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2024 AMC 12B 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 B.B. The next three 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, 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

Step kk deposits kk balls, so after step kk a total of k(k+1)2\dfrac{k(k+1)}{2} balls have been placed. Since 63⋅642=2016\dfrac{63 \cdot 64}{2} = 2016 and 64⋅652=2080,\dfrac{64 \cdot 65}{2} = 2080, ball 20242024 falls in step 64.64. The steps cycle through the bins A,B,C,D,E,A, B, C, D, E, so step kk uses position (k−1) mod 5.(k-1) \bmod 5. Here (64−1) mod 5=63 mod 5=3,(64 - 1) \bmod 5 = 63 \bmod 5 = 3, which is the fourth bin, D.D. Thus, the correct answer is D.
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Tagged: triangular number · modular arithmetic

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