AMC10 2020 B

AMC10 2020 B · Q15

AMC10 2020 B · Q15. It mainly tests Casework, Sequences in number theory (remainders patterns).

Steve wrote the digits 1, 2, 3, 4, and 5 in order repeatedly from left to right, forming a list of 10,000 digits, beginning 123451234512 . . . . He then erased every third digit from his list (that is, the 3rd, 6th, 9th, . . . digits from the left), then erased every fourth digit from the resulting list (that is, the 4th, 8th, 12th, . . . digits from the left in what remained), and then erased every fifth digit from what remained at that point. What is the sum of the three digits that were then in positions 2019, 2020, and 2021?

Steve 将数字 1, 2, 3, 4, 5 按顺序反复从左到右写成一个 10,000 个数字的列表，开头是 123451234512 ... 。然后他从列表中删除每个第 3 个数字（即从左数的第 3、6、9... 个数字），接着从剩余列表中删除每个第 4 个数字（即剩余列表从左数的第 4、8、12... 个数字），然后从那时剩下的中删除每个第 5 个数字。当时位置 2019、2020 和 2021 的三个数字之和是多少？

(A) 7 7

(B) 9 9

(D) 11 11

(E) 12 12

Answer

Correct choice: (D)

正确答案：（D）

Solution

Answer (D): The original list had period 5, with 12345 repeated. After every third number was deleted, the resulting list began 124523513412452..., which has period 10. After every fourth number was deleted from that list, the resulting list began 12423534145251312423..., which has period 15. After every fifth number was deleted from that list, the resulting list began 12425341525112425..., which has period 12. Because $2019 \bmod 12 = 3$, the numbers in positions 2019, 2020, and 2021 at that point were the same as the numbers in positions 3, 4, and 5, respectively. The requested sum is $4+2+5=11$.

答案（D）：原始数列的周期为 5，重复 12345。每删去第 3 个数后，得到的数列开头为 124523513412452...，其周期为 10。再从该数列中每删去第 4 个数后，得到的数列开头为 12423534145251312423...，其周期为 15。再从该数列中每删去第 5 个数后，得到的数列开头为 12425341525112425...，其周期为 12。因为 $2019 \bmod 12 = 3$，此时第 2019、2020、2021 位上的数字分别与第 3、4、5 位上的数字相同。所求和为 $4+2+5=11$。

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