AMC 10 · 2019 · #10

Grade 5 counting
graph-readinggraph-connectivityparitysystematic-enumeration extreme-principlesystematic-enumerationlogical-deduction ↑ Prerequisites: graph-reading
📏 Long solution 💡 4 insights 📊 Diagram
📘 View easy version →
Problem
A map shows 12 cities as a grid of 3 rows and 4 columns of dots, with 17 roads joining each dot to the dots directly beside it. Paula starts at the top-left dot, finishes at the bottom-right, and drives along exactly 13 of the roads, never driving any road twice. She may pass through the same city more than once. Count how many different routes she can drive.

Pick an answer.

(A)
0
(B)
1
(C)
2
(D)
3
(E)
4
How to solve
Strategy Extreme Principle

Tool #14 (Extreme Principle) is the crux: work out the largest number of times each city could possibly be used, add those maxima up, and compare with the number of stops 13 roads demand. The two totals turn out to be equal, so nothing is spare and every city must be used at its maximum. Tool #1 (Diagram) supplies the coordinates and road-counts that the maximum depends on. Tool #3 (Eliminate Possibilities) then kills the wrong first move and pins down which four roads are skipped. Tool #2 (Systematic List) finishes by counting the only free choices left.

1STEP 1

Put coordinates on the map

There are seventeen roads in all.

9 + 8 = 17
2STEP 2

Count the roads at each city

Count the roads meeting each city.

4 · 2 + 6 · 3 + 2 · 4 = 8 + 18 + 8 = 34 = 2 · 17
3STEP 3

How often a city can be used

Each pass uses two roads.

⌊ 2/2 ⌋ = 1, ⌊ 3/2 ⌋ = 1, ⌊ 4/2 ⌋ = 2
4STEP 4

The budget is exactly full

There is no slack at all.

14 - 2 = 12 = 2 · 1 + 6 · 1 + 2 · 2
5STEP 5

Locate the skipped roads

The skipped roads are almost fixed.

1 + 1 + 6 = 8 = 2 · (17 - 13)
6STEP 6

Fix the starting direction

One direction cannot be the start.

2 ≠ 3
7STEP 7

The rest is forced

Every remaining road is forced.

skipped = { A - (0,1), (1,2) - (2,2), (1,0) - (2,0), (3,1) - L }
8STEP 8

Read the route's shape

The route settles into one shape.

2 + 4 + 1 + 4 + 2 = 13
9STEP 9

Count the turns

Multiplying the remaining choices gives 4.

2 · 2 = 4 → (E)
Answer
4
Trace one route to confirm it really exists: A(0,2) → (1,2) → (1,1) → (1,0) → (0,0) → (0,1) → (1,1) → (2,1) → (3,1) → (3,2) → (2,2) → (2,1) → (2,0) → L(3,0). That is 13 roads with none repeated, so choice (A) 0 is out. Reversing just one loop gives a genuinely different route, so the two loops contribute 2 · 2 = 4, not 2 — choice (C) is what you get by noticing only one of the two loops. No more than 4 is possible either, because the set of driven roads was forced before any turning choice was made. The skipped-road picture is consistent too: the four skipped roads are pairwise disjoint and use up exactly the eight road-ends that had to be skipped.
💡Key takeaway

This AMC 12 problem needs only Grade 5 tools: count how many times each city could possibly be used, notice that total is exactly the number of stops 13 roads require, and the whole route is forced — only the two square loops can still be turned either way, giving 2 · 2 = 4, choice (E).

  • Put coordinates on the map
  • Count the roads at each city
  • How often can one city be used?
  • The budget is exactly full
  • Locate the four skipped roads
  • Paula cannot start downward
  • Every remaining road is forced
  • Read the shape of the route
  • Count the ways to turn