AMC 8 · 2018 · #3
Grade 4 logicProblem
Students Arn, Bob, Cyd, Dan, Eve, and Fon are arranged in that order in a circle. They start counting: Arn first, then Bob, and so forth. When the number contains a 7 as a digit (such as 47) or is a multiple of 7 that person leaves the circle and the counting continues. Who is the last one present in the circle?
Pick an answer.
AMC 8 2018 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: Six students — Arn, Bob, Cyd, Dan, Eve, Fon — stand in a circle in that order. They count off $1, 2, 3, \ldots$ around the circle starting with Arn. A student is removed the moment they say a number that is a multiple of $7$ OR that contains the digit $7$. Counting continues with the next student after each removal. Who is the very last student remaining in the circle?
Givens: Circle order: Arn, Bob, Cyd, Dan, Eve, Fon (then back to Arn); Counting starts at $1$ with Arn and continues one number per student around the circle; Elimination trigger: the spoken number is a multiple of $7$ OR contains the digit $7$; After an elimination the next student says the next number; Answer choices: (A) Arn, (B) Bob, (C) Cyd, (D) Dan, (E) Eve
Unknowns: Which student is the single survivor after $5$ eliminations
Understand
Restated: Six students — Arn, Bob, Cyd, Dan, Eve, Fon — stand in a circle in that order. They count off $1, 2, 3, \ldots$ around the circle starting with Arn. A student is removed the moment they say a number that is a multiple of $7$ OR that contains the digit $7$. Counting continues with the next student after each removal. Who is the very last student remaining in the circle?
Givens: Circle order: Arn, Bob, Cyd, Dan, Eve, Fon (then back to Arn); Counting starts at $1$ with Arn and continues one number per student around the circle; Elimination trigger: the spoken number is a multiple of $7$ OR contains the digit $7$; After an elimination the next student says the next number; Answer choices: (A) Arn, (B) Bob, (C) Cyd, (D) Dan, (E) Eve
Plan
Primary tool: #10 Create a Physical Representation
Secondary: #2 Make a Systematic List
This is a small, concrete simulation — only $6$ people and only a handful of eliminations. Tool #10 (Physical Representation) fits perfectly: lay out $6$ coins (or fingers) in a circle and physically remove one each time the count hits an unlucky number. To keep the bookkeeping clean we also use Tool #2 (Systematic List) to write the 'unlucky numbers' in order — $7, 14, 17, 21, 27, \ldots$ — so we never miss one. Tool #13 (Algebra) would be overkill here; tool #5 (Pattern) is not needed because $5$ rounds is small enough to walk through directly.
Execute — Answer: D
4.OA.B.4 Step 1 - List the unlucky numbers in order.
- A number is unlucky if it is a multiple of $7$ or contains the digit $7$.
- Going $1, 2, 3, \ldots$, the first few are $7$ (both), $14$ (multiple), $17$ (digit), $21$ (multiple), $27$ (digit).
- We only need $5$ of them because $6$ people lose one each round and the question asks for the last one left.
💡 Listing multiples of $7$ and numbers containing the digit $7$ is exactly what Grade 4 'factors and multiples' work practices.
K.G.A.1 Step 2 - Set up the physical model.
- Place $6$ coins in a circle labeled A(rn), B(ob), C(yd), D(an), E(ve), F(on).
- We will move clockwise, pointing at one coin per spoken number, and pull a coin out when the number is unlucky.
💡 Arranging objects in a ring and naming who is 'next' is the Kindergarten skill of describing positions of objects.
1.NBT.A.1 Step 3 - Round 1 — count until someone says $7$.
- Pointing in order: $1\to$ A, $2\to$ B, $3\to$ C, $4\to$ D, $5\to$ E, $6\to$ F, $7\to$ A.
- So Arn says $7$ and is removed.
- Remaining: B, C, D, E, F.
💡 Counting one number per coin around the ring is straight Grade 1 'count to 120 starting at any number'.
1.NBT.A.1 Step 4 - Round 2 — keep counting from $8$ with Bob (the next coin after the removed Arn) until $14$.
- The pointer goes $8\to$ B, $9\to$ C, $10\to$ D, $11\to$ E, $12\to$ F, $13\to$ B, $14\to$ C.
- Cyd is removed.
- Remaining: B, D, E, F.
💡 Same counting move as before — we just continue the same number line without restarting.
1.NBT.A.1 Step 5 - Round 3 — continue from $15$ with Dan (next after the removed Cyd) until $17$.
- The pointer goes $15\to$ D, $16\to$ E, $17\to$ F.
- Fon is removed.
- Remaining: B, D, E.
💡 Just three more counts — still pointing one coin per number around the shrinking ring.
K.G.A.1 Step 6 - Round 4 — continue from $18$ with the student after Fon.
- Going around: after F came A, but A is gone, so it is B.
- Counting until $21$: $18\to$ B, $19\to$ D, $20\to$ E, $21\to$ B.
- Bob is removed.
- Remaining: D, E.
💡 Skipping the already-removed coin and going to the next physical neighbor is exactly the 'next-to / beside' position language.
1.NBT.A.1 Step 7 - Round 5 — continue from $22$ with Dan (next after the removed Bob) until $27$.
- With only two coins left we alternate: $22\to$ D, $23\to$ E, $24\to$ D, $25\to$ E, $26\to$ D, $27\to$ E.
- Eve is removed.
- The only coin left in the ring is Dan, so Dan is the last one present.
- The answer is (D).
💡 When only two coins remain, counting alternates between them — a simple back-and-forth pattern any first grader can follow.
4.OA.B.4 List the unlucky numbers in order. A number is unlucky if it is a multiple of $7 K.G.A.1 Set up the physical model. Place $6$ coins in a circle labeled A(rn), B(ob), C(y 1.NBT.A.1 Round 1 — count until someone says $7$. Pointing in order: $1\to$ A, $2\to$ B, $ 1.NBT.A.1 Round 2 — keep counting from $8$ with Bob (the next coin after the removed Arn) 1.NBT.A.1 Round 3 — continue from $15$ with Dan (next after the removed Cyd) until $17$. T K.G.A.1 Round 4 — continue from $18$ with the student after Fon. Going around: after F c 1.NBT.A.1 Round 5 — continue from $22$ with Dan (next after the removed Bob) until $27$. W Review
Reasonableness: Five unlucky numbers $7, 14, 17, 21, 27$ remove exactly five of the six students, in the order Arn, Cyd, Fon, Bob, Eve. That leaves Dan as the single survivor — matches choice (D). A quick sanity check: counting all the way from $1$ to $27$ uses $27$ 'spoken numbers', and across the $5$ rounds the sums $7 + 7 + 3 + 4 + 6 = 27$ match perfectly, so no count was missed or double-counted.
Alternative: Tool #3 (Eliminate Possibilities) plus modular arithmetic: in each round with $k$ people remaining, the gap of $g$ counts lands on position $((\text{start} - 1) + g - 1) \bmod k + 1$. Round 2 has $k=5, g=7$ starting at Bob, giving position $((0)+6)\bmod 5 + 1 = 2$, the second person which is Cyd — same result. The modular shortcut confirms each elimination without re-counting.
CCSS standards used (min grade 4)
K.G.A.1Describe positions of objects using above, below, beside, in front of (Talking about who sits 'next to' whom in the circle and who becomes the next counter after a coin is pulled out.)1.NBT.A.1Count to 120 starting at any number less than 120 (Saying the numbers $1, 2, 3, \ldots, 27$ one per student around the ring without restarting after each elimination.)4.OA.B.4Find all factor pairs and recognize multiples; determine prime or composite (Identifying the multiples of $7$ ($7, 14, 21, \ldots$) that trigger an elimination, combined with the digit-$7$ check, to list the unlucky numbers in order.)
⭐ This AMC 8 problem only needs Grade 4 multiples of $7$ that you already know — the rest is just counting around a circle with coins!
⭐ This AMC 8 problem only needs Grade 4 multiples of $7$ that you already know — the rest is just counting around a circle with coins!
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