AMC 10 · 2024 · #6

Grade 4 counting

Archetype Lattice / Grid Pattern from Small Cases

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

📏 Medium solution 💡 3 insights

Problem

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
$ABC\to BAC\to BCA\to CBA.$ )

Pick an answer.

(A)

(B)

~10

(C)

~12

(D)

~15

(E)

~24

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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, $FEDCBA$?

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

#9 Solve an Easier Related Problem 3.OA.D.8 Step 1

Easier case: length $2$.
Reversing $AB$ to $BA$ takes one swap.
Record: length $2 \Rightarrow 1$ swap.

$AB \to BA$ ($1$ swap)

💡 Start at the smallest case so the counting can't go wrong.

#9 Solve an Easier Related Problem 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.

$ABC \to BAC \to BCA \to CBA$ ($3$ swaps)

💡 Using the example the problem hands you saves a step and confirms the counting style.

#9 Solve an Easier Related Problem 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.

$ABCD \to ABDC \to ADBC \to DABC \to DBAC \to DBCA \to DCBA$ ($6$ swaps)

💡 Bringing the last letter to the front always costs (length $-1$) swaps, then the rest is the same problem one size smaller.

#9 Solve an Easier Related Problem 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.

$$10 = 4 + 6$$

💡 Each new letter at the back costs (length $-1$) swaps to ship to the front, then the smaller answer takes over.

#5 Look for a Pattern 4.OA.C.5 Step 5

Look for the pattern.
Lay the answers out and watch the differences.

$$\begin{array}{c|c|c} \text{length} & \text{swaps} & \text{added} \\ \hline 2 & 1 & - \\ 3 & 3 & +2 \\ 4 & 6 & +3 \\ 5 & 10 & +4 \end{array}$$

💡 Each row adds one more than the previous row. These are the triangular numbers $1, 3, 6, 10, 15, \ldots$

#5 Look for a Pattern 4.OA.A.3 Step 6

Extend the pattern one more step to length $6$.
The next addition is $+5$, so the count is $10 + 5 = 15$.

$$\text{swaps}(6) = 10 + 5 = 15 \;\Rightarrow\; \textbf{(D)}$$

💡 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: $5 + 10 = 15$.

[1] #9 3.OA.D.8 Easier case: length $2$. Reversing $AB$ to $BA$ takes one swap. Record: length $

[2] #9 3.OA.D.8 Easier case: length $3$. The problem already shows $ABC \to BAC \to BCA \to CBA$

[3] #9 3.OA.D.8 Easier case: length $4$. Move the last letter $D$ to the front (takes $3$ swaps)

[4] #9 3.OA.D.8 Easier case: length $5$. Same plan: bring $E$ to the front in $4$ swaps, then re

[5] #5 4.OA.C.5 Look for the pattern. Lay the answers out and watch the differences.

[6] #5 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.8 Solve 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.5 Generate 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.3 Solve 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 $15$.

Problem

Toolkit + CCSS Solution

Understand

Plan

Execute — Answer: D

Review

CCSS standards used (min grade 4)

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