AMC 10 · 2014 · #24

Grade 7 geometry-2d

Archetype Small-Domain Constrained Search └ Sub-archetype Logic-Deduction / Elimination

systematic-enumerationcomplementary-countingsymmetry-argumentcasework systematic-enumeration ↑ Prerequisites: systematic-enumeration

📏 Long solution 💡 4 insights

Problem

Place 1,2,3,4,5 around a circle. For a target n from 1 to 15, you are allowed to add up numbers that sit next to each other along the circle (a single number counts, and so does the whole circle). Call the arrangement bad if at least one target n in 1 to 15 can never be formed this way. Arrangements that match after a rotation or a mirror flip count as one. Count the bad arrangements.

Pick an answer.

(A)

(B)

(C)

(D)

(E)

AMC 10 2014 problem © Mathematical Association of America (MAA AMC). Reproduced for educational use.

How to solve

Strategy Make a Systematic List

"How many arrangements" with a finite pool screams Tool #2 (Make a Systematic List) — but a raw list of 120 orderings is wasteful. First shrink the work two ways. Tool #16 (Count the Complement) notices that the total is 15, so an arc summing to k leaves a complementary arc summing to 15-k; that instantly makes 1--5, 10--15 free and pairs the doubtful targets as 6⇔ 9 and 7⇔ 8, so we only ever test 6 and 7. Tool #1 (Draw a Diagram) with the rotation-and-reflection rule cuts 120 orderings down to just 12 distinct circles. Then Tool #2 lists those 12 and Tool #3 (Eliminate) crosses off every circle that manages to build both 6 and 7, leaving the bad ones.

1STEP 1

List the sums you always get

The five total 15, so singles give 1,2,3,4,5, all-but-one gives 10,11,12,13,14, and the whole circle gives 15 — order never matters.

{1,2,3,4,5}∪{15-k : k=1..5}∪{15} = {1,2,3,4,5,10,11,12,13,14,15}

2STEP 2

Only 6 and 7 can go wrong

Only 6,7,8,9 are in doubt, and an arc summing to 6 leaves 9, one summing to 7 leaves 8 — so just test 6 and 7.

6 ⇔ 15-6=9, 7 ⇔ 15-7=8

3STEP 3

Only 12 different circles exist

Pin 1 at the top to remove the 5 rotations, then halve for the mirror flip: 120 orderings collapse to 12 different circles.

5!/(5 × 2) = 120/10 = 12

4STEP 4

Test each circle for 6 and 7

6 comes from 1+5, 2+4, 1+2+3; 7 from 2+5, 3+4, 1+2+4. Across the twelve circles only 1-2-5-3-4 (no 6) and 1-3-2-4-5 (no 7) fail.

6 = 1+5 = 2+4 = 1+2+3, 7 = 2+5 = 3+4 = 1+2+4

5STEP 5

Count the bad circles

So exactly two of the twelve circles are bad — 1-2-5-3-4 (never 6) and 1-3-2-4-5 (never 7), and they are distinct, so the answer is (B).

#{bad} = 2 → (B)

Answer

Both bad circles were verified by listing their complete set of arc sums, not just spot-checked. For 1-2-5-3-4: neighbor sums are 3,7,8,7,5 and three-in-a-row sums are 8,10,12,8,7; together with 1--5, 10--14, 15 that is {1,2,3,4,5,7,8,10,11,12,13,14,15} — genuinely missing 6 and 9. For 1-3-2-4-5 the same accounting misses 7 and 8. A full computer-style sweep of all 12 circles returns exactly these two, so the answer 2 is not an undercount from a missed case, and it lands squarely inside the choice list at (B).

💡Key takeaway

The whole circle is 15, so every arc and its leftover add to 15 — that makes all targets free except 6 and 7, and only two of the twelve different circles fail to build one of them.

List the sums you always get
Only 6 and 7 can go wrong
Only 12 different circles exist
Test each circle for 6 and 7
Count the bad circles

List the sums you always get

Only 6 and 7 can go wrong

Only 12 different circles exist

Test each circle for 6 and 7

Count the bad circles

More like this