AMC 10 · 2012 · #22

Grade 7 countingpattern
Archetype Casework by Position / Vertex Type └ Sub-archetype Count of Type
lattice-pathssystematic-enumeration systematic-enumeration ↑ Prerequisites: systematic-enumeration
📏 Medium solution 💡 3 insights 📊 Diagram
Problem
Some edges may be walked either way and some only one way, with no edge reused. Count the routes.

Pick an answer.

(A)
2112
(B)
2304
(C)
2368
(D)
2384
(E)
2400
How to solve
Strategy Make a Systematic List

The question asks "how many routes", so the goal is a complete, non-overlapping count (Tool #2, Make a Systematic List). First redraw the picture as a directed graph so the rules become mechanical: dots are vertices, diagonals are two-way edges, horizontal arrows are one-way edges (Tool #1, Draw a Diagram). The whole difficulty lives in the three left-pointing reverse arrows on the middle line, so split every route by how many of those reverse arrows it uses -- 0, 1, 2, or 3 (Tool #7, Identify Subproblems; Tool #16, Change Focus). Each case is then a clean forward count, and the four cases add up.

1STEP 1

Redraw the lattice as a graph

Redrawing shows which edges point backwards.

diagonal = two-way, horizontal = one-way (right, except 3 reverse on the midline)
2STEP 2

The reverse arrows are the whole story

Those few arrows are the whole story.

routes = N₀ + N₁ + N₂ + N₃, N_k = #{routes using exactly k reverse arrows}
3STEP 3

Case 0: no reverse arrow used

Using none of them gives 1024 routes.

N₀ = 1024
4STEP 4

Cases 1, 2, 3: reverse arrows used

Using one, two, or three adds fewer each time.

N₁ = 1024, N₂ = 320, N₃ = 32
5STEP 5

Add the four cases

Adding the cases gives 2400, choice (E).

1024 + 1024 + 320 + 32 = 2400 → (E)
Answer
2400
The four cases were built to be disjoint (they differ in how many reverse arrows are used) and exhaustive (a route must use 0,1,2, or 3 of them), so summing them is legitimate; recombining 1024+1024+320+32 independently returns 2400, matching choice (E). The total is also a smooth, highly composite number, 2400 = 2⁵ · 3 · 5², exactly the sort of value a product-and-sum of small branch counts produces. By contrast the other options carry stray large primes -- 2368 = 2⁶ · 37, 2384 = 2⁴ · 149, 2112 = 2⁶ · 3 · 11 -- which almost never arise from multiplying small choice counts, and 2304 = 2⁸ · 3² falls short of 2400. So (E) is both derived and the most structurally believable choice.
💡Key takeaway

Redraw the maze as one-way and two-way streets, notice only three backward arrows cause trouble, count routes by how many of those you use, and add: 1024+1024+320+32=2400.

  • Redraw the lattice as a graph
  • The reverse arrows are the whole story
  • Case 0: no reverse arrow used
  • Cases 1, 2, 3: reverse arrows used
  • Add the four cases

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