AMC12 2016 B

AMC12 2016 B · Q7

AMC12 2016 B · Q7. It mainly tests Patterns & sequences (misc), Powers & residues.

Josh writes the numbers 1, 2, 3, $\ldots$, 99, 100. He marks out 1, skips the next number (2), marks out 3, and continues skipping and marking out the next number to the end of his list. Then he goes back to the start of his list, marks out the first remaining number (2), skips the next number (4), marks out 6, skips 8, marks out 10, and so on to the end. Josh continues in this manner until only one number remains. What is that number?

Josh 写下数字 1, 2, 3, $\ldots$, 99, 100。他划去 1，跳过下一个数（2），划去 3，并继续按“划去一个、跳过一个”的方式直到列表末尾。然后他回到列表开头，划去剩下的第一个数（2），跳过下一个数（4），划去 6，跳过 8，划去 10，如此一直到末尾。Josh 按这种方式不断重复，直到只剩下一个数。这个数是多少？

(A) 13 13

(B) 32 32

(D) 64 64

(E) 96 96

Answer

Correct choice: (D)

正确答案：（D）

Solution

Answer (D): In the first pass Josh marks out the odd numbers 1, 3, 5, 7, $\ldots$, 99, leaving the multiples of 2: 2, 4, 6, 8, $\ldots$, 100. In the second pass Josh marks out 2, 6, 10, $\ldots$, 98, leaving the multiples of 4: 4, 8, 12, $\ldots$, 100. Similarly, in the $n$th pass Josh marks out the numbers that are not multiples of $2^n$, leaving the numbers that are multiples of $2^n$. It follows that in the 6th pass Josh marks out the numbers that are multiples of $2^5$ but not multiples of $2^6$, namely 32 and 92. This leaves 64, the only number in his original list that is a multiple of $2^6$. Thus the last number remaining is 64.

答案（D）：在第一轮中，Josh 划去奇数 $1, 3, 5, 7, \ldots, 99$，留下 2 的倍数：$2, 4, 6, 8, \ldots, 100$。在第二轮中，Josh 划去 $2, 6, 10, \ldots, 98$，留下 4 的倍数：$4, 8, 12, \ldots, 100$。类似地，在第 $n$ 轮中，Josh 划去所有不是 $2^n$ 的倍数的数，留下所有是 $2^n$ 的倍数的数。由此可知在第 6 轮中，Josh 划去那些是 $2^5$ 的倍数但不是 $2^6$ 的倍数的数，即 32 和 92。这样就只剩下 64，它是原始列表中唯一一个 $2^6$ 的倍数。因此最后剩下的数是 64。

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