find last 3 digits.......2988^687^241?

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  2988^687^241 mod 1000
  =988^687^241 mod 1000
  let 687^241 be n
  988^n mod 1000
  247^n mod 1000 * 4^n mod 1000
  first for 247^n mod 1000
  lambda(1000)=100
  so, we need 687^241 mod 100
  lambda(100)=20
  241 mod 20=1
  687^1 mod 100 = 87 mod 100 or -13 mod 100
  now,
  247^(-13) mod 1000
  multiplicative inv of 247 is 583
  583^13 mod 1000
  863
  now for 4^n mod 1000
  =(2^(2n-3) mod 125)*8
  lambda(125)=100
  So, we need 687^241 mod 100 = 87 mod 100 (already calculated)
  So n=87
  2*n-3= 171
  (2^171 mod 125)*8
  =(2^71 mod 125)*8 (Since 171 mod 100 =71)
  reduce using 2^(10*7+1) mod 125
  you'll get
  (98 mod 125)*8
  =784
  now 784*863 mod 1000
  =592(ans)
  
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