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

Concept: To find only the last k digits of a large sum, you don't need the full sum — you can compute it modulo 10^k by adding column by column (units, tens, hundreds, ...) exactly like manual addition, carrying the overflow into the next column. This works because any digit beyond the k-th place of a number contributes a multiple of 10^k, which vanishes modulo 10^k.

Application: There are 23 terms — 7, 77, 777, ..., a 23-digit number all of whose digits are 7. Compute the last three digits column by column:

1. Units column: all 23 terms contribute 7 to the units place, so the sum is 23×7 = 161, giving digit 1 with a carry of 16.

2. Tens column: only the 2nd through 23rd terms (22 terms) contribute 7 to the tens place — the 1st term is the single digit '7' and contributes 0 to tens; the sum is 22×7 + 16 (carry) = 170, giving digit 0 with a carry of 17.

3. Hundreds column: only the 3rd through 23rd terms (21 terms) contribute 7 to the hundreds place; the sum is 21×7 + 17 (carry) = 164, giving digit 4.

4. Reading the digits hundreds-tens-units gives 4, 0, 1 — so the last three digits are 401.

Cross-check: Directly computing the full 23-term sum independently also gives a number ending in ...401, confirming the column-wise result above.