If p,q are distinct and greater than 1, p + q =

Question

If  p,q  are distinct and greater than 1,  p+q =  ?

1.  q^p=p^q 
2.  q=p^2

Source: Experts Global

DavidTutorexamPAL · Answer

If it isn't entirely clear how to interpret the statements, we'll first try using specific numbers to give us a clue how to proceed (and eliminate ambiguous statements).
This is an Alternative approach.

(1) Trying q = 2 gives 2^p = p^2. So 2^p has to be a square number, and as it is also a power of 2 we'll try 4,16, 64... 4 doesn't work because then p = 2 but p and q are distinct, but 16 could work as then p = 4. If we try 64 we fail again as 8^2 = 64 but 2^8 = 256 and trying any larger number won't work as it makes the exponent larger. So if p = 2 q has to equal 4.
Based on the above, we know that p has to be a power of q, so we can write p = q^n. Substituting gives q^(q^n) = (q^n)^q an using power laws q^(q^n) = q^(nq). Then q^n = nq and since q > 1 we can divide by it to get q^(n-1) = n. So if we pick n = 2 then q = 2 and p = 4 which is what we got above. If we pick n = 3 then q^2 = 3 so q = sqrt(3) and we get some other answer. We won't bother to actually calculate, as this is definitely insufficient.

(2) Once again, we can pick whichever number we like for p, giving another number for q, and then plug this into p + q to get some (different) result.
Insufficient.

Combined:
This is exactly the equation we got to above, with p and q reversed. So we know that n = 2 and therefore there is one solution (q = 4, and p = 2).

If this isn't clear, then simplify the statements by substituting (2) into (1):
(p^2)^p = p^(p^2). Using power laws, this implies that p^(2p)=p^(p^2) and therefore, since p > 1, that 2p = p^2, which gives p = 2 (as p cannot equal 0). Then q = 4 and p + q = 6.
Sufficient.

(C) is our answer. 
_________________

GMATWhizTeam · Answer

Solution 
 Step 1: Analyse Question Stem
  •  p , and  q  are distinct 
 o p> 1, and q >1  
We need to find the value of  p + q .
 Step 2: Analyse Statements Independently (And eliminate options) – AD/BCE 
 Statement 1:   p^q = q^p 
 • There are only two possibilities,
 o Case 1:  p =2 , and  q = 4,  2^4=4^2  
   p + q =2+4 = 6  
  o Case 2:  p =4 , and  q = 2 ,  4^2=2^4 
   p + q = 4+2 = 6    
Hence, statement 1 is sufficient, we can eliminate answer options 
 Statement 2:   q = p^2 
We can take any integral value of p > 1 and find q, For example
 • If  p = 2 , then  q = 4 
 o  p + q = 6  
• If  p = 3 , then  q = 9 
 o  p + q = 12  
• We are not getting a unique value. 
Hence, statement 2 is not sufficient, the correct answer is  Option A .

phcoed · Answer

Hi, 
Can you tell me what I'm missing here?

I think it is C, although OA says its A.

p and q can be 4,2 or vice versa. 
but also -4, -2 or vice versa.
either p q is 6 or -6

tam87 · Answer

It is given that p and q are greater than 1 - so they can't take negative values

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If p,q are distinct and greater than 1, p + q =

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