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
Correct Answer: 10
Given $\mathrm{N}^{\mathrm{N}}=2^{160}=2^{5 \times 32}$
$=\left(2^{5}\right)^{32} \Rightarrow \mathrm{N}^{\mathrm{N}}=32^{32}$
$\Rightarrow \mathrm{N}=32$
$\mathrm{N}^{2}+2^{\mathrm{N}}=32^{2}+2^{32}$
$\Rightarrow\left(2^{5}\right)^{2}+2^{32}$
$\Rightarrow 2^{10}+2^{32}$
$=2^{10}\left(1+2^{22}\right)$
Or $ \mathrm{x}$ is 10
Given $\mathrm{N}^{\mathrm{N}}=2^{160}=2^{5 \times 32}$
$=\left(2^{5}\right)^{32} \Rightarrow \mathrm{N}^{\mathrm{N}}=32^{32}$
$\Rightarrow \mathrm{N}=32$
$\mathrm{N}^{2}+2^{\mathrm{N}}=32^{2}+2^{32}$
$\Rightarrow\left(2^{5}\right)^{2}+2^{32}$
$\Rightarrow 2^{10}+2^{32}$
$=2^{10}\left(1+2^{22}\right)$
Or $ \mathrm{x}$ is 10
CAT Online Course
Also Check: 841+ CAT Quant Questions with Solutions
