Challenge Problem Showdown – January 14, 2013

Tom Williams —  January 14, 2013 — 3 Comments

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
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Tom Williams

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Tom Williams is a Marketing Associate at Manhattan Prep. His background and interests are primarily within New Media and Online Marketing. As such, he spends approximately 18 hours a day in front of a computer. When he’s not interacting with Manhattan Prep students on Facebook and Twitter, Tom can be found listening to records, rooting for the Mets and Jets, and reading history books.

3 responses to Challenge Problem Showdown – January 14, 2013

  1. 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?

    • 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?

  2. Yeah you are absolutely wright !

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