If m and n are positive integers, is m^n

Question

If m and n are positive integers, is m^n < n^m?
(1) m = sqrt(n)
(2) n > 5

from 1

m = sqrt n ie: n = m^2

is m^(m^2)< m^2m , is m^m(m-2) < 1 is m(m-2)<0 is m>2.... insuff

from 2

obviously insuff

both suff

n>5, n = m^2 ( try worst case scenario n = 9 ) thus m = 3 >2...suff

Bunuel · Accepted Answer

If m and n are positive integers, is m^n < n^m?
(1) m = sqrt(n)
(2) n > 5

from 1

m = sqrt n ie: n = m^2

is m^(m^2)< m^2m , is m^m(m-2) < 1 is m(m-2)<0 is m>2.... insuff

from 2

obviously insuff

both suff

n>5, n = m^2 ( try worst case scenario n = 9 ) thus m = 3 >2...suff

shrouded1 · Answer

(1) :  m^n < n^m 
  m^{m^2} < n^m  
  (m^2)^{m^2/2} < n^m 
  n^{m^2/2} < n^m

For n,m integers and both greater than 1, this implies

m^2/2 < m 
 m(m-2) < 0

This expression is false for all m>2

Also we know for m=1 and m=2 that m^n=n^m ... so again the expression is false

So (1) is sufficient

(2) : Not sufficient as only condition on n

So I agree answer is A

The answer would be (c) if the original question was m^n <= n^m

yezz · Answer

If m and n are positive integers, is m^n < n^m?
(1) m = sqrt(n)
(2) n > 5

from 1

m = sqrt n ie: n = m^2

is m^(m^2)< m^2m , is m^m(m-2) < 1 is m(m-2)<0 is m>2.... insuff

from 2

obviously insuff

both suff

n>5, n = m^2 ( try worst case scenario n = 9 ) thus m = 3 >2...suff

Bunuel · Answer

It seems that you are right, answer should be A.

Answer to be C question shouldn't say that  m  and  n  are integers. In this case if  m=\sqrt{n}=\sqrt{2}  then  m^n=\sqrt{2}^2=2<2^{\sqrt{2}}=n^m , so (1) wouldn't be sufficient.

Also the question would be a little bit trickier in this case.

If  m=7  then  n=49  and  7^{49}>49^7 , so answer is still NO.

KarishmaB · Answer

Let's discuss the relation between m^n and n^m when m and n are positive integers. To differentiate m from n, I say m is the smaller of the two except where they are equal. Look at the pattern now: Say m = 1 n = 1: 1^1 = 1^1 n = 2: 1^2 < 2^1 n = 3: 1^3 < 3^1 and so on... Say m = 2 n = 2: 2^2 = 2^2 n = 3: 2^3 < 3^2 n = 4: 2^4 = 4^2 (m^n was less than n^m above but here they are equal so m^n should become greater from now on) n = 5: 2^5 > 5^2 n = 6: 2^6 > 6^2 You see that the difference between m^n and n^m is increasing in every step.) Say m = 3 n = 4: 3^4 > 4^3 (m^n is already greater than n^m) n = 5: 3^5 > 5^3 n = 6: 3^6 > 6^3 You see that the difference between m^n and n^m is increasing in every step.) Say m = 4 n = 5: 4^5 > 5^4 (m^n is already greater than n^m) n = 6: 4^6 > 6^4 n = 7: 4^7 > 7^4 You see that the difference between m^n and n^m is increasing in every step.) I hope you see the pattern. Let's forget about the cases where m = 1 and m = n. In every case except m = 2, m^n > n^m When m = 2, 2^3 < 3^2 and 2^4 = 4^2 . Thereafter, m^n > n^m It is a good idea to remember these relations. Let's look at the question now. Is m^n < n^m ? (1) m = \sqrt{n} First of all, both m and n are integers so n must be a perfect square. Also, m will be smaller than n except when both are equal to 1. m = 1/2/3/4/5/6 etc n = 1/4/9/16/25/36 etc If m = 1 and n = 1, we know m^n = n^m If m = 2 and n = 4, we know m^n = n^m For all other cases, we know that m^n > n^m Hence the answer to the question is 'No'. Sufficient. (2) n > 5 If m = 1, 1^6 < 6^1 If m = 2, 2^6 > 6^2 Not sufficient. Answer (A)

arnoorichenna · Answer

m^n < n^m 1) m=sqrt(n) => m^2 = n , putting this in above. m^m^2 < m^2^m => m^2m < m^2m which are equal, so NO. Sufficient. 2) n > 5 not sufficient Ans: A. ( Am I missing something...?)

EMPOWERgmatRichC · Answer

Hi All,

This DS question can be solved by TESTing VALUES.

We're told that M and N are POSITIVE INTEGERS. We're asked if M^N < N^M. This is a YES/NO question.

Fact 1: M = \sqrt{N}

IF....
N = 1
M = 1
1^1 is NOT < 1^1 and the answer to the question is NO.

N = 4
M = 2
2^4 is NOT < 4^2 and the answer to the question is NO.

N = 9
M = 3
3^9 is NOT < 9^3 and the answer to the question is NO.
This pattern will continue; the answer to the question is ALWAYS NO.
Fact 1 is SUFFICIENT.

Fact 2: N > 5

This tells us NOTHING about the value of M, so this is probably insufficient. Here's the proof.

IF...
M = 1
N = 6
1^6 < 6^1 and the answer to the question is YES.

IF....
M = 6
N = 6
6^6 is NOT < 6^6 and the answer to the question is NO.
Fact 2 is INSUFFICIENT

Final Answer:  A

GMAT assassins aren't born, they're made,
Rich

Jam2014 · Answer

if I compare with base using "n" rather than "m"

I get 
following equation after reduction :

m^n < n^m reduces to

n^(n/2) < (n)^sqrt (n)

means we need find if 
n/2 < Sqrt (n), Given n is perfect square bcs m = sqrt(n) and m is integer.

for n=1 : yes
 for n=4 : no 
for n=64 : No
for n=9 : No

then how can we answer : A is sufficient.

on other side if n>5 then
 Answer B:  is always hold. So sufficient....

I am sure I must be wrong somewhere in logic!!!

GMATinsight · Answer

Look at the highlighted steps only

n^(n/2) < (n)^sqrt (n)

for n=1 : yes  The answer is not Yes it's No here as well because  n^(n/2) will NOT be less than (n)^sqrt (n) for n=1

Hence consistent answer so SUFFICIENT

Statement 2: n>5

@n=6, and m=1, m^n will be less than n^m
@n=6, and m=2, m^n will NOT be less than n^m
NOT SUFFICIENT

I hope it helps!

MathRevolution · Answer

Forget the conventional way to solve DS questions.

We will solve this DS question using the variable approach.

Remember the relation between the Variable Approach, and Common Mistake Types 3 and 4 (A and B)[Watch lessons on our website to master these approaches and tips]

Step 1: Apply Variable Approach(VA)

Step II: After applying VA, if C is the answer, check whether the question is the key question.

StepIII: If the question is not a key question, choose C as the probable answer, but if the question is a key question, apply CMT 3 and 4 (A or B).

Step IV: If CMT3 or 4 (A or B) is applied, choose either A, B, or D.

Let's apply CMT (2), which says there should be only one answer for the condition to be sufficient. Also, this is an integer question and, therefore, we will have to apply CMT 3 and 4 (A or B).

To master the Variable Approach, visit https://www.mathrevolution.com and check our lessons and proven techniques to score high in DS questions.

Let’s apply the 3 steps suggested previously. [Watch lessons on our website to master these 3 steps]

Step 1 of the Variable Approach: Modifying and rechecking the original condition and the question.

We have to find the value Is \(m^n < n^m\) - where 'm' and 'n' are positive integers

Second and the third step of Variable Approach: From the original condition, we have 2 variables (m and n). To match the number of variables with the number of equations, we need 2 equations. Since conditions (1) and (2) will provide 1 equation each, C would most likely be the answer.

But we know that this is a key question [Integer question] and if we get an easy C as an answer, we will choose A or B.

Let’s take a look at each condition.

Condition(1) tells us that \(m = \sqrt{n}\).

=> \(m^2 = n\)

=> For n = 1 => m = 1: \(m^n = 1^1 = n^m - m^n < n^m \)- NO

=> For n = 2 => m = 4: \(m^n = 4^2 = 16 = n^m - m^n < n^m\) - NO

=> For n = 3 => m = 9: \(m^n = 9^3 > n^m = 3^9 - m^n < n^m\) - NO

Since the answer is a unique NO, the condition (1) is sufficient by CMT 1.

Condition(2) tells us that n > 5.

=> If m = 1 and n = 6: \(1^6 < 6^1 - m^n < n^m\) - YES

=> If m = 2 and n = 6: \(2^6 > 6^2 - m^n < n^m\) - NO

Since the answer is not a unique YES or a NO, the condition(2) is not sufficient by CMT 2.

Condition (1) alone is sufficient.

So, A is the correct answer.

Answer: A

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If m and n are positive integers, is m^n < n^m?

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GMAT Data Sufficiency (DS) Questions

If m and n are positive integers, is m^n < n^m?

Prep Toolkit

615 to 715 in 15 Days (Ria's Strategies)

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