in Quantitative Aptitude edited by
651 views
2 votes
2 votes
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 _______
in Quantitative Aptitude edited by
13.8k points
651 views

1 Answer

1 vote
1 vote
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 $:10$
edited by
11.6k points
Answer:

Related questions

Quick search syntax
tags tag:apple
author user:martin
title title:apple
content content:apple
exclude -tag:apple
force match +apple
views views:100
score score:10
answers answers:2
is accepted isaccepted:true
is closed isclosed:true