remainder when 15^23 + 23^23 divided by 19?

• a^n + b^n is divisible by (a+b) when n is odd
  Here 15^23 + 23^23 is divisible by (15+23) = 38 since 23 is odd
  So remainder[15^23+23^23/38] = 0...
  Since 38 is a multiple of 19, the given expression is also divisible by 19...
  Hence the remainder is 0
  
  Ans : 0
• 9 years agoHelpfull: Yes(69) No(0)
• 15^23 + 23^23 can be written as (15*1)^23+(23*1)^23
  = 1^23(15+23)
  = 15+23
  = 38
  and we know 38/19 yields a remainder of 0
• 9 years agoHelpfull: Yes(12) No(11)
• Welcome @Akib :)
• 9 years agoHelpfull: Yes(1) No(1)
• thanku @saraswathy
• 9 years agoHelpfull: Yes(0) No(1)
• 1+5=6
  1+9=10 both have remainder 0 so same for second case
  as a whole it has a 0 remaninder
• 9 years agoHelpfull: Yes(0) No(1)