If m and n are non-zero integers, is m^n > n^n? (1) |m| = n (2) m < n

aah sorry, true, A is the answer. Thanks for the tip.

Stmt 1. |m|=n if, m=-2,n will be 2 or m=2.n will be 2. The answer is a definite no! A

Went with E at first, but changed my answer to A after working it out. Good question, +1.

If m and n are non-zero integers, is m^n>n^n?

Statement 1: |m| = n

Tells us the absolute value of m equals n, so if we pick numbers we can come up with multiple solutions. For example, if m = -2 and n = 2. In this case, -2^2 = 4 and 2^2 = 4 so the answer would be no.

But, if m = -3 and n = 3. -3^3 = -27 and 3^3 = 27. In this case the answer is still no.

There isn't any more possible ways to get different values for m and n because of question stem tells us they are integers and no-zero. So there for m^n will always equal or be less than n^n. Therefore statement 1 is sufficient.

Statement 2: m < n

Picking numbers will work for this again.

Example 1. m = -2 and n =2

-2^2=2^2

Example 2. m = -3 and n =2

-3^2>2^2

Since you can have two solutions that give you two different outcomes, statement 2 is insufficient.

Ans A

Hi,
Again what can we make out of the Q stem..
m and n are non-zero integers, is m^n > n^n..
Just "there are various possiblities of value of m and n..." so lets move on to the statements

(1) |m| = n
in the given condition that m and n are non-zero integers..
m^n > n^n will never be true.. so its suff

But lets see why?
we know that n>0..
but we do not know anything about m, just that it is equal to n in numeric value..
so lets see the possiblities..

both m and n are positive..
is m^n > n^n?... NO, as both will be equal..

m is negative and n is positive...
we will have two scenarios here too when n is even integer and when n is positive integer..
when n is even, m, a -ive number to even power will make m^n = n^n..
when n is odd, LHS,m^n, will be negative, so m^n < n^n..

In each case our answer is NO..
so Sufficient A

(2) m < n
if m is -4 and n is 2.. ans will be yes..
if m is -1 and n is 1.. ans will be No..
different answers..
insuff..

ans A

St1: |m| = n --> n is positive
If m = -3, n = 3 --> m^n = -27; n^n = 27... m^n > n^n? No
If m = 3, n = 3 --> m^n = 27; n^n = 27... m^n > n^n? No
If m = -2, n = 2 --> m^n = 4; n^n = 4... m^n > n^n? No
Thus m^n is not greater than n^n. St1 is sufficient

St2: m < n
Let m = 1; n = 2 --> m^n = 1; n^n = 4... m^n > n^n? No
Let m = -2, n = -1 --> m^n = -(1/2); n^n = -1.. m^n > n^n? Yes
St2 is not sufficient.

Answer: A

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