Navvye/Olympiad-Problems-CSV huggingface dataset:
Problem: The acute angles of a right triangle are $a^{\circ}$ and $b^{\circ}$ , where $a>b$ and both $a$ and $b$ are prime numbers. What is the least possible value of $b$
$\textbf{(A)}\ 2 \qquad\textbf{(B)}\ 3 \qquad\textbf{(C)}\  5 \qquad\textbf{(D)}\ 7 \qquad\textbf{(E)}\ 11$

Solution: Since the three angles of a triangle add up to $180^{\circ}$ and one of the angles is $90^{\circ}$ because it's a right triangle, $a^{\circ} + b^{\circ} = 90^{\circ}$
The greatest prime number less than $90$ is $89$ . If $a=89^{\circ}$ , then $b=90^{\circ}-89^{\circ}=1^{\circ}$ , which is not prime.
The next greatest prime number less than $90$ is $83$ . If $a=83^{\circ}$ , then $b=7^{\circ}$ , which IS prime, so we have our answer $\boxed{7}$ ~quacker88

Answer: D 7
Problem: The acute angles of a right triangle are $a^{\circ}$ and $b^{\circ}$ , where $a>b$ and both $a$ and $b$ are prime numbers. What is the least possible value of $b$
$\textbf{(A)}\ 2 \qquad\textbf{(B)}\ 3 \qquad\textbf{(C)}\  5 \qquad\textbf{(D)}\ 7 \qquad\textbf{(E)}\ 11$

Solution: Looking at the answer choices, only $7$ and $11$ are coprime to $90$ . Testing $7$ , the smaller angle, makes the other angle $83$ which is prime, therefore our answer is $\boxed{7}$

Answer: D 7
Problem: The acute angles of a right triangle are $a^{\circ}$ and $b^{\circ}$ , where $a>b$ and both $a$ and $b$ are prime numbers. What is the least possible value of $b$
$\textbf{(A)}\ 2 \qquad\textbf{(B)}\ 3 \qquad\textbf{(C)}\  5 \qquad\textbf{(D)}\ 7 \qquad\textbf{(E)}\ 11$

Solution: It is clear that $\gcd(a,b)=1.$ By the Euclidean Algorithm, we have \[\gcd(a,b)=\gcd(a+b,b)=\gcd(90,b)=1,\] so $90$ and $b$ are relatively prime.
The least such prime number $b$ is $7,$ from which $a=90-b=83$ is also a prime number. Therefore, the answer is $\boxed{7}.$

Answer: D 7
Problem: How many distinguishable arrangements are there of $1$ brown tile, $1$ purple tile, $2$ green tiles, and $3$ yellow tiles in a row from left to right? (Tiles of the same color are indistinguishable.)
$\textbf{(A)}\ 210 \qquad\textbf{(B)}\ 420 \qquad\textbf{(C)}\  630 \qquad\textbf{(D)}\ 840 \qquad\textbf{(E)}\ 1050$

Solution: Let's first find how many possibilities there would be if they were all distinguishable, then divide out the ones we overcounted.
There are $7!$ ways to order $7$ objects. However, since there's $3!=6$ ways to switch the yellow tiles around without changing anything (since they're indistinguishable) and $2!=2$ ways to order the green tiles, we have to divide out these possibilities: $\frac{7!}{3! \cdot2}=\boxed{420}$

Answer: B 420
Problem: How many distinguishable arrangements are there of $1$ brown tile, $1$ purple tile, $2$ green tiles, and $3$ yellow tiles in a row from left to right? (Tiles of the same color are indistinguishable.)
$\textbf{(A)}\ 210 \qquad\textbf{(B)}\ 420 \qquad\textbf{(C)}\  630 \qquad\textbf{(D)}\ 840 \qquad\textbf{(E)}\ 1050$

Solution: We can choose a different frame to solve this problem. Our tile combination can be written as $Y, Y, Y, G, G, B, P.$ We can focus on $Y, Y, Y, G, G$ first, which gives us $\binom{5}{3}=10.$ Now we can insert our brown tile into this which only has $6$ choices(like $Y,B, Y, Y, G, G$ and $Y, Y, Y, G, G, B$ etc.), then insert purple tile which only has $7$ choices(like $B,Y, P, Y, Y, G, G$ and $B, Y, Y, Y, G, G, P$ etc.). Multiply them together we get $10\cdot 6\cdot 7=\boxed{420}$

Answer: B 420
Problem: How many distinguishable arrangements are there of $1$ brown tile, $1$ purple tile, $2$ green tiles, and $3$ yellow tiles in a row from left to right? (Tiles of the same color are indistinguishable.)
$\textbf{(A)}\ 210 \qquad\textbf{(B)}\ 420 \qquad\textbf{(C)}\  630 \qquad\textbf{(D)}\ 840 \qquad\textbf{(E)}\ 1050$

Solution: Let $B$ be brown, $P$ be purple, $G$ be green, and $Y$ be yellow. Then, we are just ordering $Y$ $Y$ $Y$ $G$ $G$ $B$ , and $P$ . Hence, $\frac{7!}{3! \cdot 2!} = \boxed{420}$

Answer: B 420
Problem: Driving along a highway, Megan noticed that her odometer showed $15951$ (miles). This number is a palindrome-it reads the same forward and backward. Then $2$ hours later, the odometer displayed the next higher palindrome. What was her average speed, in miles per hour, during this $2$ -hour period?
$\textbf{(A)}\ 50 \qquad\textbf{(B)}\ 55 \qquad\textbf{(C)}\ 60 \qquad\textbf{(D)}\ 65 \qquad\textbf{(E)}\ 70$

Solution: In order to get the smallest palindrome greater than $15951$ , we need to raise the middle digit. If we were to raise any of the digits after the middle, we would be forced to also raise a digit before the middle to keep it a palindrome, making it unnecessarily larger.
So we raise $9$ to the next largest value, $10$ , but obviously, that's not how place value works, so we're in the $16000$ s now. To keep this a palindrome, our number is now $16061$
So Megan drove $16061-15951=110$ miles. Since this happened over $2$ hours, she drove at $\frac{110}{2}=\boxed{55}$ mph. ~quacker88

Answer: B 55
Problem: How many positive even multiples of $3$ less than $2020$ are perfect squares?
$\textbf{(A)}\ 7 \qquad\textbf{(B)}\ 8 \qquad\textbf{(C)}\  9 \qquad\textbf{(D)}\ 10 \qquad\textbf{(E)}\ 12$

Solution: Any even multiple of $3$ is a multiple of $6$ , so we need to find multiples of $6$ that are perfect squares and less than $2020$ . Any solution that we want will be in the form $(6n)^2$ , where $n$ is a positive integer. The smallest possible value is at $n=1$ , and the largest is at $n=7$ (where the expression equals $1764$ ). Therefore, there are a total of $\boxed{7}$ possible numbers.-PCChess

Answer: A 7
Problem: How many positive even multiples of $3$ less than $2020$ are perfect squares?
$\textbf{(A)}\ 7 \qquad\textbf{(B)}\ 8 \qquad\textbf{(C)}\  9 \qquad\textbf{(D)}\ 10 \qquad\textbf{(E)}\ 12$

Solution: A even multiple square of $3$ can be represented by $3^2 \cdot 2^2 \cdot x^2$ , where $3^2$ is the multiple or $3$ and $2^2$ makes it even. Simplifying we have $36^2 \cdot x^2$ . We can divide $2020$ by $36$ (floor) and get $56$ see the result. We can then see that there are $7$ different values for $x$ . It can't be larger or else $x^2 > 56$ . And thus $\boxed{7}$

Answer: A 7
Problem: How many positive even multiples of $3$ less than $2020$ are perfect squares?
$\textbf{(A)}\ 7 \qquad\textbf{(B)}\ 8 \qquad\textbf{(C)}\  9 \qquad\textbf{(D)}\ 10 \qquad\textbf{(E)}\ 12$

Solution: It can be seen that the problem is just asking for squares that are multiples of six. Thus, all squares of multiples of six can be listed out: $6^2$ $12^2$ $18^2$ $24^2$ $30^2$ $36^2$ , and $42^2$ . Since $48^2=2196 > 2020$ , there are $\boxed{7}$ valid answers.
~airbus-a321, November 2023

Answer: A 7
Problem: Points $P$ and $Q$ lie in a plane with $PQ=8$ . How many locations for point $R$ in this plane are there such that the triangle with vertices $P$ $Q$ , and $R$ is a right triangle with area $12$ square units?
$\textbf{(A)}\ 2 \qquad\textbf{(B)}\ 4 \qquad\textbf{(C)}\  6 \qquad\textbf{(D)}\ 8 \qquad\textbf{(E)}\ 12$

Solution: Let the brackets denote areas. We are given that \[[PQR]=\frac12\cdot PQ\cdot h_R=12.\] Since $PQ=8,$ it follows that $h_R=3.$
We construct a circle with diameter $\overline{PQ}.$ All such locations for $R$ are shown below:
[asy] /* Made by MRENTHUSIASM */ size(250); pair O, P, Q, R1, R2, R3, R4, R5, R6, R7, R8, I1, I2; O = (0,0); P = (-4,0); Q = (4,0); R1 = (-4,3); R4 = (4,3); R5 = (-4,-3); R8 = (4,-3); path C; C = Circle(O,4); R3 = intersectionpoints(C,R1--R4)[0]; R2 = intersectionpoints(C,R1--R4)[1]; R6 = intersectionpoints(C,R5--R8)[0]; R7 = intersectionpoints(C,R5--R8)[1]; I1 = intersectionpoint(R2--R6,P--Q); I2 = intersectionpoint(R3--R7,P--Q); markscalefactor=0.0375; draw(rightanglemark(R1,P,Q)^^rightanglemark(R2,I1,Q)^^rightanglemark(R3,I2,P)^^rightanglemark(R4,Q,P),red); draw(Circle(O,4),dashed); draw(R1--R5^^R4--R8^^R2--R6^^R3--R7^^P--Q); dot(O,linewidth(4)); dot("$P$",P,1.5W,linewidth(4)); dot("$Q$",Q,1.5E,linewidth(4)); dot("$R_1$",R1,1.5NW,blue+linewidth(4)); dot("$R_4$",R4,1.5NE,blue+linewidth(4)); dot("$R_5$",R5,1.5SW,blue+linewidth(4)); dot("$R_8$",R8,1.5SE,blue+linewidth(4)); dot("$R_2$",R2,1.5NW,blue+linewidth(4)); dot("$R_3$",R3,1.5NE,blue+linewidth(4)); dot("$R_6$",R6,1.5SW,blue+linewidth(4)); dot("$R_7$",R7,1.5SE,blue+linewidth(4)); dot(I1,linewidth(4)); dot(I2,linewidth(4)); Label L1 = Label("$8$", align=(0,0), position=MidPoint, filltype=Fill(3,0,white)); Label L2 = Label("$3$", align=(0,0), position=MidPoint, filltype=Fill(0,3,white)); draw(P-(0,5)--Q-(0,5), L=L1, arrow=Arrows(),bar=Bars(15)); draw(R4+(2,0)--Q+(2,0), L=L2, arrow=Arrows(),bar=Bars(15)); draw(Q+(2,0)--R8+(2,0), L=L2, arrow=Arrows(),bar=Bars(15)); [/asy]
We apply casework to the right angle of $\triangle PQR:$
Together, there are $\boxed{8}$ such locations for $R.$

Answer: D 8
Problem: Points $P$ and $Q$ lie in a plane with $PQ=8$ . How many locations for point $R$ in this plane are there such that the triangle with vertices $P$ $Q$ , and $R$ is a right triangle with area $12$ square units?
$\textbf{(A)}\ 2 \qquad\textbf{(B)}\ 4 \qquad\textbf{(C)}\  6 \qquad\textbf{(D)}\ 8 \qquad\textbf{(E)}\ 12$

Solution: Let the brackets denote areas. We are given that \[[PQR]=\frac12\cdot PQ\cdot h_R=12.\] Since $PQ=8,$ it follows that $h_R=3.$
Without the loss of generality, let $P=(-4,0)$ and $Q=(4,0).$ We conclude that the $y$ -coordinate of $R$ must be $\pm3.$
We apply casework to the right angle of $\triangle PQR:$
Together, there are $2+2+4=\boxed{8}$ such locations for $R.$

Answer: D 8
Problem: How many ordered pairs of integers $(x, y)$ satisfy the equation \[x^{2020}+y^2=2y?\]
$\textbf{(A) } 1 \qquad\textbf{(B) } 2 \qquad\textbf{(C) } 3 \qquad\textbf{(D) } 4 \qquad\textbf{(E) } \text{infinitely many}$

Solution: Rearranging the terms and and completing the square for $y$ yields the result $x^{2020}+(y-1)^2=1$ . Then, notice that $x$ can only be $0$ $1$ and $-1$ because any value of $x^{2020}$ that is greater than 1 will cause the term $(y-1)^2$ to be less than $0$ , which is impossible as $y$ must be real. Therefore, plugging in the above values for $x$ gives the ordered pairs $(0,0)$ $(1,1)$ $(-1,1)$ , and $(0,2)$ gives  a  total  of $\boxed{4}$ ordered pairs.

Answer: D 4
Problem: How many ordered pairs of integers $(x, y)$ satisfy the equation \[x^{2020}+y^2=2y?\]
$\textbf{(A) } 1 \qquad\textbf{(B) } 2 \qquad\textbf{(C) } 3 \qquad\textbf{(D) } 4 \qquad\textbf{(E) } \text{infinitely many}$

Solution: Bringing all of the terms to the LHS, we see a quadratic equation \[y^2 - 2y + x^{2020} = 0\] in terms of $y$ . Applying the quadratic formula, we get \[y = \frac{2\pm\sqrt{4-4\cdot1\cdot x^{2020}}}{2}=\frac{2\pm\sqrt{4(1-x^{2020})}}{2}.\] In order for $y$ to be real, which it must be given the stipulation that we are seeking integral answers, we know that the discriminant, $4(1-x^{2020})$ must be nonnegative. Therefore, \[4(1-x^{2020}) \geq 0 \implies x^{2020} \leq 1.\] Here, we see that we must split the inequality into a compound, resulting in $-1 \leq x \leq 1$
The only integers that satisfy this are $x \in \{-1,0,1\}$ . Plugging these values back into the quadratic equation, we see that $x = \{-1,1\}$ both produce a discriminant of $0$ , meaning that there is only 1 solution for $y$ . If $x = \{0\}$ , then the discriminant is nonzero, therefore resulting in two solutions for $y$
Thus, the answer is $2 \cdot 1 + 1 \cdot 2 = \boxed{4}$

Answer: D 4
Problem: How many ordered pairs of integers $(x, y)$ satisfy the equation \[x^{2020}+y^2=2y?\]
$\textbf{(A) } 1 \qquad\textbf{(B) } 2 \qquad\textbf{(C) } 3 \qquad\textbf{(D) } 4 \qquad\textbf{(E) } \text{infinitely many}$

Solution: Set it up as a quadratic in terms of y: \[y^2-2y+x^{2020}=0\] Then the discriminant is \[\Delta = 4-4x^{2020}\] This will clearly only yield real solutions when $|x^{2020}| \leq 1$ , because the discriminant must be positive.
Then $x=-1,0,1$ . Checking each one: $-1$ and $1$ are the same when raised to the 2020th power: \[y^2-2y+1=(y-1)^2=0\] This has only has solutions $1$ , so $(\pm 1,1)$ are solutions.
Next, if $x=0$ \[y^2-2y=0 \Rightarrow y(y-2)=0\] Which has 2 solutions, so $(0,2)$ and $(0,0)$
These are the only 4 solutions, so our answer is $\boxed{4}$

Answer: D 4
Problem: How many ordered pairs of integers $(x, y)$ satisfy the equation \[x^{2020}+y^2=2y?\]
$\textbf{(A) } 1 \qquad\textbf{(B) } 2 \qquad\textbf{(C) } 3 \qquad\textbf{(D) } 4 \qquad\textbf{(E) } \text{infinitely many}$

Solution: Move the $y^2$ term to the other side to get $x^{2020}=2y-y^2 = y(2-y)$
Because $x^{2020} \geq 0$ for all $x$ , then $y(2-y) \geq 0 \Rightarrow y = 0,1,2$
If $y=0$ or $y=2$ , the right side is $0$ and therefore $x=0$
When $y=1$ , the right side become $1$ , therefore $x=1,-1$
Our solutions are $(0,2)$ $(0,0)$ $(1,1)$ $(-1,1)$ . These are the only $4$ solutions, so the answer is $\boxed{4}$

Answer: D 4
Problem: How many ordered pairs of integers $(x, y)$ satisfy the equation \[x^{2020}+y^2=2y?\]
$\textbf{(A) } 1 \qquad\textbf{(B) } 2 \qquad\textbf{(C) } 3 \qquad\textbf{(D) } 4 \qquad\textbf{(E) } \text{infinitely many}$

Solution: Since $x^{2020}$ and $y^2$ are perfect squares, they are both nonnegative. That means $y^2$ plus a nonnegative number equals $2y$ , which means $y^2 \leq 2y.$ The only possible integer values for $y$ are $0, 1, 2$
For $y=0$ $x$ can only be $0$
For $y=1$ $x^2=1$ so $x=1, -1$
For $y=2$ $x$ can only be $0$ as well.
This gives us the solutions $(0, 0)$ $(1, 1)$ $(-1, 1)$ , and $(0, 2)$ . These are the only solutions, so there is a total of $\boxed{4}$ ordered pairs.

Answer: D 4
Problem: The decimal representation of \[\dfrac{1}{20^{20}}\] consists of a string of zeros after the decimal point, followed by a $9$ and then several more digits. How many zeros are in that initial string of zeros after the decimal point?
$\textbf{(A)} \text{ 23} \qquad \textbf{(B)} \text{ 24} \qquad \textbf{(C)} \text{ 25} \qquad \textbf{(D)} \text{ 26} \qquad \textbf{(E)} \text{ 27}$

Solution: We have
\[\dfrac{1}{20^{20}} = \dfrac{1}{(10\cdot2)^{20}}=\dfrac{1}{10^{20}\cdot2^{20}}\]
Now we do some estimation. Notice that $2^{20} = 1024^2$ , which means that $2^{20}$ is a little more than $1000^2=1,000,000$ . Multiplying it with $10^{20}$ , we get that the denominator is about $1\underbrace{00\dots0}_{26 \text{ zeros}}$ . Notice that when we divide $1$ by an $n$ digit number as long as n is not a power of 10, there are $n-1$ zeros before the first nonzero digit. This means that when we divide $1$ by the $27$ digit integer $1\underbrace{00\dots0}_{26 \text{ zeros}}$ , there are $\boxed{26}$ zeros in the initial string after the decimal point. -PCChess

Answer: D 26
Problem: The decimal representation of \[\dfrac{1}{20^{20}}\] consists of a string of zeros after the decimal point, followed by a $9$ and then several more digits. How many zeros are in that initial string of zeros after the decimal point?
$\textbf{(A)} \text{ 23} \qquad \textbf{(B)} \text{ 24} \qquad \textbf{(C)} \text{ 25} \qquad \textbf{(D)} \text{ 26} \qquad \textbf{(E)} \text{ 27}$

Solution: First rewrite $\frac{1}{20^{20}}$ as $\frac{5^{20}}{10^{40}}$ . Then, we know that when we write this in decimal form, there will be 40 digits after the decimal point. Therefore, we just have to find the number of digits in ${5^{20}}$
$\log{5^{20}} = 20\log{5}$ and memming $\log{5}\approx0.69$ (alternatively use the fact that $\log{5} = 1 - \log{2}$ ), $\lfloor{20\log{5}}\rfloor+1=\lfloor{20\cdot0.69}\rfloor+1=13+1=14$ digits.
Our answer is $\boxed{26}$

Answer: D 26
Problem: The decimal representation of \[\dfrac{1}{20^{20}}\] consists of a string of zeros after the decimal point, followed by a $9$ and then several more digits. How many zeros are in that initial string of zeros after the decimal point?
$\textbf{(A)} \text{ 23} \qquad \textbf{(B)} \text{ 24} \qquad \textbf{(C)} \text{ 25} \qquad \textbf{(D)} \text{ 26} \qquad \textbf{(E)} \text{ 27}$

Solution: Just as in Solution $2,$ we rewrite $\dfrac{1}{20^{20}}$ as $\dfrac{5^{20}}{10^{40}}.$ We then calculate $5^{20}$ entirely by hand, first doing $5^5 \cdot 5^5,$ then multiplying that product by itself, resulting in $95,367,431,640,625.$ Because this is $14$ digits, after dividing this number by $10$ fourteen times, the decimal point is before the $9.$ Dividing the number again by $10$ twenty-six more times allows a string of $\boxed{26}$ zeroes to be formed. -OreoChocolate

Answer: D 26
Problem: The decimal representation of \[\dfrac{1}{20^{20}}\] consists of a string of zeros after the decimal point, followed by a $9$ and then several more digits. How many zeros are in that initial string of zeros after the decimal point?
$\textbf{(A)} \text{ 23} \qquad \textbf{(B)} \text{ 24} \qquad \textbf{(C)} \text{ 25} \qquad \textbf{(D)} \text{ 26} \qquad \textbf{(E)} \text{ 27}$

Solution: We see that $\frac{1}{20^{20}} = 9.5367432 \cdot \cdot \cdot \times 10^{-27}$ . We see that this has $27-1=26$ zeros after the decimal point before coming to $9$
Therefore, the answer is $\boxed{26}$

Answer: D 26
Problem: The decimal representation of \[\dfrac{1}{20^{20}}\] consists of a string of zeros after the decimal point, followed by a $9$ and then several more digits. How many zeros are in that initial string of zeros after the decimal point?
$\textbf{(A)} \text{ 23} \qquad \textbf{(B)} \text{ 24} \qquad \textbf{(C)} \text{ 25} \qquad \textbf{(D)} \text{ 26} \qquad \textbf{(E)} \text{ 27}$

Solution: \begin{align*}|\lceil \log \dfrac{1}{20^{20}} \rceil| &= |\lceil \log 20^{-20} \rceil| \\ &= |\lceil -20 \log(20) \rceil| \\ &= |\lceil -20(\log 10 + \log 2) \rceil| \\ &= |\lceil -20(1 + 0.301) \rceil| \\ &= |\lceil -26.02 \rceil| \\ &= |-26| \\ &= \boxed{26}

Answer: D 26
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 $2019, 2020, 2021$
$\textbf{(A)}\ 7 \qquad\textbf{(B)}\ 9 \qquad\textbf{(C)}\ 10 \qquad\textbf{(D)}\ 11 \qquad\textbf{(E)}\ 12$

Solution: Note that cycles exist initially and after each round of erasing.
Let the parentheses denote cycles. It follows that:
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 $4+2+5=\boxed{11}.$

Answer: D 11
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 $2019, 2020, 2021$
$\textbf{(A)}\ 7 \qquad\textbf{(B)}\ 9 \qquad\textbf{(C)}\ 10 \qquad\textbf{(D)}\ 11 \qquad\textbf{(E)}\ 12$

Solution: 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 $4+2+5= \boxed{11}.$

Answer: D 11
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 $2019, 2020, 2021$
$\textbf{(A)}\ 7 \qquad\textbf{(B)}\ 9 \qquad\textbf{(C)}\ 10 \qquad\textbf{(D)}\ 11 \qquad\textbf{(E)}\ 12$

Solution: 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:
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 $4+2+5=\boxed{11}.$

Answer: D 11