Difference between revisions of "2024 AMC 10B Problems/Problem 4"

Revision as of 13:42, 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?

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

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

Video Solution 1 by Pi Academy (Fast and Easy ⚡🚀)

https://youtu.be/DIl3rLQQkQQ?feature=shared

~ Pi Academy

Solution 2

~Kathan

@@ Line 35: / Line 35: @@
-==See also==
+== Solution 2 ==
-{{AMC10 box|year=2024|ab=B|num-b=3|num-a=5}}
+[[Image: 2024_AMC_12B_P04.jpeg|thumb|center|600px|]]
-{{AMC12 box|year=2024|ab=B|num-b=3|num-a=5}}
+~Kathan
-{{MAA Notice}}

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