If x = 1^1*2^2*3^3*4^4*...*98^98*99^99*100^100, how many trailing zero

A zero is the result of the product of a 5 and a 0.

In the question above, all the 5's come from the multiples of 5, like 5, 10, 15, 20 and so on . The 2's come from even numbers. We need to count only the fives as there are more 2's than 5's.

\(5^5\) contributes 5 zeroes

\(10^{10}\) contributes 10 zeroes

\(15^{15}\) contributes 15 zeroes

\(20^{20}\) contributes 20 zeroes

\(25^{25} = (5^2)^{25} = 5^{50}\) contributes 50 zeroes

So all the numbers where we have squares of 5 will give us the double number of zeroes. These numbers are 25, 50, 75 and 100...All other multiples will give us the same number of zeroes as the number itself.

So we have 5 + 10 + 15 + 20 + 50 + 30 + 35 + 40 + 45 + 100 + 55 + 60 + 65 + 70 + 150 + 80 + 85 + 90 + 95 + 200 = 1300



Option A

Arun Kumar