### 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

xandnsuch that 0 â‰¤xâ‰¤n, the functionf(_{n}x) is defined by the equationf(_{n}x) =x. The smallest value of^{n“x}nfor which the maximum off(_{n}x) occurs whenx= 4 is

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Yeah you are absolutely wright !

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?

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?