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2024 AMC 10B Problem 4

Problem 4 of 25EasierNumber Theory

Balls numbered 1,1, 2,2, 3,3, …\ldots are deposited in 55 bins, labeled A,A, B,B, C,C, D,D, and E,E, using the following procedure. Ball 11 is deposited in bin A,A, and balls 22 and 33 are deposited in bin B.B. The next 33 balls are deposited in bin C,C, the next 44 in bin D,D, and so on, cycling back to bin AA after balls are deposited in bin E.E. (For example, balls numbered 22,22, 23,23, …,\ldots, 2828 are deposited in bin BB at step 77 of this process.) In which bin is ball 20242024 deposited?

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Solution

Group gg holds gg balls, so the first gg groups swallow g(g+1)2\tfrac{g(g+1)}{2} of them. Now 63⋅642=2016\tfrac{63 \cdot 64}{2} = 2016 and 64⋅652=2080,\tfrac{64 \cdot 65}{2} = 2080, which puts ball 20242024 in group 6464 (balls 20172017 through 20802080). The bins cycle A,B,C,D,E,A, B, C, D, E, so group gg lands in bin number (g−1) mod 5.(g - 1) \bmod 5. For g=64g = 64 that’s 63 mod 5=3,63 \bmod 5 = 3, bin D.D. Therefore, the answer is D.
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Tagged: triangular number · modular arithmetic

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