Question

Number of zeroes present at the end of $36{!}^{36!}$ are

n x 36!.find n

Answer 1

Here $(36\times 35\times 34\times ..30\times ...25\times ..20\times .15\times ...10\times ..5\times ...\times 1{)}^{36!}$ In order to get zero at end of number, we need either $2\times 5=10$ or $10\times 1=10$ hence no. of 5's contributing from $36!$ is $5$(35 gives 5,25 gives 5^{ 2 },15 gives 5,5 gives 5)here every even no. contributes 2. so we don't bother about 2. no.of 10's are 3(10,20,30). so total $(2\times 5{)}^5+(10{)}^3={10}^8$ hence answer will $\left({10}^8\right)36!=8\times 36!$