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 1

After erasing every third digit, the list becomes $1245235134\ldots$ repeated. After erasing every fourth digit from this list, the list becomes $124235341452513\ldots$ repeated. Finally, after erasing every fifth digit from this list, the list becomes $124253415251\ldots$ repeated. Since this list repeats every $12$ digits and since $2019,2020,2021$ are $3,4,5$ respectively in $\pmod{12},$ we have that the $2019$ th, $2020$ th, and $2021$ st digits are the $3$ rd, $4$ th, and $5$ th digits respectively. It follows that the answer is $4+2+5= \boxed {\textbf{(D)} \text{ 11}}.$

Solution 2

As the LCM of 3, 4, and 5 is $60$ , let's look at a 60-digit block of original numbers (many will be erased by Steve). After he erases every third number (1/3), then every fourth number of what remains (1/4), then every fifth number of what remains (1/5), we are left with $\dfrac{2}{3} \cdot \dfrac{3}{4} \cdot \dfrac{4}{5} \cdot 60=24$ digits from this 60-digit block. $2019 \equiv 3 \pmod {24}, 2020 \equiv 4 \pmod {24}, 2021 \equiv 5 \pmod {24}$ . Writing out the first few digits of this sequence, we have $\underbrace{1}_{\#1}, \underbrace{2}_{\#2}, \cancel{3}, \underbrace{4}_{\#3}, \cancel{5}, \cancel{1}, \underbrace{2}_{\#4}, \cancel{3}, \cancel{4}, \underbrace{5}_{\#5}, \dots$ . Therefore, our answer is $4+2+5=\boxed{\textbf{(D) }11}$ . ~BakedPotato66

Solution 3 (Illustrations of Solutions 1 and 2)

Note that cycles exist initially and after each round of erasing.

Two solutions follow from here:

Solution 3.1 (Stepwise: Similar to Solution 1)

We will use parentheses to denote cycles. Initially, the list has a cycle of length $5:$ $(12345)=12345123451234512345\cdots.$

To find the cycle after the first round of erasing, we need a cycle of length $\mathrm{lcm}(3,5)=15$ before erasing. So, we first group $\frac{\mathrm{lcm}(3,5)}{5}=3$ copies of the current cycle into one, then erase: Work to be done.

To find the cycle after the second round of erasing, we need a cycle of length $\mathrm{lcm}(4,10)=20$ before erasing. So, we first group $\frac{\mathrm{lcm}(4,10)}{10}=2$ copies of the current cycle into one, then erase: Work to be done.

To find the cycle after the right round of erasing, we need a cycle of length $\mathrm{lcm}(5,15)=15$ before erasing. So, we first group $\frac{\mathrm{lcm}(5,15)}{15}=1$ copy of the current cycle into one, then erase: Work to be done.

I will finish shortly. No edit please.

~MRENTHUSIASM

Solution 3.2 (Direct: Similar to Solution 2)

I will finish shortly. No edit please.

~MRENTHUSIASM

Video Solution

https://youtu.be/t6yjfKXpwDs 16:40 ~IceMatrix

2020 AMC 10B (Problems • Answer Key • Resources)
Preceded by Problem 14	Followed by Problem 16
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25
All AMC 10 Problems and Solutions

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