8 + 88 + 888 + ........ + 8888 ........... 8888 . There are 21 " 8 " digits in…
2026
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 .
- A.
950
- B.
970
- C.
968
- D.
900
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Correct answer: C
Answer: 968
There are 21 terms because the last term contains 21 digits of 8. Count how many times 8 appears in each place value and add with carries.
Units place: 8 appears 21 times → 21 × 8 = 168 → write 8, carry 16.
Tens place: 8 appears 20 times → 20 × 8 = 160. Add carry 16 → 176 → write 6, carry 17.
Hundreds place: 8 appears 19 times → 19 × 8 = 152. Add carry 17 → 169 → write 9, carry 16 (to higher places).
Higher places: remaining carry does not affect the last three digits.
Reading the hundreds, tens and units gives 9, 6, 8. Hence the last three digits are 968.