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If $\text{N}$ and $x$ are positive integers such that $\text{N}^{\text{N}}=2^{160}$ and $\text{N}^{2} + 2^{\text{N}}$ is an integral multiple of $2^{x}$, then the largest possible $x$ is _______

Given that, ${N}^{N} = 2^{160}$

$\Rightarrow N^{N} = \left( 2^{10} \right)^{16}$

$\Rightarrow N^{N} = \left( 2^{5} \right)^{32}$

$\Rightarrow N^{N} = (32)^{32}$

$\therefore \; \boxed{N = 32}$

Now, $N^{2} + 2^{N} = 32^{2} + 2^{32}$

$= \left( 2^{5} \right)^{2} + 2^{32}$

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

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

Here, $N^{2} + 2^{N}$ is a integral multiple of $2^{x}.$

So, $2^{x} = 2^{10}$

$\Rightarrow \boxed{ x = 10}$

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

Correct Answer $:10$
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