What is the remainder of (16937^30)/31?

• ans. 1
  sorry for the previous soln
  11^30 which can be written as 121^15 = (124-3)^15
  thus again using binomial the rem is (-3)^15 = -(27)^5 = -(31-4)^5
  which leaves the rem as -(-4)^5 = 4^5 = (2^5)^2 = (32-1)^2
  which again leaves the rem (-1)^2 = 1
• 11 years agoHelpfull: Yes(14) No(4)
• Using Remainder Theorem :-
  16937^30/31
  ==> 11^30/31
  ==> 29^10/31
  ==> 4^5/31
  ==> 64*4*4/31
  ==> 2*4*4/31
  ==> 32/31
  Remainder is 1
• 11 years agoHelpfull: Yes(11) No(2)
• ans. 11
  ((16937)^30)/31 = (16926 + 11)^30/31
  16926 is completely divisible by 31
  thus by binomial expansion we are left with only 11^1 = 11
• 11 years agoHelpfull: Yes(2) No(9)
• 1 is the asnwer !!!
• 11 years agoHelpfull: Yes(2) No(0)
• 121^15 = (124-3)^15
  (-3)^15 = -(27)^5 = -(31-4)^5
  which leaves the rem as -(-4)^5 = 4^5 = (2^5)^2 = (32-1)^2
  which again leaves the rem (-1)^2 = 1
• 11 years agoHelpfull: Yes(1) No(0)
• answer will be :11
  
• 10 years agoHelpfull: Yes(1) No(0)
• Answeris 1.
• 11 years agoHelpfull: Yes(0) No(2)
• 1
  is the answer
• 11 years agoHelpfull: Yes(0) No(1)