AMC 10 · 2021 · #8
Easy mode Grade 5Problem
Picture a 15 by 15 grid of empty squares.
Mr. Zhou writes the numbers 1,2,3,…,225 into the squares, one number per square. He starts by writing 1 in the very middle square (row 8, column 8). Then he writes 2 in the square just to the right, 3 in the square below that, and keeps going in a clockwise spiral outward. The picture below shows the start of the spiral.
Look at the second row from the top of the grid.
What is the sum of the largest number and the smallest number in that row?
Pick an answer.
AMC 10 2021 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: Mr. Zhou writes $1$ in the middle cell (row $8$, column $8$) of a $15 \times 15$ grid and then writes $2, 3, 4, \dots, 225$ in a clockwise spiral outward — first one step east, then turning to fill each new ring. Look at the second row from the top of the grid. Find the sum of the largest number and the smallest number that appear in that row.
Givens: $15 \times 15$ grid filled with $1$ through $225 = 15^2$; $1$ sits in the center cell (row $8$, column $8$); The spiral goes clockwise; the picture shows: $1$ east to $2$, south to $3$, west to $4, 5$, north to $6, 7$, east to $8, 9, 10$, south, west, north, east — each ring adds $8k$ cells; Answer choices: (A) $367$, (B) $368$, (C) $369$, (D) $379$, (E) $380$
Unknowns: Sum of the largest and smallest numbers in row $2$ of the grid
Understand
Restated: Mr. Zhou writes $1$ in the middle cell (row $8$, column $8$) of a $15 \times 15$ grid and then writes $2, 3, 4, \dots, 225$ in a clockwise spiral outward — first one step east, then turning to fill each new ring. Look at the second row from the top of the grid. Find the sum of the largest number and the smallest number that appear in that row.
Givens: $15 \times 15$ grid filled with $1$ through $225 = 15^2$; $1$ sits in the center cell (row $8$, column $8$); The spiral goes clockwise; the picture shows: $1$ east to $2$, south to $3$, west to $4, 5$, north to $6, 7$, east to $8, 9, 10$, south, west, north, east — each ring adds $8k$ cells; Answer choices: (A) $367$, (B) $368$, (C) $369$, (D) $379$, (E) $380$
Plan
Primary tool: #9 Easier Related Problem
Secondary: #5 Look for a Pattern, #1 Draw a Diagram, #7 Identify Subproblems, #3 Eliminate Possibilities
Tool #9 (Easier Related Problem): the given diagram already solves the $7 \times 7$ sub-version — use it to learn the spiral rule, then extend to $15 \times 15$. Tool #5 (Pattern): the top-right corner of ring $k$ is the odd square $(2k-1)^2$, and each ring adds $8k$ cells along $1+ (2k-1) + 2k + 2k + 2k$ east/south/west/north/east moves. Tool #1 (Diagram): sketch the outer two rings (ring $7$ and ring $8$) to locate row $2$. Tool #7 (Subproblems): split row $2$ into 'inside ring $8$' (just the two endpoints) and 'inside ring $7$ top edge' (the middle $13$ cells), then take max and min separately.
Execute — Answer: A
4.OA.C.5 Step 1 - From the picture, the spiral's corner pattern is clear: $1 = 1^2$ at the center, $9 = 3^2$ at the top-right of the $3 \times 3$ ring, $25 = 5^2$ at the top-right of the $5 \times 5$ ring.
- In general the top-right corner of ring $k$ (the $(2k-1) \times (2k-1)$ boundary around center) is $(2k-1)^2$.
- Verify: ring $4$ corner $= 7^2 = 49$, ring $5$ corner $= 81$, ..., ring $8$ corner $= 15^2 = 225$.
💡 Grade 4 patterns: the odd squares $1, 9, 25, 49, 81, 121, 169, 225$ sit at the spiral's outer-right corners.
4.OA.C.5 Step 2 - Use the $7 \times 7$ picture to learn the move-sequence from one corner to the next.
- From $9$ at $(\text{row } 7, \text{col } 7)$ in a $7 \times 7$ embedding centered at $(\text{row } 8, \text{col } 8)$ — sorry, in the figure $9$ corresponds to the corner of the inner $3 \times 3$ block.
- The next ring fills $10$ east, $11, 12, 13$ south, $14, 15, 16, 17$ west, $18, 19, 20, 21$ north, $22, 23, 24, 25$ east.
- So from corner $(2k-1)^2$ to corner $(2k+1)^2$ the moves are $1$ east, $2k-1$ south, $2k$ west, $2k$ north, $2k$ east — total $8k$ cells in the new ring.
💡 Grade 4 pattern rule: each ring's perimeter $= 8k$, matched by the move counts.
5.OA.B.3 Step 3 - Now zoom to the outer two rings of the $15 \times 15$.
- Ring $7$ is the $13 \times 13$ boundary; its corner $169 = 13^2$ sits at row $2$, column $14$ (one row above and one column left of the outer corner).
- Working backwards: the east-step that ends at $169$ began at row $2$, column $3$, with value $158$.
- So row $2$, columns $3$ to $14$, holds values $158, 159, \dots, 169$ in order (the top edge of ring $7$).
💡 Grade 5 numerical pattern: the east-step adds $1$ per column moved.
3.NBT.A.2 Step 4 - Just before that east-step, the north-step (left column of ring $7$) ended at row $2$, column $2$.
- That step started at row $13$, column $2$ with value $146$ and ended with $146 + 11 = 157$ at row $2$, column $2$.
- So the row-$2$ entries inside ring $7$ are columns $2$ through $14$ holding $157, 158, \dots, 169$.
💡 Grade 3 sequential counting: each step in the spiral adds $1$ to the value.
3.NBT.A.2 Step 5 - Now do ring $8$ (the outer $15 \times 15$ boundary).
- It begins at $170$ right after $169$.
- The first east-step lands $170$ at row $2$, column $15$ (one step east of $169$).
- The south-step then fills rows $3, 4, \dots, 15$ at column $15$ with values $171, 172, \dots, 183$.
- After the bottom (west) and left (north) edges, the north-step ends at row $1$, column $1$ with value $211$.
- So row $2$, column $1$ is one step before $211$, i.e., $210$.
💡 Grade 3 counting on: ring $8$ has $8 \cdot 8 = 64$ cells, $170$ through $225$, placed exactly along the four edges.
3.NBT.A.2 Step 6 - Collect all of row $2$.
- Columns $1, 2, 3, \dots, 14, 15$ hold $210, 157, 158, 159, \dots, 169, 170$.
- The largest is $210$ (the outer-ring north-step neighbor of $211$) and the smallest is $157$ (the inner-ring north-step end).
- Sum: $210 + 157 = 367$.
💡 Grade 3 within-$1000$ addition: $210 + 157 = 367$ matches choice (A).
4.OA.C.5 From the picture, the spiral's corner pattern is clear: $1 = 1^2$ at the center, 4.OA.C.5 Use the $7 \times 7$ picture to learn the move-sequence from one corner to the n 5.OA.B.3 Now zoom to the outer two rings of the $15 \times 15$. Ring $7$ is the $13 \time 3.NBT.A.2 Just before that east-step, the north-step (left column of ring $7$) ended at ro 3.NBT.A.2 Now do ring $8$ (the outer $15 \times 15$ boundary). It begins at $170$ right af 3.NBT.A.2 Collect all of row $2$. Columns $1, 2, 3, \dots, 14, 15$ hold $210, 157, 158, 15 Review
Reasonableness: Sanity: every value in row $2$ should lie in $\{157, 158, \dots, 170\} \cup \{210\}$ — and indeed all these are between $1$ and $225$. The extremes are $\min = 157$ and $\max = 210$, and $367$ is in the right ballpark (between $2 \cdot 150 = 300$ and $2 \cdot 225 = 450$). Among the choices, only $(A)\ 367$ matches a clean integer split into a middle-band value plus an outer-ring value.
Alternative: Tool #2 (Systematic List) + Tool #6 (Guess and Check): write out the values in row $2$ cell-by-cell, starting from the known corner $169$ at row $2$ column $14$. Move left along the top of ring $7$ to $\min = 157$ at column $2$. Move outward to column $15$: that's $170$ (one east of $169$ in ring $8$). Move to column $1$: that's $210$ (one south of $211$ in ring $8$'s left column). Sum $210 + 157 = 367$. Same answer, just walked along the row directly.
CCSS standards used (min grade 5)
4.OA.C.5Generate a number or shape pattern following a given rule (Recognizing the odd-square corner pattern $(2k-1)^2$ for the spiral's ring corners and the $8k$ ring perimeters.)5.OA.B.3Generate two numerical patterns using two given rules and identify relationships (Tracking column position vs. value along the east-step that ends at corner $169$.)3.NBT.A.2Fluently add and subtract within 1000 (Adding $146 + 11 = 157$, $158 + 11 = 169$, $210 + 157 = 367$, etc.)
⭐ This AMC 10 problem only needs Grade 5 number patterns you already know — the spiral's corners are the odd squares $(2k-1)^2$, so ring $7$ ends at $169$ on row $2$ column $14$, the ring's top edge runs from $157$ to $169$, and ring $8$'s neighbors put $170$ at column $15$ and $210$ at column $1$. Max $+$ min $= 210 + 157 = 367$, choice (A).
⭐ This AMC 10 problem only needs Grade 5 number patterns you already know — the spiral's corners are the odd squares $(2k-1)^2$, so ring $7$ ends at $169$ on row $2$ column $14$, the ring's top edge runs from $157$ to $169$, and ring $8$'s neighbors put $170$ at column $15$ and $210$ at column $1$. Max $+$ min $= 210 + 157 = 367$, choice (A).
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