AMC 8 · 2017 · #15
Easy mode Grade 4Problem
Picture a grid of letters and numbers. In the very middle, there is an . Around the are 's, then 's further out, then 's at the edges.
You want to spell out by walking from letter to letter. The rules are:
- Start at the 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 , your second step on a , and your third step on an .
How many different paths spell this way?
Pick an answer.
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
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.
💡 Just describing positions like "above, below, left, right" of A is a Kindergarten geometry idea.
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.
💡 Spotting that the same count of 3 repeats at every M is a Grade 4 "find the repeating rule" pattern observation.
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.
💡 Confirming that the count of 2 holds for both types of C is another Grade 4 pattern-rule check.
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 $\textbf{(D)}$.
💡 Multiplying the choices at each step is exactly the Grade 3 "number of groups times number per group" multiplication idea.
K.G.A.1 Step 1 — Count moves from A to an adjacent M. Drawing the grid, the central A ha 4.OA.C.5 Step 2 — From each M, count adjacent C's that continue the path. Take the M dire 4.OA.C.5 Step 3 — From each C, count adjacent 8's. Check both kinds of C: the C at a corn 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.1Describe 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.3Solve 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.5Generate 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!