Find the last digit of 1^3 + 2^3 + .............99^3

1) 0
2) 1
3) 2
4) 5
5) 8

• 1) 0 is correct ans
  n^2(n+1)^2/4 is formula for this qs
• 8 years agoHelpfull: Yes(0) No(0)
• answer is 0.
  since 1^3+2^3+.....9^3=45
  therefore last digit of 1^3+2^3+3^3+...........+99^3=0 (45*10=450)
  
• 8 years agoHelpfull: Yes(0) No(0)
• ANS is 0
  By using Arithmeti Progression
  (1^3+2^3+3^3+4^3+......+n^3) =1/4*n^2*(n+1)^2
  so here n=99
  1/4*99^2*100^2=24502500
  from above last digit i s 0
• 8 years agoHelpfull: Yes(0) No(0)
• 1^3+2^3+3^3+4^3+...................+9^3 repeat 10 times if we consider only last digit.....
  and this unit digit of cube result is a sum of 1 to 9=45
  this will 10 times means 45*10=450
  so the last digit is 0.(ans)
• 6 years agoHelpfull: Yes(0) No(0)
• is the answer 0
• 5 years agoHelpfull: Yes(0) No(0)
• 0 is the correct answer for it
  use formulae (n*(n+1))^2/4
• 5 years agoHelpfull: Yes(0) No(0)
• 1^3=1
  2^3=8
  3^3=7(last digit)
  4^3=4
  5^3=5
  6^3=6
  7^3=3
  8^3=2
  9^3=9
  10^3=0
  
  Sum of all last digit= 0
  
  so we can conclude one thing
  11^3=1
  12^3=8
  .
  .
  .
  .
  .
  .
  99^3=9
  
  so by this
  for every 10 numbers addition, it is 0
  
  So final answer = ZERO ........... i.e. 0
• 4 years agoHelpfull: Yes(0) No(0)
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