AMC 8 · 2017 · #15

Easy mode Grade 4
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Problem

Picture a grid of letters and numbers. In the very middle, there is an AA. Around the AA are MM's, then CC's further out, then 88's at the edges.

You want to spell out AMC8\text{AMC8} by walking from letter to letter. The rules are:

  • Start at the AA in the middle.
  • Each step moves to a square right next to the current one (up, down, left, or right — not diagonal).
  • Your first step must land on an MM, your second step on a CC, and your third step on an 88.

How many different paths spell AMC8\text{AMC8} this way?

(A) 8(B) 9(C) 12(D) 24(E) 36\textbf{(A) }8\qquad\textbf{(B) }9\qquad\textbf{(C) }12\qquad\textbf{(D) }24\qquad\textbf{(E) }36

Pick an answer.

(A)
8
(B)
9
(C)
12
(D)
24
(E)
36
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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

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) $36$

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!

⭐ 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!