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Remainder problem
by ketav sharma - Monday, 19 September 2011, 02:00 PM
  2^2001/2001 remainder ?
With solution

Re: Remainder problem
by sankar a - Friday, 23 September 2011, 01:12 AM
  2001=3*23*29

2^2001 mod 3 =2
2^2001 mod 23=12
2^2001 mod 29=14

Solve the following equations using Chinese Remainder Theorem
3x+2=23y+12=29z+14

we get N= 2001k + 449

Remainder is 449 tongueout tongueout