(16937)^30 is divisible by 31

• Ans is:1
  (16937)^30
  16937/31 Remainder is 11
  11^30/31
  11^n unit digits are 1,1,1,1,...n+1
  So 11^30/31=Ans is 1..
• 10 years agoHelpfull: Yes(12) No(3)
• fermats theorem remainder would be 1
  
• 10 years agoHelpfull: Yes(4) No(1)
• the remainder is 18
• 10 years agoHelpfull: Yes(2) No(3)
• reminder is 1 ,using fermets theorem
• 10 years agoHelpfull: Yes(2) No(1)
• 16937^30/31
  [(546*31)+11]^30/31
  11^30/31
  if a^(r-1)/r then remainder is 1
  so 11^30/31 remainder is 1
• 10 years agoHelpfull: Yes(2) No(3)
• unit digit 7, 7^30 so last two digit is 49, 49-31= 18 so ans 18
• 10 years agoHelpfull: Yes(1) No(1)
• 16937/31= remainder 11
  (11^30)=(121^15)
  ((124-3)/31=remainder -3
  3^15=27^5
  27^5=(31-4)^5
  remainder=4^5=64*16
  (63+1)*16
  answer is 16
  
• 10 years agoHelpfull: Yes(1) No(1)
• why u have subtracted 49 @appu
  
• 10 years agoHelpfull: Yes(0) No(1)
• khusboo........apply ur formula on 6^2/3.....
  
• 10 years agoHelpfull: Yes(0) No(0)
• eulers formula
  euler no. of prime no is no.-1
  so
  
  31-1=30
  in such cases remainder will be one always
• 10 years agoHelpfull: Yes(0) No(0)