Difference between revisions of "2024 AMC 10B Problems/Problem 4"
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+ | == Solution 2 == | ||
+ | [[Image: 2024_AMC_12B_P04.jpeg|thumb|center|600px|]] | ||
+ | ~Kathan | ||
==Video Solution 1 by Pi Academy (Fast and Easy ⚡🚀)== | ==Video Solution 1 by Pi Academy (Fast and Easy ⚡🚀)== | ||
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~ Pi Academy | ~ Pi Academy | ||
+ | == Video Solution by Daily Dose of Math == | ||
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+ | https://youtu.be/GsXiQWPowoE | ||
− | + | ~Thesmartgreekmathdude | |
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− | ~ | ||
==See also== | ==See also== |
Revision as of 15:01, 16 November 2024
- The following problem is from both the 2024 AMC 10B #4 and 2024 AMC 12B #4, so both problems redirect to this page.
Contents
[hide]Problem
Balls numbered 1, 2, 3, ... are deposited in 5 bins, labeled A, B, C, D, and E, using the following procedure. Ball 1 is deposited in bin A, and balls 2 and 3 are deposited in bin B. The next 3 balls are deposited in bin C, the next 4 in bin D, and so on, cycling back to bin A after balls are deposited in bin E. (For example, balls numbered 22, 23, ..., 28 are deposited in bin B at step 7 of this process.) In which bin is ball 2024 deposited?
Solution 1
Consider the triangular array of numbers: .
The numbers in a row congruent to will be in bucket A. Similarly, the numbers in a row congruent to will be in buckets B, C, D, and E respectively. Note that the row ends with the triangle number, .
We must find values of , that make close to .
Trying we find that . Since, will be the last ball in row , ball will be in row . Since , ball will be placed in bucket .
~numerophile
Solution 2
~Kathan
Video Solution 1 by Pi Academy (Fast and Easy ⚡🚀)
https://youtu.be/DIl3rLQQkQQ?feature=shared
~ Pi Academy
Video Solution by Daily Dose of Math
~Thesmartgreekmathdude
See also
2024 AMC 10B (Problems • Answer Key • Resources) | ||
Preceded by Problem 3 |
Followed by Problem 5 | |
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 |
2024 AMC 12B (Problems • Answer Key • Resources) | |
Preceded by Problem 3 |
Followed by Problem 5 |
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 12 Problems and Solutions |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.