Thank you for using the timer!
We noticed you are actually not timing your practice. Click the START button first next time you use the timer.
There are many benefits to timing your practice, including:
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
Hello from the GMAT Club BumpBot!
Thanks to another GMAT Club member, I have just discovered this valuable topic, yet it had no discussion for over a year. I am now bumping it up - doing my job. I think you may find it valuable (esp those replies with Kudos).
Want to see all other topics I dig out? Follow me (click follow button on profile). You will receive a summary of all topics I bump in your profile area as well as via email. _________________