Question 14

# 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

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