If N and x are positive integers such that $$N^{N}$$ = $$2^{160}\ and \ N{^2} + 2^{N}\ $$ is an integral multi

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 the largest possible x is

Accepted Answer

It is given that N N N^{N} N N = 2 160 2^{160} 2 160 We can rewrite the equation as N N N^{N} N N = ( 2 5 ) 160 / 5 (2^5)^{160/5} ( 2 5 ) 160/5 = 32 32 32^{32} 3 2 32 ⇒ \Rightarrow ⇒ N = 32 N 2 + 2 N N{^2} + 2^{N} N 2 + 2 N = 32 2 + 2 32 = 2 10 + 2 32 = 2 10 ∗ ( 1 + 2 22 ) 32^2+2^{32}=2^{10}+2^{32}=

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