AMC 10 · 2007 · #19
Grade 7 geometry-2dThe wheel shown is spun twice, and the randomly determined numbers opposite the pointer are recorded. The first number is divided by 4, and the second number is divided by 5. The first remainder designates a column, and the second remainder designates a row on the checkerboard shown. What is the probability that the pair of numbers designates a shaded square?
Pick an answer.
AMC 10 2007 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 wheel with the numbers $1, 2, 3, 6, 7, 9$ is spun twice, each number equally likely. The first result is divided by $4$; its remainder picks a column ($1$, $2$, or $3$). The second result is divided by $5$; its remainder picks a row ($1$, $2$, $3$, or $4$). Find the probability that the chosen column-and-row square on the checkerboard is shaded.
Givens: The wheel shows six equally likely numbers: $1, 2, 3, 6, 7, 9$; First spin $\div 4$: the remainder is the column number ($1, 2, 3$); Second spin $\div 5$: the remainder is the row number ($1, 2, 3, 4$); The board is $3$ columns by $4$ rows, shaded in a checkerboard pattern: shaded squares are $(\text{col},\text{row}) = (1,1),(3,1),(2,2),(1,3),(3,3),(2,4)$; Answer choices: (A) $\frac{1}{3}$, (B) $\frac{4}{9}$, (C) $\frac{1}{2}$, (D) $\frac{5}{9}$, (E) $\frac{2}{3}$
Unknowns: The probability that the designated square is shaded
Understand
Restated: A wheel with the numbers $1, 2, 3, 6, 7, 9$ is spun twice, each number equally likely. The first result is divided by $4$; its remainder picks a column ($1$, $2$, or $3$). The second result is divided by $5$; its remainder picks a row ($1$, $2$, $3$, or $4$). Find the probability that the chosen column-and-row square on the checkerboard is shaded.
Givens: The wheel shows six equally likely numbers: $1, 2, 3, 6, 7, 9$; First spin $\div 4$: the remainder is the column number ($1, 2, 3$); Second spin $\div 5$: the remainder is the row number ($1, 2, 3, 4$); The board is $3$ columns by $4$ rows, shaded in a checkerboard pattern: shaded squares are $(\text{col},\text{row}) = (1,1),(3,1),(2,2),(1,3),(3,3),(2,4)$; Answer choices: (A) $\frac{1}{3}$, (B) $\frac{4}{9}$, (C) $\frac{1}{2}$, (D) $\frac{5}{9}$, (E) $\frac{2}{3}$
Plan
Primary tool: #2 Make a Systematic List
Secondary: #1 Draw a Diagram, #7 Identify Subproblems
Tool #2 (Make a Systematic List) — the column and row are each a remainder, so first list what remainder every wheel number gives. This turns a vague spinner into a clean table of how likely each column and each row is. Tool #1 (Draw a Diagram) — read the six shaded squares straight off the board as (column, row) pairs so we know exactly which outcomes count as a win. Tool #7 (Identify Subproblems) — because the two spins are independent, the probability of any one square is (its column's probability) times (its row's probability); sum that over the six shaded squares.
Execute — Answer: C
4.NBT.B.6 Step 1 List column remainders (÷4)
- Divide each wheel number by $4$ and record the remainder — this is the column.
- $1\to1$, $2\to2$, $3\to3$, $6\to2$, $9\to1$, and $7\to3$.
- So remainder $1$ comes from $\{1,9\}$, remainder $2$ from $\{2,6\}$, remainder $3$ from $\{3,7\}$ — two numbers each.
- Every column is therefore equally likely.
💡 Sorting the six numbers by their remainder mod $4$ shows each column is fed by exactly two numbers, so the columns tie.
4.NBT.B.6 Step 2 List row remainders (÷5)
- Now divide each wheel number by $5$ and record the remainder — this is the row.
- $1\to1$, $6\to1$; $2\to2$, $7\to2$; $3\to3$; $9\to4$.
- So row $1$ comes from $\{1,6\}$ and row $2$ from $\{2,7\}$ (two numbers each), while row $3$ comes only from $\{3\}$ and row $4$ only from $\{9\}$ (one number each).
- The rows are not equally likely.
💡 Mod $5$ splits the six numbers unevenly, so the top two rows are half as likely as the bottom two.
7.SP.C.7 Step 3 Read the shaded squares
- Read the shaded squares off the board as (column, row) pairs.
- They are $(1,1),(3,1)$ in row $1$; $(2,2)$ in row $2$; $(1,3),(3,3)$ in row $3$; and $(2,4)$ in row $4$ — six shaded squares forming a checkerboard.
- Group them by row so we can reuse the row probabilities: rows $1$ and $2$ (the likely rows) hold three shaded squares, and rows $3$ and $4$ (the unlikely rows) hold the other three.
💡 Grouping wins by row lets us pair each square with the row probability it depends on.
7.SP.C.8 Step 4 Probability from the likely rows
- Because the two spins are independent, each square's probability is its column probability times its row probability.
- Every column is $\frac{1}{3}$.
- For the three shaded squares in rows $1$ and $2$, the row probability is $\frac{1}{3}$, so each such square has probability $\frac{1}{3}\cdot\frac{1}{3}=\frac{1}{9}$.
💡 Independent spins multiply, and the three same-strength squares add up to one-third.
7.SP.C.8 Step 5 Probability from the unlikely rows, then total
- For the three shaded squares in rows $3$ and $4$, the column probability is still $\frac{1}{3}$ but the row probability is only $\frac{1}{6}$, so each has probability $\frac{1}{3}\cdot\frac{1}{6}=\frac{1}{18}$.
- Adding the two groups gives the total probability of landing on a shaded square.
- That equals $\frac{1}{2}$, which is choice (C).
💡 The weaker squares contribute half as much each, and the two groups together make exactly one half.
4.NBT.B.6 Divide each wheel number by $4$ and record the remainder — this is the column. $ 4.NBT.B.6 Now divide each wheel number by $5$ and record the remainder — this is the row. 7.SP.C.7 Read the shaded squares off the board as (column, row) pairs. They are $(1,1),(3 7.SP.C.8 Because the two spins are independent, each square's probability is its column p 7.SP.C.8 For the three shaded squares in rows $3$ and $4$, the column probability is stil Review
Reasonableness: The answer $\frac{1}{2}$ is believable: the board is a perfect checkerboard, so exactly half of the twelve squares are shaded, and a quick symmetry argument confirms the weighting does not break that balance. Every column is equally likely, so within any single row the shaded and unshaded squares are hit with matching total weight. Because each row's three squares split evenly between shaded and unshaded across the checkerboard, no row can tip the balance, and the whole board stays at one-half regardless of the uneven row probabilities. That rules out the lopsided choices like (A) $\frac{1}{3}$ or (E) $\frac{2}{3}$, and the exact sum lands on (C).
Alternative: Change focus with Tool #16: instead of summing six products, note that in each row the shaded squares and the unshaded squares are hit with equal probability (columns are uniform, and every row of this checkerboard has its shaded/unshaded squares symmetric under the column distribution). So shaded and unshaded are equally likely overall, forcing the probability to be exactly $\frac{1}{2}$ — no arithmetic needed.
CCSS standards used (min grade 7)
4.NBT.B.6Find whole-number quotients and remainders with up to four-digit dividends (Dividing each wheel number by 4 and by 5 to get the column and row remainders.)7.SP.C.7Develop probability models and use them to find probabilities of events (Turning the remainder counts into probabilities for each column and each row.)7.SP.C.8Find probabilities of compound events using organized lists, tables, and simulation (Multiplying independent column and row probabilities and summing over the shaded squares.)
⭐ Sort the spinner numbers by their remainders to see how likely each column and row is, then multiply column-chance by row-chance for every shaded square and add — the checkerboard's balance makes it exactly $\frac{1}{2}$.
⭐ Sort the spinner numbers by their remainders to see how likely each column and row is, then multiply column-chance by row-chance for every shaded square and add — the checkerboard's balance makes it exactly $\frac{1}{2}$.
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