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

Concept: When several numbers made of one repeated digit (8, 88, 888, …) are added together, don't try to add the huge total directly — add column by column (units, tens, hundreds, …), exactly like ordinary column addition. At any given place value, only the terms that actually have a digit there contribute 8 to that column's sum, and any extra tens are carried forward into the next column.

Application: The series has 24 terms: 8, 88, 888, …, up to a 24-digit term. Build the last three digits of the sum from the units column upward.

1. Units place: every one of the 24 terms has a units digit of 8, so this column totals 24 × 8 = 192. Write down 2 and carry 19 into the tens column.

2. Tens place: only the terms with two or more digits contribute an 8 here — that is all terms except the first ("8" itself), i.e. 23 terms. Adding the carry, this column totals 23 × 8 + 19 = 184 + 19 = 203. Write down 3 and carry 20 into the hundreds column.

3. Hundreds place: only the terms with three or more digits contribute an 8 here — that is all terms except the first two ("8" and "88"), i.e. 22 terms. Adding the carry, this column totals 22 × 8 + 20 = 176 + 20 = 196. Write down 6.

Reading the digits found from the hundreds column down to the units column gives 6, 3, 2 — so the last three digits of the sum are 632.

Cross-check: Check the column-and-carry method on a short version of the same series before trusting it for 24 terms. For just 4 terms: 8 + 88 + 888 + 8888 = 9872 by direct addition. Using the same column logic — units: 4 × 8 = 32 → digit 2, carry 3; tens: 3 × 8 + 3 = 27 → digit 7, carry 2; hundreds: 2 × 8 + 2 = 18 → digit 8, carry 1; thousands: 1 × 8 + 1 = 9 — reproduces 9872 exactly. This confirms the method is reliable, so applying the same reasoning through all 24 terms correctly gives 632.