gmatusa2010 wrote:If m and n are positive integers, is m^n < n^m?
(1) m = root(n)
(2) n > 5

m^n<n^m ==> m^n-n^m < 0 --(I)

Statement 1: m = root(n) ==> n = m^2 -- (II)
From (I) and (II), m^(m^2) - m^2m < 0 -- (III)
Since m is positive, if m = 1, then (III) =0
if m = 2, then (III) = 0
if m > 2, then (III) > 0 -- (IV)

Hence, Statement 1 alone is not sufficient.

Statement 2: n > 5. Since we dont have any information about m, nor its relation with n, Statement 2 alone is not sufficient.

However, Statement I and Statement II together is sufficient, since based on (IV), the equation is always > 0 for all m > 2==> n > 4.

Hence, answer is C.

gmatusa2010 wrote:If m and n are positive integers, is m^n < n^m?
(1) m = root(n)
(2) n > 5

Its obvious 1&2 alone is not sufficient, try 1&2 BOTH

Using 1: root(n)^n < n^root(n)
n^(n/2) < n^(root(n))
n/2<root(n)
=> n/2-root(n) <0
=> root(n)/2(root(n)-2)<0
=>root(n)<2
=>n<4

Using 2, n>5 so combine says, answer in no. Suff