AMC10 2011 A

AMC10 2011 A · Q23

AMC10 2011 A · Q23. It mainly tests Arithmetic sequences basics, Patterns & sequences (misc).

Seven students count from 1 to 1000 as follows: [description of skipping middles in groups of 3 iteratively]. What number does George say?

七个学生按以下方式从1数到1000： [每3个数跳过中间的一个，迭代进行]。乔治说的是什么数字？

(A) 37 37

(B) 242 242

(D) 728 728

(E) 998 998

Answer

Correct choice: (C)

正确答案：（C）

Solution

Answer (C): After each person counts, the numbers left for the next person form an arithmetic progression. For example, Alice leaves all of the numbers $2, 5, 8, 11, 14, \ldots, 2 + 3\cdot 332$ for Barbara. If a student leaves the progression $a, a + d, a + 2d, a + 3d, a + 4d, \ldots$, then the next student leaves the progression $a + d, (a + d) + 3d, (a + d) + 6d, \ldots$. This implies that in the following table, each number in the third column is three times the previous entry in the third column, and each entry in the second column is the sum of the two entries in the row above: \[ \begin{array}{|c|c|c|} \hline \text{Left for} & \text{First Term} & \text{Common Difference} \\ \hline \text{Alice} & 1 & 1 \\ \text{Barbara} & 2 & 3 \\ \text{Candice} & 5 & 9 \\ \text{Debbie} & 14 & 27 \\ \text{Eliza} & 41 & 81 \\ \text{Fatima} & 122 & 243 \\ \text{George} & 365 & 729 \\ \hline \end{array} \] George is left with the single term $365$.

答案（C）：每个人数完之后，留给下一个人的数字构成一个等差数列。例如，Alice 给 Barbara 留下所有这些数：$2, 5, 8, 11, 14, \ldots, 2 + 3\cdot 332$。如果某个学生留下的等差数列为 $a, a + d, a + 2d, a + 3d, a + 4d, \ldots$，那么下一个学生留下的等差数列为 $a + d, (a + d) + 3d, (a + d) + 6d, \ldots$。这意味着在下表中，第三列的每个数都是第三列上一行对应数的 3 倍，而第二列的每一项都是其上一行中第二列与第三列两项之和： \[ \begin{array}{|c|c|c|} \hline \text{留给} & \text{首项} & \text{公差} \\ \hline \text{Alice} & 1 & 1 \\ \text{Barbara} & 2 & 3 \\ \text{Candice} & 5 & 9 \\ \text{Debbie} & 14 & 27 \\ \text{Eliza} & 41 & 81 \\ \text{Fatima} & 122 & 243 \\ \text{George} & 365 & 729 \\ \hline \end{array} \] George 最后只剩下单独的一项 $365$。

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