AMC 8 · 2005 · #20
Easy mode Grade 5Problem
Picture a circle with equally spaced points around it, labeled through in clockwise order, just like a clock.
Alice and Bob both start on point .
Each turn:
- Alice moves points clockwise.
- Bob moves points counterclockwise.
The game ends as soon as they land on the same point after a turn. How many turns does this take?
Pick an answer.
Toolkit + CCSS Solution
Understand
Restated: A circle has $12$ equally-spaced points numbered $1$ through $12$. Alice and Bob both start at point $12$. Each turn, Alice moves $5$ points clockwise and Bob moves $9$ points counterclockwise. After how many turns do they first land on the same point?
Givens: $12$ points evenly spaced around a circle, numbered $1$ to $12$ clockwise; Both start at point $12$; Each turn: Alice moves $+5$ (clockwise), Bob moves $-9$ (counterclockwise); Answer choices: (A) $6$, (B) $8$, (C) $12$, (D) $14$, (E) $24$
Unknowns: The smallest number of turns after which Alice and Bob are on the same point
Understand
Restated: A circle has $12$ equally-spaced points numbered $1$ through $12$. Alice and Bob both start at point $12$. Each turn, Alice moves $5$ points clockwise and Bob moves $9$ points counterclockwise. After how many turns do they first land on the same point?
Givens: $12$ points evenly spaced around a circle, numbered $1$ to $12$ clockwise; Both start at point $12$; Each turn: Alice moves $+5$ (clockwise), Bob moves $-9$ (counterclockwise); Answer choices: (A) $6$, (B) $8$, (C) $12$, (D) $14$, (E) $24$
Plan
Primary tool: #2 Make a Systematic List
Secondary: #5 Look for a Pattern
The answer choices are small ($6, 8, 12, 14, 24$), and after each turn both positions are easy to update by adding or subtracting on a $12$-point clock. That is the classic setup for Tool #2 (Systematic List): build a table of (turn, Alice's point, Bob's point) and watch for the first row where they agree. Tool #5 (Look for a Pattern) backs it up — each turn Alice gains $5$ points and Bob loses $9$, so the gap between them changes by a fixed amount every turn, and a constant change makes the wrap-around easy to predict instead of recomputing from scratch.
Execute — Answer: A
4.OA.C.5 Step 1 - Set up the table.
- Both start at point $12$, so call that turn $0$.
- Each turn Alice's point goes up by $5$, wrapping back from $12$ to $1$ when she passes it; Bob's point goes down by $9$, wrapping from $1$ back to $12$ when he passes it.
- We will list the positions turn by turn until they match.
💡 Grade 4 "generate a number pattern from a rule" — each player follows a simple add-or-subtract rule on the clock.
5.OA.B.3 Step 2 - Compute Alice's positions.
- Starting from $12$, add $5$ each turn and wrap (treat point $12$ as the same as $0$ when adding):
💡 Grade 5 "generate two numerical patterns using two given rules" — Alice's column is one of the two patterns we will compare.
5.OA.B.3 Step 3 - Compute Bob's positions.
- Starting from $12$, subtract $9$ each turn and wrap (after passing point $1$, the next counterclockwise point is $12$, then $11$, and so on):
💡 Same Grade 5 idea — Bob's column is the second pattern. Notice his positions cycle every $4$ turns: $3, 6, 9, 12, 3, 6, \ldots$.
5.OA.B.3 Step 4 - Compare the two columns turn by turn.
- The two players land on the same point the first time the rows match.
💡 Lining the two patterns side by side is exactly the Grade 5 standard's payoff — the first matching row gives the answer.
4.OA.C.5 Step 5 - Read the answer from the table.
- The first turn on which Alice and Bob share a point is turn $6$ (both at point $6$).
💡 A systematic list ends when the searched-for row appears — here, the first "same" row.
4.OA.C.5 Set up the table. Both start at point $12$, so call that turn $0$. Each turn Ali 5.OA.B.3 Compute Alice's positions. Starting from $12$, add $5$ each turn and wrap (treat 5.OA.B.3 Compute Bob's positions. Starting from $12$, subtract $9$ each turn and wrap (af 5.OA.B.3 Compare the two columns turn by turn. The two players land on the same point the 4.OA.C.5 Read the answer from the table. The first turn on which Alice and Bob share a po Review
Reasonableness: Quick sanity check on the meeting point. Alice's total clockwise travel after $6$ turns is $5 \times 6 = 30$ points. On a $12$-point circle, $30 = 2 \times 12 + 6$, so she ends $6$ points clockwise of her start at $12$ — that lands on point $6$. Bob's total counterclockwise travel is $9 \times 6 = 54$ points; $54 = 4 \times 12 + 6$, so he ends $6$ points counterclockwise of $12$ — which is also point $6$ (counting $11, 10, 9, 8, 7, 6$). Both land on $6$, confirming turn $6$ and answer (A). None of the smaller turns ($1$ through $5$) produced a match, so $6$ really is the first.
Alternative: Tool #5 (Look for a Pattern) on the gap: each turn the gap between Alice and Bob (measured clockwise from Alice to Bob) changes by $5 + 9 = 14$ points, which on a $12$-point circle is the same as a $2$-point shift. Starting with gap $0$, the gap after $k$ turns is $2k \pmod{12}$. They meet again when the gap returns to $0$, so the smallest $k > 0$ with $2k$ a multiple of $12$ is $k = 6$, giving answer (A).
CCSS standards used (min grade 5)
4.OA.C.5Generate a number or shape pattern that follows a given rule (Building Alice's and Bob's position sequences from the rules "add $5$ on a $12$-point clock" and "subtract $9$ on a $12$-point clock.")5.OA.B.3Generate two numerical patterns using two given rules; form ordered pairs and compare them (Lining Alice's and Bob's position columns side by side and scanning for the first turn where the two patterns produce the same point.)
⭐ On any "when do they meet on a circle?" question, a side-by-side table of each player's position usually finds the answer in fewer turns than the answer choices suggest — here it takes just $6$ rows to see Alice and Bob both land on point $6$.
⭐ On any "when do they meet on a circle?" question, a side-by-side table of each player's position usually finds the answer in fewer turns than the answer choices suggest — here it takes just $6$ rows to see Alice and Bob both land on point $6$.