AMC 10 · 2013 · #22

Grade 7 counting
Archetype Casework by Position / Vertex Type └ Sub-archetype Vertex or Element Role
systematic-enumerationpermutations-basicdivisibility-rules systematic-enumerationcasework ↑ Prerequisites: permutations-basicdivisibility-rules
📏 Medium solution 💡 3 insights 📊 Diagram
Problem
A regular octagon ABCDEFGH has center J. Each of the four lines AJE, BJF, CJG, DJH passes through two opposite vertices and the center J. Place the digits 1 through 9, each used exactly once, on the eight vertices and the center so that the three numbers on each of the four lines have the same sum. In how many ways can this be done?

Pick an answer.

(A)
384
(B)
576
(C)
1152
(D)
1680
(E)
3456

AMC 10 2013 problem © Mathematical Association of America (MAA AMC). Reproduced for educational use.

How to solve
Strategy Make a Systematic List

This is a 'how many ways' counting question, so the plan is to pin down the structure, then multiply independent choices. First name the common line sum with a variable to get an equation linking it to the center digit; that equation limits the center to just a few digits. For each allowed center digit, the leftover eight digits pair up into equal-sum pairs; then I count how many ways those pairs can be placed on the four lines and flipped end for end. Multiplying the independent counts gives the total.

1STEP 1

Add all four line sums

Let S be the common line sum. Adding all four counts every vertex once and the center J four times: 4S = 45 + 3J.

4S = (1+2+…+9) + 3J = 45 + 3J
2STEP 2

Find which center digits work

For 4S to be a multiple of 4, so must 45 + 3J — testing the digits, that happens only for J = 1, 5, or 9.

45 + 3J ≡ 0 (mod 4) → J ∈ {1, 5, 9}
3STEP 3

Pair the leftover eight digits

With J fixed, the two vertices on each line must share one target sum — 11, 10, or 9 — and the eight leftovers pair up in exactly one way.

pair sum = S - J = (45 - J)/4; J=1→ 11, J=5→ 10, J=9→ 9
4STEP 4

Place and flip the four pairs

Assign the four pairs to the four lines in 4! = 24 ways and flip each line independently in 2⁴ = 16 ways: 24 × 16 = 384.

4! × 2⁴ = 24 × 16 = 384
5STEP 5

Combine the three center choices

The three center digits give disjoint cases, so 3 × 384 = 1152, choice (C).

3 × 384 = 1152
Answer
1152
The total 1152 factors as 3 x 24 x 16, matching the three independent stages: 3 center choices, 4! ways to assign pairs to lines, and 2⁴ ways to flip pairs end for end. It sits sensibly between the choices: 384 (C's answer for a single center) is exactly one center's count, so forgetting to multiply by 3 would land on (A); 1152 is 384 tripled, confirming (C). The value is comfortably under the total 9! arrangements, as expected once the equal-sum constraint prunes most placements.
💡Key takeaway

The center digit sits on every line, so only 1, 5, or 9 can balance the four sums; after that the leftovers pair up in one way and you just count how to place and flip them, giving 3 x 24 x 16 = 1152.

  • Add all four line sums
  • Find which center digits work
  • Pair the leftover eight digits
  • Place and flip the four pairs
  • Combine the three center choices

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