Competition · AMC preparation · step 4 of 4
AMC 10 · 2012B · #25
Grade 7 counting
Pick an answer.
AMC 10 2012 problem © Mathematical Association of America (MAA AMC). Reproduced for educational use.
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.
Redraw the lattice as a graph
Dots are points, diagonals are two-way edges, horizontal arrows one-way; only three arrows, all on the middle line, point left.
Turning a busy picture into 'two-way lines and one-way lines' makes the movement rules something you can check edge by edge.
5.G.A.2Draw A DiagramThe reverse arrows are the whole story
Only those three reverse arrows let the bug back up, so split every route by how many it uses: 0, 1, 2, or 3 — four disjoint groups.
One awkward feature -- the backward arrows -- controls everything, so organizing the count around it is the cleanest split.
7.SP.C.8Identify SubproblemsCase 0: no reverse arrow used
Ignoring the reverse arrows, multiply the choices at each vertical bundle of arrows across the lattice: 1024 forward-only routes.
Ways-to-reach a point add up from the points before it, so one careful pass across the lattice counts every forward route at once.
7.SP.C.8Make A Systematic ListCases 1, 2, 3: reverse arrows used
A reverse step blocks re-crossing the middle line, so the detour is forced: 1024 routes use one, 320 two, 32 all three.
Once the bug commits to a backward step, the no-reuse rule forces most of the rest of that route, so each reverse arrow adds only a limited, countable set of detours.
4.OA.A.3Make A Systematic ListAdd the four cases
Disjoint and complete cases just add: 1024 + 1024 + 320 + 32 = 2400, which is choice (E).
Disjoint cases that cover everything just add, with nothing counted twice and nothing missed.
Cases that never overlap and cover everything simply add, with nothing counted twice or missed.
▸ Why?
Separate cases have no shared routes, so their counts can be added directly.
▸ Why?
Together the cases make up every possible route, so their total is the whole count.
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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