Question

If $500!=2^m\times$ an integer, then

• A. $m=494$
• B. $m=496$
• C. It is equivalent to no. of$n$is $400!=2nx$an integer
• D. $m=500C2$

Answer 1

The correct option is A

$m=494$

Explanation for the correct option:

Given, $500!=2^m\times$ an integer

In general, $\left|\frac{k}{2}\right|$ integers less than or equal to$k$are even, such that some are divisible by $4$, some by $8$and so on.

Therefore, we need to find the total by adding the values $\left|\frac{k}{2}\right|+\left|\frac{k}{4}\right|+...$

The highest power of $2$ not greater than $500$ is $2^8$

Therefore,

$\Rightarrow \frac{500}{2}+\frac{500}{4}+\frac{500}{8}+\frac{500}{16}+\frac{500}{32}+\frac{500}{64}+\frac{500}{2\times 64}+\frac{500}{256}$

$\Rightarrow 250+125+62+31+15+7+3+1=494$

Hence . correct option is (A).