Latest M4maths Aptitude Question SOLUTION: upto how many integers we have to multiply 1*2*3*4*5*...... to make 31 zeros at the end

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(#M40034876) M4MATHS Numerical Ability question 8-4-2014 puzzle Keep an EYE

upto how many integers we have to multiply 1*2*3*4*5*......
to make 31 zeros at the end

Asked In M4maths Ghanshyam Gulati (10 years ago)

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arnold  (10 years ago)

we can do following
if the number is close to multiple of 5
for example
take 101!
the closest multiple is 100
divide 100 by 5 we get 20 multiple, now we divide 100 by 5^2 we get 4 multiple
thus we have 24 trailing zeros in 101 !
similarly, for 125!
125 divide by 5= 25 multiple
125 divide by 5^2=5 multiple
125 divide by 5^3=1 multiple
thus total we have 31 trailing zeros
in short, for trailing zeros
n!= n/5+n/(5^2)+ n/(5^3)+.....

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sruthi  (10 years ago)

the integers should multiply and give product as 125
so
125/5=25
25/5=5
5/5=1
25+5+1=31
31 zeroes will be there

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