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Question 80
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
Correct Answer: 10
Solution
It is given that $$N^{N}$$ = $$2^{160}$$
We can rewrite the equation as $$N^{N}$$ = $$(2^5)^{160/5}$$ = $$32^{32}$$
$$\Rightarrow$$ N = 32
$$N{^2} + 2^{N}$$ = $$32^2+2^{32}=2^{10}+2^{32}=2^{10}*(1+2^{22})$$
Hence, we can say that $$N{^2} + 2^{N}$$ can be divided by $$2^{10}$$
Therefore, x$$_{max}$$ = 10
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