Difference between revisions of "2024 AMC 10B Problems/Problem 4"
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==Solution 1== | ==Solution 1== | ||
− | + | Consider the triangular array of numbers: | |
+ | <cmath>1</cmath> | ||
+ | <cmath>2, 3</cmath> | ||
+ | <cmath>4, 5, 6</cmath> | ||
+ | <cmath>7, 8, 9, 10</cmath> | ||
+ | <cmath>11, 12, 13, 14, 15</cmath> | ||
+ | <cmath>\vdots</cmath>. | ||
+ | |||
+ | The numbers in a row congruent to <math>1 \bmod{5}</math> will be in bucket A. Similarly, the numbers in a row congruent to <math>2, 3, 4, 0 \bmod{5}</math> will be in buckets B, C, D, and E respectively. Note that the <math>n^\text{th}</math> row ends with the <math>n^\text{th}</math> triangle number, <math>\frac{n(n+1)}{2}</math>. | ||
+ | |||
+ | We must find values of <math>n</math>, that make <math>\frac{n(n+1)}{2}</math> close to <math>2024</math>. | ||
+ | <cmath>\frac{n(n+1)}{2} \approx 2024</cmath> | ||
+ | <cmath>n(n+1) \approx 4048</cmath> | ||
+ | <cmath>n^2 \approx 4048</cmath> | ||
+ | <cmath>n \approx 63</cmath> | ||
+ | |||
+ | Trying <math>n = 63</math> we find that <math>\frac{n(n+1)}{2} = 2016</math>. Since, <math>2016</math> will be the last ball in row <math>63</math>, ball <math>2024</math> will be in row <math>64</math>. Since <math>64 \equiv 4 \bmod{5}</math>, ball <math>2024</math> will be placed in bucket <math>\boxed{\text{D. } D}</math>. | ||
+ | |||
+ | ~numerophile | ||
+ | |||
+ | == Solution 2 == | ||
+ | [[Image: 2024_AMC_12B_P04.jpeg|thumb|center|600px|]] | ||
+ | ~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 == | ||
+ | |||
+ | https://youtu.be/GsXiQWPowoE | ||
+ | |||
+ | ~Thesmartgreekmathdude | ||
==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.