What is the last 2 digits of
(101*102*103*197*198*199)

• if we divide (101*102*103*197*198*199) by 100 then remainder gives its last two digit
  
  rem(101*102*103*197*198*199/100)
  =rem(1*2*3*(-3)*(-2)*(-1))
  =rem(-36)
  =rem(100-36)
  =rem(64)
  so, last two digit = 64
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• Answer will be 64
  
  For finding last 2 digit of a number which is obtained from the multiplication of numbers , use the remainder theorem.
• 10 years agoHelpfull: Yes(0) No(2)
• 101*102*103*197*198*199
  =(100+1)(100+2)(100+3)(200-3)(200-2)(200-1)
  =(1)(2)(3)(-3)(-2)(-1)
  = -36
  = 100-36
  = 64 =>Ans
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• 24.......we consider the last 2 digits for every number and then multiply to obtain every last 2 numbers.....
• 10 years agoHelpfull: Yes(0) No(4)
• dividing each no. by 100 and then taking only the remainder, we get
  1*2*3*(-3)*(-2)*(-1)=-36
  therefore the last 2 digit is (100-36)=64
  ans:- 64
• 10 years agoHelpfull: Yes(0) No(0)