Difference between revisions of "2024 AMC 10A Problems/Problem 6"
MRENTHUSIASM (talk | contribs) |
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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 16:53, 8 November 2024
Problem
What is the minimum number of successive swaps of adjacent letters in the string that are needed to change the string to
(For example,
swaps are required to change
to
one such sequence of swaps is
)
Solution
Procedurally, it takes:
swaps for
to move to the sixth spot, giving
swaps for
to move to the fifth spot, giving
swaps for
to move to the fourth spot, giving
swaps for
to move to the third spot, giving
swap for
to move to the second spot (so
becomes the first spot), giving
Together, the answer is
~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.