If a=1^2, b=2^3, c=3^4, d=4^5...........z=26^27,then how many zeros would be present in the end in the product of all alpahabets?
For this we need to calculate the number of 5's in the product which will combine with 2's to give a zero in the end. We need not calculate the number of 2's because they are much more than no. of 5's.
5^6 has 6 fives.
10^11 is 2^11*5^11 which gives 11 fives.
15^16 is 3^16*5^16 which gives 16 fives.
20^21 is 4^21*5^21 which gives 21 fives.
25^26 is 5^52 which has 52 fives.
So total number of zeroes in the end is 6+11+16+21+52 = 106
