what is the remainder of (16937^30)/31

• 16937=16926+11,now 16926 is completely divisible.. So Wat remains is (11^30)/31
  Which is (11^6)^5./31
  11^6 gives 4 as remainder..so 4^5/31 is remaining...which gives 1 as remainder.
  
  
  ANS IS 1
• 10 years agoHelpfull: Yes(27) No(12)
• a^(r-1)/r]reminder=1....
  
• 9 years agoHelpfull: Yes(22) No(3)
• a^(p-1)/p have remainder 1 if
  p=prime no.
  31 is prime no. so ans is 1
• 8 years agoHelpfull: Yes(14) No(2)
• the last digit is comming as 7,9,3,9,3,1 if we do (16937)^1 (16937)^2 (16937)^3....and this seq is repeatng for evry (7)^6 ,.in 30 5 times 6 comes sothe last digit will be (16937)^30 is 1 .. reaminder will be 1
  
• 10 years agoHelpfull: Yes(2) No(8)
• Reminder 1
  According to formula a^p-1/p = rem 1 where a and p are co prime
• 8 years agoHelpfull: Yes(2) No(0)
• i was able to make out quotient please help me
  16937^30 = 1.3834^90
  now let log (1.3834^90)/31 = log x
  then 90log 1.3834 = log x + log 31
  then 12.6852 = log x + log 31
  12.6852 - log 31 + log x
  11.1939 = log x
  x = ant log 11.1939
  x = 1.5629* 10^11
  NOW HOW TO GET REMAINDER
• 10 years agoHelpfull: Yes(0) No(17)
• if a=2 & r=6 then remainder of
  2^5/6 this is not 1.explain why.
• 8 years agoHelpfull: Yes(0) No(4)