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a tough one

by sana.noor » Sun May 05, 2013 6:53 pm
If m and n are positive integers, is m^n < n^m?
(1) m = under root (n)
(2) n > 5

C

source: jeff sackmann total gmat math
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by hemant_rajput » Sun May 05, 2013 8:47 pm
sana.noor wrote:If m and n are positive integers, is m^n < n^m?
(1) m = under root (n)
(2) n > 5

C

source: jeff sackmann total gmat math
1. m = (n ^ 1/2) lets take n = 1

m^n = 1 ^ 1 and n^m = 1 ^ 1

for n = 9

m^n = 3 ^ 9 and n^m = 9 ^ 3. m^n > n^m

for n = 16

m^n = 4 ^ 16 and n^m = 16 ^ 4. m^n > n^m

So I think A is sufficient to say that m^n will never be less than n^m.

2. n> 5 can be 6

2^ 6 > 6^2

3^ 6 > 6^3

and so on

so this is also sufficient.

IMO D.
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by Atekihcan » Sun May 05, 2013 9:36 pm
sana.noor wrote:If m and n are positive integers, is m^n < n^m?

(1) m = under root (n)
(2) n > 5
Statement 1: As m and n are positive integers, m = √n = n^(1/2) means m < n
So, m^n = (n^(1/2))^n = n^(n/2) and, n^m = n^(√n)

Now, we need to compare n^(n/2) and n^(√n), i.e. n/2 and √n
Note that, n/2 = √n for n = 4 and for n = 1, n/2 < √n.
So, for n < 4, n/2 < √n and for n > 4, n/2 > √n

As m and n are positive integers and m = √n, possible values of n are 1, 4, 9,... etc.
Now, for n = 1, m = 1, m^n = n^m = 1
And, for n = 4, m = 2, m^n = n^m = 16
The rest of the possible values of n are all greater than 4. So, for them, n/2 is always greater than √n, i.e. m^n is always greater than n^m.

In other words, m^n ≥ n^m. i.e. m^n will be never less than n^m.
So, statement 1 is sufficient.

Statement 2: Let us assume n = 6 > 5
Now, if m = 1, m^n = 1^6 = 1 < n^m = 6^1 = 6
And, if m = 2, m^n = 2^6 = 64 > n^m = 6^2 = 36

So, statement 2 is not sufficient.

Answer : A
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