2020 AMC 10 B

Complete problem set with solutions and individual problem pages

Problem 15 Medium

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 \ldots. He then erased every third digit from his list (that is, the 3rd, Gth, 9th, .. .digits from the left), then erased every fourth digit from the resulting list (that is, the 4 th, 8 th, 12 th, . . , 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 the positions 2019,2020,2021?(2020 AMC 10B, Question #15)

  • A.

    7

  • B.

    9

  • C.

    10

  • D.

    11

  • E.

    12

Answer:D

After erasing every third digit, the list becomes 1245235134 \ldots. repeated. After erasing every fourth digit from this list, the list becomes 124235341452513 . . repeated. Finally, after erasing every fifth digit from this list, the list becomes 24253415251 . . repeated. Since this list repeats every 12 digits and since 2019,2020,2020 are 3,4,5 respectively in (\bmod 12), we have that the 2019 th,2020 th, and 2021st digits are the 3 rd, 4 th, and 5 th digits respectively. It follows that the answer is 4+2+5=|(\text{D}) \perp 1|.