7 + 77 + 777 + ... up to 23 terms (the 23rd term consists of 23 sevens). What…

Concept: To find the last three digits (the remainder on dividing by 1000) of a sum of many numbers, add the terms column by column — units, tens, hundreds — exactly as in ordinary long addition, carrying between columns; the full sum is never needed. A term contributes to a column only if it actually has a digit there — a term shorter than that column position contributes nothing to it.

Application:

1. Every term of this series is 7 times a repunit — a number made only of 1s. The k-th term (which has k sevens) equals 7 × Rk.

2. So the whole sum equals 7 × (R1 + R2 + … + R23). Finding the last three digits of the whole sum reduces to finding the last three digits of R1 + R2 + … + R23, then multiplying by 7.

3. Units column: every one of the 23 repunits has a 1 in the units place, so the units-column total is 23 — write 3, carry 2.

4. Tens column: only repunits with 2 or more digits (R2 through R23 — 22 of them) have a 1 in the tens place; adding the carry gives 22 + 2 = 24 — write 4, carry 2.

5. Hundreds column: only repunits with 3 or more digits (R3 through R23 — 21 of them) have a 1 in the hundreds place; adding the carry gives 21 + 2 = 23 — write 3, carry 2 (any further carry only affects digits beyond the hundreds place, which do not matter here).

6. So R1 + R2 + … + R23 ends in …343. Multiplying by 7: 7 × 343 = 2401, so the whole sum ends in 401.

Cross-check: Test the same column-addition method on a short version of the series, where the full sum can be checked directly. For 3 terms: R1 + R2 + R3 = 1 + 11 + 111 = 123, and 7 × 123 = 861 — which is indeed 7 + 77 + 777 computed directly. The method reproduces the direct sum exactly, confirming it can safely be extended to 23 terms without ever writing out the full 23-digit number.

So the last three digits of the sum are 401.