what is remainder of (16937^30)/31

• Its according to the Fermat's Theorem which says that the remainder of a^(n-1)/n will be 1 only if n is a prime number. Hence the answer of this question will be 1. :)
• 8 years agoHelpfull: Yes(19) No(0)
• 16937^31-1/31=1
  as a^p-1/p=1
• 8 years agoHelpfull: Yes(5) No(0)
• According to the Fermat's Theorem
  remainder of a^(n-1)/n will be 1 (n=prime no.)
  so......the ans is "1".....
• 8 years agoHelpfull: Yes(1) No(0)
• 1 will be the answer according to fermat's theorem..
• 8 years agoHelpfull: Yes(0) No(1)
• There is a formula for this remainder problem"R(a^m/b)=R(a/b)m
  (16937/31)^30,the ans will be 1.
• 8 years agoHelpfull: Yes(0) No(0)
• 6^2+(r-2)^2 = r^2
  r=20
  
• 8 years agoHelpfull: Yes(0) No(0)