2024 AMC 10A Problems/Problem 6

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

$\textbf{(A)}~6\qquad\textbf{(B)}~10\qquad\textbf{(C)}~12\qquad\textbf{(D)}~15\qquad\textbf{(E)}~24$

Solution 1 (Analysis)

Procedurally, it takes:

$5$ swaps for $A$ to move to the sixth spot, giving $BCDEFA.$

$4$ swaps for $B$ to move to the fifth spot, giving $CDEFBA.$

$3$ swaps for $C$ to move to the fourth spot, giving $DEFCBA.$

$2$ swaps for $D$ to move to the third spot, giving $EFDCBA.$

$1$ swap for $E$ to move to the second spot (so $F$ becomes the first spot), giving $FEDCBA.$

Together, the answer is $5+4+3+2+1=\boxed{\textbf{(D)}~15}.$

~MRENTHUSIASM

Solution 2 (Recursive Approach)

We can proceed by a recursive tactic on the number of letters in the string.

Looking at the string $A$ , there are $0$ moves needed to change it to the string $A$

Then, there is $1$ move to change $AB$ to $BA$ .

Similarly, there is $3$ moves needed for three letters (said in the problem).

There are $6$ moves needed to change $ABCD$ to $DCBA$ .

We see a pattern of $0,1,3,6,...$ . We notice that the difference between consecutive terms is increasing by $1$ , so in the same way, for $5$ letters, we would need $10$ moves, and for $6$ , we would need $\boxed{\textbf{(D)}~15}$ moves.

Thinking why, when we start making these moves, we see that for a string of length $n$ , it takes $n-1$ moves to move the last letter to the front. After, we get a string that will be changed identically to a string of length $n-1$ . This works in our pattern above and is another way to think about the problem!

~world123

Note:

The sequence consists of [url=https://artofproblemsolving.com/wiki/index.php/Triangular_number?srsltid=AfmBOorS0aNYW5Ipg2yd_Xe-Y5_hOBKasngGeiehAL098VYyZuMQVulA]triangular numbers[/url] shifted a term up (as it starts with $0$ on term $1$ and $1$ on term $2$ .) Thus, our explicit formula is $\dfrac{(n-1)(n+1-1)}{2} = \dfrac{(n)(n-1)}{2}$ and as $n = 6$ in this case ( $6$ letters), our answer is $\dfrac{(6)(6-1)}{2} = \boxed{\textbf{(D)}~15}$ ~ NSAoPS

Solution 3 (Solution 2 Done Fast)

We can see that the most efficient way to change $ABCDEF$ to $FEDCBA$ is the same as changing $ABCDE$ to $EBCDA$ and then moving $F$ to the front in $5$ moves. Similarly, this would carry on downwards, where to change $ABCDE$ to $EBCDA$ would be to change $ABCD$ to $DCBA$ and move $E$ $4$ times to the front. This pattern will carry on until $AB$ to $BA$ would be $1$ , and $A$ to $A$ would be $0$ . The answer would be $0(A)+1(B)+2(C)+3(D)+4(E)+5(F)$ , which is $\boxed{\textbf{(D)}~15}$ moves.

~Moonwatcher22

Video Solution by Pi Academy

https://youtu.be/6qYaJsgqkbs?si=K2Ebwqg-Ro8Yqoiv

Video Solution 1 by Power Solve

https://youtu.be/j-37jvqzhrg?si=ieBRx0-CUihcKttE&t=616

Video Solution by Daily Dose of Math

https://youtu.be/-EFTk2pBFug

~Thesmartgreekmathdude

2024 AMC 10A (Problems • Answer Key • Resources)
Preceded by Problem 5	Followed by Problem 7
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25
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