Question

If $10!=2^p\cdot 3^q\cdot 5^r\cdot 7^s$, then

• A. $2q=p$
• B. $pqrs=64$
• C. number of divisors of $10!$ is $270$
• D. number of ways of putting $10!$ as a product of two natural numbers is $135$

Answer 1

Answer: (A), (B), (C), (D)

The correct options are
A $2q=p$
B $pqrs=64$
C number of divisors of $10!$ is $270$
D number of ways of putting $10!$ as a product of two natural numbers is $135$

Exponent of $2$ is
$\left[\frac{10}{2}\right]+\left[\frac{10}{2^2}\right]+\left[\frac{10}{2^3}\right]=5+2+1=8$

Exponent of $3$ is
$\left[\frac{10}{3}\right]+\left[\frac{10}{3^2}\right]=3+1=4$

Exponent of $5$ is
$\left[\frac{10}{5}\right]=2$

Exponent of $7$ is
$\left[\frac{10}{7}\right]=1$
The number of divisors of $10!$ is $(8+1)(4+1)(2+1)(1+1)=270$.

The number of ways of putting $N$ as a product of two natural numbers is $\frac{270}{2}=135$.