Competition · AMC preparation · step 4 of 4
AMC 8 · 2018 · #3
Grade 4 logicPick an answer.
AMC 8 2018 problem © Mathematical Association of America (MAA AMC). Reproduced for educational use.
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, … — 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.
List the unlucky numbers
A number is 'unlucky' if it's a multiple of 7 or contains a 7; the first five are 7, 14, 17, 21, 27 — enough to remove five of six.
Listing multiples of 7 and numbers containing the digit 7 is exactly what Grade 4 'factors and multiples' work practices.
4.OA.B.4Make A Systematic ListSeat everyone in a circle
Model it physically: set 6 coins in a ring — A(rn), B(ob), C(yd), D(an), E(ve), F(on) — and pull one out on each unlucky number.
Arranging objects in a ring and naming who is 'next' is the Kindergarten skill of describing positions of objects.
K.G.A.1Create A Physical RepresentationCount round 1
Round 1: 1→A, 2→B, 3→C, 4→D, 5→E, 6→F, 7→A — Arn says 7 and leaves; B, C, D, E, F remain.
Counting one number per coin around the ring is straight Grade 1 'count to 120 starting at any number'.
1.NBT.A.1Create A Physical RepresentationCount round 2
Round 2: continue from Bob: 8→B, 9→C, 10→D, 11→E, 12→F, 13→B, 14→C — Cyd leaves; B, D, E, F remain.
Same counting move as before — we just continue the same number line without restarting.
When the count continues from 8 with Bob and runs up to the unlucky number 14, the student who says 14 — and so is removed — is Cyd.
▸ Why?
Bob says 8, and reaching 14 takes 6 more counts around the ring, so 14 lands 6 turns after Bob.
▸ Why?
From 8 up to 14 the extra numbers 9, 10, 11, 12, 13, 14 are counted, and each number is exactly one student's turn, so it is exactly 6 steps forward.
▸ Why?
The 5 students still in the ring take turns in a fixed repeating order, so 6 steps is one full loop of 5 back to Bob plus 1 more step, which lands on Cyd.
▸ Why?
Going around the fixed ring of 5 students, 5 steps bring the count right back to Bob, where this round's count began.
▸ Why?
The 6 steps split with no leftover into 5 — the full loop back to Bob — plus 1, and that 1 extra step moves from Bob to the very next student, Cyd.
Count round 3
Round 3: from Dan — 15→D, 16→E, 17→F — Fon says 17 and leaves; B, D, E remain.
Just three more counts — still pointing one coin per number around the shrinking ring.
1.NBT.A.1Create A Physical RepresentationCount round 4
Round 4: after Fon skip the out Arn to Bob — 18→B, 19→D, 20→E, 21→B — Bob leaves; D, E remain.
Skipping the already-removed coin and going to the next physical neighbor is exactly the 'next-to / beside' position language.
K.G.A.1Create A Physical RepresentationCount round 5
Round 5: only Dan and Eve left, alternating 22→D, 23→E, 24→D, 25→E, 26→D, 27→E — Eve leaves, so Dan is the last one; answer (D).
When only two coins remain, counting alternates between them — a simple back-and-forth pattern any first grader can follow.
1.NBT.A.1Create A Physical RepresentationThis AMC 8 problem only needs Grade 4 multiples of 7 that you already know — the rest is just counting around a circle with coins!
- List the unlucky numbers
- Seat everyone in a circle
- Count round 1
- Count round 2
- Count round 3
- Count round 4
- Count round 5
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