Art of Problem Solving
Art of Problem Solving
AoPS Online
Math texts, online classes, and more
for students in grades 5-12.
Visit AoPS Online ‚
Books for Grades 5-12 Online Courses
Beast Academy
Engaging math books and online learning
for students ages 6-13.
Visit Beast Academy ‚
Books for Ages 6-13 Beast Academy Online
AoPS Academy
Small live classes for advanced math
and language arts learners in grades 2-12.
Visit AoPS Academy ‚
Find a Physical Campus Visit the Virtual Campus

2024 AMC 10A Problems/Problem 6: Difference between revisions

← Older editNewer edit →
MRENTHUSIASM (talk | contribs)
editor
9,986 edits
No edit summary
MRENTHUSIASM (talk | contribs)
editor
9,986 edits
No edit summary
Line 8: Line 8:
Procedurally, it takes:
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>
Together, the answer is <math>5+4+3+2+1=\boxed{\textbf{(D)}~15}.</math>

Revision as of 15:53, 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

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

These problems are copyrighted © by the Mathematical Association of America, as part of the American Mathematics Competitions. Error creating thumbnail: File missing

Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=2024_AMC_10A_Problems/Problem_6&oldid=231835"