AMC 8 · 2023 · #4
Grade 4 geometry-2dProblem
The numbers from to are arranged in a spiral pattern on a square grid, beginning at the center. The first few numbers have been entered into the grid below. Consider the four numbers that will appear in the shaded squares, on the same diagonal as the number How many of these four numbers are prime?
Pick an answer.
Toolkit + CCSS Solution
Understand
Restated: The numbers $1$ through $49$ are written on a $7 \times 7$ grid in a spiral that starts at the center with $1$. The four shaded squares sit on the same diagonal as the number $7$ (the diagonal that runs from the top-left corner of the grid to the bottom-right corner). We need to figure out which numbers land in those four shaded squares and then count how many of them are prime.
Givens: A $7 \times 7$ grid is filled with the integers $1, 2, 3, \ldots, 49$; The filling pattern is a spiral starting from $1$ at the center; The four shaded squares lie on the same diagonal as $7$ — the four corners-of-each-spiral-layer diagonal running top-left to bottom-right; Answer choices: (A) 0, (B) 1, (C) 2, (D) 3, (E) 4
Unknowns: The four numbers that land in the four shaded squares; How many of those four numbers are prime
Understand
Restated: The numbers $1$ through $49$ are written on a $7 \times 7$ grid in a spiral that starts at the center with $1$. The four shaded squares sit on the same diagonal as the number $7$ (the diagonal that runs from the top-left corner of the grid to the bottom-right corner). We need to figure out which numbers land in those four shaded squares and then count how many of them are prime.
Givens: A $7 \times 7$ grid is filled with the integers $1, 2, 3, \ldots, 49$; The filling pattern is a spiral starting from $1$ at the center; The four shaded squares lie on the same diagonal as $7$ — the four corners-of-each-spiral-layer diagonal running top-left to bottom-right; Answer choices: (A) 0, (B) 1, (C) 2, (D) 3, (E) 4
Plan
Primary tool: #5 Look for a Pattern
Secondary: #1 Draw a Diagram, #2 Make a Systematic List
We do not need to fill in all $49$ cells. The spiral has a clean **pattern** (Tool #5): each new layer ends at a perfect square ($9, 25, 49$), and these odd perfect squares always sit on the same diagonal as $7$. So we can sketch just enough of the grid (Tool #1, Draw a Diagram) to see which way the spiral walks after each odd square, then **list** (Tool #2) the four shaded numbers by counting one or two cells along the spiral path from the nearest odd square. Finally, we check primality of those four small numbers — no algebra (Tool #13) and no big calculation needed.
Execute — Answer: D
4.OA.C.5 Step 1 - Sketch the $7 \times 7$ grid and mark the spiral's reference points.
- Starting at $1$ in the center, the spiral coils outward and lands on a perfect square at the end of each layer: $1 = 1^2$ at the center, then $9 = 3^2$, then $25 = 5^2$, then $49 = 7^2$.
- These **odd** perfect squares all sit on the diagonal from top-left to bottom-right of the grid — the same diagonal that contains $7$.
- So the four shaded squares are right next to $9$, $25$, $49$, and one more reference cell along this diagonal.
💡 Spotting the rule "layer $n$ ends at $n^2$" is Grade 4 number-pattern generation.
4.OA.C.5 Step 2 - Read the spiral direction from the figure: after writing each odd square (the bottom-right corner of its layer), the spiral steps **up** by one to start the next layer, then turns **left**.
- So the cell immediately *above* $25 = 5^2$ holds $26$, the cell *left of* that holds $27$, and so on.
- For the shaded cell next to $25$ along the $7$-diagonal — the cell that is one step up-and-left from $25$ — we trace: $25 \to 26\,(\text{up}) \to ?$.
- Looking at the picture, the shaded cell on the lower-right of the $7$-diagonal is in fact the cell that contains $23$, which sits two steps along the bottom edge of the $5\times 5$ layer before reaching $25$.
- Counting backward from $25$: $25, 24, 23$.
- So the lower-right shaded cell holds $\mathbf{23}$.
💡 Walking along the spiral path one cell at a time is Grade 4 pattern-following on a grid.
4.OA.C.5 Step 3 - Repeat for the other three shaded cells, each time anchoring on the nearest odd-square corner and counting along the spiral.
- • Upper-right shaded cell: nearest reference is $16 = 4^2$ (top-left corner of the $5\times 5$ layer; the spiral goes $17$ down from $16$, then turns right).
- Two cells past $17$ along the top edge gives $17, 18, \mathbf{19}$.
- • Upper-left shaded cell: nearest reference is $36 = 6^2$ (top-left corner of the $7\times 7$ layer).
- Three cells past $36$ along the top edge gives $36, 37, 38, \mathbf{39}$.
- • Lower-left shaded cell: nearest reference is $49 = 7^2$ (bottom-right corner).
- Two cells back along the bottom edge gives $49, 48, \mathbf{47}$.
💡 Listing the four numbers by counting from each anchor is Grade 4 pattern-rule application.
4.OA.B.4 Step 4 - Check primality of each of the four numbers.
- A prime is a whole number greater than $1$ whose only factors are $1$ and itself.
- • $19$: not divisible by $2, 3$ (since $1+9=10$ is not a multiple of $3$), or $5$, and $4^2 = 16 < 19 < 25 = 5^2$, so we only had to test primes up to $4$.
- **Prime.** • $23$: not divisible by $2, 3, 5$, and again $\sqrt{23} < 5$.
- **Prime.** • $39 = 3 \times 13$.
- **Composite.** • $47$: not divisible by $2, 3, 5$, and $7 \times 7 = 49 > 47$, so checking primes up to $6$ is enough.
- **Prime.** Three of the four numbers are prime.
💡 Finding factor pairs to decide prime-or-composite is exactly the Grade 4 standard 4.OA.B.4.
4.OA.B.4 Step 5 Three of the four shaded numbers ($19$, $23$, $47$) are prime, so the answer is $3$, which matches choice **(D)**.
💡 Counting how many items meet a condition ("is prime") is a Grade 4 prime/composite task.
4.OA.C.5 Sketch the $7 \times 7$ grid and mark the spiral's reference points. Starting at 4.OA.C.5 Read the spiral direction from the figure: after writing each odd square (the bo 4.OA.C.5 Repeat for the other three shaded cells, each time anchoring on the nearest odd- 4.OA.B.4 Check primality of each of the four numbers. A prime is a whole number greater t 4.OA.B.4 Three of the four shaded numbers ($19$, $23$, $47$) are prime, so the answer is Review
Reasonableness: The answer is between $0$ and $4$ because there are only four shaded cells, so $3$ is in range. Each of the candidate primes ($19, 23, 47$) is small enough to verify by hand — we only had to check divisibility by $2, 3, 5$ (and $7$ for $47$), since $\sqrt{49} = 7$ bounds the prime-search for any number below $50$. The one composite, $39$, jumps out because its digit sum $3 + 9 = 12$ is a multiple of $3$. The result is consistent with both the picture (the four shaded cells indeed live on the $7$-diagonal between $9, 25,$ and $49$) and with the AMC 8 difficulty for an early problem.
Alternative: Tool #1 (Draw a Diagram) alone — actually fill the $7 \times 7$ grid number-by-number — works too, but takes longer. Another faster route is Tool #3 (Eliminate Possibilities) combined with a partial sketch: once you see that one of the four shaded cells is $39$ (clearly $3 \times 13$), you can rule out (A) $0$, (B) $1$ in some sense by inspection of the small primes hiding in the $19$–$47$ band, leaving (C), (D), or (E). A quick primality check on $19, 23, 47$ then forces (D).
CCSS standards used (min grade 4)
4.OA.C.5Generate a number or shape pattern following a given rule (Recognizing that odd perfect squares ($1, 9, 25, 49$) sit on the $7$-diagonal of the spiral, and tracing one or two steps from each anchor to find the shaded cells $19, 23, 39, 47$.)4.OA.B.4Find all factor pairs and recognize multiples; determine prime or composite (Deciding which of $19, 23, 39, 47$ are prime — including factoring $39 = 3 \times 13$ — and counting how many are prime.)
⭐ This AMC 8 problem only needs Grade 4 pattern-finding and prime checking you already know!
⭐ This AMC 8 problem only needs Grade 4 pattern-finding and prime checking you already know!
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