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(#M40016278) MBA question number systems, progressions Keep an EYE

Find the number of zeros at the end of the product of the factorials of the first 25 natural numbers.

1) 54
2) 55
3) 56
4) 75
5) 76

Asked In MBA (12 years ago)

Unsolved Read Solution (3)

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ayush  (9 years ago)

the solution is very easy... we just have to see that number of 5 that comes in any number factorial...
as from 1! to 4! their is no 5 in any factorial, 0*4=0
from 5! to 9! their is one 5 in every factorial ,1*5=5
from 10! to1 4! their is two 5 in every factorial,2*5=10
from 15! to 19! their is three 5 in every factorial,3*5=15
from 20! to 24! their is four 5 in every factorial,4*5=20
but when 25! comes their are six fives as the number 25 will give u 2 fives, 6
sum all answer = 56

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Rupesh  (9 years ago)

25! = 6 zero
20! to 25! each have 4 zero so total zero= 5*4 = 20
15! to 19 each have 3 zero so total zero = 5*3 = 15
10! to 14! each have 2 zero so total zero = 5*2 =10
5! to 9! each have 1 zero so total zero = 5*1 = 5
total sum = 56

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vyshnav  (9 years ago)

3
2*27=one 2
5*11=one 5
7*8=three 2's
25*3= two*5's
19*4=two*2's

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