### A special implication: Suppose you were to find the number of zeroes in the value of the following factorial values

- While solving 45!, 46!, 47!, 48!, 49!. Notice the number of zeroes in each of the cases will be equal to 10.
- It is not difficult to understand that the number of fives in any of these factorials is equal to 10. The number of zeroes will only change at 50! (It will become 12).
- In fact, this will be true for all factorial values between two consecutive products of 5.
- Thus, 50!, 51!, 52!, 53! And 54! will have 12 zeroes (since they all have 12 fives). Similarly, 55!, 56!, 57!, 58! And 59! will each have 13 zeroes.
- While there are 10 zeroes in 49! there are directly 12 zeroes in 50!. This means that there is no value of a factorial which will give 11 zeroes. This occurs because to get 50! we multiply the value of 49! by 50. When you do so, the result is that we introduce two 5’s in the product. Hence, the number of zeroes jumps by two.
- Note : at 124! you will get 24 + 4 fi 28 zeroes. At 125! you will get 25 + 5 + 1 = 31 zeroes. (A jump of 3 zeroes.)

Q.

n! has 23 zeroes. What is the maximum possible value of n?

Q.

n! has 13 zeroes. The highest and least values of n are?

- 59 and 55
- 59 and 55
- 59 and 55
- 59 and 55

Q.

Find the number of zeroes in the product 11+22+33+...+4949

Q.

Find the number of zeroes in: 1001×992×983×974×……….×1100

Q.

Find the number of zeroes in: 1 1! × 2 2! × 3 3! × 4 4! × 5 5! × ……..10 10!

- 5! * 10!
- 5! + 10!
- 5! - 10!
- 5! / 10!