Re: Define A = The sum of digits of the number 7^100 and B = The sum of
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15 Nov 2021, 14:43
This question is not, as suggested above, a 650-700 level question. It's more like a 900-level problem, and it's not something you'd ever see on the GMAT. The solutions above are incomplete; they answer the question assuming we'll be summing digits until we get down to a single-digit number. That is what will end up happening here, but I don't think that's obvious just reading the question.
When you divide a number by 9, the remainder you get will always be equal to the remainder you get when you divide the sum of the number's digits by 9. So, for example, if you divide 3725 by 9, the remainder you'd get will be equal to the remainder you get when you divide 3+7+2+5 = 17 by 9, so it will be 8. That's a fairly straightforward consequence of the divisibility test for 9 (where you sum the digits to test if a number is divisible by 9), just adapted to remainders instead of divisibility.
If we now consider powers of 7, as in this question, the number 7^3 = 343 will give us a remainder of 1 when we divide by 9 (which you can see by summing its digits, or noticing 342 is a multiple of 9, and 343 = 342 + 1). If you know modular/remainder arithmetic (which is never tested in this way on the GMAT,) you'll know that we can 'multiply remainders' if we're always dividing by a fixed number, so if the remainder is 1 when we divide 7^3 by 9, the remainder will be 1^33 = 1 when we divide (7^3)^33 = 7^99 by 9, and when we divide 7^100 = (7)(7^99) by 9, the remainder will be (7)(1) = 7.
As a consequence of the second paragraph above, since the remainder is 7 when we divide 7^100 by 9, when we sum the digits of 7^100 (to get the number "A" described in the question), we must also get a number with a remainder of 7 when we divide by 9. There's no way to work out precisely what A is equal to here without a computer, but we do know we're adding the digits of 7^100, which is less than 10^100, a 101-digit number. So we're adding fewer than 101 digits, each of which will be at most 9, and the value of A cannot possibly be bigger than 900. When we add the digits of A to get the number called "B", we'll be adding three digits that are at most 9, so we'll get a number less than 27, and finally when we add the digits of this number to finally answer the question, we're adding a tens digit that is at most '2' to a units digit, so we must get something 11 or less. But we also must get something with a remainder of 7 when we divide by 9, and the only such number is 7 itself, so that must be the answer to the question.
You could technically answer the question more quickly just by estimating how large the answer can be (it needs to be very small because we add the digits of our numbers so many times, and only one answer choice is reasonable; note that E is clearly wrong since the answer is obviously calculable somehow) but then you're not proving the answer is what it is, only eliminating the answers that can't be right. Regardless, this is not even remotely close to resembling a real GMAT question, even if a couple of the concepts in the solution might be useful on rare occasions.