Find the remainder when 83^261 is divided by 17

• 83^261 mod 17
  
  (17*5-2)^261 % 17
  (-2)^261 % 17 [Since 17*5%17 = 0]
  Now 2 nd 17 are co-primes... So we can use Euler's theorem...
  17 => 17^1 [a^p*b^q*..]
  E[17]=16 [ Since E(N)=N*(1-1/a)*(1-1/b)...]
  
  Therefore we can write
  [(-2^16)^16 * (-2)^5] % 17
  1^16 * (-32) % 17 [Since Rem[M^E(N)/N] = 1]
  -32 % 17
  (-32+17) % 17 [Since rem can't be negative]
  -15 % 17
  (-15+17) % 17
  2 % 17
  
  Hence reminder is 2...
  
  Ans : 2
• 10 years agoHelpfull: Yes(15) No(1)
• 83^261 mod 17
  
  (17*5-2)^261 % 17
  (-2)^261 % 17 [Since 17*5%17 = 0]
  Now 2 nd 17 are co-primes... So we can use Euler's theorem...
  17 => 17^1 [a^p*b^q*..]
  E[17]=16 [ Since E(N)=N*(1-1/a)*(1-1/b)...]
  
  Therefore we can write
  [(-2)^16 * (-2)^16 * (-2)^5] % 17
  1 * 1 * (-32) % 17 [Since Rem[M^E(N)/N] = 1]
  -32 % 17
  (-32+17) % 17 [Since rem can't be negative]
  -15 % 17
  (-15+17) % 17
  2 % 17
  
  Hence reminder is 2...
  
  Ans : 2
• 10 years agoHelpfull: Yes(1) No(1)
• 83^261 mod 17
  solution-
  83^4 contain unit digit is 1;
  so 83^(261/4) mod 17;
  now 83^(1) Write only reminder on power;
  (261/4=65 is quotient and 1 is reminder)
  now only 83/17=4 quotient and 15 is reminder
  ans = 15 ;
  
• 10 years agoHelpfull: Yes(0) No(2)
• 83^261%17
  (17*5-2)^261 % 17
  (-2)^261 % 17 [Since 17*5%17 = 0]
  now 261%4=1 we can write 261 as 4n+1
  so that
  (-2)^1/17
  =-2+17=15
  so remainder will be 15
• 10 years agoHelpfull: Yes(0) No(2)
• if 261 is divided by 4 then remainder will be 1
  so, 83^261 became 83^1
  now 83^1 is divided by 17
  then remainder will be 15
  
• 10 years agoHelpfull: Yes(0) No(2)