Difference between revisions of "2024 AMC 10A Problems/Problem 6"

Revision as of 18:41, 9 November 2024

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 triangular numbers 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

Solution 4 (If you don't notice anything)

Simply, just blindly doing it, the most straightforward way is to shift F all the way back. From ABCDEF:

ABCDFE -> ABCFDE -> ABFCDE -> AFBCDE -> FABCDE

For E:

FABCED -> FABECD -> FAEBCD -> FEABCD

For D:

FEABDC -> FEADBC -> FEDABC

For C:

FEDACB -> FEDCAB

For B:

FEDCBA

That's it, you don't need to do it with A. Still $\boxed{\textbf{(D)}~15}$ moves.

~pepper2831

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

@@ Line 1: / Line 1: @@
-The length of your computer screen is x ft. The height is y ft. What is the probability that a sneaky user (hint, like you) can view the AMC 10A Problem 6 2024 BEFORE the actual test starts?
+== Problem ==
+What is the minimum number of successive swaps of adjacent letters in the string <math>ABCDEF</math> that are needed to change the string to <math>FEDCBA?</math> (For example, <math>3</math> swaps are required to change <math>ABC</math> to <math>CBA;</math> one such sequence of swaps is
+<math>ABC\to BAC\to BCA\to CBA.</math>)
-A: 0% B: 0.00% C: -1000% D: 500% * -1 E: pi ^ 0 * -1
+<math>\textbf{(A)}~6\qquad\textbf{(B)}~10\qquad\textbf{(C)}~12\qquad\textbf{(D)}~15\qquad\textbf{(E)}~24</math>
+== Solution 1 (Analysis) ==
+Procedurally, it takes:
+* <math>5</math> swaps for <math>A</math> to move to the sixth spot, giving <math>BCDEFA.</math>
+* <math>4</math> swaps for <math>B</math> to move to the fifth spot, giving <math>CDEFBA.</math>
+* <math>3</math> swaps for <math>C</math> to move to the fourth spot, giving <math>DEFCBA.</math>
+* <math>2</math> swaps for <math>D</math> to move to the third spot, giving <math>EFDCBA.</math>
+* <math>1</math> swap for <math>E</math> to move to the second spot (so <math>F</math> becomes the first spot), giving <math>FEDCBA.</math>
+Together, the answer is <math>5+4+3+2+1=\boxed{\textbf{(D)}~15}.</math>
+~MRENTHUSIASM
+== Solution 2 (Recursive Approach)==
+We can proceed by a recursive tactic on the number of letters in the string.
+Looking at the string <math>A</math>, there are <math>0</math> moves needed to change it to the string <math>A</math>
+Then, there is <math>1</math> move to change <math>AB</math> to <math>BA</math>.
+Similarly, there is <math>3</math> moves needed for three letters (said in the problem).
+There are <math>6</math> moves needed to change <math>ABCD</math> to <math>DCBA</math>.
+We see a pattern of <math>0,1,3,6,...</math>. We notice that the difference between consecutive terms is increasing by <math>1</math>, so in the same way, for <math>5</math> letters, we would need <math>10</math> moves, and for <math>6</math>, we would need <math>\boxed{\textbf{(D)}~15}</math> moves.
+Thinking why, when we start making these moves, we see that for a string of length <math>n</math>, it takes <math>n-1</math> moves to move the last letter to the front. After, we get a string that will be changed identically to a string of length <math>n-1</math>. This works in our pattern above and is another way to think about the problem!
+~world123
+Note:
+The sequence consists of triangular numbers shifted a term up (as it starts with <math>0</math> on term <math>1</math> and <math>1</math> on term <math>2</math>.)
+Thus, our explicit formula is <cmath>\dfrac{(n-1)(n+1-1)}{2} = \dfrac{(n)(n-1)}{2}</cmath> and as <math>n = 6</math> in this case (<math>6</math> letters), our answer is <math>\dfrac{(6)(6-1)}{2} = \boxed{\textbf{(D)}~15}</math> ~ NSAoPS
+== Solution 3 (Solution 2 Done Fast)==
+We can see that the most efficient way to change <math>ABCDEF</math> to <math>FEDCBA</math> is the same as changing <math>ABCDE</math> to <math>EBCDA</math> and then moving <math>F</math> to the front in <math>5</math> moves. Similarly, this would carry on downwards, where to change <math>ABCDE</math> to <math>EBCDA</math> would be to change <math>ABCD</math> to <math>DCBA</math> and move <math>E</math> <math>4</math> times to the front. This pattern will carry on until <math>AB</math> to <math>BA</math> would be <math>1</math>, and <math>A</math> to <math>A</math> would be <math>0</math>. The answer would be <math>0(A)+1(B)+2(C)+3(D)+4(E)+5(F)</math>, which is <math>\boxed{\textbf{(D)}~15}</math> moves.
+~Moonwatcher22
+== Solution 4 (If you don't notice anything)==
+Simply, just blindly doing it, the most straightforward way is to shift F all the way back. From ABCDEF:
+ABCDFE -> ABCFDE -> ABFCDE -> AFBCDE -> FABCDE
+For E:
+FABCED -> FABECD -> FAEBCD -> FEABCD
+For D:
+FEABDC -> FEADBC -> FEDABC
+For C:
+FEDACB -> FEDCAB
+For B:
+FEDCBA
+That's it, you don't need to do it with A. Still <math>\boxed{\textbf{(D)}~15}</math> moves.
+~pepper2831
+== 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
+==See also==
+{{AMC10 box|year=2024|ab=A|num-b=5|num-a=7}}
+{{MAA Notice}}

2024 AMC 10A (Problems • Answer Key • Resources)
Preceded by Problem 5	Followed by Problem 7
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