Question

How many zeroes will be there at the end of the expression $N=7\times 14\times 21\times ...\times 777?$

• A. 24
• B. 25
• C. 26
• D. None of these

Answer 1

The correct option is C 26

$N=7\times 14\times 21\times ...\times 777$
Method 1
In this expression, every fifth term is a multiple of 5.
Now, there are 111 terms in the expression.
Therefore, number of $5s=\left(111/5\right)+\left(111/25\right)=22+4=26$

Method 2
$N=7\times 14\times 21\times ...\times 777=(7\times 1)\times (7\times 2)\times (7\times 3)...\times (7\times 111)=7^{111}\times (1\times 2\times 3\times ...\times 111)=7^{111}\times 111!$
Number of zeroes in $111!=\left(111/5\right)+\left(111/5^2\right)$
= 22 + 4 = 26