Each letter in the addition sentence below represents a ofdifferent digit. What are the values A, B and C? AA+BB+CC=BAC

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David Shabazi answered

Values for A, B, and C have to be within the 1-9 range, only single digits.

In AA+BB+CC=BAC, I know that whichever three numbers I pick for A, B, and C, they must be at least 100, since BAC represents three digits. So I can't assign 1, 2, and 3 for B, A, and C respectively, because that would only give me 66 and would not be the correct answer. The sum of A, B, and C would have to be AT LEAST 10 in order to get a valid (not necessarily correct) answer, because in that way, we'll have an answer that's three digits instead of two. So instead of 3, let's make it 7, since 1 + 2 + 7 = 10. I also know that I have to assign the 1 to B, because the answer will be in the 100-range, and B is the starting number in BAC.

So let's say B = 1, A = 2, and C = 7. Let's plug that in.

AA+BB+CC=BAC

22+11+77=110

Unfortunately, that's not right, because the values we assigned for A, B, and C don't match up with what we got for BAC (which is 110). Let's bump up the 7 to an 8.

AA+BB+CC=BAC

22+11+88=121.

Nope, still not right. But let's think about this. Right now, we have the correct value for B, which is 1, as well as A, which is 2, but not for C. Right now, we assigned 8 for C, but we currently have it at 1. So it's 7 units away from being 8, what can we do? Well, we can increase the value for A by 7 units, because if we set A = 9, that will change the C value from 1 to 8, and the A value from 2 to 9, so everything should work out. Let's try it:

A = 9, B = 1, C = 8

AA+BB+CC=BAC

99+11+88 = 198

And there you go. So you mainly have to experiment with it, but you can also get some information and clues to what values you can use by just studying the equation and seeing what's going on. Like knowing that you can't use the numbers 1, 2, and 3 in this problem because adding 11, 22, and 33 together wouldn't give you a three digit number for BAC. (Truthfully, I was a little puzzled at first when I saw this question, but it turned out to actually be pretty easy to solve.)

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