Question

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 $3$ rd, $6$ th, $9$ th, $\ldots$ digits from the left), then erased every fourth digit from the resulting list (that is, the $4$ th, $8$ th, $12$ th, $\ldots$ 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 $2021?$

$7$

$9$

$10$

$11$

$12$

Answer: D

Concepts:pattern recognition process simulation least common multiple

Difficulty rating: 1660

Solution:

Start with the repeating block $12345.$ Deleting every third digit repeats over $\operatorname{lcm}(3,5)=15$ original positions: $\begin{aligned} &123451234512345 \\ &\quad \longrightarrow 1245235134, \end{aligned}$ so the new period has length $10.$

Deleting every fourth digit from this period repeats over $\operatorname{lcm}(4,10)=20$ positions: $\begin{aligned} &12452351341245235134 \\ &\quad \longrightarrow 124235341452513, \end{aligned}$ so the new period has length $15.$

Deleting every fifth digit from this period gives $\begin{aligned} &124235341452513 \\ &\quad \longrightarrow 124253415251, \end{aligned}$ which has period $12.$ Since $2019\equiv 3\pmod{12},$ the digits in positions $2019,$ $2020,$ and $2021$ are the 3rd, 4th, and 5th digits of this period: $4,2,5.$ Their sum is $11.$

Thus, the correct answer is D .

2020 AMC 10B Problem 15

Answer: D

Difficulty rating: 1660

Solution:

Problem 15 in Other Years