Find the last 3 digit of 2988^687

• (2988%1000)^687
  988^687%1000
  (988^684)*(988^3)
  (988%1000)^3
  (976144%1000)*988
  (144*988)%1000
  142272%1000
  272
  ans=272
• 11 years agoHelpfull: Yes(12) No(18)
• dividing a no. by 1000,remainder is the last three digit of number
  2988^687/1000
  =(3*1000-12)^687/1000
  =(-12)^687/1000
  ={(-12)^3}^229/1000
  =(-1728)^229/1000
  ={(-1728)^2}^114*(-1728)/1000
  =(2985984)^114*(-1728)/1000
  =(984)^114*(-1728)/1000
  =(-16)^114*(-1728)/1000
  ={(-16)^3}^38*(-1728)/1000
  =(-4096)^38*(-1728)/1000
  =(-96)^38*(-1728)/1000
  =(9216)^19*(-728)/1000
  =(216)^19*272/1000
  =(46656)^9*216*272/1000
  =(656^9)*58752/1000
  =(656^3)^3*752/1000 (consider last 3 or 4 digit of product)
  =(416)^3*752/1000
  =296*752/1000
  =592/1000
  =rem(592)
• 11 years agoHelpfull: Yes(10) No(1)
• 988^2=976144%1000=144
  988^4=144^2%1000=736
  988^8=736^2%1000=696
  988^16=696^2%1000=416
  988^32=416^2%1000=056
  988^64=056^2%1000=136
  988^128=136^2%1000=496
  988^256=496^2%1000=016
  988^512=016^2%1000=256
  
  988^687=988^512.988^128.988^32.988^8.988^4.988^2.988
  
  988^687=256.496.56.696.736.144.988
  
  988^687=592
  
  i think 592 is answer it's correct or not ?
  
• 11 years agoHelpfull: Yes(6) No(6)
• (..8)^(2)=...4 ie. a num ending with 8 raised to the power of 2 gives a num ending with 4 (since 8^2 is 64).similarly 8^3=512..ends with 2.
  8^4=4096 ends with 6..
  8^5=32768..again this series repeats...the cycle is 4...
  hence 687%4=3
  8 6 2..the ans is 2
• 11 years agoHelpfull: Yes(3) No(11)
• 142272%1000
  272
  ans=272
• 11 years agoHelpfull: Yes(2) No(1)
• koi k pass simple explanation nahi hai kyaa...?
• 11 years agoHelpfull: Yes(2) No(0)
• this is for last digit alone
• 11 years agoHelpfull: Yes(1) No(4)
• 272 is the answer
• 11 years agoHelpfull: Yes(1) No(7)
• 592 is the correct answer
  
• 11 years agoHelpfull: Yes(1) No(1)
• last to digit is 88
  88^687=== 687%4==3 is remainder
  88^3===472 answer
• 11 years agoHelpfull: Yes(0) No(0)
• 687/4=remainder 3,
  for remainder 3 base unit=(base unit)^3
  i.e. 8^3=512
• 11 years agoHelpfull: Yes(0) No(0)
• the formula for this problem is
  (x+1)^n = 1+nx+n(n-1)*x^2 therefore the ans will be
  
  (2987+1)^687= 1 + 069 + 641*169
  = 070 + 329
  = 399 answer
• 10 years agoHelpfull: Yes(0) No(0)