AMC 8 · 2007 · #18

Grade 5 number-theory
Archetype Lattice / Grid Pattern from Small Cases └ Sub-archetype Periodic Placement Formula
multi-digit-arithmeticplace-valuepattern-recognitionunits-digit-tracking pattern-recognitionidentify-subproblemseasier-related-problem ↑ Prerequisites: multi-digit-arithmeticplace-value
📏 Short solution 💡 2 insights
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Problem
Multiply two 99-digit numbers — N₁ = 303,030,303,…,030,303 and N₂ = 505,050,505,…,050,505. Call the thousands digit of the product A and the units digit B. Find A + B.

Pick an answer.

(A)
3
(B)
5
(C)
6
(D)
8
(E)
10

AMC 8 2007 problem © Mathematical Association of America (MAA AMC). Reproduced for educational use.

How to solve
Strategy Change Focus / Count the Complement

Tool #16 (Change Focus) is the key move: instead of computing the whole 198-digit product, focus only on what the question actually needs — the last four digits. Standard column multiplication shows that the last four digits of a product depend only on the last four digits of each factor; everything to the left feeds into higher places and never comes back down. Tool #9 (Easier Problem) confirms the shortcut by trying a baby version (much shorter 303… and 505… numbers) and checking that the last four digits of the product are unchanged.

1STEP 1

Ignore the whole product: the thousands and units digits sit in its last four places, so only each factor's last four digits reach them.

last 4 digits of N₁ × N₂ = last 4 digits of (last 4 digits of N₁) × (last 4 digits of N₂)
2STEP 2

Read each factor's last four digits: N₁ ends 0303, N₂ ends 0505 — so we multiply 303 by 505.

last 4 of N₁ = 0303 = 303, last 4 of N₂ = 0505 = 505
3STEP 3

Multiply the two small numbers by partial products to get 153015.

303 × 505 = 300 × 500 + 300 × 5 + 3 × 500 + 3 × 5 = 150000 + 1500 + 1500 + 15 = 153015
4STEP 4

Its last four digits are 3015, so the thousands digit A = 3, the units digit B = 5, and A + B = 8.

last 4 of 153015 = 3015 → A = 3, B = 5 → A + B = 8 → (D)
Answer
8
Sanity check with shorter versions: take N₁' = 30303 and N₂' = 50505 (the same repeating patterns, but only 5 digits each). Then N₁' × N₂' = 30303 × 50505 = 1,530,458,415, whose last four digits are 8415 — wait, that does not match. Recompute carefully: 30303 × 50505 = 30303 × 50000 + 30303 × 505 = 1,515,150,000 + 15,303,015 = 1,530,453,015. The last four digits are 3015, exactly the same 3015 we obtained from 303 × 505. So the thousands digit is 3, the units digit is 5, and A + B = 8, matching answer (D). The reason the longer factors give the same last four digits is precisely the place-value shortcut from step 1.
💡Key takeaway

When you only need the last few digits of a giant product, throw away every digit to the left — Grade 5 place value reduces this AMC 8 problem to a single tidy multiplication.

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