We invite you to test your GMAT knowledge for a chance to win! Each week, we will post a new Challenge Problem for you to attempt. If you submit the correct answer, you will be entered into that week’s drawing for a free Manhattan GMAT Prep item. Tell your friends to get out their scrap paper and start solving!

Here is this week’s problem:

For positive integers k and n, the “k-power remainder of n” is defined as r in the following equation:

n = k^w + r, where w is the largest integer such that r is not negative. For instance, the 3-power remainder of 13 is 4, since 13 = 3² + 4. In terms of k and w, what is the largest possible value of r that satisfies the given conditions?

To see the answer choices, and to submit your answer, visit our Challenge Problem Showdown page on our site.

Because of the error in last week’s Challenge Problem, we do not feel that it would be fair to announce a winner for last week’s problem. Therefore, we will be awarding TWO winners for this week’s problem. Now that you have double the chance to win, you have no excuse not to enter!

Discuss this week’s problem with like-minded GMAT takers on our Facebook page.

The weekly winner, drawn from among all the correct submissions, will receive One Year of Access to our Challenge Problem Archive, AND the OG Archer, AND Our Six Computer Adaptive Tests ($92 value).

Other Posts You May Like:

• Challenge Problem Showdown – May 14th, 2012
• Challenge Problem Showdown – May 7th, 2012
• Challenge Problem Showdown – April 30th, 2012