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

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10 — Easy First pin down $N$ from $N^N=2^{160}$ by writing $2^{160}$ as a perfect &ldquo;something to its own power.&rdquo; Then factor out the largest power of $2$ from $N^2+2^{N}$; the leftover factor must be odd. 1 Find $N$. We need $N^N=2^{160}$. Since $32=2^5$, $\;32^{32}=(2^5)^{32}=2^{160}$, matching $N^N$ with $N=32$. $$\begin{aligned} &N^N = 2^{160} = (2^5)^{32} = 32^{32}\ &\Rightarrow\ N = 32 \end{aligned}$$ 2 Form $N^2+2^{N}$. Substitute $N=32$ and write everything in powers of $2$. $$\beg

CAT 2018 Slot 2 — QA Question 26

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