last three digit of 2988^678

• (2988)^687= ((2^2)*(3^2)*83)^(6*113)
  
  = ((2^12)*(3^18)*(83^6))^113
  = (2^1356)*(3^2034)*(..69^113)
  
  now powers of 69 repeat itself after 69^10
  
  so (..6)*(..9)*(..09)=486
• 11 years agoHelpfull: Yes(0) No(5)
• This is a tough question.
  Last three digits of an expression is nothing but the remainder when we divide the given expression by 1000.
  So 29886781000=(−12)6781000=126781000
  We write 1000 as 8 ×125. So we separately find the remainders by dividing the given number by 8 and 125.
  Given expression is clearly divisible by 8. Also
  Euler totient theorem says [aϕ(N)N]Remainder=1
  Here ϕ(N) is the number of co primes of a number N. In this case ϕ(125) = 100.
  So (12100)6.1278125=16.1278125
  Now 1278=478.378=(1+15)39.(10−1)39
  (1+15)39.(−1)39(1−10)39
  - (1+39C1.15+39C2.152.... ). (1+39C1.10+39C2.102−39C3.103....
  If you observe clearly, 153 and 103 are 15 multiples. So we can omit the remaining terms.
  -(1+39.15 + 39.19.152 ).(1-39.10 + 39.19.102)
  By writing 39 as 40-1 we can simplify this equation fast.
  -(1+(40-1).15 + (40-1).19.152).(1-390 + (40-1).19.100
  -(1+100 - 15 - 19.152 ).(1-140 - 19.100)
  -(101 - 15 - (20-1).152).(1-140+100)
  -(101-15+100).(-39)
  -(61).39
  -2379 = 4
  So the remainder is 4.
  The original expression is in the format of N = 8K = 125 L + 4
  The minimum number satisfies this condition is L = 4.
  So 125 (4) + 4 = 504
  So answer is 504
  
• 9 years agoHelpfull: Yes(0) No(0)