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Difference between revisions of "2024 AMC 10B Problems/Problem 4"
 
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{{duplicate|[[2024 AMC 10B Problems/Problem 2|2024 AMC 10B #4]] and [[2024 AMC 12B Problems/Problem 2|2024 AMC 12B #4]]}}
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{{duplicate|[[2024 AMC 10B Problems/Problem 4|2024 AMC 10B #4]] and [[2024 AMC 12B Problems/Problem 4|2024 AMC 12B #4]]}}
  
 
==Problem==
 
==Problem==
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==Solution 1==
 
==Solution 1==
  
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Ball D, 63*64/2=2016.
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Consider the triangular array of numbers:
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<cmath>1</cmath>
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<cmath>2, 3</cmath>
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<cmath>4, 5, 6</cmath>
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<cmath>7, 8, 9, 10</cmath>
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<cmath>11, 12, 13, 14, 15</cmath>
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<cmath>\vdots</cmath>.
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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>.
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We must find values of <math>n</math>, that make <math>\frac{n(n+1)}{2}</math> close to <math>2024</math>.
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<cmath>\frac{n(n+1)}{2} \approx 2024</cmath>
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<cmath>n(n+1) \approx 4048</cmath>
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<cmath>n^2 \approx 4048</cmath>
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<cmath>n \approx 63</cmath>
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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>.
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~numerophile
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== Solution 2 ==
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[[Image: 2024_AMC_12B_P04.jpeg|thumb|center|600px|]]
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~Kathan
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==Video Solution 1 by Pi Academy (Fast and Easy ⚡🚀)==
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https://youtu.be/DIl3rLQQkQQ?feature=shared
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~ Pi Academy
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== Video Solution by Daily Dose of Math ==
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https://youtu.be/GsXiQWPowoE
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~Thesmartgreekmathdude
  
 
==See also==
 
==See also==

Latest 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.

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?

$\textbf{(A) } A \qquad\textbf{(B) } B \qquad\textbf{(C) } C \qquad\textbf{(D) } D \qquad\textbf{(E) } E$

Solution 1

Consider the triangular array of numbers: \[1\] \[2, 3\] \[4, 5, 6\] \[7, 8, 9, 10\] \[11, 12, 13, 14, 15\] \[\vdots\].

The numbers in a row congruent to $1 \bmod{5}$ will be in bucket A. Similarly, the numbers in a row congruent to $2, 3, 4, 0 \bmod{5}$ will be in buckets B, C, D, and E respectively. Note that the $n^\text{th}$ row ends with the $n^\text{th}$ triangle number, $\frac{n(n+1)}{2}$.

We must find values of $n$, that make $\frac{n(n+1)}{2}$ close to $2024$. \[\frac{n(n+1)}{2} \approx 2024\] \[n(n+1) \approx 4048\] \[n^2 \approx 4048\] \[n \approx 63\]

Trying $n = 63$ we find that $\frac{n(n+1)}{2} = 2016$. Since, $2016$ will be the last ball in row $63$, ball $2024$ will be in row $64$. Since $64 \equiv 4 \bmod{5}$, ball $2024$ will be placed in bucket $\boxed{\text{D. } D}$.

~numerophile

Solution 2

2024 AMC 12B P04.jpeg

~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

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. AMC logo.png

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