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

(Recursive Solution)
(→‎Solution 2)
Line 26: Line 26:
  
 
We can proceed by a recursive tactic on the number of letters in the string.
 
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>
 
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>.
 
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).
 
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>.
 
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{15}</math> moves.
 
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{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!
 
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
 
~world123
  

Revision as of 18:08, 8 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 $ABC\rightarrow BAC\rightarrow BCA\rightarrow CBA.$)

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

Solution

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

Note: This is my first time writing a solution, so please feel free to edit this and change it, thank you in advance! (Work in progress)

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 $\boxed{15}$ 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

See also

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
All AMC 10 Problems and Solutions

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