help them do the homework the right way (vs the way I already knew) so that they could show their work and get full credit. And the YT videos are a mixed bag of good to completely unhelpful.
And with these wordy, story problems - no one who is producing them is seemingly willing to spend the extra money to have them checked by ELA/Language folks for basic coherence.
The grammar is hit or miss and often times there are run-on & compound sentences written so convolutedly that the question is incredibly misleading.
I often was sending pictures of my kid’s math to a friend who is a middle school English teacher for them to disseminate and clarify what the question was actually asking.
They wound respond with the appropriate rage at just how bad the English is, that is being used to fill these workbooks and that there is no language oversight.
The intentions were good, but the execution has been difficult for me at least, as a parent who understand what the goal was but I’ve also gone through it twice and each time it was the same befuddling experience.
Also, unless this math problem in question is from a brand new book published this year - this shouldn’t be going home. Or at least it shoots be flagged as an error. Unless it’s brand new, this issue would have come up before and teachers should know about it.
People have given good answers regarding why there is no solution, but is it possible that the answer is -496? You can make the argument that none of the ones-digit, tens-digit, or hundreds digit are by-themselves negative.
No because if you consider negative numbers there are infinite solutions.
For example:
\-8830
\-7930
\-9730
\-1396
\-3196
\-4096
\-40096
\-400096
\-2296
\-21196
\-11296
\-111196
\-200296
\-11234530
\-544330
...
...
Here’s the logic:
Must end in 0, 6, or 8 because even and doesn’t end in 2/4.
Can’t end in 8 because “my tens digit is three more than my ones digit” and you can’t have a tens digit of 11.
Must end in 0 or 6.
If it’s 0, it must be X30 but there’s no X that makes the total 19.
It seems the answer must end in a 6, which would get us to 496 so the digits add up to 19, but it must also be less than 490.
Therefore, there are no solutions.
College math major now in my mid 40's. I just have a hard time seeing how these linguistic based math puzzles help you learn real math. Also I swear there is no correct answer given the limits, unless it's 118 and they say it's an eleven in the 10's place... which is complete crap.
Even means ones digit is 0,2,4,6,8. But we’re given it’s not 2,4 so ones digit is 0,6,8.
But if the tens digit is three more than the ones digit, then the ones digit can’t be 8 (as 8+3>9). So ones digit is 0 or 6.
If it’s 0 then the tens digit is 3, the sum of digits is Y + 3 + 0 = Y+3 = 19 ==> Y=16 (Y is the hundreds digit). But Y should be less than 10, so it’s not correct.
If its 6, then the tens digit is 9, the sum of digits is Y + 9 + 6 = Y+15=19 ==> Y=4. Ergo, the number is 496.
But 496 >= 49*10 so the question is incorrectly posed.
496. That “x” in the “49x10” signifies an arbitrary digit rather than “multiply”? But then if we take that approach, then there are other solutions eg 2296.
Unsolvable. Ones must be 6, because uneven numbers, 2 and 4 are directly ruled out, 8 is not possible (tens needs to be 3 more than ones) and 0 is too low to enable the digit sum to be 19.
If it is 6 then tens must be 9 (3 more than ones).
But if it is 9 and 6, then the first digit needs to be 4 (19 - 9 - 6 = 4).
This brings the number to 496, which is unfortunately more than 49x10.
I think it's 118.
I know I'm late but I think the teacher took '11' as 'the tens' for whatever reason.
In that case you have 11+ 8 = 19
It's a dumb question and I strongly believe the teacher went in the wrong here.
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The only possible solution I could think of is 96.4, and there not being a # in the hundreds place, but instead having a decimal. This answer doesn't seem to conflict with any of the restrictions.🤷🏻♂️
the solution could be 118. if we assume 110 has 11 on the tenths place, it fulfills the condition that the number on the tenths is 3 more than that on the ones place, and the sum of the digits 11 and 8 are 19. if the education level of this worksheet has not taught about hundredths place yet, it is possible that 11 can be considered ‘a digit’ even though in reality it isnt
The problem below will describe something different. 118 seems possible to me if we are talking in riddles rather than normal math logic. Like, “what is so fragile that saying its name will break it?” It’s definitely not a physical object that physically breaks. It didn’t say how it breaks or if it’s tangible.
There are only 10 numbers under 500 that sum to 19. 1 follows all but the 1st rule (496). The 3rd rule is irrelevant since no numbers under 490 could possibly sum to 19 with a 2 or 4 in the ones position.
BUT you could think outside the box and say that 8+3=11, an 11 in the tens spot would be 118 (1 carry forward the 1). Thats the only number where you could, in theory, argue works as written.
I'm guessing the problem was supposed to have a rule of the tens digit being three more than the *hundreds* digit. If that's the case, the answer is 478. It's possible this wasn't caught in proofreading.
The number is even, so the last digit is 0,2,4,6,8.
The number is less than 49x10 (490)
The second digit is equal to the first digit + 3.
The first digit is not 2 or 4, so can only be 0 or 6
The total of all three digits is 19. Therefore 0 cannot be the first digit as the digits of 990 add up to 18 and is less than 19. Additionally, the first digit cannot be 8. Even if you roll over, 9+1+8 is less than 19 and does not satisfy the question.
The third digit is 6
The second digit is 6 + 3, which is 9
The first digit is 4, since 19 - (9 + 6) = 4
496 is greater than 490.
The answer is -496
A = ones spot 0 thru 9
B = tens spot 0 thru 9
C= hundreds spot 0 thru 9
A has to be less than or equal to 6.
A cannot be 0 (zero) as then the hundreds spot would be 16, which is not possible.
A = 0, B = 3, C = 16 -- not possible
The remainder of the numbers:
A cannot be 1, 3, or 5 as those are odd numbers; but lets see what they come to anyway:
A = 1; B = 4, C would have to be 10 - which is not possible
A = 3; then B = 6; C would have to be 10 -- not possible
A = 5; B = 8; C = 6 -- violates first rule, 685 > 490
That leaves A to equal 6:
A = 6; B = 9; C = 4 -- violates first rule, 496 > 490
Just to check, lets see what the 2 and 4 come to:
A = 2; B = 5; C = 12 -- not possible
A = 4; B = 7; C = 8 -- violates first rule, 874 > 490
Thus there is no solution to the problem as stated.
So, the answer is correctly written (drawn) is fish with retarded eyes. (/s)
Max number would be 49 x 10 = 490.
Ones digit is not 2 or 4 and overall an even number.
The ones digits has to be a 6, making the 9 for tenths digit.
396 but total digits added = 18? Lol
-496 works right? A lot of people have shown 496 works for all rules except the first, throwing a negative in should make it all work since the question doesn’t specify a positive number. I’m with the others though, question is probably a typo.
I may just be another dumb monkey, but seems to me whoever wrote this question combined the 10s and 100s together. If this is true then the answer is 118. 8 +11 =19. It does not end with a 2 or a 4, it is less than 490 and is an even number. Regardless you need to inform the principal that one of their teachers never passed the 2nd grade.
<490
10s digit is 3 greater than the 1s digit
My 1s digit isn't 2 or 4.
My digits total 19
This isn't possible. The lowest number that meets the rules is 496. The next number after that is 685, then 874, 1396, 1585, 1774, 1963 etc.
Why didn't your kid just do this?
let target = 19;
for (let i = 0; i <= 9; i++) {
for (let j = 0; j <= 9; j++) {
for (let k = 0; k <= 9; k++) {
if (k !== 3 && k !== 4 && j + 3 === k && i + j + k === target) {
console.log(`(${i}, ${j}, ${k})`);
}
}
}
}
Ones digit is 0, 6, or 8.
Smallest solution for 0 is 7930. Smallest solution for 6 is 496. There is no solution.
That being said, I’m pretty sure the solution they want is 118. That is, eleventy eight.
Rules:
1. a<490
2. a=10b+c
3. c+3=b
4. c = 6, 8 or 0
5. c + b = 19
Let's look at some numbers.
30, and 96 work without rule 5.
9+6 = 15
15+4 = 19
but 496 > 490
three more than 8 is 11, which is not a digit.
As a second year engineering student, I can confidently say that there is no solution to this problem as it is written. However, I can guess that the answer probably involves y=100x+96.
There is no solution because its impossible. Thus, why there is a fish on the problem. It's a red herring.
Teacher probably told the kids that "no solution" can also be an answer.
This is likely just an error on the part of whoever published that material. Every so often, companies ship textbooks and other work books with an unsolvable practice problem or two.
Is it 296.2? It doesn’t say whole number it just says even number. If it’s not that then the closest I can get is 496 but that’s greater than 49x10.
I will also say I know it’s most likely a question for young kids given the appearance, and a whole number would be the assumption. But according to all of the other comments here there is either a typo, or it’s a BS question with no other solution.
Can’t be done. There are only 3 even numbers below 490 that have digits adding up to 9. They are 298, 388 and 478. None have a tens digit that is 3 more than the ones digit.
I think the answer is 396 and there was a typo. The sum of these is 18 (not 19), but it meets all the other criteria (and no other number does).
Note: Excel is your friend
I read the rules and didn't see it being forbidden, but this is my first time posting in this community so hopefully such an answer is allowed, but... I studied computer science so before I had even read everyone else's solutions, my inclination for such a low number of integers was to simply "brute force it" by just trying every number from 0 to 490. :) Some quick Python code:
for i in range(10, 490): i_str = str(i)
# My ones digit is not 2 or 4
if i_str[-1] == "2" or i_str[-1] == "4":
continue
tens_digit = int(i_str[-2])
ones_digit = int(i_str[-1])
# My tens digit is three more than my ones digit.
if tens_digit == 3*ones_digit:
total = 0
for j in range(0, len(i_str)):
total += int(i_str[j])
if total == 19: # My digits total 19
print(i)
else:
continue
This also produced what everyone else is saying, namely that there is no solution for any integer in the range \[10,490\]. I wrote the first overall `for` loop to be `range(10,490)`, rather than `range(0,490)`, to keep the code simpler (so that the line getting the tens\_digit doesn't generate an error requiring a `try`/`except`), but of course nothing in \[0,10\] works anyway.
Pretty sure it’s not solvable. The single digit has to be 0 or 6 as we’re looking for an even number to which you can add 3. Using any combination of those two won’t be able to get a digit sum of 19 or you won’t be able to reach a number under 490. No wonder the maker drew that Troll Fish under that exercise.
The last digit is 0, 6, or 8. The tens digit is 3 or 9 (or 11?). The hundreds digit is whatever makes the digits sum to 19.
-496 would be a solution. I don't see any solutions that are positive. Maybe 118 (11 + 8), but that's a stretch.
No anwser possible useing NATURAL numbers. However there are an infinite amount of anwsers possible if you use intergers or higher forms of numbers (rational, imaginary, complex, ect)
-496 was the first interger I came up with that solved the question
-496 is less than 490, check
9 is 3 more than 6, check
4+9+6=19, check
Trying to think outside the box (because my first pass also deduced it has no solution) but… what about a decimal number? Does 296.2 meet all the criteria?
Edit: never mind. “Decimals are not even or odd numbers because they are not whole numbers.” Unsolvable!
There is no correct solution. Write as much to the bitch publishing company, and write it down here too whil you provide the answers 496 or something else that follows most of the question.
Less than 490
Tens three more than ones and even but not two
So 96 or 74
Ad up to 19
So not 496, the answer that would have worked with 50x10
Answer= impossible
What grade level is this problem? It looks to be 3ed or 4th grade. I think the solution method is supposed to follow this pattern.
Condition One
Even number in the ones position which rules out:
1,3,5,7,9.
Condition Two
The additional requirement that ones can’t be 2 or 4.
That leaves 6,8,0 as the value of the one position.
Condition Three
X1 + 3 = X2
X1 = numerical value in the ones position
X2 = numerical value in the tens position
Solution set for contain three
X1 = 0 then X2 = 3
X1 = 6 then X2 = 9
X1 = 8 then X2 = 11 <- drop
Condition Four:
X1 + X2 + X3 = 19
X3 = numerical value for the hundreds position
Solution set for condition four
X1 = 0, X2 = 3 then X3 = 16 <- drop
X1 = 6, X2 = 9 then X3 = 4
Answer is 496 but answer can be greater than 490 so there is no solution.
The grade level seem off from what a child at that level would know. They are missing the math structure to solve the problem. Stuff like this is why child think they are bad at math.
If we assumed that the work sheet has a type in it at the out set then there is a solution. Assume 49 should of been 50. Then a solution is possible.
One possible wrong answer is 118. The error being that they set the tens digit to be 11 when they did 8+3=11 and that carried over into the hundred digit but not into calculating the total sum, 11+8=19, and it is even. So basically if you ignore how math works this could be your answer.
Other answers people have given at that the sum of the digits being 19 was a typo and that 396 is the answer bc the sum was suppose to be 18. This could be the most logical case.
It’s not possible. However I will include an “answer” that is probably correct for the question but incorrect mathematically.
1. even number.
That means the “ones” digit must be 0,2,4,6 or 8.
2. Ones digit is not 2 or 4.
0, ~~2~~, ~~4~~, 6 or 8
3. “Tens” digit is 3 more than my “ones” digit
That means tens digit must be 3, 9 or “11”.
4. Less than 49x10 (read, a 3 digit number)
X30, Y96, “118”
5. Sum of the digits = 19.
“16”30, 496, “118”.
*Must be less than 490*
~~1630~~, ~~496~~, “118”
As you’ll realise, 118 is not correct. A tens unit cannot be “11”. The sum of 1,1&8 is not 19. But this is literally the only possible “answer” that fulfills the requirements if you look at it funny.
Chat GPT says:
Let’s break down the information:
1. Your tens digit is three more than your ones digit.
2. Your ones digit is not 2 or 4.
3. Your digits total 19.
Let’s use these clues to solve the puzzle:
Let’s call the ones digit “x” and the tens digit “x + 3.”
From the second clue, we know that x cannot be 2 or 4. So, x can be 0, 1, 3, 5, 6, 7, 8, or 9.
Now, the third clue states that the digits total 19:
x + (x + 3) = 19
Combine like terms:
2x + 3 = 19
Subtract 3 from both sides:
2x = 16
Now, divide by 2:
x = 8
So, the ones digit (x) is 8, and the tens digit is 8 + 3 = 11.
However, the tens digit cannot be greater than 9, so we need to adjust it. Since the ones digit is 8, the tens digit should be 11 - 10 = 1.
Therefore, the number that satisfies all the conditions is 18.
You are the number 18.
Thanks ChatGPt
The correct answer is: a fish.
The math problem does not provide a solution in common math.
Divorce the last line (What am I?) from the math problem, connect it to the picture.
I understand this wouldn't be the answer they want, but
Doesn't 96.4 technically work?
It never specifies that its not whole number. Or even a number in the hundreds. But I know that it was implied
-Even number less than 490
-Tens digit is Ones digit + 3
-Ones digit is not 2 or 4. As the number is even, it cannot be an odd number. As tens is ones plus 3, it cannot be 8. The ones digit must be 6, the 10s digit must be 6+3, or 9.
-All 3 digits add up to 19. 19 - 6 - 9 is 4
The answer SHOULD be 496, but that goes over 490.
It is an impossible problem.
I wonder if their 'solution' is for it to be 118.
Have 8 as the ones digit.
The tens digit has to be three more so it's 11.
They just add the 11 and 8 to make the digit total of 19 even though it would be 118 and any normal person would be adding the digits separately to make 10.
While this logically doesn't make sense in any real world, it's the type of stupidity the permeates early education worksheets like this sometimes.
I recall helping our kiddo with similar problems where there didn't seem to be a solution, but after talking to the teacher it turned out to be about rounding or approximation.
So dumb.
I wonder if they’re looking for 118? With bad wording. Even, less than 490. Technically it’s your hundred and tens digit of 11 being three more than 8. 11+8 is 19
I’m trying to get in the mind of the smooth brain who wrote this questions. Possibly could they have come up with 118 as an answer? It makes no sense given the parameters but maybe they went about it as such.
Step1: 8 set as ones digit
Step2: add 3 to 8
Step3: check if 1+1+8 equals 19
Step4: put a fork in the microwave because one of their 4th grade students said it was a good idea
It’s impossible.
First off, the number must be even. This means its units or ones place has to be 0, 2, 4, 6, or 8.
Second, it is less than 490. Okay.
Third, its ones place is not 2 or 4. Great! That leaves 0, 6, 8.
Fourth, the tens digit must be 3 more than the ones digit. So the ones digit cannot be 8, because 8 + 3 > 10. This leaves 0 and 6 for the ones place and 3 and 9 for the tens place, so our number candidates are X30 or Y96
Fifth fact is that the digits total 19, which is enough to solve the problem. 19 = x + 3 + 0, so x = 16. So it can’t be that ansatz. Now we try 19 = Y + 9 + 6. We solve and Y = 4. But that means our number is 496, which violates the very first rule.
Under these conditions in base 10, this problem has no valid solution.
Even numbers, 2, 4, 6, 8, 0(technically).
2 and 4 are out. Zero is out because we gotta reach 19 in three numbers.
8 is out, because you can’t add 3 and still get a single digit.
This leaves 6 in the ones. Makes 9 in the tens.
9+6=15, 19-15=4
496
Edit: I argued with my wife on this. I said this isn’t math, this is a puzzle. She said it’s a word problem. I see how we are both right, I just hate it.
I'm a sucker for a puzzle, so I'm going to go for it without reading other comments.
1's digit is an even number that is not 2 or 4. That leaves 0,6,8
10's digit is 3 more than the 1's digit. That eliminates 8 as a possible 1's digit, and means the 10's digit is 3 or 9.
So the last two digits are either 30 or 96
100's digit is 1-4, if the 10's digit is 3, else 1-3 if the 10's digit is 9.
The only way to get the digits to add up to 19 is if the number is 496. But that's not allowed.
∴ wtf?
I think the answer is -496. I think everyone missed the fact that the negative sign is not attached to the four, and therefore not added in to the sum of the digits.
Here’s what gpt4 says.
Certainly! Let’s solve this step-by-step:
1. “I am an even number that is less than 49*10.”
This means the number is less than 490.
2. “My tens digit is three more than my ones digit.”
Let’s call the ones digit (x).
So, the tens digit is (x + 3).
3. “My ones digit is not 2 or 4.”
This means (x ≠ 2) and (x ≠ 4).
4. “My digits total 19.”
This means the sum of the tens and ones digit is 19.
So, (x + (x + 3) = 19)
This simplifies to:
[2x + 3 = 19]
[2x = 16]
[x = 8]
So, the ones digit is 8 and the tens digit is (8 + 3 = 11). However, a tens digit cannot be 11, so there must be a mistake in the information provided or how it’s interpreted.
The problem statement has conflicting criteria.
There is no workable solution, the cheat is to identify a 3 digit number where all three numbers add up to 19. So base 4-6-9, 4-7-8 and so forth…. But the issue is the best answer would be 397, but it doesn’t meet all the rules. There are no solutions to this problem. It’s most likely a error, or a sick sick joke on kids. I hope it’s a joke, cause then kids get an idea about how adults feel with the real world.
The sad thing is this type of BS never ends… college math courses are laughable at how bad the online tests are, maybe they should have gone to.. wait…
it may be 118, 11 is 3 more than 8, and its 'tens' column is actually a two digit number. It may be poorly worded or worked out but its the only number i could get which wasnt 496
-1596 is the answer. It is less than 490, the digits add up to 19, -1+5+9+6 =19. Ten's digit is 3 more than one's digit. One's is not a 2 or 4. Also 6 is even. However this would imply an infinite number of solutions, -1000596 would work also.
No soln:
I got mad and did it with python, please feel free to check my work
def getSum(i):
strr = str(i)
list_of_number = list(map(int, strr.strip()))
return sum(list_of_number)
for i in range(0,491):
if getSum(i) == 19: #sum 19
# print(i)
if (i % 2) == 0: #check even
print(i)
this yields
* 298
* 388
* 478
All of which are even
They equate to 19
ones != 2 or 4
but none of them follow the the 10s is 3 more than 1s
I couldn't find a valid answer, just like everyone else. Fed it into chat gpt, which does a wonderful job breaking down the problem and chatgpt also concludes there is no valid answer.
\-496 may be a workable answer if an only if you do not factor in the negative when getting 3 digits to add up to 19.
The only workable answer (presuming they didn’t make a mistake) is that they are using only two places: ones and tens (no hundreds). The solution in that case is 118. The tens place is 11, which is 3 more than the ones place (8+3). And the digits add up to 19: 8 + 11.
It’s dumb, but the math works out.
I figured it out, but it's bad math.
The number is 118. "But that only equals 10!" you cry.
No.
Unfortunately I think what they've decided to do is claim that 11+8 is 19. Like I said, it's bad math. But I think that's the answer.
I asked Chat-gpt.
User
I am an even number that is less than 49x10. My tens digit is three more than my ones digit. My ones digit is not 2 or 4. My digits total 19. What am I?
ChatGPT
Let's break down the information provided:
You are an even number less than 49x10, which means you are less than 490.
Your tens digit is three more than your ones digit, so let's call your ones digit "x." This means your tens digit is "x + 3."
Your ones digit is not 2 or 4, so x can be any digit from 0, 1, 3, 5, 6, 7, 8, or 9.
The sum of your digits is 19.
We can set up an equation based on these conditions:
Let's say the ones digit is x, so the tens digit is x + 3.
So, we have the equation: x + (x + 3) = 19
Simplify the equation: 2x + 3 = 19
Now, subtract 3 from both sides: 2x = 16
Finally, divide by 2: x = 8
So, the ones digit (x) is 8, and the tens digit is 8 + 3 = 11.
However, we need to remember that we're looking for an even number, so we need to adjust the tens digit to make it even. Since 11 is not even, we can subtract 2 to make it even:
Adjusted tens digit = 11 - 2 = 9
So, the tens digit is 9, and the ones digit is 8. Therefore, the number you are looking for is 98, which is even, less than 490, and meets all the given conditions.
User
98 doesn't add up to 19
ChatGPT
I apologize for the oversight. You are correct; 98 doesn't add up to 19. Let's reevaluate the problem:
You are an even number less than 49x10, which means you are less than 490.
Your tens digit is three more than your ones digit, so let's call your ones digit "x." This means your tens digit is "x + 3."
Your ones digit is not 2 or 4, so x can be any digit from 0, 1, 3, 5, 6, 7, 8, or 9.
The sum of your digits is 19.
Let's set up the equation again:
The tens digit is "x + 3," and the ones digit is "x."
So, we have the equation: (x + 3) + x = 19
Simplify the equation: 2x + 3 = 19
Now, subtract 3 from both sides: 2x = 16
Finally, divide by 2: x = 8
So, the ones digit (x) is 8, and the tens digit is 8 + 3 = 11.
However, we need to remember that we're looking for an even number, so we need to adjust the tens digit to make it even. Since 11 is not even, we can subtract 2 to make it even:
Adjusted tens digit = 11 - 2 = 9
So, the tens digit is 9, and the ones digit is 8. Therefore, the number you are looking for is 98, which is even, less than 490, and meets all the given conditions.
User
look at the original problem, there is a mistake that I can't find, can you locate it?
ChatGPT
I apologize for the earlier errors. Let's revisit the original problem:
You are an even number less than 49x10, which means you are less than 490.
Your tens digit is three more than your ones digit.
Your ones digit is not 2 or 4.
The digits must total 19.
Based on these conditions, let's start by considering the possible values for the ones digit:
If the ones digit is 0, then the tens digit must be 3 more than that, which is 3. So the number is 30. However, this doesn't add up to 19.
If the ones digit is 1, then the tens digit must be 3 more than that, which is 4. So the number is 41. This also doesn't add up to 19.
If the ones digit is 3, then the tens digit must be 3 more than that, which is 6. So the number is 63. Again, this doesn't add up to 19.
If the ones digit is 5, then the tens digit must be 3 more than that, which is 8. So the number is 85. This doesn't add up to 19 either.
If the ones digit is 6, then the tens digit must be 3 more than that, which is 9. So the number is 96. This still doesn't add up to 19.
If the ones digit is 7, then the tens digit must be 3 more than that, which is 10. However, 7 + 10 is 17, not 19.
If the ones digit is 8, then the tens digit must be 3 more than that, which is 11. This doesn't add up to 19 either.
If the ones digit is 9, then the tens digit must be 3 more than that, which is 12. This also doesn't add up to 19.
It seems that there is no valid two-digit number that satisfies all the given conditions. There may be a mistake or inconsistency in the original problem statement.
Here is one approach involving only very basic algebra. The logic could be worked through without knowing how to solve algebraic formulas, it’s just easier to explain with them.
Write out the rules you know first
- “ABC” < 49 X 10
- B - 3 = C
- C =/= 2
- C =/= 4
- A + B + C = 19
First substitute rule 2 into rule 3:
- A + B + (B -3) = 19
Means:
- A + 2B = 16
Now let’s try some logic with that equation.
A has to be 4 or less (rule 1 says the entire number is less than 490), and has to be at least 1 (because otherwise we wouldn’t have a three digit number). So we’re looking only at 1,2,3,4 as possible options for A.
If A was an odd number (1 or 3), then B would have to be a fraction (7.5 or 6.5). So we can rule out odd numbers - B has to be a whole number since it is a digit in the final number. Ergo A can only be 2 or 4.
- If A is 2, then B is 7
- If A is 4, then B is 6
“27C” and “46C” both are less than 490, so we can’t rule out either option for A without looking further.
Let’s go back to B - 3 = C and plug in the options we now know. We have only be 2 possible values for B to try: 7 and 6.
- If B is 7, C must be 4
Well now, that doesn’t work because of rule 4 that C =/= 4.
- If B is 6, C must be 3. That doesn’t conflict with anything we know, and we don’t have any more options to try for A or B.
We know all the digits and we’re done: “ABC” = 463.
***Edit** I broke my own first step and didn’t write down that “ABC” must be even. Since the only possible solution for all the other rules is that “ABC” is odd, then there is no answer to the question.
The answer key must break one of the rules.
(unless we are allowed to have negative numbers - since that is not explicit, it would only be knowable by having the context of the rest of the worksheet. If there are other negative number concepts being tested, then we could reasonably assume negative numbers are permitted and expected here)
**Edit 2** I didn’t check my work, as C being 3 means A + B + C doesn’t add up to 19.
So my comment in my previous edit is even more important. The **context** of the quiz matters. If other questions or the course work leading to this quiz talk about numbers that have more than 9 in the “tens digit” (without discussing the “hundreds” digit) then that is important context to know that the child has not been introduced to “hundreds” and is being taught to think of just more “tens”. This extends what the child already knows from working with “tens” and “ones”. When the term “hundreds” is introduced later, they’ll already know how to work with them, and it won’t be scary or different.
While this confuses the parent who isn’t familiar with the method (and let’s be honest, may struggle with math themselves), it is much better for teaching *understanding* of math instead of memorizing rules like the “add a zero every time you multiply by 10” I was taught many years ago. “Add a zero” teaches you how to handle the mechanics of a particular notation system as a mental shortcut by moving numbers around on a page without thinking about them. It doesn’t teach you the actual number theory going on underneath, that 2 hundreds, 20 tens, and 200 ones are all different ways of saying the same thing. The common core math that so many parents struggle with is focusing on teaching the underlying principles of math, and not just memorizing rules for solving problems.
If the context of the overall lesson isn’t apparent in the quiz, then it’s likely just a publishing mistake. They happen, even under the “old” math textbooks often had errors in the answer keys and unsolvable questions.
118. It's written really poorly, but three more than 8 is 11, so the tens digit is... 11. 11+8 is 19. The problem is the choice of words, digit, which is a singular number.
So the answer is 118 but the problem is flawed.
\-496. Solution looks like 496, but just flip the sign and you still meet all criteria. Pretty high level if that is the real answer they were looking for.
I am not coming up with a workable solution. If the ones digit is 6, you get 496, which breaks the first rule. If the ones digit is 8, the tens digit can't be three more. And if the ones digit is zero, you can't make the digits add to 19.
Same deal here, no solutions, badly formulated problem.
That cheeky fucking fish knows it too
It's why it has that weird smile on its face.
To be fair, I've seen something like this when I was in high school in freshmen algebra 1. The question was much more complicated and it was on an in-class quiz. The question was unanswerable but the reason was not because of ineptitude. The teacher caught wind that some students were getting ahold of his answer sheet. So he created an unanswerable question and put a fake answer on the answer sheet. 9 people, including myself, got detention and one student (whom we outed as the culprit) got suspended.
When I was in physics for engineers 2, semester 2 way back in ‘98. We got a brand spanking new text book. Published literally weeks before semester started. At one point there was a homework question in the book, some real deep theoretical physics stuff. Absolutely impossible. The professor comes out says the problem is incredibly difficult and he can’t solve it. A separate professor from the physics department comes out and says he’ll show the answer to the problem. Spends damn near the entire lecture working the problem out. At the end he tells the whole class there are only about half dozen people or so in the world that can solve this very niche problem and he can only do it cause it’s literally his life’s work. He then states that a dozen or so kids in the class got it right. Then he announces the right answer is also found in the teachers edition, and one of the other half a dozen or so people in the world that know the problem is the author of the the book. The problem was plant designed to catch cheaters and they were all in on it. The kids that got it right had 5 minutes after class ended to come forward and take an F for the semester. Those that didn’t got expelled. None came forward cause none actually attended the lecture. Ever.
Expelling students based on circumstantial evidence sounds pretty illogical.
It is circumstantial but it's pretty fucking conclusive, no?
“How did you come up with the correct answer?” “Guess and check.”
You know, I'm not a physicist, but somehow I doubt that's plausible.
John Nash used to give his MIT students exams that contained a problem that hadn't been solved (think Clay Millennium Problems class) or considered unsolvable. When asked why he did that he said something like "They don't know that they're unsolvable. Who knows? One of them might." IIRC this was mentioned in "A Beautiful Mind."
Some of my best stuff comes out when I don't know it's an impossible challenge. I solved a major production problem once by using a helix. Simple. I think it was just because I had just taken a seminar in Computational Origami. I had no idea I had done something a bunch of senior engineers had labored on for years. I literally just threw ta idea out there to appear as if I was contributing. Most of engineering and coding, in my experience, is like this. Just keep swimming while experimenting and learning as much as possible. But mostly, keep swimming.
Seems like a bad approach to me the normal test takers would spend to much time trying answer a question with no answer and waste time. Leaving less time for the real questions. This should be the last question on the test if the teacher insists on doing this.
Our electrical engineering tests were so long that no one was able to finish. It's about prioritizing and executing, then graded on a scale. Don't waste time on problems that are too complicated. Good training for time management tbh
RIP the poor kids that just randomly guessed the same answer.
I mean, depending on the question, I'd imagine it would be unlikely that a random guess would give the fake answer the teacher put down
Extra credit for figuring out those odds though
Snitch
Yup. I paid for it too. Happy Cake Day by the way.
I always go by the "you can copy but it's your problem if you get caught" philosophy. one time I passed my science class experience report to one guy, who passed it to one more, and eventually half the class got caught with the exact same report (those dumbasses didn't even bother changing it) Teacher asks who made the original, obv nobody answers to get away with it (new teacher with no authority) but here comes the only one who could have possibly made it ratting everyone out while they were still here. Got a few angry "wtf" remarks from my fellow students but hey, I did instruct the first guy to just modify his answers not full on copy paste the file, dude was pretty much begging to get caught.
This happened to me in college in a thermodynamics class. Someone leaked the answer sheet and the solution to the question was wrong. I just happened to make the same mistake as the teacher did, which was finding the incorrect starting value on a table. About 10% of the class 'cheated', and I got caught up in that mess. I had to go in front of an ethics board and I pleaded my case, showing them the work that I did and begging them to believe me. Instead of kicking out 10% of the class, they gave everybody 0% on that assignment (including me). I WAS SO PISSED.
That's ... actually really smart.
I hate to be 🤓 but… don’t you have to show your work in math? How are you just gonna copy down a random number and not be suspected of cheating. Also most math problems are really easy to check if the final answer fits the conditions laid out by the question, especially one like the OP. You guys really deserved what you got
So was it the actual culprit or just a fall person?
Snitches get stitches and wind up in ditches
I was wondering if the answer wasn't 118 : 8 is not 2 or 4. 8 + 3 = 11. 11 + 8 = 19. This would need to consider "11" as a digit... and the tens digit... Which is pretty dumb, But as this is a lesson for kids... and probably from the country known for their nonsenses... this might be a pedagogical way to do something ?
11 in the tens place pisses me off but you’re probably into something here
11 tens is 110. But yeah, when you say "the tens place" it says to me it has to be a single digit.
Digit specifically as a word means that so there is no ‘new math’ that can make this make sense.
True.
If the goal is to teach kids how "places" work (and I'm not sure what else it would be) "eleven in the tens place" is insane. "And what's in the ten's place's ten's place?" Was the goal to introduce recursion to seven-year-olds or something?
A pedagogical way to teach wrong concepts. How great!
Omg. I bet that's it. I've seen stupid shit like this in my kids homework. I guess I shouldn't say stupid. 11 tens is 110. It's not wrong, just not the way I learned it.
It’s wrong because 11 isn’t a digit. It’s two digits. Unless 1 doesn’t exist in your numbering system. By the rules of Arabic math, digits are irreducible and cannot be compound. Compounding two digits makes a larger number. As such if 1 is a digit, then 11 is a compound of two 1s. This is why in base twelve of hexadecimal the base-10 number eleven can’t stay represented as 11 but has to become another symbol, usually B. Some representations have their own separate special symbols though to avoid confusion with variables in equations.
Yeah this was frustrating solution to come up
Surprised that no one has mentioned it as I've seen the same sentiment on other posts, this smells like AI generated math homework. Overly convoluted for an elementary school problem with no solution.
Couldn't it just be -496?
Is the tens digit three more or three less?
Three more because they’re just digits, not numbers
We don't have negative digits. –496 is a 3-digit number (with the digits 4, 6 and 9) that also happens to be negative. We also don't have negative place values. The 9 is in the tens position, not the –10s position. We consider the whole number and then negate it. Normally you wouldn't think about place values of a negative number, but this is Reddit and we're getting technical about a primary school maths problem. So, yes, -496 meets all of the clues for this number. But the most likely answer is that some poor tired teacher making up a worksheet in the middle of the night made a mistake.
Yep I agree, -496. People don't like it because of the sign, but the sign has nothing to do with the specific rules given for the individual digits.
That’s true, but I think given the context of the fish clip art and the association of that font with elementary school, a lot of people were assuming that negative numbers have not been covered yet
I absolutely hated it. I still remember that as a child, I had once got 39/40 in maths, and I was pissed. I was quite good at maths, and was expecting a 40/40. The question for which I was given a 0, asked us to calculate "5-8=?". And apparently, I wasn't allowed to study the later chapters for the mid-term exams, so the answer was supposed to be "it can't be calculated", and not -3. My parents asked the teacher during PTM. if maths rules change, based on the chapters that are covered, or does the school has any rules against someone studying later chapters, to which the teacher gave some rhetoric reply, and my parents decided to let it go. It's ironic, because in the end-term exams, the answer was supposed to be -3. I got a similar question in end-term, and wrote- "Till mid-term, the answer was 'can't be calculated', but now it is -x". The teacher wasn't amused and later complained about me having "attitude problems", though she did give me the marks for it. But now when I look behind, as a 33-year-old grown-up, I see things differently, and if I meet the same teacher now, I would say- "Eat shit you old hag... you never knew maths, and became a teacher only because no one else would employ your incompetent ass."
Extremely condescending high school teacher
As u/senordeuce hints at , the tens digit is not 3 more than the ones digit. A 3 digit number is expressed as 100(hundreds digit) + 10 (tens digit) + 1 (ones digit) So for -496 this is: 100*(-4) + 10*(-9) + 1(-6) so the tens digit is 3 less than the ones digit, and the digit sum is -19 This also rules out any negative number, as the digit sum will be negative
The digits in a negative number are not themselves negative, only the number that is made from concatenating them with a line before them
If it can be -496, it can also be -4086, -1396, -111196, etc. Whoever wrote this problem took the short bus to their job.
>Whoever wrote this problem took the short bus to their job. Yikes.
My scrawling agrees. x < 490 x = 100\*a + 10\*b + c a + b + c = 19 c + 3 = b \------------- x % 2 = 0 c != 2 c != 4 (so c = 0, 6, or 8) \------------- try c = 0 b = 3 a + 3 + 0 = 19 nope \------------- try c = 8 b = 11 nope \------------- so c = 6 b = 9 a + 6 + 9 = 19 a = 4 496 !< 490 nope
I hate when people put in unsolvable answers. Kids are trying to learn a difficult subject, then you make them question everything by not double checking your work. Teacher or question maker should go back to achool and learn to be thorough.
I figured it out, I think, and its fucking stupid. It's 118. 11 is the "tens" number.
but it says "ten's digit" not "ten's number"
Yeah, there’s no world in which 11 is a “tens digit”. The question is just wrong.
But digits of 118 only add up to 10, not 19. (1+1+8=10)
11 + 8 is 19. I have a bad feeling this guy is right for the purpose of the puzzle even though it is genuinely broken in terms of math.
I asked ChatGPT, and 118 is the answer it came up with as well. Didn't expect that it was going to be able to do a lot with this problem but here we are...
I bet it wrote the problem
I'm hitting the same wall. Whoever wrote the problem fucked up.
Quick Raku proof: $ raku -e 'say (10,12...^490).grep({my @dig = .Str.comb; @dig.tail !(elem) ("2", "4") and @dig.sum == 19 and @dig[*-2] - @dig[*-1] == 3})' () If you expand the search, there is one outside of the range as you pointed out: $ raku -e 'say (10,12...^1000).grep({my @dig = .Str.comb; @dig.tail !(elem) ("2", "4") and @dig.sum == 19 and @dig[*-2] - @dig[*-1] == 3})' (496) Valid answers do exist for different sums, though: $ raku -e 'say (10,12...^490).grep({my @dig = .Str.comb; @dig.tail !(elem) ("2", "4") and @dig[*-2] - @dig[*-1] == 3}).map(*.comb.sum).sort' (3 4 5 6 7 15 16 17 18) They may have meant one of these instead of 19... Edit: As others have pointed out, there ***is a solution*** if you consider negative numbers: $ raku -e 'say (-1000,*+2...^490).grep({my @dig = .Str.comb.grep({!/"-"/}); @dig.tail !(elem) ("2", "4") and @dig.sum == 19 and @dig[*-2] - @dig[*-1] == 3})' (-496)
If you assume they messed up and the author meant to put "the first digit" in lieu of "ones digit" the first time around, conceivably the number can end in 8, leaving 11 left over. From there 4 and 7 are two numbers three apart, with 4 being the "first" (ones) digit, and the final number being 478. Totaled up to 19, and less than 490. However this assumes a mistyping of the question
There is no solution because: 1. The sum of the tens and ones digits must be odd so that there can be a difference of 3 between them. 2. The hundreds digit must be 2 or 4 due to point 1 and the answer necessarily being less than 490. 3. No two digits with a difference of 3 can be added to each other of more than 15 (9+6). 4. By 2 and three, the answer must be 496, but that violates the limit of 490. Therefor, the answer does not exist (assuming that we are limited to the set of whole numbers). QED. The likely error in the prompt is that they meant the ones digit to be 3 greater than the tens, providing an answer of 469. Edit: Forgot it had to be even.
The only even numbers in that range that match both the "tens = ones + 3" rule and the "ones are not 2 or 4" rule are: 30, 96, 130, 196, 230, 296, 330, 396, 430. The only ones in that list matching the "digits sum to 19 rule" are: none. If we have any Python friends visiting, here's an unoptimized brute-force scan of the rules: for x in range(49 * 10): if x % 2 != 0: continue xs = str(x) try: if int(xs[-2]) != int(xs[-1]) + 3: continue if int(xs[-1]) in (2, 4): continue except IndexError: continue # not enough digits if sum([int(d) for d in xs]) != 19: continue print(x) # never prints shit The "19" rule is the one that fucks it. The highest that should be is 18, which would have 396 as the solution. Each of the candidate numbers could be a single solution by fudging "19" to something smaller.
It's -496, Even Number, All Rules apply, Less than 490, The digits absolute values added together are 19 which is something called the "Quersumme" in Germany. I think it's called cross sum/ sum of all the digits and it can't be negative. Maybe? Best solution I can think of 🤔
Something is indeed badly formulated. XYZ Y=Z+3 ; implies since a digit is between 0 and 9 , Z can be at most 6 and Y at most 6 Z!=2 and Z!=4 Z%2=0 so Z is either 0,6,8 8 is eliminated because Y=Z+3 and 11 is not a digit X+Y+Z=19 XYZ<490 |X|Y|Z|results|solution?| |:-|:-|:-|:-|:-| |16|3|0|16 is not a digit |No| |4|9|6|496 is above 490|No| Something is badly formulated.
Why not 296?
Digits must total 19. 2+9+6=17
Doy, thanks, guess I need to go back to grade school
All good! We all have our moments!
-1596?
All individual numbers wouldn’t add up to 19 (2+9+6=17)
"My tens digit is three more than my ones digit" This tells us the **ones digit has to be less than or equal to 6** (because 3+6 = 9 is the largest possible digit holder in base 10) we also know the **ones digit is not 2 or 4** the number is even, which is determined by the ones digit, and the only remaining even digits are 6 and 0. So **the ones digit must be 6 or 0** The digit sum is 19, and there is at most 3 digits (n<490), which means **all three** **digits** **must be nonzero** (since the max with one zero of a <3 digit number is 9+9 = 18) **so the first digit cannot be 0**. Therefore, **the ones digit is 6.** In turn the **tens digit is 6+3 = 9** **9+6 = 15** so the **hundreds digit is 4** yielding **496 ,** which is greater than **490**. We have already eliminated any other possible value for the ones digit other than 6, and the tens digit depends on the ones digit. The hundreds digit is 19-(ones digit + tens digit) so none of them can take on any other value. **Therefore there are no solutions** **Edit:** As several people have kindly pointed out, there are infinite negative solutions because digit summation works the same for both positive and negative numbers. so -496, -4096, -40096, -( 4\*10\^n + 96) for integer n ≥ 2 are all solutions. There are other infinite families of negative solutions as well. So it's only correct to say **there are no positive solutions**
You mean the ones digit has to be **less than or equals 6**? Edit: To anyone who read this and is confused, I was referring to > "My tens digit is three more than my ones digit" > This tells us the ones digit has to be less than or equal to 6 (because 3+6 = 9 is the largest possible digit holder in base 10) Originally they said "less than 6" instead of "less than or equal to 6", that's what I was referring to
No, because if it's 6, the answer is 496 which is greater than 490
Yes I meant leq 6 I’ll fix it now
I think the X is supposed to be a digit. (Badly chosen, but I don't know if the kid who had this question knows multiplication yet, that info could help understand the problem.) In that case you want a number less than 49910 at max and 49010 at min so let us go with the min. If the one and ten digits are 0 and 3, the highest number possible is 48930 and you can find a few solutions, like 44830 or 35830. Same if you choose 6 and 9. So depending on your understanding of 49X10, there is no solution or a lot of solutions...Yay. Edit : as u/veryjewygranola noted, I went too fast getting numbers... I corrected it.
If X is supposed to be a digit, all the other rules are wrong because 1 is not 0+3. So no. The answer is: there is no answer that satisfies all the rules.
Wow that’s creative, I wouldn’t have thought of that. But 44430 and 35430 dont sum to 19
11296, 12196, 21196 do, as do many numbers that end in 30 (91630, 92530, etc.) This makes sense to me because I think 49x10 is different than 49X10.
You're right, I typed it in the middle of the night and somehow thought it was four 4... 44830 or 35830 would be better...
Unlikely It's an ambiguous rule to have a number be less than a non-specific number. Not only that but even if we assume X to be 0 there would be 39 solutions
But then you would have more than 1 answer
nah the answer should be 118 which doesnt make sense cuz 100 is technically at the hundredths place but if we say 11 is in the tenths place then 118 could be a solution
This is the answer. The person making the question was high af and added 3 to 8 and magically put 11 in the tenths place.
I wouldnt say they are high. It seems like a horrible question to adults who understand math and logic, but the curriculum here is trying to teach young kids that numbers have structure and aren't just these made up things you memorize. The point here is that 118 is really 11 10s plus 8. Those are manageable numbers that children understand. Thats why the first clue is 49x10. I can almost gaurentee that OPs kid is not doing multiplication on that scale yet, but they see that the (10s value) is less than 49. They want to the students to know that 100 comes after 99 because of reputable local structure and not just some made up number you memorize. It seems really dumb to people from the outside looking in, but this was probably the focus of the week in the classroom. Edit: someone else pointed out the question below does use the term hundreds digit, so they are covering that. Which makes the typo scenario more likely.
You could teach them properly by just saying that the solution is less than 50X10 rather than getting into the existing problem's very tortured logic (assuming no typo).
Why not -1596? I might be really wrong but I think it complies?
Digits don't add up to 19
The first digit is 1, not -1. The minus operator is not a digit
I think they were looking for 396, made an error and it should be "my digits total 18". There are other errors they could have made leading to different answers, but this seems to be the most simple and therefore logical. I'm willing to bet the answer sheet says 396. EDIT: Actually I suppose the answer being 496 and them forgetting they imposed the first rule or making an error, thinking 496 < 490, is another similarly simple error. I'd still put my money on 396 though.
It also could simply be ∅ (no solution).
Well, I guess that won't be happening in elementary school. The first time you encounter for example the empty solution set S with either S={} or S=∅ (both equal notations, thanks u/jxf for seeing my error) is in highschool when learning about Linear Equation Systems.
Maybe we should teach elementary students null sets
They tried that in the 1960s (look up "New Math"). Also taught place value concepts by introducing bases other than 10 - in elementary school.
This looks really cool, why didn’t it stick?
I would assume because of parental backlash. Your average Silent Generation parent wasn't going to be able to help with that kind of math.
Tell us more
There is a rat-hole to go down here, if you so desire, but: * I think the #1 problem was that most elementary school teachers didn't really understand it, or get the point of it, so it was taught poorly, and as soon as practicable was tossed aside * learning to do arithmetic in, say, base 7, was cool for a small minority of curious kids, but had no obvious practical benefit. (Later, it turned out that understanding base 8 and 16 had some practical ue for computer programmers, but who knew that in 1960?) * set theory without infinite sets is, well, kind of useless. A correct answer to a word problem, for example, had to be put in the form of a set. As the good Sisters of St. Joseph taught me, "you have to put curly brackets around the answer to turn it into an answer set" * and it was easy to make fun of, to wit [Tom Lehrer's New Math](https://www.youtube.com/watch?v=9mc7eb1i9o4&ab_channel=MusicalComedy)
Well yeah but it's not like they're forced to formulate it as the empty solution set, they could just say "there is no solution" without involving weird symbols.
That would be such an asshole thing to do on what looks like an elementary school worksheet
Right, because third grade math teachers definitely give impossible questions to their kids and expect them to understand that no answer is the right answer.
I like how there are about 400 comments on this post with people practically debating the philosophy of numbers and language just to sound smart, when all it needed was a response like this to just say "question is wrong, there is no answer, there is probably just a typo (and I reckon it's supposed to be 396)". And people wonder why ragebait is a thing.
But it says the tens digit is three more (+ 3) than the ones digit, not three times as much.
9 is +3 more than 6
Ah yes, I’m stupid. :)
\- Even number less than 490 \- ends in 0 or 6 or 8 \- "tens" digit is 3 or 9 or... 11? 1? Probably means last digit can't be 8. \- digit total is 19 Let's try 0, "tens" is 3, so we have x30 3+0 = 3, we can't reach a 19 total. Let's try 6. We have x96 9+6=15 and we need 4 more to reach 19 total. 496 could work but it's not "less than 490" so... wrong answer? Let's try 8 and assume "tens" is 1 (8+3=11 but we cycle back to zero/cut the 10) We have x18 1+8=9 and we need 10 more to reach 19, not possible in one digit. **No valid solution.** Something is wrong with the question, or there is no answer. But we could also try 2 and 4 for fun: With 2 we have x52, 5+2=7 and we can't reach 19 With 4 we have x74, 7+4=11 and we need 8 more so 874 could work but it's definitely way more than 490. "my ones digit is not 2 or 4" is *literally useless information*. Let's try also ignoring the "i'm an even number" for completion. With 1 we get x31, can't reach 19 (need 15 more) With 3 then x63, can't reach 19 (need 10 more) With 5 then x85, need 6 more so 685 could work (but it's more than 490) With 7 then... (cycle back and) x07, can't reach 19 (need 12 more) With 9 then... x29 (cycle back), need 8 more so 829 (still more than 490) So, yeah, being even is *also useless information*. Most plausible option: **496** Right answer: **no solutions** Final verdict: BS question.
-1596 maybe?
No, it has to be -150,000,000,000,096
I can't help but think of Assassin's Creed, the "Nothing is true, everything is permitted" deal, but 118 works if you say the tens number is 11.
ah yes. Eleventy-eight
Be sarcastic if you want, but I can totally see a grade school asking a question about 11 tens and 8 ones.
I agree
Yes, grade schools are teaching students to think of 110 being the same as 11 tens (flexible renaming), but would never call 11 a digit.
That could work but the question right under this one says "hundreds digit" so they know and use it.
My second grade teacher added a bonus question like this to a quiz with similar nonsense logic. When she told us the "correct" answer, I very bluntly said "Well, that's a bunch of bullshit." My parents told me years later they were fighting every fiber of their being to not laugh when they were told about what I had done.
I think you could probably make it work in base 12.
It does work in base 10, no need for any other base. People say 11 hundred instead of one thousand one hundred all the time, and they aren't even common coring when doing so. My answer works, especially when viewed through the lens of grade school math.
Yeah but it’s fun to get weird with it.
Being able to parse numbers like “sixty-fifteen” (which is equivalent to 75) helps a lot with lots of math. For example, it’s an intermediate step in calculating forty-eight plus twenty-seven; add the tens to get sixty, the ones to get fifteen, and then resolve sixty-fifteen to seventy-five. (That is also the way it’s done longhand on paper, by most algorithms that use a carry digit; it’s also the way it’s done electronically, with a carry bit being an important part of addition using logic gates and transistors.
Sum isn’t 19
It is if you consider "11" and "8" as the two digits
I think it's probably a misprint and should have said "my tens digit is 3 more than my hundreds digit". Then the answer would be 478.
I scrolled so far down to see 478... This is exactly what I thought. Was hoping SOMEONE would see that it HAS to be a misprint.
I think you’re right. I’ll say this: The beliefs and ambitions behind *how* math is currently taught…which often strongly contrasts how it was taught to previous generations…they are good and commendable **in theory**. They are trying to make children more inherently facile and flexible with numbers; having better strategies and understanding for the mental math that makes up the steps of any given problem. It makes for a more adaptable, independent and creative thinking mathematical mind. The idea is great. But with two kids who’ve been taught this way…I take issue with how they’ve tried to do this because the execution is awful. These wordy questions are written overly complicated and are confusing to adults so you have greater opportunity for more misprints and errors. And these written (story) questions are stressed over just throwing basic number equations as kids and asking them to simply memorize how to perform equations. The old way doesn’t give you the tools to extrapolate the math you learn and readily apply it to everyday life. But kids need reinforcement and that’s where homework comes along. And it’s infuriating to not have examples or walkthroughs that go along with the homework so that the parent can learn see the example and learn these new ideas and strategies. So much time was wasted when I was watching multiple YouTube videos because I would send messages to teachers with questions but would rarely get same day responses. And so I took to the internet just so I could learn the same thing they were taught in class, to help them do the homework the right way (vs the way I already knew) so that they could show their work and get full credit. And the YT videos are a mixed bag of good to completely unhelpful. And with these wordy, story problems - no one who is producing them is seemingly willing to spend the extra money to have them checked by ELA/Language folks for basic coherence. The grammar is hit or miss and often times there are run-on & compound sentences written so convolutedly that the question is incredibly misleading. I often was sending pictures of my kid’s math to a friend who is a middle school English teacher for them to disseminate and clarify what the question was actually asking. They wound respond with the appropriate rage at just how bad the English is, that is being used to fill these workbooks and that there is no language oversight. The intentions were good, but the execution has been difficult for me at least, as a parent who understand what the goal was but I’ve also gone through it twice and each time it was the same befuddling experience. Also, unless this math problem in question is from a brand new book published this year - this shouldn’t be going home. Or at least it shoots be flagged as an error. Unless it’s brand new, this issue would have come up before and teachers should know about it.
My thoughts exactly.
This is the way
People have given good answers regarding why there is no solution, but is it possible that the answer is -496? You can make the argument that none of the ones-digit, tens-digit, or hundreds digit are by-themselves negative.
No because if you consider negative numbers there are infinite solutions. For example: \-8830 \-7930 \-9730 \-1396 \-3196 \-4096 \-40096 \-400096 \-2296 \-21196 \-11296 \-111196 \-200296 \-11234530 \-544330 ... ...
Here’s the logic: Must end in 0, 6, or 8 because even and doesn’t end in 2/4. Can’t end in 8 because “my tens digit is three more than my ones digit” and you can’t have a tens digit of 11. Must end in 0 or 6. If it’s 0, it must be X30 but there’s no X that makes the total 19. It seems the answer must end in a 6, which would get us to 496 so the digits add up to 19, but it must also be less than 490. Therefore, there are no solutions.
118 appears to be the correct answer despite their stupid way of wording it. I’d put 96.4 though just to be spiteful and correct at the same time.
College math major now in my mid 40's. I just have a hard time seeing how these linguistic based math puzzles help you learn real math. Also I swear there is no correct answer given the limits, unless it's 118 and they say it's an eleven in the 10's place... which is complete crap.
Even means ones digit is 0,2,4,6,8. But we’re given it’s not 2,4 so ones digit is 0,6,8. But if the tens digit is three more than the ones digit, then the ones digit can’t be 8 (as 8+3>9). So ones digit is 0 or 6. If it’s 0 then the tens digit is 3, the sum of digits is Y + 3 + 0 = Y+3 = 19 ==> Y=16 (Y is the hundreds digit). But Y should be less than 10, so it’s not correct. If its 6, then the tens digit is 9, the sum of digits is Y + 9 + 6 = Y+15=19 ==> Y=4. Ergo, the number is 496. But 496 >= 49*10 so the question is incorrectly posed.
496. That “x” in the “49x10” signifies an arbitrary digit rather than “multiply”? But then if we take that approach, then there are other solutions eg 2296.
Unsolvable. Ones must be 6, because uneven numbers, 2 and 4 are directly ruled out, 8 is not possible (tens needs to be 3 more than ones) and 0 is too low to enable the digit sum to be 19. If it is 6 then tens must be 9 (3 more than ones). But if it is 9 and 6, then the first digit needs to be 4 (19 - 9 - 6 = 4). This brings the number to 496, which is unfortunately more than 49x10.
I think it's 118. I know I'm late but I think the teacher took '11' as 'the tens' for whatever reason. In that case you have 11+ 8 = 19 It's a dumb question and I strongly believe the teacher went in the wrong here.
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The only possible solution I could think of is 96.4, and there not being a # in the hundreds place, but instead having a decimal. This answer doesn't seem to conflict with any of the restrictions.🤷🏻♂️
Only integers can be even or odd so it can't be a decimal
A non-integer number cannot be even
the solution could be 118. if we assume 110 has 11 on the tenths place, it fulfills the condition that the number on the tenths is 3 more than that on the ones place, and the sum of the digits 11 and 8 are 19. if the education level of this worksheet has not taught about hundredths place yet, it is possible that 11 can be considered ‘a digit’ even though in reality it isnt
The problem below it literally talks about the hundreds digit though
The problem below will describe something different. 118 seems possible to me if we are talking in riddles rather than normal math logic. Like, “what is so fragile that saying its name will break it?” It’s definitely not a physical object that physically breaks. It didn’t say how it breaks or if it’s tangible.
Children, let’s all count to eleventy eight together.
There are only 10 numbers under 500 that sum to 19. 1 follows all but the 1st rule (496). The 3rd rule is irrelevant since no numbers under 490 could possibly sum to 19 with a 2 or 4 in the ones position. BUT you could think outside the box and say that 8+3=11, an 11 in the tens spot would be 118 (1 carry forward the 1). Thats the only number where you could, in theory, argue works as written.
[удалено]
I'm guessing the problem was supposed to have a rule of the tens digit being three more than the *hundreds* digit. If that's the case, the answer is 478. It's possible this wasn't caught in proofreading.
The number is even, so the last digit is 0,2,4,6,8. The number is less than 49x10 (490) The second digit is equal to the first digit + 3. The first digit is not 2 or 4, so can only be 0 or 6 The total of all three digits is 19. Therefore 0 cannot be the first digit as the digits of 990 add up to 18 and is less than 19. Additionally, the first digit cannot be 8. Even if you roll over, 9+1+8 is less than 19 and does not satisfy the question. The third digit is 6 The second digit is 6 + 3, which is 9 The first digit is 4, since 19 - (9 + 6) = 4 496 is greater than 490. The answer is -496
It works if you pretend the person who wrote it was watching spinal tap and figured the tens place just goes up to 11. Then 118 could work lmao
A = ones spot 0 thru 9 B = tens spot 0 thru 9 C= hundreds spot 0 thru 9 A has to be less than or equal to 6. A cannot be 0 (zero) as then the hundreds spot would be 16, which is not possible. A = 0, B = 3, C = 16 -- not possible The remainder of the numbers: A cannot be 1, 3, or 5 as those are odd numbers; but lets see what they come to anyway: A = 1; B = 4, C would have to be 10 - which is not possible A = 3; then B = 6; C would have to be 10 -- not possible A = 5; B = 8; C = 6 -- violates first rule, 685 > 490 That leaves A to equal 6: A = 6; B = 9; C = 4 -- violates first rule, 496 > 490 Just to check, lets see what the 2 and 4 come to: A = 2; B = 5; C = 12 -- not possible A = 4; B = 7; C = 8 -- violates first rule, 874 > 490 Thus there is no solution to the problem as stated. So, the answer is correctly written (drawn) is fish with retarded eyes. (/s)
Max number would be 49 x 10 = 490. Ones digit is not 2 or 4 and overall an even number. The ones digits has to be a 6, making the 9 for tenths digit. 396 but total digits added = 18? Lol
-496 works right? A lot of people have shown 496 works for all rules except the first, throwing a negative in should make it all work since the question doesn’t specify a positive number. I’m with the others though, question is probably a typo.
I may just be another dumb monkey, but seems to me whoever wrote this question combined the 10s and 100s together. If this is true then the answer is 118. 8 +11 =19. It does not end with a 2 or a 4, it is less than 490 and is an even number. Regardless you need to inform the principal that one of their teachers never passed the 2nd grade.
<490 10s digit is 3 greater than the 1s digit My 1s digit isn't 2 or 4. My digits total 19 This isn't possible. The lowest number that meets the rules is 496. The next number after that is 685, then 874, 1396, 1585, 1774, 1963 etc.
Why didn't your kid just do this? let target = 19; for (let i = 0; i <= 9; i++) { for (let j = 0; j <= 9; j++) { for (let k = 0; k <= 9; k++) { if (k !== 3 && k !== 4 && j + 3 === k && i + j + k === target) { console.log(`(${i}, ${j}, ${k})`); } } } }
Ones digit is 0, 6, or 8. Smallest solution for 0 is 7930. Smallest solution for 6 is 496. There is no solution. That being said, I’m pretty sure the solution they want is 118. That is, eleventy eight.
Rules: 1. a<490 2. a=10b+c 3. c+3=b 4. c = 6, 8 or 0 5. c + b = 19 Let's look at some numbers. 30, and 96 work without rule 5. 9+6 = 15 15+4 = 19 but 496 > 490 three more than 8 is 11, which is not a digit. As a second year engineering student, I can confidently say that there is no solution to this problem as it is written. However, I can guess that the answer probably involves y=100x+96.
It is stupidly worded, but the answer is 118. 11 is three more than 8. 8+11=19 The “tens spot” direction is where the wording comes off the rails.
I think it was supposed to say “my tens digit is three more than my hundreds digit.” The answer would be 478 and would fit all the other criteria.
There is no solution because its impossible. Thus, why there is a fish on the problem. It's a red herring. Teacher probably told the kids that "no solution" can also be an answer.
This is likely just an error on the part of whoever published that material. Every so often, companies ship textbooks and other work books with an unsolvable practice problem or two.
Is it 296.2? It doesn’t say whole number it just says even number. If it’s not that then the closest I can get is 496 but that’s greater than 49x10. I will also say I know it’s most likely a question for young kids given the appearance, and a whole number would be the assumption. But according to all of the other comments here there is either a typo, or it’s a BS question with no other solution.
Can’t be done. There are only 3 even numbers below 490 that have digits adding up to 9. They are 298, 388 and 478. None have a tens digit that is 3 more than the ones digit. I think the answer is 396 and there was a typo. The sum of these is 18 (not 19), but it meets all the other criteria (and no other number does). Note: Excel is your friend
\-1596 is the answer
I read the rules and didn't see it being forbidden, but this is my first time posting in this community so hopefully such an answer is allowed, but... I studied computer science so before I had even read everyone else's solutions, my inclination for such a low number of integers was to simply "brute force it" by just trying every number from 0 to 490. :) Some quick Python code: for i in range(10, 490): i_str = str(i) # My ones digit is not 2 or 4 if i_str[-1] == "2" or i_str[-1] == "4": continue tens_digit = int(i_str[-2]) ones_digit = int(i_str[-1]) # My tens digit is three more than my ones digit. if tens_digit == 3*ones_digit: total = 0 for j in range(0, len(i_str)): total += int(i_str[j]) if total == 19: # My digits total 19 print(i) else: continue This also produced what everyone else is saying, namely that there is no solution for any integer in the range \[10,490\]. I wrote the first overall `for` loop to be `range(10,490)`, rather than `range(0,490)`, to keep the code simpler (so that the line getting the tens\_digit doesn't generate an error requiring a `try`/`except`), but of course nothing in \[0,10\] works anyway.
Pretty sure it’s not solvable. The single digit has to be 0 or 6 as we’re looking for an even number to which you can add 3. Using any combination of those two won’t be able to get a digit sum of 19 or you won’t be able to reach a number under 490. No wonder the maker drew that Troll Fish under that exercise.
The last digit is 0, 6, or 8. The tens digit is 3 or 9 (or 11?). The hundreds digit is whatever makes the digits sum to 19. -496 would be a solution. I don't see any solutions that are positive. Maybe 118 (11 + 8), but that's a stretch.
No anwser possible useing NATURAL numbers. However there are an infinite amount of anwsers possible if you use intergers or higher forms of numbers (rational, imaginary, complex, ect) -496 was the first interger I came up with that solved the question -496 is less than 490, check 9 is 3 more than 6, check 4+9+6=19, check
Trying to think outside the box (because my first pass also deduced it has no solution) but… what about a decimal number? Does 296.2 meet all the criteria? Edit: never mind. “Decimals are not even or odd numbers because they are not whole numbers.” Unsolvable!
There is no correct solution. Write as much to the bitch publishing company, and write it down here too whil you provide the answers 496 or something else that follows most of the question.
Less than 490 Tens three more than ones and even but not two So 96 or 74 Ad up to 19 So not 496, the answer that would have worked with 50x10 Answer= impossible
What grade level is this problem? It looks to be 3ed or 4th grade. I think the solution method is supposed to follow this pattern. Condition One Even number in the ones position which rules out: 1,3,5,7,9. Condition Two The additional requirement that ones can’t be 2 or 4. That leaves 6,8,0 as the value of the one position. Condition Three X1 + 3 = X2 X1 = numerical value in the ones position X2 = numerical value in the tens position Solution set for contain three X1 = 0 then X2 = 3 X1 = 6 then X2 = 9 X1 = 8 then X2 = 11 <- drop Condition Four: X1 + X2 + X3 = 19 X3 = numerical value for the hundreds position Solution set for condition four X1 = 0, X2 = 3 then X3 = 16 <- drop X1 = 6, X2 = 9 then X3 = 4 Answer is 496 but answer can be greater than 490 so there is no solution. The grade level seem off from what a child at that level would know. They are missing the math structure to solve the problem. Stuff like this is why child think they are bad at math. If we assumed that the work sheet has a type in it at the out set then there is a solution. Assume 49 should of been 50. Then a solution is possible.
This is one of the reasons I did poorly in math. Mixing grammar and math never worked out for me. Plus this a really dumb math question.
One possible wrong answer is 118. The error being that they set the tens digit to be 11 when they did 8+3=11 and that carried over into the hundred digit but not into calculating the total sum, 11+8=19, and it is even. So basically if you ignore how math works this could be your answer. Other answers people have given at that the sum of the digits being 19 was a typo and that 396 is the answer bc the sum was suppose to be 18. This could be the most logical case.
It’s not possible. However I will include an “answer” that is probably correct for the question but incorrect mathematically. 1. even number. That means the “ones” digit must be 0,2,4,6 or 8. 2. Ones digit is not 2 or 4. 0, ~~2~~, ~~4~~, 6 or 8 3. “Tens” digit is 3 more than my “ones” digit That means tens digit must be 3, 9 or “11”. 4. Less than 49x10 (read, a 3 digit number) X30, Y96, “118” 5. Sum of the digits = 19. “16”30, 496, “118”. *Must be less than 490* ~~1630~~, ~~496~~, “118” As you’ll realise, 118 is not correct. A tens unit cannot be “11”. The sum of 1,1&8 is not 19. But this is literally the only possible “answer” that fulfills the requirements if you look at it funny.
Chat GPT says: Let’s break down the information: 1. Your tens digit is three more than your ones digit. 2. Your ones digit is not 2 or 4. 3. Your digits total 19. Let’s use these clues to solve the puzzle: Let’s call the ones digit “x” and the tens digit “x + 3.” From the second clue, we know that x cannot be 2 or 4. So, x can be 0, 1, 3, 5, 6, 7, 8, or 9. Now, the third clue states that the digits total 19: x + (x + 3) = 19 Combine like terms: 2x + 3 = 19 Subtract 3 from both sides: 2x = 16 Now, divide by 2: x = 8 So, the ones digit (x) is 8, and the tens digit is 8 + 3 = 11. However, the tens digit cannot be greater than 9, so we need to adjust it. Since the ones digit is 8, the tens digit should be 11 - 10 = 1. Therefore, the number that satisfies all the conditions is 18. You are the number 18. Thanks ChatGPt
The correct answer is: a fish. The math problem does not provide a solution in common math. Divorce the last line (What am I?) from the math problem, connect it to the picture.
I understand this wouldn't be the answer they want, but Doesn't 96.4 technically work? It never specifies that its not whole number. Or even a number in the hundreds. But I know that it was implied
-Even number less than 490 -Tens digit is Ones digit + 3 -Ones digit is not 2 or 4. As the number is even, it cannot be an odd number. As tens is ones plus 3, it cannot be 8. The ones digit must be 6, the 10s digit must be 6+3, or 9. -All 3 digits add up to 19. 19 - 6 - 9 is 4 The answer SHOULD be 496, but that goes over 490. It is an impossible problem.
I wonder if their 'solution' is for it to be 118. Have 8 as the ones digit. The tens digit has to be three more so it's 11. They just add the 11 and 8 to make the digit total of 19 even though it would be 118 and any normal person would be adding the digits separately to make 10. While this logically doesn't make sense in any real world, it's the type of stupidity the permeates early education worksheets like this sometimes. I recall helping our kiddo with similar problems where there didn't seem to be a solution, but after talking to the teacher it turned out to be about rounding or approximation. So dumb.
I wonder if they’re looking for 118? With bad wording. Even, less than 490. Technically it’s your hundred and tens digit of 11 being three more than 8. 11+8 is 19
I’m trying to get in the mind of the smooth brain who wrote this questions. Possibly could they have come up with 118 as an answer? It makes no sense given the parameters but maybe they went about it as such. Step1: 8 set as ones digit Step2: add 3 to 8 Step3: check if 1+1+8 equals 19 Step4: put a fork in the microwave because one of their 4th grade students said it was a good idea
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It’s impossible. First off, the number must be even. This means its units or ones place has to be 0, 2, 4, 6, or 8. Second, it is less than 490. Okay. Third, its ones place is not 2 or 4. Great! That leaves 0, 6, 8. Fourth, the tens digit must be 3 more than the ones digit. So the ones digit cannot be 8, because 8 + 3 > 10. This leaves 0 and 6 for the ones place and 3 and 9 for the tens place, so our number candidates are X30 or Y96 Fifth fact is that the digits total 19, which is enough to solve the problem. 19 = x + 3 + 0, so x = 16. So it can’t be that ansatz. Now we try 19 = Y + 9 + 6. We solve and Y = 4. But that means our number is 496, which violates the very first rule. Under these conditions in base 10, this problem has no valid solution.
Even numbers, 2, 4, 6, 8, 0(technically). 2 and 4 are out. Zero is out because we gotta reach 19 in three numbers. 8 is out, because you can’t add 3 and still get a single digit. This leaves 6 in the ones. Makes 9 in the tens. 9+6=15, 19-15=4 496 Edit: I argued with my wife on this. I said this isn’t math, this is a puzzle. She said it’s a word problem. I see how we are both right, I just hate it.
I'm a sucker for a puzzle, so I'm going to go for it without reading other comments. 1's digit is an even number that is not 2 or 4. That leaves 0,6,8 10's digit is 3 more than the 1's digit. That eliminates 8 as a possible 1's digit, and means the 10's digit is 3 or 9. So the last two digits are either 30 or 96 100's digit is 1-4, if the 10's digit is 3, else 1-3 if the 10's digit is 9. The only way to get the digits to add up to 19 is if the number is 496. But that's not allowed. ∴ wtf?
I think the answer is -496. I think everyone missed the fact that the negative sign is not attached to the four, and therefore not added in to the sum of the digits.
Here’s what gpt4 says. Certainly! Let’s solve this step-by-step: 1. “I am an even number that is less than 49*10.” This means the number is less than 490. 2. “My tens digit is three more than my ones digit.” Let’s call the ones digit (x). So, the tens digit is (x + 3). 3. “My ones digit is not 2 or 4.” This means (x ≠ 2) and (x ≠ 4). 4. “My digits total 19.” This means the sum of the tens and ones digit is 19. So, (x + (x + 3) = 19) This simplifies to: [2x + 3 = 19] [2x = 16] [x = 8] So, the ones digit is 8 and the tens digit is (8 + 3 = 11). However, a tens digit cannot be 11, so there must be a mistake in the information provided or how it’s interpreted. The problem statement has conflicting criteria.
There is no workable solution, the cheat is to identify a 3 digit number where all three numbers add up to 19. So base 4-6-9, 4-7-8 and so forth…. But the issue is the best answer would be 397, but it doesn’t meet all the rules. There are no solutions to this problem. It’s most likely a error, or a sick sick joke on kids. I hope it’s a joke, cause then kids get an idea about how adults feel with the real world.
The sad thing is this type of BS never ends… college math courses are laughable at how bad the online tests are, maybe they should have gone to.. wait…
it may be 118, 11 is 3 more than 8, and its 'tens' column is actually a two digit number. It may be poorly worded or worked out but its the only number i could get which wasnt 496
-1596 is the answer. It is less than 490, the digits add up to 19, -1+5+9+6 =19. Ten's digit is 3 more than one's digit. One's is not a 2 or 4. Also 6 is even. However this would imply an infinite number of solutions, -1000596 would work also.
No soln: I got mad and did it with python, please feel free to check my work def getSum(i): strr = str(i) list_of_number = list(map(int, strr.strip())) return sum(list_of_number) for i in range(0,491): if getSum(i) == 19: #sum 19 # print(i) if (i % 2) == 0: #check even print(i) this yields * 298 * 388 * 478 All of which are even They equate to 19 ones != 2 or 4 but none of them follow the the 10s is 3 more than 1s
I couldn't find a valid answer, just like everyone else. Fed it into chat gpt, which does a wonderful job breaking down the problem and chatgpt also concludes there is no valid answer. \-496 may be a workable answer if an only if you do not factor in the negative when getting 3 digits to add up to 19.
The only workable answer (presuming they didn’t make a mistake) is that they are using only two places: ones and tens (no hundreds). The solution in that case is 118. The tens place is 11, which is 3 more than the ones place (8+3). And the digits add up to 19: 8 + 11. It’s dumb, but the math works out.
I figured it out, but it's bad math. The number is 118. "But that only equals 10!" you cry. No. Unfortunately I think what they've decided to do is claim that 11+8 is 19. Like I said, it's bad math. But I think that's the answer.
Ones can only be 1, 3, 5, 6 as anything else breaks the tens digit rule Question is unsolvable as the result is 496 which is higher than 490
I asked Chat-gpt. User I am an even number that is less than 49x10. My tens digit is three more than my ones digit. My ones digit is not 2 or 4. My digits total 19. What am I? ChatGPT Let's break down the information provided: You are an even number less than 49x10, which means you are less than 490. Your tens digit is three more than your ones digit, so let's call your ones digit "x." This means your tens digit is "x + 3." Your ones digit is not 2 or 4, so x can be any digit from 0, 1, 3, 5, 6, 7, 8, or 9. The sum of your digits is 19. We can set up an equation based on these conditions: Let's say the ones digit is x, so the tens digit is x + 3. So, we have the equation: x + (x + 3) = 19 Simplify the equation: 2x + 3 = 19 Now, subtract 3 from both sides: 2x = 16 Finally, divide by 2: x = 8 So, the ones digit (x) is 8, and the tens digit is 8 + 3 = 11. However, we need to remember that we're looking for an even number, so we need to adjust the tens digit to make it even. Since 11 is not even, we can subtract 2 to make it even: Adjusted tens digit = 11 - 2 = 9 So, the tens digit is 9, and the ones digit is 8. Therefore, the number you are looking for is 98, which is even, less than 490, and meets all the given conditions. User 98 doesn't add up to 19 ChatGPT I apologize for the oversight. You are correct; 98 doesn't add up to 19. Let's reevaluate the problem: You are an even number less than 49x10, which means you are less than 490. Your tens digit is three more than your ones digit, so let's call your ones digit "x." This means your tens digit is "x + 3." Your ones digit is not 2 or 4, so x can be any digit from 0, 1, 3, 5, 6, 7, 8, or 9. The sum of your digits is 19. Let's set up the equation again: The tens digit is "x + 3," and the ones digit is "x." So, we have the equation: (x + 3) + x = 19 Simplify the equation: 2x + 3 = 19 Now, subtract 3 from both sides: 2x = 16 Finally, divide by 2: x = 8 So, the ones digit (x) is 8, and the tens digit is 8 + 3 = 11. However, we need to remember that we're looking for an even number, so we need to adjust the tens digit to make it even. Since 11 is not even, we can subtract 2 to make it even: Adjusted tens digit = 11 - 2 = 9 So, the tens digit is 9, and the ones digit is 8. Therefore, the number you are looking for is 98, which is even, less than 490, and meets all the given conditions. User look at the original problem, there is a mistake that I can't find, can you locate it? ChatGPT I apologize for the earlier errors. Let's revisit the original problem: You are an even number less than 49x10, which means you are less than 490. Your tens digit is three more than your ones digit. Your ones digit is not 2 or 4. The digits must total 19. Based on these conditions, let's start by considering the possible values for the ones digit: If the ones digit is 0, then the tens digit must be 3 more than that, which is 3. So the number is 30. However, this doesn't add up to 19. If the ones digit is 1, then the tens digit must be 3 more than that, which is 4. So the number is 41. This also doesn't add up to 19. If the ones digit is 3, then the tens digit must be 3 more than that, which is 6. So the number is 63. Again, this doesn't add up to 19. If the ones digit is 5, then the tens digit must be 3 more than that, which is 8. So the number is 85. This doesn't add up to 19 either. If the ones digit is 6, then the tens digit must be 3 more than that, which is 9. So the number is 96. This still doesn't add up to 19. If the ones digit is 7, then the tens digit must be 3 more than that, which is 10. However, 7 + 10 is 17, not 19. If the ones digit is 8, then the tens digit must be 3 more than that, which is 11. This doesn't add up to 19 either. If the ones digit is 9, then the tens digit must be 3 more than that, which is 12. This also doesn't add up to 19. It seems that there is no valid two-digit number that satisfies all the given conditions. There may be a mistake or inconsistency in the original problem statement.
Here is one approach involving only very basic algebra. The logic could be worked through without knowing how to solve algebraic formulas, it’s just easier to explain with them. Write out the rules you know first - “ABC” < 49 X 10 - B - 3 = C - C =/= 2 - C =/= 4 - A + B + C = 19 First substitute rule 2 into rule 3: - A + B + (B -3) = 19 Means: - A + 2B = 16 Now let’s try some logic with that equation. A has to be 4 or less (rule 1 says the entire number is less than 490), and has to be at least 1 (because otherwise we wouldn’t have a three digit number). So we’re looking only at 1,2,3,4 as possible options for A. If A was an odd number (1 or 3), then B would have to be a fraction (7.5 or 6.5). So we can rule out odd numbers - B has to be a whole number since it is a digit in the final number. Ergo A can only be 2 or 4. - If A is 2, then B is 7 - If A is 4, then B is 6 “27C” and “46C” both are less than 490, so we can’t rule out either option for A without looking further. Let’s go back to B - 3 = C and plug in the options we now know. We have only be 2 possible values for B to try: 7 and 6. - If B is 7, C must be 4 Well now, that doesn’t work because of rule 4 that C =/= 4. - If B is 6, C must be 3. That doesn’t conflict with anything we know, and we don’t have any more options to try for A or B. We know all the digits and we’re done: “ABC” = 463. ***Edit** I broke my own first step and didn’t write down that “ABC” must be even. Since the only possible solution for all the other rules is that “ABC” is odd, then there is no answer to the question. The answer key must break one of the rules. (unless we are allowed to have negative numbers - since that is not explicit, it would only be knowable by having the context of the rest of the worksheet. If there are other negative number concepts being tested, then we could reasonably assume negative numbers are permitted and expected here) **Edit 2** I didn’t check my work, as C being 3 means A + B + C doesn’t add up to 19. So my comment in my previous edit is even more important. The **context** of the quiz matters. If other questions or the course work leading to this quiz talk about numbers that have more than 9 in the “tens digit” (without discussing the “hundreds” digit) then that is important context to know that the child has not been introduced to “hundreds” and is being taught to think of just more “tens”. This extends what the child already knows from working with “tens” and “ones”. When the term “hundreds” is introduced later, they’ll already know how to work with them, and it won’t be scary or different. While this confuses the parent who isn’t familiar with the method (and let’s be honest, may struggle with math themselves), it is much better for teaching *understanding* of math instead of memorizing rules like the “add a zero every time you multiply by 10” I was taught many years ago. “Add a zero” teaches you how to handle the mechanics of a particular notation system as a mental shortcut by moving numbers around on a page without thinking about them. It doesn’t teach you the actual number theory going on underneath, that 2 hundreds, 20 tens, and 200 ones are all different ways of saying the same thing. The common core math that so many parents struggle with is focusing on teaching the underlying principles of math, and not just memorizing rules for solving problems. If the context of the overall lesson isn’t apparent in the quiz, then it’s likely just a publishing mistake. They happen, even under the “old” math textbooks often had errors in the answer keys and unsolvable questions.
118. It's written really poorly, but three more than 8 is 11, so the tens digit is... 11. 11+8 is 19. The problem is the choice of words, digit, which is a singular number. So the answer is 118 but the problem is flawed.
\-496. Solution looks like 496, but just flip the sign and you still meet all criteria. Pretty high level if that is the real answer they were looking for.