Thanks guys for the reply...
The explanation given is as follows and it is from manhattan gmat cat. I am still wondering why google gives such as huge answer for 60!.
If an integer ends in 0, then that integer is divisible by 10; when that integer is divided by 10, the final 0 disappears. Therefore, the number of 0's at the end of an integer is exactly the number of times that integer is divisible by 10.
Since 10 = 2 × 5, this number (the number of times the integer is divisible by 10) is the smaller of the powers to which 2 and 5 are raised in the integer's prime factorization. For instance, an integer whose prime factorization contains 28 × 57 and an integer whose prime factorization contains 27 × 58 will both end with seven 0's.
In a factorial, such as 60!, there are many more 2's than 5's in the prime factorization: the factorial product has more numbers contributing 2's than 5's, and more of those numbers contribute repeated factors. Therefore, the question can be rephrased: How many 5's are in the prime factorization of (60!) ?
60! is the product of all the integers from 1 to 60, inclusive. Each of the twelve multiples of 5 in this range - 5, 10, 15, ..., 60 - contributes a 5 to the prime factorization. Additionally, a second 5 is contributed by each of the two multiples of 25 (25 and 50). Therefore, the prime factorization of 60! contains fourteen 5's, and so 60! will end with fourteen 0's.
