Latest Elitmus Aptitude Question SOLUTION: Find the quotient when total no. Of divisors of 16! is divided by right most non-zero digit of 15!

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(#M40157785) ELITMUS Numerical Ability question Keep an EYE

Find the quotient when total no. Of divisors of 16! is divided by right most non-zero digit of 15!

Asked In Elitmus Sniper (7 years ago)

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abdullah imran  (7 years ago)

16!= (2^4) * (3*5) * (2*7)* (13)*(2^2*3) * (11)*(2*5)*(3^2)*(2^3)*(7)*(2*3)*(5)*(2^2)*1
=(2^15)*(3^6)*(5^3)*(7^2)*(11)*(13)
so number of divisor= (15+1)*(6+1)*(3+1)*(2+1)*(1+1)*(1+1)
=16*7*4*3*2*2
=5376

The generalized formula to find the non zero digit of N = [N/5]!×2^[N/5]×[N/5]R!
(see http://www.campusgate.co.in/2013/10/finding-right-most-non-zero-digit-of.html )
=[15/5]! * 2 ^[15/5] * 0!
=48
=8(unit digit)

So Required Anwser=5376/8= 672

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pp  (7 years ago)

first find the factors of 16!
and then divide with non zero digit of 15!
factors of 16!
prime no below 16 are 2 3 5 7 11 13
now highest power of 2 in 16! is 15
3is 6,5 is 3,7 is 2,11,is 1,13 is 1
so no of factors is 16*7*4*3*2*2=5376
now last digit in 15!
it can be written as 2^11*3^6*5^3*7^2*11^1*13*1
5^3*2^3 gives 0 so remaining is 2^3 =8

therefore 5376/8=672

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sambeet mandal  (7 years ago)

how did you get the number of divisors?

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