2020 AMC 10B Problem 15

Problem 15 of 25IntermediateNumber Theory

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?

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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

3

rd,

4

th, and

5

th digits of this period:

4,2,5.

Their sum is

11.

Thus, the correct answer is D .

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