8+88+888+.....+8888........8888. There are 21 “8” digits in the last term of the series.
Find the last three digits of the sum.

• 968 is the answer.
  
  As their is 21 no of 8 will add in first summation then 20 and then 19.
  so 21*8=168 from that 8 will come 16 as a carry
  and 20*8=160 , 160 + 16 (carry from first term) = 176 and from that 6 will come 17 as a carry
  at last 19*8=152 , 152+17=169 , so from that 9 will come
  and answer become 968
• 9 years agoHelpfull: Yes(47) No(3)
• Simply add
  In Unit Digit (8+8+8+....21times)=21*8=168
  unit digit =8, carry =16
  In tens Digit (8+8+8+....20times)=20*8=160+(16)=176
  unit digit =6, carry =17
  In hundred Digit (8+8+8+....19times)=19*8=152+17=169
  unit digit =9, carry =16
  so last three digits= 968
• 9 years agoHelpfull: Yes(29) No(0)
• there are 21 8's at unit digit which results to (21*8=168) 8 at the unit digit of the sum and 16 is the carry for tens digit number. there are 20 8's at tens digit that results to (20*8=160) 160+16(from the carry)=6 at the tens digit and now 17 is the carry. similarly for 100th place (19*8)results to 152+17 and digit is 9.
  hence answer is 968.
• 9 years agoHelpfull: Yes(4) No(0)
• answer is 968
• 9 years agoHelpfull: Yes(3) No(0)
• 21th term- 8888......888
  20th term- 8888.......88
  19th term- 8888.......8
  --------------------------------
  568
  
• 9 years agoHelpfull: Yes(0) No(9)
• 21*8= 168
  21*7+16=163
  21*6+16= 142
  so the last three terms are 238
• 9 years agoHelpfull: Yes(0) No(5)