2020 AMC 10B Problems/Problem 15

Problem

skibidi toilet

Solution 1 (Simulation)

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

Let the parentheses denote cycles. It follows that:

Initially, the list has cycles of length $5:$ $(12345)=12345123451234512345\cdots.$

To find one cycle after the first round of erasing, we need one cycle of length $\operatorname{lcm}(3,5)=15$ before erasing. So, we first group $\frac{15}{5}=3$ copies of the current cycle into one, then erase: $\begin{align*} (12345)&\longrightarrow(123451234512345) \\ &\longrightarrow(12\cancel{3}45\cancel{1}23\cancel{4}51\cancel{2}34\cancel{5}) \\ &\longrightarrow(1245235134). \end{align*}$ As a quick confirmation, one cycle should have length $15\cdot\left(1-\frac{1}{3}\right)=10$ at this point.

To find one cycle after the second round of erasing, we need one cycle of length $\operatorname{lcm}(4,10)=20$ before erasing. So, we first group $\frac{20}{10}=2$ copies of the current cycle into one, then erase: $\begin{align*} (1245235134)&\longrightarrow(12452351341245235134) \\ &\longrightarrow(124\cancel{5}235\cancel{1}341\cancel{2}452\cancel{3}513\cancel{4}) \\ &\longrightarrow(124235341452513). \end{align*}$ As a quick confirmation, one cycle should have length $20\cdot\left(1-\frac{1}{4}\right)=15$ at this point.

To find one cycle after the third round of erasing, we need one cycle of length $\operatorname{lcm}(5,15)=15$ before erasing. We already have it here, so we erase: $\begin{align*} (124235341452513)&\longrightarrow(1242\cancel{3}5341\cancel{4}5251\cancel{3}) \\ &\longrightarrow(124253415251). \end{align*}$ As a quick confirmation, one cycle should have length $15\cdot\left(1-\frac{1}{5}\right)=12$ at this point.

Since $2019,2020,2021$ are congruent to $3,4,5$ modulo $12,$ respectively, the three digits in the final positions $2019,2020,2021$ are $4,2,5,$ respectively: $(12\underline{425}3415251).$ Therefore, the answer is $4+2+5=\boxed{\textbf{(D)}\ 11}.$

~MRENTHUSIASM (inspired by TheBeautyofMath)

Solution 2 (Simulation)

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 $2019,2020,2021$ 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 $4+2+5= \boxed {\textbf{(D)}\ 11}.$

Solution 3 (Analysis)

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

We will consider one cycle after all three rounds of erasing. Suppose this cycle has length $L$ before any round of erasing. It follows that:

Initially, one cycle has length $5,$ from which $L$ must be divisible by $5.$

After the first round of erasing, one cycle has length $L\left(1-\frac13\right)=\frac23L,$ from which $L$ must be divisible by $3.$

After the second round of erasing, one cycle has length $L\left(1-\frac13\right)\left(1-\frac14\right)=\frac12L,$ from which $L$ must be divisible by $2.$

After the third round of erasing, one cycle has length $L\left(1-\frac13\right)\left(1-\frac14\right)\left(1-\frac15\right)=\frac25L,$ from which $L$ must be divisible by $5.$

The least such positive integer $L$ is $\operatorname{lcm}(5,3,2,5)=30.$ So, there is a repeating pattern for every $30$ digits on the original list. As shown below, the digits erased in the first, second, and third rounds are colored in red, yellow, and green, respectively: $[asy] /* Made by MRENTHUSIASM */ size(20cm); for (real i=2.5; i<30; i+=3) { fill((i-0.5,0)--(i-0.5,4)--(i+0.5,4)--(i+0.5,0)--cycle,red); } for (real i=4.5; i<30; i+=6) { fill((i-0.5,0)--(i-0.5,4)--(i+0.5,4)--(i+0.5,0)--cycle,yellow); } fill((7,0)--(7,4)--(8,4)--(8,0)--cycle,green); fill((18,0)--(18,4)--(19,4)--(19,0)--cycle,green); fill((27,0)--(27,4)--(28,4)--(28,0)--cycle,green); for (real i=1; i<5; ++i) { for (real j=0; j<30; ++j) { label("$"+string(1+j%5)+"$",(j+0.5,i-0.5)); } } for (real i=0; i<30; ++i) { label("$\vdots$",(i+0.5,-1/3)); } add(grid(30,4,linewidth(1.25))); [/asy]$ As indicated by the white squares, each group of $30$ digits on the original list has $\frac25\cdot30=12$ digits remaining on the final list.

Since $2019,2020,2021$ are congruent to $3,4,5$ modulo $12,$ respectively, the three digits in the final positions $2019,2020,2021$ are $4,2,5,$ respectively: $[asy] /* Made by MRENTHUSIASM */ size(20cm); for (real i=2.5; i<30; i+=3) { fill((i-0.5,0)--(i-0.5,1)--(i+0.5,1)--(i+0.5,0)--cycle,red); } for (real i=4.5; i<30; i+=6) { fill((i-0.5,0)--(i-0.5,1)--(i+0.5,1)--(i+0.5,0)--cycle,yellow); } fill((7,0)--(7,1)--(8,1)--(8,0)--cycle,green); fill((18,0)--(18,1)--(19,1)--(19,0)--cycle,green); fill((27,0)--(27,1)--(28,1)--(28,0)--cycle,green); for (real j=0; j<30; ++j) { label("$"+string(1+j%5)+"$",(j+0.5,0.5)); } draw((3.5,-1.25)--(3.5,-0.2),linewidth(1.25),EndArrow); draw((6.5,-1.25)--(6.5,-0.2),linewidth(1.25),EndArrow); draw((9.5,-1.25)--(9.5,-0.2),linewidth(1.25),EndArrow); add(grid(30,1,linewidth(1.25))); [/asy]$ Therefore, the answer is $4+2+5=\boxed{\textbf{(D)}\ 11}.$

~MRENTHUSIASM

Solution 4 (Analysis)

As the LCM of $3$ , $4$ , and $5$ is $60$ , let us look at a $60$ -digit block of original numbers (many will be erased by Steve). After he erases every third number $\left(\dfrac{1}{3}\right)$ , then every fourth number of what remains $\left(\dfrac{1}{4}\right)$ , then every fifth number of what remains $\left(\dfrac{1}{5}\right)$ , 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 5

Lemma: Given a sequence $a_1, a_2, a_3, \cdots$ , and an positive integer $k>2$ . If we erase every $k$ th item in this sequence, and we name $b_1, b_2, b_3, \cdots$ as the remaining sequence. Then we have $b_{(k-1)m+1}=a_{km+1}, b_{(k-1)m+2}=a_{km+2}, \cdots, b_{(k-1)m+k-1}=a_{km+k-1}.$

Proof: For $a_{km+j}$ with some $j, 1\le j\le k-1$ , we will have $m$ items removed before this item, so it becomes $b_{(k-1)m+j}$ in the new sequence. Hence, we have $b_{(k-1)m+j}=a_{km+j}$ .

If we start with $a_1, a_2, a_3, \cdots,$ and let $b_1, b_2, \cdots$ be the sequence after removing all the indexes that are multiples of $3$ . Then, we have $b_{2n+1}=a_{3n+1},b_{2n+2}=a_{3n+2}$ .

Similarly, if we start with $b_1, b_2, b_3, \cdots,$ and remove all multiples of $4$ , and get $c_1, c_2, \cdots$ We have $c_{3n+1}=b_{4n+1},c_{3n+2}=b_{4n+2}, c_{3n+3}=b_{4n+3}$ .

If we start with $c_1, c_2, \cdots$ and $d_1, d_2, \cdots$ are remove all multiples of $5$ , we get $d_{4n+1}=c_{5n+1}, d_{4n+2}=c_{5n+2}, d_{4n+3}=c_{5n+3}, d_{4n+4}=c_{5n+4}.$ Therefore, $\begin{align*} d_{2019}&=d_{4\cdot504+3}=c_{5\cdot 504+3}=c_{2523}=c_{3\cdot 840+3}=b_{4\cdot 840+3}=b_{3363}=b_{2\cdot 1681+1}=a_{3\cdot 1681+1}=a_{5044}=a_4=4, \\ d_{2020}&=d_{4\cdot504+4}=c_{5\cdot 504+4}=c_{2524}=c_{3\cdot841+1}=b_{4\cdot841+1}=b_{3365}=b_{2\cdot1682+1}=a_{3\cdot 1682+1}=a_{5047}=a_2=2, \\ d_{2021}&=d_{4\cdot505+1}=c_{5\cdot505+1}=c_{2526}=c_{3\cdot841+3}=b_{4\cdot841+3}=b_{3367}=b_{2\cdot1683+1}=a_{3\cdot1683+1}=a_{5050}=a_5=5, \end{align*}$ and the answer is $4+2+5=\boxed{\textbf{(D)}\ 11}$ .

~szhangmath

Video Solution (HOW TO CRITICALLY THINK!!!)

https://youtu.be/yDX4h6YSY8Q

~Education, the Study of Everything

Video Solution

https://youtu.be/t6yjfKXpwDs?t=1004

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