Question

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

Accepted Answer

10. Given $N^N = 2^{160}$. Let $N = 2^a$. $$(2^a)^{(2^a)} = 2^{a \cdot 2^a} = 2^{160}$$ $$a \cdot 2^a = 160 = 5 	imes 32 = 5 	imes 2^5$$ Thus, $a = 5$. So $N = 2^5 = 32$. Now we find $N^2 + 2^N = (2^5)^2 + 2^{32} = 2^{10} + 2^{32} = 2^{10}(1 + 2^{22})$. Since $1 + 2^{22}$ is an odd integer, the highest power of 2 dividing $N^2 + 2^N$ is $2^{10}$. Therefore, the largest possible $x$ is 10.

Largest exponent dividing expression

Explanation

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