AMC 8 · 2020 · #21
Easy mode Grade 5Problem
Picture a checkerboard with 64 squares, colored black and white in the usual alternating pattern. One white square in the bottom row is marked P, and one white square in the top row is marked Q.
A marker starts on square P. Each step, the marker moves up one row, landing on one of the white squares that touches its current square at a corner.
How many different 7-step paths take the marker from P to Q? (The sample path is shown in the figure.)
Pick an answer.
AMC 8 2020 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 checkerboard has $64$ alternating black and white squares. A marker starts on white square $P$ in the bottom row. Each step moves the marker diagonally up-left or up-right to an adjoining white square in the row above. The marker needs to reach white square $Q$ in the top row using exactly $7$ steps. How many distinct $7$-step paths from $P$ to $Q$ are there?
Givens: An $8 \times 8$ checkerboard with alternating black and white squares; Marker starts at $P$ (white square in the bottom row); Target square $Q$ is in the top row (white); Each step moves the marker to a white square diagonally above (up-left or up-right) and must stay on the board; Exactly $7$ steps are taken (bottom row to top row); Answer choices: (A) $28$, (B) $30$, (C) $32$, (D) $33$, (E) $35$
Unknowns: The number of distinct $7$-step paths from $P$ to $Q$
Understand
Restated: A checkerboard has $64$ alternating black and white squares. A marker starts on white square $P$ in the bottom row. Each step moves the marker diagonally up-left or up-right to an adjoining white square in the row above. The marker needs to reach white square $Q$ in the top row using exactly $7$ steps. How many distinct $7$-step paths from $P$ to $Q$ are there?
Givens: An $8 \times 8$ checkerboard with alternating black and white squares; Marker starts at $P$ (white square in the bottom row); Target square $Q$ is in the top row (white); Each step moves the marker to a white square diagonally above (up-left or up-right) and must stay on the board; Exactly $7$ steps are taken (bottom row to top row); Answer choices: (A) $28$, (B) $30$, (C) $32$, (D) $33$, (E) $35$
Plan
Primary tool: #1 Draw a Diagram
Secondary: #5 Look for a Pattern, #7 Identify Subproblems
Path-counting on a grid is the textbook job for Tool #1 (Draw a Diagram): redraw the board with $P$ at the bottom and write the number of paths into each reachable white square, working one row at a time. Tool #5 (Look for a Pattern) shows up because the count at each square is the sum of the counts at the two diagonal squares below it — exactly the Pascal's-triangle pattern, modified at the edges where one diagonal falls off the board. Tool #7 (Identify Subproblems) makes the bookkeeping clean: "how many ways to reach $(r,c)$" is solved by adding the answers to the two smaller subproblems "how many ways to reach $(r-1, c-1)$" and "how many ways to reach $(r-1, c+1)$".
Execute — Answer: A
5.G.A.1 Step 1 - Set up a coordinate system: label columns $0$-$7$ left to right and rows $0$-$7$ bottom to top.
- From the figure, $P$ is at (row $0$, col $5$) and $Q$ is at (row $7$, col $6$).
- Confirm that every legal move goes from a white square to a white square diagonally one row up; the marker can never land on a black square or step sideways.
💡 Setting up rows and columns on the board is the Grade 5 coordinate-grid idea: every white square gets an address $(r, c)$.
4.OA.C.5 Step 2 - State the counting rule.
- Let $N(r, c)$ be the number of $r$-step paths from $P$ to the white square at $(r, c)$.
- A white square at $(r, c)$ is reachable only from $(r-1, c-1)$ or $(r-1, c+1)$ (and only when those squares are on the board), so $N(r, c) = N(r-1, c-1) + N(r-1, c+1)$.
- Seed the recursion with $N(0, 5) = 1$ (there is exactly one way to be at $P$ to start).
💡 Splitting "reach $(r, c)$" into the two smaller subproblems "reach the two diagonal neighbors below" is the Grade 4 "generate a pattern from a rule" move.
4.OA.C.5 Step 3 - Fill in the counts row by row, using the rule from Step 2.
- Row $1$ from $P$: $(1, 4) = 1$ and $(1, 6) = 1$.
- Row $2$: $(2, 3) = 1$, $(2, 5) = 1 + 1 = 2$, $(2, 7) = 1$.
- Row $3$: $(3, 2) = 1$, $(3, 4) = 1 + 2 = 3$, $(3, 6) = 2 + 1 = 3$ (the square at $(3, 8)$ is off the board, so nothing is added there).
💡 Each new number is the sum of the two diagonal numbers right below it — the same Pascal's-triangle pattern from Grade 4.
4.OA.C.5 Step 4 - Continue the row-by-row sums to row $6$, watching the right edge of the board (column $8$ doesn't exist, so squares like $(4, 7)$ inherit only the count from $(3, 6)$).
- Row $4$: $(4, 3) = 1 + 3 = 4$, $(4, 5) = 3 + 3 = 6$, $(4, 7) = 3$.
- Row $5$: $(5, 4) = 4 + 6 = 10$, $(5, 6) = 6 + 3 = 9$.
- Row $6$: $(6, 5) = 10 + 9 = 19$, $(6, 7) = 9$.
💡 Writing the running total into each white square keeps the bookkeeping visual, with edge squares getting only one contribution.
4.NBT.B.4 Step 5 - Compute $N(7, 6) = N(6, 5) + N(6, 7) = 19 + 9 = 28$.
- So there are exactly $28$ distinct $7$-step paths from $P$ to $Q$, which is choice (A).
💡 Adding the two contributions $19 + 9 = 28$ is a Grade 4 multi-digit addition.
5.G.A.1 Set up a coordinate system: label columns $0$-$7$ left to right and rows $0$-$7$ 4.OA.C.5 State the counting rule. Let $N(r, c)$ be the number of $r$-step paths from $P$ 4.OA.C.5 Fill in the counts row by row, using the rule from Step 2. Row $1$ from $P$: $(1 4.OA.C.5 Continue the row-by-row sums to row $6$, watching the right edge of the board (c 4.NBT.B.4 Compute $N(7, 6) = N(6, 5) + N(6, 7) = 19 + 9 = 28$. So there are exactly $28$ d Review
Reasonableness: Sanity-check with the unconstrained version: without any edge or board limits, a $7$-step random walk that ends up shifted by exactly $+1$ to the right (from column $5$ to column $6$) needs $4$ right-steps and $3$ left-steps, giving $\binom{7}{4} = 35$ paths. Our answer $28$ is a bit smaller, which makes sense because the right edge of the board cuts off some paths (a marker reaching $(4, 7)$ on the right edge only has one continuation instead of two). The leak is exactly $35 - 28 = 7$ paths killed by the edge, which is a believably small correction.
Alternative: Tool #16 (Change Focus / Complement) gives the slick alternative above: count $\binom{7}{4} = 35$ free paths first, then subtract the paths that would step off the right edge using the reflection principle. For students who haven't met binomial coefficients yet, Tool #2 (Systematic List) also works — enumerate all $7$-step sequences of L/R, throw out the ones that go off-board, and count what's left.
CCSS standards used (min grade 5)
5.G.A.1Use a pair of perpendicular number lines forming a coordinate system (Labeling rows $0$-$7$ and columns $0$-$7$ on the board so every white square has an address $(r, c)$, locating $P = (0, 5)$ and $Q = (7, 6)$.)4.OA.C.5Generate a number or shape pattern following a given rule (Applying the rule $N(r, c) = N(r-1, c-1) + N(r-1, c+1)$ row by row to build the path-count pattern (a Pascal's-triangle style growth).)4.NBT.B.4Fluently add and subtract multi-digit whole numbers (Adding small whole numbers like $10 + 9 = 19$ and $19 + 9 = 28$ during the row-by-row fill.)
⭐ This AMC 8 problem only needs Grade 5 coordinate grids and the Grade 4 "add the two numbers below" Pascal's-triangle pattern you already know!
⭐ This AMC 8 problem only needs Grade 5 coordinate grids and the Grade 4 "add the two numbers below" Pascal's-triangle pattern you already know!
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