Difference between revisions of "2020 AMC 10B Problems/Problem 15"
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Steve wrote the digits <math>1</math>, <math>2</math>, <math>3</math>, <math>4</math>, and <math>5</math> in order repeatedly from left to right, forming a list of <math>10,000</math> digits, beginning <math>123451234512\ldots.</math> He then erased every third digit from his list (that is, the <math>3</math>rd, <math>6</math>th, <math>9</math>th, <math>\ldots</math> digits from the left), then erased every fourth digit from the resulting list (that is, the <math>4</math>th, <math>8</math>th, <math>12</math>th, <math>\ldots</math> 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 <math>2019, 2020, 2021</math>? | Steve wrote the digits <math>1</math>, <math>2</math>, <math>3</math>, <math>4</math>, and <math>5</math> in order repeatedly from left to right, forming a list of <math>10,000</math> digits, beginning <math>123451234512\ldots.</math> He then erased every third digit from his list (that is, the <math>3</math>rd, <math>6</math>th, <math>9</math>th, <math>\ldots</math> digits from the left), then erased every fourth digit from the resulting list (that is, the <math>4</math>th, <math>8</math>th, <math>12</math>th, <math>\ldots</math> 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 <math>2019, 2020, 2021</math>? | ||
− | <math>\textbf{(A)} \ | + | <math>\textbf{(A)}\ 7 \qquad\textbf{(B)}\ 9 \qquad\textbf{(C)}\ 10 \qquad\textbf{(D)}\ 11 \qquad\textbf{(E)}\ 12</math> |
− | ==Solution 1 | + | ==Solution 1 (Simulation)== |
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Note that cycles exist initially and after each round of erasing. | Note that cycles exist initially and after each round of erasing. | ||
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Let the parentheses denote cycles. It follows that: | Let the parentheses denote cycles. It follows that: | ||
<ol start="0" style="margin-left: 1.5em;"> | <ol start="0" style="margin-left: 1.5em;"> | ||
<li>Initially, the list has cycles of length <math>5:</math> <cmath>(12345)=12345123451234512345\cdots.</cmath></li><p> | <li>Initially, the list has cycles of length <math>5:</math> <cmath>(12345)=12345123451234512345\cdots.</cmath></li><p> | ||
− | <li>To find one cycle after the first round of erasing, we need one cycle of length <math>\ | + | <li>To find one cycle after the first round of erasing, we need one cycle of length <math>\operatorname{lcm}(3,5)=15</math> before erasing. So, we first group <math>\frac{15}{5}=3</math> copies of the current cycle into one, then erase: |
<cmath>\begin{align*} | <cmath>\begin{align*} | ||
(12345)&\longrightarrow(123451234512345) \\ | (12345)&\longrightarrow(123451234512345) \\ | ||
Line 28: | Line 17: | ||
&\longrightarrow(1245235134). | &\longrightarrow(1245235134). | ||
\end{align*}</cmath> | \end{align*}</cmath> | ||
− | As a quick | + | As a quick confirmation, one cycle should have length <math>15\cdot\left(1-\frac{1}{3}\right)=10</math> at this point.</li><p> |
− | <li>To find one cycle after the second round of erasing, we need one cycle of length <math>\ | + | <li>To find one cycle after the second round of erasing, we need one cycle of length <math>\operatorname{lcm}(4,10)=20</math> before erasing. So, we first group <math>\frac{20}{10}=2</math> copies of the current cycle into one, then erase: |
<cmath>\begin{align*} | <cmath>\begin{align*} | ||
(1245235134)&\longrightarrow(12452351341245235134) \\ | (1245235134)&\longrightarrow(12452351341245235134) \\ | ||
Line 35: | Line 24: | ||
&\longrightarrow(124235341452513). | &\longrightarrow(124235341452513). | ||
\end{align*}</cmath> | \end{align*}</cmath> | ||
− | As a quick | + | As a quick confirmation, one cycle should have length <math>20\cdot\left(1-\frac{1}{4}\right)=15</math> at this point. |
</li><p> | </li><p> | ||
− | <li>To find one cycle after the third round of erasing, we need one cycle of length <math>\ | + | <li>To find one cycle after the third round of erasing, we need one cycle of length <math>\operatorname{lcm}(5,15)=15</math> before erasing. We already have it here, so we erase: |
<cmath>\begin{align*} | <cmath>\begin{align*} | ||
(124235341452513)&\longrightarrow(1242\cancel{3}5341\cancel{4}5251\cancel{3}) \\ | (124235341452513)&\longrightarrow(1242\cancel{3}5341\cancel{4}5251\cancel{3}) \\ | ||
&\longrightarrow(124253415251). | &\longrightarrow(124253415251). | ||
\end{align*}</cmath> | \end{align*}</cmath> | ||
− | As a quick | + | As a quick confirmation, one cycle should have length <math>15\cdot\left(1-\frac{1}{5}\right)=12</math> at this point.</b></li><p> |
</ol> | </ol> | ||
+ | Since <math>2019,2020,2021</math> are congruent to <math>3,4,5</math> modulo <math>12,</math> respectively, the three digits in the final positions <math>2019,2020,2021</math> are <math>4,2,5,</math> respectively: <cmath>(12\underline{425}3415251).</cmath> Therefore, the answer is <math>4+2+5=\boxed{\textbf{(D)}\ 11}.</math> | ||
− | + | ~MRENTHUSIASM (inspired by TheBeautyofMath) | |
− | + | ==Solution 2 (Simulation)== | |
+ | After erasing every third digit, the list becomes <math>1245235134\ldots</math> repeated. After erasing every fourth digit from this list, the list becomes <math>124235341452513\ldots</math> repeated. Finally, after erasing every fifth digit from this list, the list becomes <math>124253415251\ldots</math> repeated. Since this list repeats every <math>12</math> digits and since <math>2019,2020,2021</math> are <math>3,4,5</math> respectively | ||
+ | in <math>\pmod{12},</math> we have that the <math>2019</math>th, <math>2020</math>th, and <math>2021</math>st digits are the <math>3</math>rd, <math>4</math>th, and <math>5</math>th digits respectively. It follows that the answer is <math>4+2+5= \boxed {\textbf{(D)}\ 11}.</math> | ||
+ | |||
+ | ==Solution 3 (Analysis)== | ||
+ | Note that cycles exist initially and after each round of erasing. | ||
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We will consider one cycle after all three rounds of erasing. Suppose this cycle has length <math>L</math> before any round of erasing. It follows that: | We will consider one cycle after all three rounds of erasing. Suppose this cycle has length <math>L</math> before any round of erasing. It follows that: | ||
<ol start="0" style="margin-left: 1.5em;"> | <ol start="0" style="margin-left: 1.5em;"> | ||
Line 57: | Line 51: | ||
<li>After the third round of erasing, one cycle has length <math>L\left(1-\frac13\right)\left(1-\frac14\right)\left(1-\frac15\right)=\frac25L,</math> from which <math>L</math> must be divisible by <math>5.</math></li><p> | <li>After the third round of erasing, one cycle has length <math>L\left(1-\frac13\right)\left(1-\frac14\right)\left(1-\frac15\right)=\frac25L,</math> from which <math>L</math> must be divisible by <math>5.</math></li><p> | ||
</ol> | </ol> | ||
− | The least such positive integer <math>L</math> is <math>\ | + | The least such positive integer <math>L</math> is <math>\operatorname{lcm}(5,3,2,5)=30.</math> So, there is a repeating pattern for every <math>30</math> 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> | <asy> | ||
/* Made by MRENTHUSIASM */ | /* Made by MRENTHUSIASM */ | ||
Line 63: | Line 57: | ||
for (real i=2.5; i<30; i+=3) { | for (real i=2.5; i<30; i+=3) { | ||
− | fill((i-0.5,0)--(i-0.5, | + | 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) { | for (real i=4.5; i<30; i+=6) { | ||
− | fill((i-0.5,0)--(i-0.5, | + | fill((i-0.5,0)--(i-0.5,4)--(i+0.5,4)--(i+0.5,0)--cycle,yellow); |
} | } | ||
− | fill((7,0)--(7, | + | fill((7,0)--(7,4)--(8,4)--(8,0)--cycle,green); |
− | fill((18,0)--(18, | + | fill((18,0)--(18,4)--(19,4)--(19,0)--cycle,green); |
− | fill((27,0)--(27, | + | fill((27,0)--(27,4)--(28,4)--(28,0)--cycle,green); |
for (real i=1; i<5; ++i) { | for (real i=1; i<5; ++i) { | ||
for (real j=0; j<30; ++j) { | for (real j=0; j<30; ++j) { | ||
− | label("$"+string(1+j%5)+"$",(j+0.5,i | + | label("$"+string(1+j%5)+"$",(j+0.5,i-0.5)); |
} | } | ||
} | } | ||
for (real i=0; i<30; ++i) { | for (real i=0; i<30; ++i) { | ||
− | label("$\vdots$",(i+0.5, | + | label("$\vdots$",(i+0.5,-1/3)); |
} | } | ||
− | add(grid(30, | + | add(grid(30,4,linewidth(1.25))); |
</asy> | </asy> | ||
− | As indicated by the white squares, each group has <math>\frac25\cdot30=12</math> digits remaining. | + | As indicated by the white squares, each group of <math>30</math> digits on the original list has <math>\frac25\cdot30=12</math> digits remaining on the final list. |
Since <math>2019,2020,2021</math> are congruent to <math>3,4,5</math> modulo <math>12,</math> respectively, the three digits in the final positions <math>2019,2020,2021</math> are <math>4,2,5,</math> respectively: | Since <math>2019,2020,2021</math> are congruent to <math>3,4,5</math> modulo <math>12,</math> respectively, the three digits in the final positions <math>2019,2020,2021</math> are <math>4,2,5,</math> respectively: | ||
Line 115: | Line 109: | ||
add(grid(30,1,linewidth(1.25))); | add(grid(30,1,linewidth(1.25))); | ||
</asy> | </asy> | ||
− | Therefore, the answer is <math>4+2+5=\boxed{\textbf{(D)} \ | + | Therefore, the answer is <math>4+2+5=\boxed{\textbf{(D)}\ 11}.</math> |
~MRENTHUSIASM | ~MRENTHUSIASM | ||
+ | |||
+ | ==Solution 4 (Analysis)== | ||
+ | As the LCM of <math>3</math>, <math>4</math>, and <math>5</math> is <math>60</math>, let's look at a <math>60</math>-digit block of original numbers (many will be erased by Steve). After he erases every third number (<math>1/3</math>), then every fourth number of what remains (<math>1/4</math>), then every fifth number of what remains (<math>1/5</math>), we are left with <math>\dfrac{2}{3} \cdot \dfrac{3}{4} \cdot \dfrac{4}{5} \cdot 60=24</math> digits from this <math>60</math>-digit block. <math>2019 \equiv 3 \pmod {24}, 2020 \equiv 4 \pmod {24}, 2021 \equiv 5 \pmod {24}</math>. Writing out the first few digits of this sequence, we have <math>\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</math>. Therefore, our answer is <math>4+2+5=\boxed{\textbf{(D)}\ 11}</math>. | ||
+ | |||
+ | ~BakedPotato66 | ||
==Video Solution== | ==Video Solution== |
Latest revision as of 22:34, 31 August 2021
Contents
Problem
Steve wrote the digits , , , , and in order repeatedly from left to right, forming a list of digits, beginning He then erased every third digit from his list (that is, the rd, th, th, digits from the left), then erased every fourth digit from the resulting list (that is, the th, th, th, 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 ?
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
- To find one cycle after the first round of erasing, we need one cycle of length before erasing. So, we first group copies of the current cycle into one, then erase: As a quick confirmation, one cycle should have length at this point.
- To find one cycle after the second round of erasing, we need one cycle of length before erasing. So, we first group copies of the current cycle into one, then erase: As a quick confirmation, one cycle should have length at this point.
- To find one cycle after the third round of erasing, we need one cycle of length before erasing. We already have it here, so we erase: As a quick confirmation, one cycle should have length at this point.
Since are congruent to modulo respectively, the three digits in the final positions are respectively: Therefore, the answer is
~MRENTHUSIASM (inspired by TheBeautyofMath)
Solution 2 (Simulation)
After erasing every third digit, the list becomes repeated. After erasing every fourth digit from this list, the list becomes repeated. Finally, after erasing every fifth digit from this list, the list becomes repeated. Since this list repeats every digits and since are respectively in we have that the th, th, and st digits are the rd, th, and th digits respectively. It follows that the answer is
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 before any round of erasing. It follows that:
- Initially, one cycle has length from which must be divisible by
- After the first round of erasing, one cycle has length from which must be divisible by
- After the second round of erasing, one cycle has length from which must be divisible by
- After the third round of erasing, one cycle has length from which must be divisible by
The least such positive integer is So, there is a repeating pattern for every 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: As indicated by the white squares, each group of digits on the original list has digits remaining on the final list.
Since are congruent to modulo respectively, the three digits in the final positions are respectively: Therefore, the answer is
~MRENTHUSIASM
Solution 4 (Analysis)
As the LCM of , , and is , let's look at a -digit block of original numbers (many will be erased by Steve). After he erases every third number (), then every fourth number of what remains (), then every fifth number of what remains (), we are left with digits from this -digit block. . Writing out the first few digits of this sequence, we have . Therefore, our answer is .
~BakedPotato66
Video Solution
https://youtu.be/t6yjfKXpwDs 16:40 ~IceMatrix
See Also
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 |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.