Find last three digits of 2988^687

• (2988^687)/1000 leaves remainder as the last 3 digits of result
  this can be = 988^(4*171+3)=988^3 by cyclicity
  this leaves 272 as last 3 digits
  
• 11 years agoHelpfull: Yes(5) No(14)
• please give a proper solution
• 11 years agoHelpfull: Yes(2) No(1)
• unit digit of 2988 is 8
  The cyclicity of 2,3,7,8 is 4
  The cyclicity of 0,1,5,6 is the no. itself
  The cyclicity of 4,9 is 2
  To check whether the number is divisible by 4 we have to divide the last two digit by 4. Likewise in 687 we have to divide 87 by 4 we get the remainder 3.So we have to put 3 as a power of unit digit i.e. 8(in base number)
  8^3=432 therefore the ans is 432
• 11 years agoHelpfull: Yes(0) No(8)
• @chandra lekha
  Cycle is not repeating at power of 4 right?
  then how u taken as multiple of 4 and all...
  pls explain...
• 11 years agoHelpfull: Yes(0) No(0)
• pls explain properly
• 11 years agoHelpfull: Yes(0) No(0)
• @durga 8^3=512 not 432
  so,ans:512
• 11 years agoHelpfull: Yes(0) No(0)
• mine came 592 :p
• 11 years agoHelpfull: Yes(0) No(0)
• (2988^687)/1000 leaves remainder as the last 3 digits of result
  this can be = 988^(4*171+3)=988^3 by cyclicity
  this leaves 272 as last 3 digits
• 11 years agoHelpfull: Yes(0) No(0)
• ans is 504
  
• 8 years agoHelpfull: Yes(0) No(0)