Common sense:
For every factor of 10, there will be one trailing zero similarly for every factor of 5 there will be one trailing zero (as 5*2 =10, and there are enough number of 2′s). But we have to take care of repetitions.
10, 20,…., 90 = 9 zeros
100 = 2 zeros
5, 15, 25……95 = 10 zeros
and 1 extra 5 in each of 25, 50 and 75 = 3 zeros
so total 9+2+10+3 = 24 zeros.
Short formula:
total 0s = (100/5) + (100/25) + (100/125) + ... = 24
Common sense:
For every factor of 10, there will be one trailing zero similarly for every factor of 5 there will be one trailing zero (as 5*2 =10, and there are enough number of 2′s). But we have to take care of repetitions.
10, 20,…., 90 = 9 zeros
100 = 2 zeros
5, 15, 25……95 = 10 zeros
and 1 extra 5 in each of 25, 50 and 75 = 3 zeros
so total 9+2+10+3 = 24 zeros.
Short formula:
total 0s = (100/5) + (100/25) + (100/125) + ... = 24
Common sense:
Reply 1For every factor of 10, there will be one trailing zero similarly for every factor of 5 there will be one trailing zero (as 5*2 =10, and there are enough number of 2′s). But we have to take care of repetitions.
10, 20,…., 90 = 9 zeros
100 = 2 zeros
5, 15, 25……95 = 10 zeros
and 1 extra 5 in each of 25, 50 and 75 = 3 zeros
so total 9+2+10+3 = 24 zeros.
Short formula:
total 0s = (100/5) + (100/25) + (100/125) + ... = 24