If N and x are positive integers such that N^N = 2160 and N^2 + 2^N

Question

If N and x are positive integers such that  N^N   = 2^{160} , and  N^2  +  2^N  is an integral multiple of  2^x , then find the largest possible value of integer x.

A. 2
B. 8
C. 10
D. 16
E. 20

chetan2u · Answer

The question is wrong the way it is written. You cannot have a positive integer N that would satisfy  N^N  = 2160.

As we talk of power of 2 at various places, the question should be  N^N   = 2^{160} .

Editing the question accordingly. Please let me know if it was meant to be something else.

chetan2u · Answer

N^N   = 2^{160} 
As N is a positive integer, it has to have a base of 2, so let  N=2^a 
So  (2^x)^{2^x}   = 2^{160} 
 2^{x*2^x}   = 2^{160} 
Equating the powers
 x*2^x=160=5*2^5 
x=5 and  N=2^5=32

N^2  +  2^N = 32^2+2^{32}=(2^5)^2+2^{32}=2^{10}+2^{32}=2^{10}(1+2^{22}) 
Therefore, the maximum value of x=10

C

Bunuel · Answer

Discussed here: if-n-and-x-are-positive-integers-such-that-n-n-2-160-and-n-2-2-n-320442.html?fl=similar

If N and x are positive integers such that N^N = 2160 and N^2 + 2^N

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