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Re: GMAT prep DS- Remainder [#permalink]
29 Jun 2010, 18:33
This post received KUDOS
bunuel's solution is certainly the quickest way to solve. (Of course, it would be (k+1)^3 = k^3 + 3k^2 + 3k + 1, and since each of the "k" terms is divisible by k, the remainder is 1).
But if this deduction/concept didn't jump out at you, you could also solve fairly quickly by picking numbers. If we pick different integer values for k and always end up with the same remainder for n/k, then we can trust that (1) is sufficient. On the other hand, if ever we get different remainder values, we know at once (1) is insufficient:
(1): let k=2. Then, n = (2+1)^3 = 27. 27/2 leaves a remainder of 1. ----let k=3. Then, n = (3+1)^3 = 64. 63 is clearly divisible by 3. Thus, 64/3 leaves a remainder of 1 again. ----let k=4. Then, n = (4+1)^3 = 125. 124 is clearly divisible by 4. Thus, 125/4 leaves a remainder of 1 again.
Re: What is the remainder when the positive integer n is divided [#permalink]
03 Jan 2015, 09:47
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