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Digits

by yellowho » Fri Mar 11, 2011 12:36 am
I don't understand the answer here, 28+82 seems to work so does 37,73 etc...
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by sanju09 » Fri Mar 11, 2011 1:20 am
yellowho wrote:I don't understand the answer here, 28+82 seems to work so does 37,73 etc...

Remember this for ever and a day that the sum of a two digit number and the number obtained by reversing its digits is always divisible by 11 that can never exceed 198 or the hundreds' if any, will only be 1.

Statement 1 shows that such a sum has its tens and hundreds as same; A is definitely equal to 1. And if A = 1, and 11C is divisible by 11, C can only be zero. Since B + A or B + 1 ends in a 0, B must be [spoiler]9

Statement 1 alone is sufficient.[/spoiler]
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by Stuart@KaplanGMAT » Fri Mar 11, 2011 3:17 am
yellowho wrote:I don't understand the answer here, 28+82 seems to work so does 37,73 etc...
Hi,

you have to remember that all of the "A"s have to be the same.

So, when you add 28+82, you get 110. However, by choosing 28 and 82, you've chosen A=2 and B=8. Since the sum should be "AAC", and since A=2, 110 does NOT equal "AAC".

The max sum of two 2 digit numbers will be 198 (99 + 99), so (1) tells us that A, the hundreds digit of our sum, equals 1. Once we realize that A=1, the only possible value for B that will make A and B sum to a 3 digit number beginning in "11" (i.e. "AA") is 9.

You can test it out:

19 + 91 = 110... works!

18 + 81 = 99... fail!

17 + 71 = 88... fail!

The sum keeps getting smaller, so no need to test any other values of A and B.
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