The last digit of the expression 4+92+43+94+45+96+…….+499+9100 is ?

• 4+92+43+94+45+96+…….+499+9100
  = 4+(92+43+94+45+96+…….+499+9100)
  = 4+(92+94+96+……+9100)+(43+45+47+...+499)
  = 4 + [4505/2 *(92+9100)] + [229/2 *(43+499)]
  = 4 + 4505*4596 + 229*271
  => unit digit = 4 + 0 + 9 = 3
• 10 years agoHelpfull: Yes(55) No(3)
• 41+43+45+......499+92+94+96+......9100
  for 41+43+45
  l=a+(n-1)*d
  499=41+(n-1)2
  n=230
  s=n(a+l)/2
  s=230(41+499)/2=621
  similarly
  for 92+94+96
  s=last two digit is 10
  then 21+10=31
  ans=31
• 10 years agoHelpfull: Yes(0) No(7)
• in alakh solution
  
  4 is not added??
  so
  4+21+10=5 will be the ans
  
• 10 years agoHelpfull: Yes(0) No(2)
• easy one
  ans is 3
  
  
  
• 10 years agoHelpfull: Yes(0) No(2)