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.
Choose the correct choice from the following choices
A) 958  B) 968  C) 978  D) 988  E) 999
As their is 21 no of 8 will add in first summation then 20 and then 19.   so 21*8= 
<gadget id="calculator">21*8</gadget>
<output>168</output>
168 from that 8 will come 16 as a carry   and 20*8= 
<gadget id="calculator">20*8</gadget>
<output>160</output>
160 , 160 + 16 (carry from first term)= 176 and from that 6 will come 17 as a carry   at last  19*8= 
<gadget id="calculator"> 19*8</gadget>
<output>152</output>
152 , 152+17=169 , so from that 9 will come   and answer become 968   ANSWER:968
<result>B</result>
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.