Q 137. 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.
Answers
The question is very simple.
Since it is mentioned that there are 28 8s in the last term of the series. That means there are 28 terms in total because there's a single 8 in the first term.
So, the ones place of the sum will be ones digit of (28 times 8) i.e. ones digit of 224 i.e. 4 and 22 as carry that has to be added in the tens place of the sum.
The tens place will be ones digit of [22+ ( 27 times 8)] = ones digit of (22 + 216) = ones digit of 238 i.e. 8 amd the 23 again carry forwarded to the left side for hundreds place.
Hundreds place of the sum will be ones digit of [23 + (26 times 8)] = ones digit of (23 + 208) = ones digit of 231 i.e. 1.
Therefore, last three digits of the sum will be 184.
Answer:986
Step-by-step explanation:
21*8=168 carry 16
20*8=160+16=176 carry 17
19*8=152+17=169
the value is 968