Latest CAT Aptitude Question SOLUTION: find the number of consecutive zero at the end of 1!*2!*3!*4!*5!*-------*50!

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(#M40152631) CAT Numerical Ability question Keep an EYE

find the number of consecutive zero at the end of 1!*2!*3!*4!*5!*-------*50!

Asked In CAT pratik (8 years ago)

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NITESH KUMAR SINGH  (8 years ago)

262 zeroes
no of zeroes in 50!=12
no of zeroes in 49!=10
no of zeroes in 48!=10
.
.
.no of zeroes in in 45!=10
similarly
no of zeroes in 49! to 45!=10*5
no of zeroes in 44! to 40!=9*5
no of zeroes in in 35! to 39!=8*5
.
.
no of zeroes in20! to 24!=4*5 as no 25 has 2 multiple of 5
.
.
no of zeroes in 5! to 9!=1*5
total no= 12 + ((10*11)/2)*5-25=262

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Appala Naidu Gollu  (8 years ago)

12 zeros
short cut
50/5=10
10/5=2
total 10+2= 12 zeros

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ekta  (8 years ago)

appala can u pls explain ur trick

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tarshika biswas  (8 years ago)

We can clearly see that in the above expression, the no. of 2's is greater than the no. of 5's.. So we need to count the no. of 5's here.
then we get, 5! * 10! * 15! * 20! * 25! * 30! * 35! * 40! * 45! * 50!
= (5/5) * (10/5) * (15/5) * (20/5) * ((25/5) + (25/25)) * ((30/5) + (30/25)) * ((35/5) + (35/25)) * ((40/5) + (40/25)) * ((45/5) + (45/25)) * ((50/5) + (50/25))
= 1 * 2 * 3 * 4 * 6 * 7 * 8 * 9 * 10 * 12 = 8709120 No.s of zeros will be there... but I think there will be '+' instead of '*' in the question

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