AMC 8 · 2018 · #3

Grade 4 logic

Archetype Branching State Enumeration └ Sub-archetype Rule-Governed Process Simulation

modular-arithmeticsystematic-enumerationlogical-deduction systematic-enumerationcasework ↑ Prerequisites: multi-digit-arithmeticdivisibility-rules

📏 Long solution 💡 3 insights

📘 View easy version →

Problem

Six students — Arn, Bob, Cyd, Dan, Eve, Fon — stand in a circle in that order. They count off 1, 2, 3, … 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?

Pick an answer.

(A)

$text{Arn}$

(B)

$text{Bob}$

(C)

$text{Cyd}$

(D)

$text{Dan}$

(E)

$text{Eve}$

AMC 8 2018 problem © Mathematical Association of America (MAA AMC). Reproduced for educational use.

How to solve

Strategy Create a Physical Representation

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.

1STEP 1

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.

Unlucky numbers: 7, 14, 17, 21, 27, …

2STEP 2

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.

Circle: A → B → C → D → E → F → A

3STEP 3

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.

1 A, 2 B, 3 C, 4 D, 5 E, 6 F, 7 A ×

4STEP 4

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.

8 B, 9 C, 10 D, 11 E, 12 F, 13 B, 14 C ×

5STEP 5

Round 3: from Dan — 15→D, 16→E, 17→F — Fon says 17 and leaves; B, D, E remain.

15 D, 16 E, 17 F ×

6STEP 6

Round 4: after Fon skip the out Arn to Bob — 18→B, 19→D, 20→E, 21→B — Bob leaves; D, E remain.

18 B, 19 D, 20 E, 21 B ×

7STEP 7

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).

22 D, 23 E, 24 D, 25 E, 26 D, 27 E × → (D) Dan

Answer

text{Dan}

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.

💡Key takeaway

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!

More like this