Latest other Aptitude Question SOLUTION: find last non-zero digit in 2012!

• Placement Papers
• /
• other
• /
• Numerical Ability
• /
• Number System

other Question Papers:

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

find last non-zero digit in 2012!

Asked In other Harsh Poddar (9 years ago)

Unsolved Read Solution (3)

Is this Puzzle helpful?   (0)   (0) Submit Your Solution Number System

Read Solution (3) :

rakesh  (9 years ago)

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

Is this solution Helpfull? Yes (1) | No   

rakesh  (9 years ago)

ans :8

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

Is this solution Helpfull? Yes | No (1)  

sreekanth  (9 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.

Is this solution Helpfull? Yes | No   

Please Login to Read Solution.