ENDING ZEROES

$7\times14\times21\times\cdots\times777=7^{111}\times1\times2\times3\times\cdots\times111=7^{111}\times111!$ For each zero at the end of $N$ , there are one 2 and one 5 in the list of the factors of $N$ . Since the number of 2's in the factor is much bigger than that of 5's there, we can ignore 2.

The number of 5's in $111!$ $\begin{array}{l} =\left\lfloor\frac{111}5\right\rfloor+\left\lfloor\frac{111}{25}\right\rfloor\\ =22+4\\ =26 \end{array}$

Since $7^{111}$ contains no 5's, it is ignored.

Therefore there are 26 zeros at the end of N.

simple again : group it as (7X1)X(7X2)X(7X3)... (7X111) now if we take out all the 7s , we get 7e111 X (111 !(factorial)) now the answer depends on 111 factorial and we find out the number of 5s in 111 factorial as out of 2s and 5s [for number of 0s] number of 5s is the restricting factor, so floor of 111/5 +111/25 =26