find last non-zero digit in 2012!

• =(8)^210*2011*2012
  =4*1*2=8
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• ans :8
  
  ....6*2011*2012=...8
  
• 10 years agoHelpfull: Yes(0) No(1)
• 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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