### Challenge Problem Showdown – January 14, 2013

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 â‰¤ x â‰¤ 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

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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?