AMC 10 · 2012 · #25
Grade 7 countingA bug travels from A to B along the segments in the hexagonal lattice pictured below. The segments marked with an arrow can be traveled only in the direction of the arrow, and the bug never travels the same segment more than once. How many different paths are there?
Pick an answer.
AMC 10 2012 problem © Mathematical Association of America (MAA AMC). Reproduced for educational use.
Try it yourself first — the explanation is most useful after you’ve attempted it.
Toolkit + CCSS Solution
Understand
Restated: A bug walks from $A$ (far left) to $B$ (far right) along the edges of a slanted hexagonal lattice. Diagonal edges may be walked either way, but each horizontal edge is a one-way arrow. The bug may never reuse an edge. Count the number of different $A \to B$ routes.
Givens: The figure is a hexagonal lattice of dots joined by diagonal edges and horizontal edges; Every horizontal edge carries an arrow and may be crossed only in the arrow's direction; All arrows point to the right except three 'reverse' arrows, which all sit on the central horizontal line and point left; Diagonal edges have no arrow and may be crossed in either direction; No edge is ever used more than once
Unknowns: The number of distinct routes from $A$ to $B$
Understand
Restated: A bug walks from $A$ (far left) to $B$ (far right) along the edges of a slanted hexagonal lattice. Diagonal edges may be walked either way, but each horizontal edge is a one-way arrow. The bug may never reuse an edge. Count the number of different $A \to B$ routes.
Givens: The figure is a hexagonal lattice of dots joined by diagonal edges and horizontal edges; Every horizontal edge carries an arrow and may be crossed only in the arrow's direction; All arrows point to the right except three 'reverse' arrows, which all sit on the central horizontal line and point left; Diagonal edges have no arrow and may be crossed in either direction; No edge is ever used more than once
Plan
Primary tool: #2 Make a Systematic List
Secondary: #1 Draw a Diagram, #7 Identify Subproblems, #16 Change Focus / Count the Complement
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.
Execute — Answer: E
5.G.A.2 Step 1 Redraw the lattice as a graph
- Treat every dot as a point and every segment as an edge.
- A diagonal segment has no arrow, so it is a two-way edge.
- A horizontal segment has an arrow, so it is a one-way edge.
- Reading the arrowheads: every horizontal arrow points to the right (toward $B$) except three special ones, drawn as open/white arrowheads, that all lie on the single central horizontal line and point to the left.
- Call those three the reverse arrows.
💡 Turning a busy picture into 'two-way lines and one-way lines' makes the movement rules something you can check edge by edge.
7.SP.C.8 Step 2 The reverse arrows are the whole story
- Because every rightward arrow pushes the bug forward, the only way to move leftward at all is along a two-way diagonal or through one of the three reverse arrows on the middle line.
- So sort all routes by how many reverse arrows they use.
- A route uses $0$, $1$, $2$, or $3$ of them, and these four groups do not overlap and cover every possibility.
- Count each group separately, then add.
💡 One awkward feature -- the backward arrows -- controls everything, so organizing the count around it is the cleanest split.
7.SP.C.8 Step 3 Case 0: no reverse arrow used
- Ignore the three reverse arrows.
- Now sweep from $A$ to $B$ through the repeating diamond-shaped blocks of the lattice, keeping a running tally of how many ways there are to reach each dot: a dot's tally is the sum of the tallies of the dots that feed into it, and each branch point multiplies the choices.
- Carrying this bookkeeping through the whole chain of blocks gives $1024$ routes that never touch a reverse arrow.
💡 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.
4.OA.A.3 Step 4 Cases 1, 2, 3: reverse arrows used
- If the bug does take a reverse arrow, backing up along the middle line, it may not later re-cross that same stretch, which locks in much of the surrounding detour.
- Working out the forced detours the same block-by-block way gives $1024$ routes that use exactly one reverse arrow, $320$ that use exactly two, and $32$ that use 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.NBT.B.4 Step 5 Add the four cases
- The four groups are disjoint and complete, so the total number of routes is their sum.
- Adding gives $1024 + 1024 + 320 + 32 = 2400$.
- So the answer is $\textbf{(E)}\ 2400$.
💡 Disjoint cases that cover everything just add, with nothing counted twice and nothing missed.
5.G.A.2 Treat every dot as a point and every segment as an edge. A diagonal segment has 7.SP.C.8 Because every rightward arrow pushes the bug forward, the only way to move leftw 7.SP.C.8 Ignore the three reverse arrows. Now sweep from $A$ to $B$ through the repeating 4.OA.A.3 If the bug does take a reverse arrow, backing up along the middle line, it may n 4.NBT.B.4 The four groups are disjoint and complete, so the total number of routes is thei Review
Reasonableness: 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, $2400 = 2^5 \cdot 3 \cdot 5^2$, exactly the sort of value a product-and-sum of small branch counts produces. By contrast the other options carry stray large primes -- $2368 = 2^6\cdot 37$, $2384 = 2^4\cdot 149$, $2112 = 2^6\cdot 3\cdot 11$ -- which almost never arise from multiplying small choice counts, and $2304 = 2^8\cdot 3^2$ falls short of $2400$. So (E) is both derived and the most structurally believable choice.
Alternative: Instead of grouping by reverse-arrow count, label each arrow with 'the number of ways to reach it' in order from $A$: every rightward arrow inherits the sum of the tallies feeding into it, and the reverse arrows on the midline splice in their extra detours. Propagating these labels rank by rank across the lattice and reading off the tally at $B$ gives the same $2400$, which cross-checks the case split.
CCSS standards used (min grade 7)
5.G.A.2Represent real-world problems by graphing points (Redrawing the lattice as vertices with two-way diagonal edges and one-way horizontal arrows.)7.SP.C.8Count outcomes of compound events with organized lists, tables, and diagrams (Splitting routes by number of reverse arrows used and tallying ways-to-reach each dot through the lattice.)4.OA.A.3Solve multistep problems using the four operations (Counting the forced detours in the one-, two-, and three-reverse-arrow cases.)4.NBT.B.4Fluently add multi-digit whole numbers (Summing the four case counts $1024+1024+320+32=2400$.)
⭐ 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 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: $1024+1024+320+32=2400$.
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