8+88+888+8888…. There are 21 “8”digits in the last term of the series. Find last three digits of sum.
a.458
b.648
c.658
d.568

• 968
  
  There is one 8 in first term, two 8s in second term, so twenty one 8s in 21st term.
  While summing all terms of this series, we observe that, there are 21, 20 and 19 number of 8s in units, tens and hundredth place respectively.
  So sum of all units digits=21x8=168, with carry 16 to tens place.
  Sum of all tens digits=20x8=160 + 16 carry=176, so carry 17 to hundredth place.
  Sum of all hundredth digits=19x8=152 + 17 carry=169, so carry 16 to thousand place.
  Thus last 3 digits( hundred, tens, unit, ) of sum are 968
• 6 years agoHelpfull: Yes(15) No(0)
• i donot even get 968 in options
• 6 years agoHelpfull: Yes(1) No(0)
• 568 option d
• 6 years agoHelpfull: Yes(1) No(1)
• Ans 648
  
  For last digit 21×8=168. Insert 8 carry 16
  For last 2 nd digit 21*8=168+ cqrry
  168+16=184 insert 4. Carry 18
  Last 3rd 21*8=168+18=186
  Insert 6 carry 18
  
  3rd&&2nd&&last
  648
• 6 years agoHelpfull: Yes(1) No(5)
• 8(1+11+111+...+111... 11)
  =8*(1*1,1*2,1*3,....,1*19,1*20,1*21)
  =8*(...121)last three digits
  =.... 968 last three digits
• 5 years agoHelpfull: Yes(1) No(0)
• Answer is 968 . But the option given here is wrong.!!!
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• d
  21 terms i.e. last digit of the sum is (21*8)= 8 and 16 in hand.
  second last digit is (20*8)= 6 and 17 in hand
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• option d .568
• 6 years agoHelpfull: Yes(0) No(2)
• given that find last 3 digits sum
  we are adding numbers so
  take last 3 digits sum
  888
  888
  888
  ans 568(last 3 digits)
• 6 years agoHelpfull: Yes(0) No(0)
• 968
  for last digit (8*21)=168 ,8 remains,16 carry
  for 2nd last digit 16+(8*20)=176 ,6 remains ,17 carry
  for 3rd last digit 17+(8*19)=169 ,9 remains ,16 carry
  so final ans is ............968.............
• 5 years agoHelpfull: Yes(0) No(0)