Challenge Problem Showdown – January 14, 2013


challenge problem
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 all non-negative integers x and n such that 0 ≤ x ≤ n, the function fn(x) is defined by the equation fn(x) = xn“x. The smallest value of n for which the maximum of fn(x) occurs when x = 4 is

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

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 GMAT Navigator, AND Our Six Computer Adaptive Tests ($92 value).

  1. Ziya January 16, 2013 at 2:16 pm

    Yeah you are absolutely wright !

  2. Rohan January 15, 2013 at 12:11 am

    I too agree with salman, since x<=n and both are +ve, Fn(4) = 4^(n-4) will keep increasing as n increases or in other word.. MAX(Fn(x)) is not constrained. Are we missing something?

  3. Salman January 14, 2013 at 2:52 pm

    I don’t follow the question. What do u mean by “The smallest value of n for which the maximum of f (x) occurs when x = 4″. If x and n are both positive, and n is larger than x (which is 4), then as n increases, f(x) will also increase. So there isn’t any max value. Am I misreading the question?