AMC 10 · 2024 · #6

Grade 4 counting

Archetype Lattice / Grid Pattern from Small Cases └ Sub-archetype Figure-N Growth Sequence

pattern-recognitionsystematic-enumerationsequences-arithmetic easier-related-problempattern-recognitionoptimization-counting ↑ Prerequisites: systematic-enumerationmulti-digit-arithmetic

📏 Medium solution 💡 3 insights

Problem

Starting from the string ABCDEF, you may swap any two adjacent letters at a time. What is the smallest number of such swaps that turns the string into its reverse, FEDCBA?

Pick an answer.

(A)

(B)

~10

(C)

~12

(D)

~15

(E)

~24

AMC 10 2024 problem © Mathematical Association of America (MAA AMC). Reproduced for educational use.

How to solve

Strategy Solve an Easier Related Problem

A length-6 reversal is too big to swap-and-count from scratch with confidence. Tool #9 (Solve an Easier Related Problem) says: shrink the string. The problem already tells us ABC (length 3) takes 3 swaps; we can quickly do length 2, 4, and 5 ourselves. Once those small cases are in hand, Tool #5 (Look for a Pattern) lines them up — 1, 3, 6, 10, … — and the rule (each new length adds the next whole number) gives the length-6 answer without any algebra.

1STEP 1

Easier case, length 2: reversing AB to BA takes just 1 swap.

AB → BA (1 swap)

2STEP 2

Easier case, length 3: the problem's own ABC → BAC → BCA → CBA is already 3 swaps.

ABC → BAC → BCA → CBA (3 swaps)

3STEP 3

Easier case, length 4: send D to the front (3 swaps), then reverse the leftover ABC (3 more) → 6 swaps.

ABCD → ABDC → ADBC → DABC → DBAC → DBCA → DCBA (6 swaps)

4STEP 4

Easier case, length 5: ship E to the front (4 swaps), then reverse ABCD (6 swaps) → 10 swaps.

10 = 4 + 6

5STEP 5

Line the counts up — 1, 3, 6, 10 — and each step adds one more than the last: the triangular numbers.

length & swaps & added ; 2 & 1 & - ; 3 & 3 & +2 ; 4 & 6 & +3 ; 5 & 10 & +4

6STEP 6

Extend one rung to length 6: the next jump is +5, so the count is 10 + 5 = 15 → (D).

swaps(6) = 10 + 5 = 15 → (D)

Answer

~15

Cross-check with a different counting style: how many pairs of letters end up in opposite order? In ABCDEF every pair is in alphabetical order; in FEDCBA every pair is in reverse order. So every pair must cross exactly once during the swaps, and each adjacent swap flips exactly one pair. The number of pairs from 6 letters is

\frac{(6 · 5)}{2}

= 15, so the minimum number of swaps is exactly 15. Matches (D). Also, (A) 6 and (B) 10 are too small (length-5 alone takes 10), (E) 24 is wastefully large, and (C) 12 has no natural counting story — only (D) lands on the triangular-number rung.

💡Key takeaway

When the string feels too long, shrink it first. Lengths 2, 3, 4, 5 need 1, 3, 6, 10 swaps — the next stair adds 5, so length 6 needs 15.

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