8 + 88 + 888 + 8888 + ...... up to 24 terms. What are the last three digits of…

Concept: In column (place-value) addition, the digit contributed to each place -- units, tens, hundreds, ... -- by every term is summed independently. The digit finally written at a place is (that column's own sum + carry coming in from the place to its right) mod 10, and the carry passed on to the next place is that total divided by 10 (integer part). This lets us find only the last few digits of a huge sum without evaluating the whole number, provided we count, for each place, only the terms that actually have a digit there.

Application:

1. The 24 terms are 8, 88, 888, 8888, ..., up to a 24-digit term -- every term is made only of the digit 8. All 24 terms have a units digit of 8. Only terms with 2 or more digits (23 of the 24 terms -- from 88 up to the 24-digit term) have a tens digit of 8. Only terms with 3 or more digits (22 of the 24 terms -- from 888 up to the 24-digit term) have a hundreds digit of 8.

2. Units place: 24 terms contribute 8 each: 24 x 8 = 192. Units digit = 2, carry = 19.

3. Tens place: 23 terms contribute 8 each: 23 x 8 = 184. Add the incoming carry: 184 + 19 = 203. Tens digit = 3, carry = 20.

4. Hundreds place: 22 terms contribute 8 each: 22 x 8 = 176. Add the incoming carry: 176 + 20 = 196. Hundreds digit = 6.

5. Reading the digits found in order (hundreds, tens, units) gives the last three digits of the sum: 632.

Cross-check: each term is 8 times a string of ones (8 = 8x1, 88 = 8x11, 888 = 8x111, ...), so the whole sum equals 8 times the sum of the first 24 repunits. Adding the 24 terms directly and reducing the running total modulo 1000 at every step gives the same last three digits, confirming the column-wise result independently.

Answer: 632