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 $280$
• D. Number of ways of putting $10!$ as a product of two natural numbers is $135$

Answer 1

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

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

Highest power of $2$: $\frac{10}{2}=5,\frac{5}{2}=2.5->2,\frac{2}{2}=1\Rightarrow 5+2+1=8$
Thus, $p=8$
Highest power of $3$: $\frac{10}{3}=3.3->3,\frac{3}{3}=1\Rightarrow 3+1=4$
Thus, $q=4$
Highest power of $5$: $\frac{10}{5}=2$
Thus, $r=2$
Highest power of $7$: $\frac{10}{7}=1$
Thus, $s=1$
Thus, $p\times q\times r\times s=64$
Now, $10!$ has prime factors $2,3,5$ and $7$. Hence, number of factors $=$ $(p+1)(q+1)(r+1)(s+1)=270$
Thus, number of ways of writing $10!$ as a product of $2$ factors $=$ $\frac{270}{2}=135$
Hence, (a)(b) and (d) are correct.