AMC 8 · 2024 · #14

Grade 2 geometry-2d
Archetype Path Length Comparison & Shortest Path └ Sub-archetype Shortest Route in Weighted Network
systematic-enumerationmulti-digit-arithmetic shortest-pathidentify-subproblems ↑ Prerequisites: multi-digit-arithmeticmental-arithmetic
📏 Medium solution 💡 4 insights 📊 Diagram
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Problem
Six towns A, M, C, X, Y, Z are connected by one-way roads with given distances (km). Using only these roads (in the direction of the arrows), we want the shortest total distance from A to Z in km.

Pick an answer.

(A)
28
(B)
29
(C)
30
(D)
31
(E)
32

AMC 8 2024 problem © Mathematical Association of America (MAA AMC). Reproduced for educational use.

How to solve
Strategy Draw a Diagram

This is a classic map problem about positions and connections, so the first move is Tool #1 — get the towns, arrows, and distances laid out in a clean picture. Then break the big question "A→ Z" into smaller subproblems: the shortest distance from A to each intermediate town (Tool #7). That lets us reuse partial sums instead of re-adding the same numbers. Finally, at the last step we list all routes into Z exhaustively (Tool #2) and pick the smallest, then check it against the answer choices with Tool #3.

1STEP 1

Organize the arrows: only A→X and A→M leave A, and only M→Z, Y→Z, C→Z enter Z, so every route ends with one final hop into Z.

A → X, A → M, M → Z, Y → Z, C → Z
2STEP 2

First layer: A→X = 5 (only one way in). For M, compare direct A→M = 8 with the detour A→X→M = 5+2 = 7, so the shortest to M is 7.

A→ X: 5. A→ M: min(8, 5+2) = min(8, 7) = 7
3STEP 3

Next layer, reusing those sums: A→Y = min(15, 13) = 13 and A→C = min(21, 18) = 18.

A→ Y: min(5+10, 7+6) = min(15, 13) = 13. A→ C: min(7+14, 13+5) = min(21, 18) = 18
4STEP 4

Last hop into Z: compare via M (7+25 = 32), via Y (13+17 = 30), via C (18+10 = 28); the smallest is 28.

min(7+25, 13+17, 18+10) = min(32, 30, 28) = 28
5STEP 5

Match 28 to the choices: it is exactly (A). The values 30 and 32 are the longer routes, and 29 and 31 never occur as a path total.

28 → (A)
Answer
28
Re-add the optimal route A → X → M → Y → C → Z in one shot: 5 + 2 + 6 + 5 + 10 = 28, matching the step-by-step accumulation exactly. The route starts on the cheapest arrow out of A (A→ X = 5), visits every intermediate town along the way, and ends on the cheapest arrow into Z (C→ Z = 10) — so 28 also feels right by size. The fact that C→ Z = 10 (the shortest single arrow into Z) appears in our winning route is a nice sanity check.
💡Key takeaway

This AMC 8 problem only needs Grade 2 addition within 100 and two-digit comparison you already know!

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