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(#M40039634) OTHER Numerical Ability question solve Keep an EYE

find last non-zero digit in 2012!

Asked In other Harsh Poddar (10 years ago)

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rakesh  (10 years ago)

=(8)^210*2011*2012
=4*1*2=8

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rakesh  (10 years ago)

ans :8

....6*2011*2012=...8

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sreekanth  (10 years ago)

write the num as (5*a)+b.then the formula for rightmost nonzero digit in num! is rightmost(2^a * a! * b!)
lets solve the above problem
2012=5*402 + 2
so a=402 and b=2
rightmost digit in 2012!=rightmst(2^402 * 402! * 2!)
we know rightmost digit of 2^n is 2 if n%4=1, 4 if n%4=2, 8 if n%4=3 , 6 if n%4 =0.so rightmost digit in 2^402 is 4 and that of 2! is 2.
so,rightmost digit in 2012!=rightmst(4 * 402! * 2)=rightmst( 402! * 8)--->I
apply same rule for 402!=5*80 + 2=rightmst(2^80 * 80! * 2)
=rightmst(6* 80! * 2)=rightmst( 80! * 12)----->II
same for 80!=5*16 + 0
right most of 80!=rightmst(2^16 * 16! * 0!)=rightmst(4 * (2^3 * 3! * 1!) * 2)IV
using IV in I
rightmost digit in 2012!=rightmst(4 * 8 * 2)=rightmst(64)=4!!!Thats the answer.

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