Q). If X = ( 16^3 + 17^3 + 18^3 + 19^3 ), then X divided by 70 leaves a remainder of
Option
1) 0
2) 1
3) 69
4) 35

• X = ( 16^3 + 17^3 + 18^3 + 19^ 3 )
  
  (16^3+19^3) is divisible by 16+19=35
  (17^3+18^3) is divisible by 17+18=35
  
  so,X = ( 16^3 + 17^3 + 18^3 + 19^ 3 ) has a factor 35
  
  also, X = ( 16^3 + 17^3 + 18^3 + 19^ 3 )is an even no. so, X has a factor 2.
  [since, E+O+E+O=E]
  
  so, X is completely divisible by 70 & leaves a remainder 0
  
  1) 0
• 10 years agoHelpfull: Yes(36) No(0)
• . x = 163 + 173 + 183 + 193
  = (163 + 193) + (173 + 183)
  = (16 + 19)(162 + 16 × 19 + 192) + (17 + 18)(172 + 17 × 18 + 182)
  = 35 × (an odd number) + 35 × (another odd number) = 35 × (an even number)
  = 35 × (2k) … (k is a positive integer)
  ∴ x = 70k
  ∴ x is divisible by 70.
  Remainder when x is divided by 70 = 0
  Hence, option 1
• 10 years agoHelpfull: Yes(2) No(1)
• rakesh y did u take 16 nd 19 as a pair.....y not 16 nd 17...pls explain
• 10 years agoHelpfull: Yes(1) No(3)
• we know sum of cubes is given by the formula [n(n+1)/2]^2.
  therfore sum of the above series will be [19*20/2]^2-[15*16/2]^2=21700 which is divisible by 70 giving a remainder 0
• 10 years agoHelpfull: Yes(1) No(1)
• Last two digit of all the 4 terms respectively =36,33,32,39
  36+33+32+39=140%70=0 (ans)
• 10 years agoHelpfull: Yes(1) No(0)
• none of these
• 10 years agoHelpfull: Yes(0) No(0)