AMC10 2024 B

AMC10 2024 B · Q4

AMC10 2024 B · Q4. It mainly tests Arithmetic sequences basics, Basic counting (rules of product/sum).

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?

编号为1、2、3、...的小球被放入标记为A、B、C、D、E的5个箱子中，使用以下程序。小球1放入箱子A，小球2和3放入箱子B。接下来的3个小球放入箱子C，接下来的4个放入箱子D，依此类推，在放入箱子E后循环回到箱子A。（例如，第7步将编号22、23、...、28的小球放入箱子B。）小球2024放入哪个箱子？

(A) A A

(B) B B

(D) D D

(E) E E

Answer

Correct choice: (D)

正确答案：（D）

Solution

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 $\boxed{\text{D. } D}$.

考虑三角形数组： \[1\] \[2, 3\] \[4, 5, 6\] \[7, 8, 9, 10\] \[11, 12, 13, 14, 15\] \[\vdots\]。模5余1的行放入桶A。类似地，模5余2、3、4、0的行分别放入桶B、C、D、E。第 $n$ 行以第 $n$ 个三角数 $\frac{n(n+1)}{2}$ 结束。需找到使 $\frac{n(n+1)}{2}$ 接近2024的 $n$。 \[\frac{n(n+1)}{2} \approx 2024\] \[n(n+1) \approx 4048\] \[n^2 \approx 4048\] \[n \approx 63\] 试 $n = 63$，得 $\frac{n(n+1)}{2} = 2016$。2016是第63行的最后一个球，小球2024在第64行。由于 $64 \equiv 4 \pmod{5}$，小球2024放入桶 $\boxed{\text{D. } D}$。

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