If p and q are two nonzero integers, is p^q q^p ?

Question

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If p and q are two nonzero integers, is p^q > q^p ?

(1) p = √q 
(2) p ≠ 1

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yashikaaggarwal · Answer

To find: p^q > q^p ?

(1) p = √q
Case1: p = 1 = q
1 = 1 {Incorrect}

Case2: p = 2 , q = 4
2^4 = 4^2 
16 = 16 {Incorrect}

Case3: p = 3 , q = 9
3^9 > 9^3
19683 > 729 {correct}
{Insufficient}

(2) p ≠ 1
Case1: p = 2 , q = 4
2^4 = 4^2 
16 = 16 {Incorrect}

Case2: p = 3 , q = 9
3^9 > 9^3
19683 > 729 {correct}

Case3: p = 4 , q = 16
4^16 > 16^4 {correct}
{Insufficient}

Combining both :
Case1: p = 2 , q = 4
2^4 = 4^2 
16 = 16 {Incorrect}

Case2: p = 3 , q = 9
3^9 > 9^3
19683 > 729 {correct}
Both have many 2 & 3 integer common
(Insufficient)

IMO E

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QuantMadeEasy · Answer

Step 1: Understanding statement 1 alone 
Squaring both sides
 p^2  = q
Using values of p and q
When p=1, q = 1, then value of  p^q  = 1 is equal to value of  q^p  =1
When p = 3, q = 9, then value of  p^q = 3^9  is greater than  q^p =9^3= 3^6 
Insufficient

Step 2: Understanding statement 2 alone 
Again using values of p and q
When p=2, q=2, then value of  p^q  = 4 is equal to value of  q^p  =4
When p = 3, q = 9, then value of  p^q = 3^9  is greater than  q^p =9^3= 3^6 
Insufficient

Step 3: Clubbing statement 1 and 2 
When p=2, q=4, then value of  p^q = 2^4  = 16 is equal to value of  q^p =4^2  = 16
When p = 3, q = 9, then value of  p^q  =  3^9  is greater than  q^p =9^3= 3^6 
Insufficient

IMO E

pankulbansal · Answer

Answer: E

According to Statement (1) , 
 p=\sqrt{q} 
Plugging in values for the variables:
If p=1 and q=1, then  p^q=1^1  is equal to  q^p=1^1 , so the answer is "No"
If p=2 and q=4, then  p^q=2^4  is equal to  q^p=4^2 , so the answer is still "No"
If p=3 and q=9, then  p^q=3^9  is greater than  q^p=9^3 , so the answer is finally "Yes"
 No definite answer, hence (1) is not sufficient

According to Statement (2) ,
 p≠1 
Although you cannot use p=q=1, you can still use the second plug-in, p=2 and q=4, and the third plug-in, p=3 and q=9
 Thus, there is no definite answer. Hence (2) is also not sufficient

Both Statements Together ,
You can use the same plug-ins as you used for Statement (2)
 Thus, there is still no definite answer, so both statements are together not sufficient
Hence answer is E

GMATinsight · Answer

Answer: Option E

Please check the video for the step-by-step solution.

GMATinsight's Solution

https://www.youtube.com/watch?v=D5MVXdlDAnY

HoneyLemon · Answer

Is it  E  ? 
If p and q are two nonzero integers, is p^q > q^p ?

(1) p = √q
 -- p can be positive or -ve .. but q is always positive .. so not sufficient . 
(2) p ≠ 1
 -- P and q can be anything .. so not sufficient .

Combining .. still p can be positive or -ve .. but q is always positive .. so not sufficient . 
Hence E .

minustark · Answer

If p and q are two nonzero integers, is p^q > q^p ?

(1) p = √q
(2) p ≠ 1

1) When p = q = 1, the condition doesn't hold true. Again, when p =2, q = 4, the two sides are also equal. Lets take p =3, q =9, then 3^9 > 9^3. Two different options. Not sufficienT.

2) Still not sufficient as we can see from the discussion about statement 1.

Together, for p =2 and p = 3, we are getting two different answers. Not sufficient.

E is the answer.

MayankSingh · Answer

IMO E

(1) p = √q
 Square root is a non-negative function. Therefore , p = +Ve integer
q=p^2

p^q = (p)^p^2 & q^p= p^2p

For p^q > q^p , (p)^p^2 > p^2p => p^2 > 2p => p<0 or p>2 , since p +ve integer, p>2
Since , no information about p, so   insufficient

(2) p ≠ 1
 No information about p & q values, so   insufficient.

Together,   p ≠ 1, p= +ve integer , so p>1 --But P=2 (only integer) doesn't satisfy, except that all other satisfy. 
  In Sufficient

exc4libur · Answer

(1) insufic
pq=(2,4): 2^4>4^2, 16=16, no
pq=(4,16): 4^16>16^4, yes

(2) insufic
pq=(2,4): 2^4>4^2, 16=16, no
pq=(4,16): 4^16>16^4, yes

(1/2) insufic
pq=(2,4): 2^4>4^2, 16=16, no
pq=(4,16): 4^16>16^4, yes

Ans (E)

bumpbot · Answer

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If p and q are two nonzero integers, is p^q > q^p ?

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