2020 AMC 10B Problems/Problem 15

Problem

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 the positions $2019, 2020, 2021$ ?

$\textbf{(A)} \text{ 7} \qquad \textbf{(B)} \text{ 9} \qquad \textbf{(C)} \text{ 10} \qquad \textbf{(D)} \text{ 11} \qquad \textbf{(E)} \text{ 12}$

Solution

Bash it out until you find a pattern. In the end, you'll see that the $2019^{th}$ , $2020^{th}$ , and the $2021^{st}$ numbers are $2, 4, 5$ , giving the answer $2+4+5= \boxed {D)11}$ ~DragonWarrior123

Video Solution

https://youtu.be/t6yjfKXpwDs

~IceMatrix

2020 AMC 10B (Problems • Answer Key • Resources)
Preceded by Problem 14	Followed by Problem 16
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2020 AMC 10B Problems/Problem 15

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Problem

Solution

Video Solution

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