Question

How many zeros will be there at the end of the product $2{!}^{2!}\times 4{!}^{4!}\times 6{!}^{6!}\times 8{!}^{8!}\times 10{!}^{10!}$ ?

• A. $10!+6!$
• B. $2(10!)$
• C. $10!+8!+6!$
• D. $6!+8!+2(10!)$

Answer 1

The correct option is D $6!+8!+2(10!)$

Zero containing terms are $6!$ i.e. it has $5!$ in it i.e. $1$ zero

$8!\to 1$ more

and $10!$ will have $2$ zero.

Total zero is $6!+8!+2(10!)$

Hence, the answer is $6!+8!+2(10!).$