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 | ||
==See also== | ==See also== | ||
Revision as of 03:42, 14 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.
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:
![\[1\]](http://latex.artofproblemsolving.com/3/4/c/34c083dd559771e8de90399d77a028e339c7a7f7.png)
![\[2, 3\]](http://latex.artofproblemsolving.com/8/a/e/8ae07c1f25096946a7cff66784d7924f47a3abc4.png)
![\[4, 5, 6\]](http://latex.artofproblemsolving.com/f/f/8/ff8dcb8d3ac03f8b2c48bce6582d624c389ee296.png)
![\[7, 8, 9, 10\]](http://latex.artofproblemsolving.com/4/4/d/44d97050255659ad9ce8a081f51cd6cd2174ca91.png)
![\[11, 12, 13, 14, 15\]](http://latex.artofproblemsolving.com/d/8/8/d8845f6caa2fc968b31e676455daed892ffbc89b.png)
![\[\vdots\]](http://latex.artofproblemsolving.com/7/6/4/76403adfdd35fb846675aaeaac92daec680a4fba.png)
.
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 
.
![\[\frac{n(n+1)}{2} \approx 2024\]](http://latex.artofproblemsolving.com/4/a/7/4a78f43250219436b82de09ab2b2ad958b732b4c.png)
![\[n(n+1) \approx 4048\]](http://latex.artofproblemsolving.com/e/a/7/ea704a22c5ad8d061f2052674da8fa083ea6b43b.png)
![\[n^2 \approx 4048\]](http://latex.artofproblemsolving.com/b/e/8/be8be73ee09c4442f9c33f5fd76323896980467a.png)
![\[n \approx 63\]](http://latex.artofproblemsolving.com/5/9/4/5946edb647f7db76bf3f643331c2dc46e0ba69f0.png)
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
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. 
