find the remainder when 59^73^5! is divided by 37?
find the reminder when 6^65^56 is divided by 43?
find the reminder when 20!+20^23 divided by 23?

• 59^73^5! / 37 => 5! = 120 (divisible by 4) => 59^73^1 / 37
  73 when divided by 4 gives remainder 1.
  => 59^1/37
  => remainder is 22
• 10 years agoHelpfull: Yes(6) No(1)
• can any one please explian the above 3 answers...
• 10 years agoHelpfull: Yes(2) No(0)
• Any number raised to the power 36 would give remainder as 1 when divied by 37.....so find the remainder of 73^5!=73^120 with respect to 36...it gives a rainder of 1...and 59 gives remainder 22 when divided by 37.....So solving it u get remainder as 22
• 10 years agoHelpfull: Yes(1) No(1)
• 3) we know that (p-2)!/p remainder 1, when p is prime and greater than 5
  hence (23-2)!/23 remainder 1
  ie 21!/23 remainder 1
  now we can write 21! as 21x20!
  so,
  21x20!/23 remainder 1
  ie, 21/23 x 20!/23 remainder 1
  now, for, 21/23 remainder 21 or -2
  now see that if a is he remainder for 20!/23, then a x -2 should be 24 for total remainder o be 1.. (as for 24/23 remainder 1) so a =-12
  so 20!/23 remainder is -12 or11
  
  now the second part..
  20^23/23
  for 20/23 remainder is 20 or -3, hence remainder -3^23
  for each -3^3/23 yields a remainder -4 (-27/23)
  we can write -3^23/23 as (-3^3)^7 x -3^2
  ie, -4^7 x-3^2/23
  now each -4^3/23 yields a remainder of -5,
  so for -4^7/23 yields a remainder -5^2 x -4
  and -4^7 x -3^2/23 yields a remainder -5^2x-4x-3^2 ie 25x-4x9 ie 100
  again 100/23 yeilds a remainder 8
  
  previously from 1st part we got a remainder 11, and 2nd part 8..
  hence total remainder 19
• 10 years agoHelpfull: Yes(1) No(1)
• 59^73^5! Div by 37 reminder is 12
  6^65^56 div by 43 reminder is 6
  20!+20^23 reminder is 0
• 10 years agoHelpfull: Yes(0) No(9)
• can u please explain the answer
• 10 years agoHelpfull: Yes(0) No(0)