When mind becomes dim.

For each factor $10$ , there will be a terminating zero; that is, for every pair factors of $5$ and $2$ , there will be a terminating zero. We must, therefore, find the number of factors $5$ and the numbers of factors $2$ in $100!$ . Then count which is lower. If the number of factors $5$ is lower than the number of factors $2$ , then the numbers of factors $5$ is the number of terminating zeros. The factor $5$ is present once in each of $5,10,15,20,30,35,40,45,55,60,65,70,80,85,90,$ and $95$ ; $16$ factors. And twice in each of $25,50,75$ and $100$ ; $8$ factors. A total of $24$ factors. The factor $2$ is present once in each $2,4,10...98$ and so forth. A total of $97$ factors. Of these factors, only $24$ are needed to pair with the $24$ factors of $5$ . Therefore, the number of terminating zeros of $100!$ is $24$ .

Since 2 is prime . Power of 2 in 100! =(100/2)+(100/2^2)+(100/2^3)+(100/2^4)+(100/2^5)+(100/2^6)+(100/2^7)+(100/2^8).... =50+25+12.5+6.25+3.1+1.5+(0....) =97 Power of 5 in 100! =(100/5)+(100/5^2)+(100/5^3)+..... =20+4+0.8=24 No. of 2's =97,no. of 5's=24 To get zero's we need both 5's and 2's in the product. Therefore no. of zeroes =24