AMC 8 · 2017 · #15

Grade 4 counting

Archetype Branching State Enumeration └ Sub-archetype Constrained Sequence or Path Count

permutations-basicsystematic-enumerationspatial-visualizationpattern-recognition tree-enumerationidentify-subproblemspattern-recognition ↑ Prerequisites: systematic-enumeration

📏 Medium solution 💡 3 insights 📊 Diagram

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Problem

In the arrangement of letters and numerals below, by how many different paths can one spell AMC8? Beginning at the A in the middle, a path only allows moves from one letter to an adjacent (above, below, left, or right, but not diagonal) letter. One example of such a path is traced in the picture.

Pick an answer.

(A)

(B)

(C)

(D)

(E)

AMC 8 2017 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.

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Toolkit + CCSS Solution

Understand

Restated: Letters and numerals are arranged in a diamond-shaped grid with the letter A in the center. Starting at the central A, we move only to horizontally or vertically adjacent cells (no diagonals) and want to spell out A-M-C-8. How many different 4-step paths spell AMC8?

Givens: Central cell is A; surrounding letters are arranged as in the figure; Allowed moves: up, down, left, or right to an adjacent cell only; Each step must land on the next required letter: A $\to$ M $\to$ C $\to$ 8; Answer choices: (A) $8$, (B) $9$, (C) $12$, (D) $24$, (E) $36$

Unknowns: The total number of distinct paths that spell AMC8 starting from the central A

Understand

Plan

Primary tool: #1 Draw a Diagram

Secondary: #7 Identify Subproblems, #5 Look for a Pattern

The problem is purely spatial — we are walking on a grid — so Tool #1 (Draw a Diagram) is the natural starting point. With the picture in hand we use Tool #7 (Identify Subproblems) to split the 3-move walk into three independent counting questions: A $\to$ M, then M $\to$ C, then C $\to$ 8. Tool #5 (Look for a Pattern) helps once we notice that every M has the same number of valid C-neighbors and every C has the same number of valid 8-neighbors, so we can multiply the three small counts instead of enumerating all paths.

Execute — Answer: D

#1 Draw a Diagram K.G.A.1 Step 1

Step 1 — Count moves from A to an adjacent M.
Drawing the grid, the central A has exactly four orthogonal neighbors: one M above, one below, one to the left, and one to the right.
So there are $4$ choices for the first move.

$$\text{A} \to \text{M}: \; 4 \text{ choices}$$

💡 Just describing positions like "above, below, left, right" of A is a Kindergarten geometry idea.

#5 Look for a Pattern 4.OA.C.5 Step 2

Step 2 — From each M, count adjacent C's that continue the path.
Take the M directly above A as a sample: its four orthogonal neighbors are A (below — already used), and three C's (left, above, right).
So there are $3$ valid C choices from that M.
The same is true for the M below, left, and right of A by the grid's symmetry.

$$\text{M} \to \text{C}: \; 3 \text{ choices from every M}$$

💡 Spotting that the same count of 3 repeats at every M is a Grade 4 "find the repeating rule" pattern observation.

#5 Look for a Pattern 4.OA.C.5 Step 3

Step 3 — From each C, count adjacent 8's.
Check both kinds of C: the C at a corner of the inner diamond (next to A's diagonal) is adjacent to two M's and two 8's, and the C at a tip of the outer diamond is adjacent to one M and two 8's.
Either way the number of adjacent 8's is exactly $2$, so each C gives $2$ choices for the last move.

$$\text{C} \to 8: \; 2 \text{ choices from every C}$$

💡 Confirming that the count of 2 holds for both types of C is another Grade 4 pattern-rule check.

#7 Identify Subproblems 3.OA.A.3 Step 4

Step 4 — Combine the three independent move-counts using the multiplication principle (a Grade 3 idea: total arrangements = product of choices at each independent stage).
The total number of A-M-C-8 paths is $4 \times 3 \times 2 = 24$, which is answer choice $\textbf{(D)}$.

$$4 \times 3 \times 2 = 24 \;\Rightarrow\; \textbf{(D)}$$

💡 Multiplying the choices at each step is exactly the Grade 3 "number of groups times number per group" multiplication idea.

[1] #1 K.G.A.1 Step 1 — Count moves from A to an adjacent M. Drawing the grid, the central A ha

[2] #5 4.OA.C.5 Step 2 — From each M, count adjacent C's that continue the path. Take the M dire

[3] #5 4.OA.C.5 Step 3 — From each C, count adjacent 8's. Check both kinds of C: the C at a corn

[4] #7 3.OA.A.3 Step 4 — Combine the three independent move-counts using the multiplication prin

Review

Reasonableness: Does $24$ make sense? The path has $3$ moves; at the worst there are $4$ options at each move, giving an upper bound of $4 \times 4 \times 4 = 64$. Our count of $24$ is comfortably below that, and it matches the most-symmetric-looking choice (D). Sanity-checking by symmetry: paths starting up = paths starting down = paths starting left = paths starting right. Starting up, the M-above-A has $3$ C-choices and each C has $2$ eight-choices, giving $3 \times 2 = 6$ paths per starting direction, then $6 \times 4 = 24$ total. Same answer — consistent.

Alternative: Tool #2 (Make a Systematic List) would also work: order the four directions out of A (up, right, down, left), and for each branch list the three C's and the two 8's per C. That produces $4 \times 3 \times 2 = 24$ labeled paths and lets you double-check that no path repeats a cell.

CCSS standards used (min grade 4)

K.G.A.1 Describe positions of objects using above, below, beside, in front of (Identifying that the central A has four orthogonal neighbors (above, below, left, right), all of which happen to be M.)
3.OA.A.3 Solve multiplication and division word problems within 100 (Combining the per-step choice counts with the multiplication principle: $4 \times 3 \times 2 = 24$.)
4.OA.C.5 Generate a number or shape pattern following a given rule (Spotting that every M produces the same $3$ C-options and every C produces the same $2$ eight-options, so the same rule repeats at every cell of the same letter.)

⭐ This AMC 8 problem only needs Grade 4 pattern-spotting plus the simple Grade 3 idea that the total number of paths is the product of the choices at each step!

Problem

Toolkit + CCSS Solution

Understand

Plan

Execute — Answer: D

Review

CCSS standards used (min grade 4)

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