If AA + BB + CC = ABC, What Are A, B and C?

If AA + BB + CC = ABC, what are the values of A, B, and C? All the digits are distinct and positive. Watch the video for a solution.

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[Selicre]

You can also solve this algebraically, and it works out to be very neat:
11(a+b+c) = 100a + 10b + c
89a = 10c + b
As all the variables are integers and in the range of [1, 9], a can only be 1, and so the right-hand expression turns into "cb" as a number, which gives you the solution of 198.

[patrickwienhoft]

AA means 10*A + A = 11*A
So the "actual" equation is 11A + 11B + 11C = 100A + 10B + C.
11A + 11B + 11C = 11(A+B+C) is divisible by 11 and the result is a 3 digit number.
A 3 digit number that is divisible by 11 has the property that the first and third digit add up to the 2nd digit, so A + C = B.
You can put that in the equatoin to get
22A + 22C = 110A + 11C
11C = 88A
C = 8A
As all numbers are smaller than 10, so 8A needs to be smaller than 10 as well. So A = 1. C = 8 then. As A + C = B, B = 9.
11 + 99 + 88 = 198.

edit: after watching your solution, I feel mine is more elegant. I feel smart now, haha :D

[philippepoulin]

Easy.
Units: If A + B + C = xC, then A + B = 10.
Tens: If A + B + C + 1 = xB, then C + 1 = B
Then the hundreds leftover is 1, so A is 1. If A is 1, then B is 9 (1+9=10), which means C is 8 (8+1=9).

[justaskurtinaitis]

I just wrote equation like this: 11A+11B+11C=100A+10B+C
From there its quite easy to conclude the answer.

[kamoroso]

My solution:
10a+a+10b+b+10c+c=100a+10b+c
11a+11b+11c=100a+10b+c
b+10c=89a
a=1
b=9
c=8

[notoriouswhitemoth]

I figured it out from the other end - starting with the fact that each summand is a multiple of 11, and that, as such, the sum has to be a multiple of 11. Given that, I started listing three-digit multiples of eleven, with an emphasis on those with three unique digits.

[BatPhil]

You don't need to consider 2 options - start with the units:

A + B = 10 (no other possibility if it is to generate C in the units column)
B = C + 1 (from the carry over to the tens column)

So, A + B + C + 1 < 20 gives A = 1 (it can't be more than 19 - think about it!)
This then leads to B = 9 and C = 8

Quite simple really and can be done in your head.

[cloverpepsi]

A=1, B=9, C= 8.
Here we go-
A+B+C= C with something leftover. By this logic, C in the second row MUST be smaller than B. You must have different values because each letter is a different number. You have A leftover in the hundreds, so that means you had A leftover in the tens. Therefore, C+A=B.
Using these two equations, we can extrapolate that A+B+C = A//C (// means concatenation.) Subtract C from each side, A+B= 10A. Subtract a from each side, B= 9A
Here's what's been established.
C+A=B,
B=9A.
Therefore, C+A=9A, C=8A.
The only multiples of 8 and 9 between 1 and 9 are, well, 8 and 9. Therefore, C=8 and B=9. And, obviously, 1=A.
Plug this in, you get
11
99
+. 88

198

Perfect!

[calcustom]

I found the max possible (99, 88, 77 = 264) then realized the only possible way to get 9's and 8's in the final number was to change the 77 to 11. I planned to take that logic down for each number, but luckily that worked immediately.

[JohnLeePettimoreIII]

I got this one. I don't always figure out the solutions, but i still enjoy learning these new ways of arriving at answers.

[DaKnightsofawesome]

well, if the last digit must be c, then that means that a + b = 10. Since we know that the second column is a + b + c + 1 >= ab, we know that 11 + c = ab. So 20 >= ab >= 11. there's only one number in that range where the digits add to ten: 19. so a = 1 and b = 9. now we know that 11 + c = 19 so c = 19 - 11 so c = 8. test it out: 11 + 99 + 88 = 198.

[potato-hjnm]

I completely mksunderstood this I thought you had to times them together.

[AREmrys]

I varied the procedure slightly in the third step: Knowing that A+B+C leads to C in the right column and at the same time to B in the middle column, we can conclude that B = C+1, as we only have to add the carryover from the right column. That gives us the two triples 11/99/88 and 22/88/77, and by checking them we see that only the first one is valid.

[iWantToDetonate]

The "logical" approach to solve A is much easier. If we just assume B and C are equal to 8 and 9 (sums up to 17), for A being able to be 2 it has to be 3 at least, which is impossible. So we can rule out the case A=2 without even looking at the other numbers. Now we know that A+B+C=AB is actually 1+B+C+1=10+B. Subtracting B and 2 gives us C=8. Solving for B is now trivial.

[dominikstepien]

My solution
11A+11B+11C=100A+10B+C
A+B=10 (so last number stay C)
A+C=9 (B will stay if it is equal to ten, but there is 1 overflow)
And after calculation you get answer A=1, B=9 and C=8.

[erinzoretich]

I liked the video but thought your explanation was too confusing. Why are we even working through the possibility of carrying a 2? Many other commenters have great solutions that use better algebra then I would have thought of. But here was my logic:

Because C passes right down to the sum, A + B must add together and end in 0, so it must be that A + B = 10. Two different numbers valued 9 or less cannot ever equal 20 or more. Because we now know that A + B is equal to 10, A + B + C must be equal to or greater than 10, so it's absolutely going to carry something over to the next column, the tens column.

Now, three different numbers valued 9 or less can absolutely equal more than 20. We could be carrying a 1 or a 2 here. BUT! We already know A + B + C = C (all units) from the ones column. We can surmise that if C were large enough to make A + B + C equal to 20 or more, it wouldn't be the UNIT that would be left over in the ones place of that very same sum. The only two possibilities for a three number sum are 9 + 8 + 7 = 24, and 8 + 7 + 6 = 21. In neither of those does the ones place of the sum have the same unit as ANY of the possible addends. We're carrying a 1, not a 2.

Next, the tens column of our problem now essentially reads "1 + A + B + C = B". You can simplify that to 1 + A + C = 0 or (we're in units, remember) 1 + A + C = 10, further to A + C = 9.

We just figured out A + C = 9. Looking at our tens and hundreds column, there's no way even with the carried 1 from the ones into the tens column those numbers can equal 20 or more. A + C = 9, remember? 9 + (carried) 1 + B cannot equal 20 or more. B is a unit, and can only be 9 at most!

Let's go ahead and assume A = 1 and see how it all plays out.

A + B = 10
1 + B = 10
B = 9

A + C = 9
1 + C = 10
C = 8

Go back, does it work?

Yup! 11 + 99 + 88 equals exactly 198. That wasn't so hard after all.

[ngarcia]

Also, since 28C is guaranteed to be greater than 264 (the highest possible sum you worked out in the beginning), you know A = 2 and B = 8 is not a solution.

[Coldheart]

I got A = 1 straight away. Going by the units column, A + B + C = xC, subtract C from both sides gives us A + B = x0, and the largest number ending in 0 possible is 10. Which immediately gives B = 9. It saved me working out an impossible solution.

[NeatNit]

This was an especially easy/simple one (nothing wrong with that).

If we start with the units digit, A + B + C = C. That means that A + B = 0, or 10, or 20 etc. But two single-digit non-zero numbers can only do 10.
Therefore: A + B = 10
Now the tens. We carry over the 10 from the units, and we have: A + B + C + 1 = B. Replace A+B with 10 and we get: 10 + C + 1 = B, but the 10 is irrelevent.
Therefore: B = C + 1
As for the hundreds, we already know that A+B+C+1 gave us a single 1 to carry over.
Therefore: A = 1
We have A now so we can use it in the previous equations:
1 + B = 10
Therefore: B = 9
9 = C + 1
Therefore: C = 8

And we got the correct (and only) solution, 198, without any guesswork and IMO in a much more straightforward way than yours.

I was honestly expecting a catch or a different solution (different digits) to turn up in the video, since that happens to me often with your videos.

[Huntracony]

If you tell me something is a math problem, I assume "AA" means A^2, not 11A.