# CAT 2018 Quant Questions

Question:
If N and x are positive integers such that NN = 2160 and N2 + 2N is an integral multiple of 2x, then the largest possible x is

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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