Question

According to Fermat's theorem, $n^p-n=M\left(p\right)=a$ is a multiple of $p$, when $p$ is a prime number and $n\in N$.
It is also known that the highest index of power of a prime number $p$ which divides $n!$ is equal to $\left[\frac{n}{p}\right]+\left[{\frac{n}{p}}^2\right]+\left[{\frac{n}{p}}^3\right]+\dots$,
where $\left[x\right]=$ greatest integer less than or equal to $x$.

The highest power of $7$ which divides $1000!$ is

• A. $7^{164}$
• B. $7^{162}$
• C. $7^{167}$
• D. $7^{142}$

Answer 1

The correct option is A $7^{164}$

The highest power of $7$ in $1000!$ will be
$=\left[\frac{1000}{7}\right]+\left[\frac{1000}{7^2}\right]+\left[\frac{1000}{7^3}\right]$
$=\left[142.8\right]+\left[20.408\right]+\left[2.915\right]$
$=142+20+2$
$=164$
Hence $7^{164}$ perfectly divides $1000!$.