find the remainder when 1^39 + 2^39 + 3^39 + 4^39+ . . . . . +12^39 when divided by 39

• Fermat's Little Theorem: a^p = amod(p) where p is prime
  x^39 = (x^3)^13 = (x^3)mod(13) ... 13 is prime not 39
  thus, sum of (x^3)mod(13) where x goes from 1-12
  Modulo operator property: amod(n)+bmod(n) = [(a+b)mod(n)]mod(n) ... quite logical too
  sum = (sum of x^3 from 1-12)mod(13)
  sum = [(12)(12+1)(2*12+1)/6] mod(13) = 26*25mod(13) = 0
• 3 years agoHelpfull: Yes(0) No(0)