swati007 wrote:

How many zeros exist in the end of the product of alphabets(a-z) where a = 4^4, b = 5^5, c = 6^6, ....z = 29^29?

a) 99

b) 100

c) 101

d)102

e)103

I think, It is 100.

A zero is created when 2 is multiplied by 5. So we should find out how many multiplications of 2 and 5 are possible in the expression.

Obviously 2 will occur more frequently than does 5. So number of 5 in the above expression will determine the number of zeros.

5 can be found in following terms

5^5 = 5 times

10^10 = 10 times

15^15 = 15 times

20^20 = 20 times

25^25 = 50 times

Total = 100 times. So 100 zeros should be there.

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