Find the remainder when number represented by 22334 raised to power (1^2+2^2 +......+66^2) is divided by 5

• 1^2+2^2+.....+66^2=n*(n+1)*(2n+1)/6=98021
  22334^98021/5
  rem of 22334/5=4
  hence 4^98021/5
  =4..ans
• 11 years agoHelpfull: Yes(18) No(1)
• answer is 4
  
  we know a number with units digit as 4 raised to any power has units digit 4 or 6
  so when we divide the number by 5 we have to check units digit
  if units digit is 4 remainder is 4 and if units digit is 6 remainder is 1
  
  now 1^2+2^2+.....+66^2 has 33 even terms i.e 2,4,6 etc whose square is always even
  so when all these even terms are added we get another even term
  this expression has 33 odd terms whose square is always odd
  so when these 33 odd terms are squared and added we get sum of 33 odd terms which is odd
  
  so we see that there is an addition of even and odd terms which give the overall result as odd
  so the number is raised to odd power so its unit digit is 4
  as number whose unit digit is 4 when this number is raised to odd power we get 4 as unit digit
  for ex 4^1 = 4, 4^3 = 64 etc
  
  so as the unit digit of the number is 4 we get the remainder as 5
  
• 11 years agoHelpfull: Yes(7) No(1)
• ans may be 4
• 11 years agoHelpfull: Yes(1) No(0)
• check mail
• 11 years agoHelpfull: Yes(0) No(6)