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

Given: $N^N = 2^{160}$ and we need to find the largest possible $x$ such that $2^x$ divides $N^2 + 2^N$.

Since $N$ is a positive integer and $N^N = 2^{160}$, we need $N$ to be a power of $2$.

Let $N = 2^k$ for some positive integer $k$.

Then: $(2^k)^{2^k} = 2^{160}$

This gives us: $2^{k \cdot 2^k} = 2^{160}$

Therefore: $k \cdot 2^k = 160$

Checking small values of $k$:

• If $k = 5$: $5 \cdot 2^5 = 5 \cdot 32 = 160$ ✓

So $N = 2^5 = 32$.

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Now we calculate $N^2 + 2^N$:

$N^2 = 32^2 = 1024 = 2^{10}$

$2^N = 2^{32}$

Therefore: $N^2 + 2^N = 2^{10} + 2^{32}$

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To find the highest power of $2$ that divides this expression, we factor out the smaller power:

$N^2 + 2^N = 2^{10} + 2^{32}$

$= 2^{10}(1 + 2^{22})$

Since $2^{22}$ is even and $1$ is odd, $(1 + 2^{22})$ is odd and contains no factors of $2$.

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Therefore, the highest power of $2$ that divides $N^2 + 2^N$ is exactly $2^{10}$.

The largest possible value of $x$ is $10$.