AMC 10 · 2024 · #6
Grade 4 countingProblem
What is the minimum number of successive swaps of adjacent letters in the string ABCDEF that are needed to change the string to FEDCBA? (For example, 3 swaps are required to change ABC to CBA; one such sequence of swaps is
ABC→BAC→BCA→CBA.)
Pick an answer.
AMC 10 2024 problem © Mathematical Association of America (MAA AMC). Reproduced for educational use.
Try it yourself first — the explanation is most useful after you’ve attempted it.
Toolkit + CCSS Solution
Understand
Restated: 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, $FEDCBA$?
Givens: Start string: $ABCDEF$ (length $6$); Target string: $FEDCBA$ (the reverse); Each move swaps two letters that sit next to each other; Worked example: $ABC \to BAC \to BCA \to CBA$ takes $3$ swaps; Answer choices: (A) $6$, (B) $10$, (C) $12$, (D) $15$, (E) $24$
Unknowns: The minimum number of adjacent swaps to reverse a string of $6$ letters
Understand
Restated: 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, $FEDCBA$?
Givens: Start string: $ABCDEF$ (length $6$); Target string: $FEDCBA$ (the reverse); Each move swaps two letters that sit next to each other; Worked example: $ABC \to BAC \to BCA \to CBA$ takes $3$ swaps; Answer choices: (A) $6$, (B) $10$, (C) $12$, (D) $15$, (E) $24$
Plan
Primary tool: #9 Solve an Easier Related Problem
Secondary: #5 Look for a Pattern
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 — $1, 3, 6, 10, \ldots$ — and the rule (each new length adds the next whole number) gives the length-$6$ answer without any algebra.
Execute — Answer: D
3.OA.D.8 Step 1 - Easier case: length $2$.
- Reversing $AB$ to $BA$ takes one swap.
- Record: length $2 \Rightarrow 1$ swap.
💡 Start at the smallest case so the counting can't go wrong.
3.OA.D.8 Step 2 - Easier case: length $3$.
- The problem already shows $ABC \to BAC \to BCA \to CBA$, which is $3$ swaps.
- Record: length $3 \Rightarrow 3$ swaps.
💡 Using the example the problem hands you saves a step and confirms the counting style.
3.OA.D.8 Step 3 - Easier case: length $4$.
- Move the last letter $D$ to the front (takes $3$ swaps), then reverse the leftover $ABC$ inside (takes $3$ more swaps from the length-$3$ case).
- Total: $3 + 3 = 6$ swaps.
💡 Bringing the last letter to the front always costs (length $-1$) swaps, then the rest is the same problem one size smaller.
3.OA.D.8 Step 4 - Easier case: length $5$.
- Same plan: bring $E$ to the front in $4$ swaps, then reverse $ABCD$ in $6$ swaps.
- Total: $4 + 6 = 10$ swaps.
💡 Each new letter at the back costs (length $-1$) swaps to ship to the front, then the smaller answer takes over.
4.OA.C.5 Step 5 - Look for the pattern.
- Lay the answers out and watch the differences.
💡 Each row adds one more than the previous row. These are the triangular numbers $1, 3, 6, 10, 15, \ldots$
4.OA.A.3 Step 6 - Extend the pattern one more step to length $6$.
- The next addition is $+5$, so the count is $10 + 5 = 15$.
💡 The same rule (bring the last letter to the front in $5$ swaps, then reverse the front $5$ letters in $10$ swaps) gives the same total: $5 + 10 = 15$.
3.OA.D.8 Easier case: length $2$. Reversing $AB$ to $BA$ takes one swap. Record: length $ 3.OA.D.8 Easier case: length $3$. The problem already shows $ABC \to BAC \to BCA \to CBA$ 3.OA.D.8 Easier case: length $4$. Move the last letter $D$ to the front (takes $3$ swaps) 3.OA.D.8 Easier case: length $5$. Same plan: bring $E$ to the front in $4$ swaps, then re 4.OA.C.5 Look for the pattern. Lay the answers out and watch the differences. 4.OA.A.3 Extend the pattern one more step to length $6$. The next addition is $+5$, so th Review
Reasonableness: 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 $\dfrac{6 \cdot 5}{2} = 15$, 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.
Alternative: Tool #5 (Look for a Pattern) reading the same data straight off as a formula: lengths $2, 3, 4, 5, 6$ give $1, 3, 6, 10, 15$, the triangular numbers $T_{n-1} = \dfrac{(n-1)n}{2}$. Plug in $n = 6$: $T_5 = \dfrac{5 \cdot 6}{2} = 15$, again (D). This is the same idea as the pair-counting check, just stated as a closed form.
CCSS standards used (min grade 4)
3.OA.D.8Solve two-step word problems using the four operations (Counting swaps in the small cases (length $2$, $3$, $4$, $5$) by breaking each reversal into "bring the last letter to the front, then reverse the rest.")4.OA.C.5Generate a number pattern that follows a given rule and identify apparent features (Lining up the small-case answers $1, 3, 6, 10$ and noticing each next term adds one more than the previous — the triangular-number rule.)4.OA.A.3Solve multistep word problems with whole numbers using the four operations (Extending the pattern by one step: $10 + 5 = 15$ to finish at length $6$.)
⭐ 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 $15$.
⭐ 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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