Navvye/Olympiad-Problems-CSV huggingface dataset:
Problem: $\emph{triangular number}$ is a positive integer that can be expressed in the form $t_n = 1+2+3+\cdots+n$ , for some positive integer $n$ . The three smallest triangular numbers that are also perfect squares are $t_1 = 1 = 1^2$ $t_8 = 36 = 6^2$ , and $t_{49} = 1225 = 35^2$ . What is the sum of the digits of the fourth smallest triangular number that is also a perfect square?
$\textbf{(A) } 6 \qquad \textbf{(B) } 9 \qquad \textbf{(C) } 12 \qquad \textbf{(D) } 18 \qquad \textbf{(E) } 27$

Solution: We have $t_n = \frac{n (n+1)}{2}$ .
If $t_n$ is a perfect square, then it can be written as $\frac{n (n+1)}{2} = k^2$ ,
where $k$ is a positive integer.
Thus, $n (n+1) = 2 k^2$ . Rearranging, we get $(2n+1)^2-2(2k)^2=1$ , a Pell equation. So $\frac{2n+1}{2k}$ must be a truncation of the continued fraction for $\sqrt{2}$
\begin{eqnarray*} 1+\frac12&=&\frac{2\cdot1+1}{2\cdot1}\\ 1+\frac1{2+\frac1{2+\frac12}}&=&\frac{2\cdot8+1}{2\cdot6}\\ 1+\frac1{2+\frac1{2+\frac1{2+\frac1{2+\frac12}}}}&=&\frac{2\cdot49+1}{2\cdot35}\\ 1+\frac1{2+\frac1{2+\frac1{2+\frac1{2+\frac1{2+\frac1{2+\frac12}}}}}}&=&\frac{2\cdot288+1}{2\cdot204} \end{eqnarray*}
Therefore, $t_{288} = \frac{288\cdot289}{2} = 204^2 = 41616$ , so the answer is $4+1+6+1+6=\boxed{18}$

Answer: D 18
Problem: $\emph{triangular number}$ is a positive integer that can be expressed in the form $t_n = 1+2+3+\cdots+n$ , for some positive integer $n$ . The three smallest triangular numbers that are also perfect squares are $t_1 = 1 = 1^2$ $t_8 = 36 = 6^2$ , and $t_{49} = 1225 = 35^2$ . What is the sum of the digits of the fourth smallest triangular number that is also a perfect square?
$\textbf{(A) } 6 \qquad \textbf{(B) } 9 \qquad \textbf{(C) } 12 \qquad \textbf{(D) } 18 \qquad \textbf{(E) } 27$

Solution: As mentioned above, $t_n = \frac{n (n+1)}{2}$ . If $t_n$ is a perfect square, one of two things must occur when the fraction is split into a product. Either $\frac{n}{2}$ and $n+1$ must both be squares, or $n$ and $\frac{n+1}{2}$ must both be squares, and thus the search for the next perfect square triangular number can be narrowed down by testing values of $n$ that are close to or are perfect squares. After some work, we reach $n = 288$ $1$ less than $289$ , and $t_{288} = \frac{288\cdot289}{2} = 144 \cdot 289 = 41616$ . This product is a perfect square, and thus the sum of the digits of the fourth smallest perfect square triangular number is therefore $4+1+6+1+6=\boxed{18}$

Answer: D 18
Problem: $\emph{triangular number}$ is a positive integer that can be expressed in the form $t_n = 1+2+3+\cdots+n$ , for some positive integer $n$ . The three smallest triangular numbers that are also perfect squares are $t_1 = 1 = 1^2$ $t_8 = 36 = 6^2$ , and $t_{49} = 1225 = 35^2$ . What is the sum of the digits of the fourth smallest triangular number that is also a perfect square?
$\textbf{(A) } 6 \qquad \textbf{(B) } 9 \qquad \textbf{(C) } 12 \qquad \textbf{(D) } 18 \qquad \textbf{(E) } 27$

Solution: According to the problem, we want to find integer $p$ such $\frac{n(n+1)}{2}=p^2$ , after expanding, we have $n^2+n=2p^2, 4n^2+4n=8p^2, (2n+1)^2-8p^2=1$ , we call $2n+1=q$ , the equation becomes $q^2-8p^2=1$ , obviously $(q,p)=(3,1)$ is the elementary solution for this pell equation, thus the forth smallest solution set $q_4+2\sqrt{2}p_4=(3+2\sqrt{2})^4=577+408\sqrt{2}$ , which indicates $p=204, p^2=41616$ leads to $\boxed{18}$

Answer:  18
Problem: $\emph{triangular number}$ is a positive integer that can be expressed in the form $t_n = 1+2+3+\cdots+n$ , for some positive integer $n$ . The three smallest triangular numbers that are also perfect squares are $t_1 = 1 = 1^2$ $t_8 = 36 = 6^2$ , and $t_{49} = 1225 = 35^2$ . What is the sum of the digits of the fourth smallest triangular number that is also a perfect square?
$\textbf{(A) } 6 \qquad \textbf{(B) } 9 \qquad \textbf{(C) } 12 \qquad \textbf{(D) } 18 \qquad \textbf{(E) } 27$

Solution: If $n \choose 2$ is a square, then ${(2n-1)^2 \choose 2}$ is also a square. We can prove this quite simply:
\[{(2n-1)^2 \choose 2}\] \[= \frac{(2n-1)^2 \cdot ((2n-1)^2 - 1)}{2}\] \[= \frac{(2n-1)^2 \cdot (2n \cdot (2n - 2))}{2}\] \[= (2n-1)^2 \cdot 4{n \choose 2}.\]
Therefore, ${(2 \cdot 9 - 1)^2 \choose 2}$ is a square. Note that $T_n = {n+1 \choose 2}$ . We can easily check all smaller possibilities using a bit of casework, and they don't work. Our solution is thus ${289 \choose 2} = 204^2 = 41616$ , and so the answer is $\boxed{18}$

Answer:  18
Problem: $\emph{triangular number}$ is a positive integer that can be expressed in the form $t_n = 1+2+3+\cdots+n$ , for some positive integer $n$ . The three smallest triangular numbers that are also perfect squares are $t_1 = 1 = 1^2$ $t_8 = 36 = 6^2$ , and $t_{49} = 1225 = 35^2$ . What is the sum of the digits of the fourth smallest triangular number that is also a perfect square?
$\textbf{(A) } 6 \qquad \textbf{(B) } 9 \qquad \textbf{(C) } 12 \qquad \textbf{(D) } 18 \qquad \textbf{(E) } 27$

Solution: We want to find integer $n_i$ and $m_i$ such that $t_{n_i} =\frac{n_i (n_i + 1)}{2}=m_i^2, n_0 = 0.$
We use the formula $\sqrt{n_{i+1}} = \sqrt{2n_i} + \sqrt{n_i + 1}$ and get
\[n_1 = ( \sqrt{2n_0} + \sqrt{n_0 + 1})^2 = (0+1)^2 = 1,\] \[n_2= ( \sqrt{2n_1} + \sqrt{n_1 + 1})^2 = ( \sqrt{2} + \sqrt{1 + 1})^2 = 8,\] \[n_3= ( \sqrt{2n_2} + \sqrt{n_2 + 1})^2 = ( \sqrt{16} + \sqrt{8 + 1})^2 = 49,\] \[n_4 = ( \sqrt{2n_3} + \sqrt{n_3 + 1})^2 = ( \sqrt{98} + \sqrt{49 + 1})^2 =  ((7 + 5)\sqrt{2})^2 = 288,\] \[n_5 = ( \sqrt{2n_4} + \sqrt{n_4 + 1})^2 = ( \sqrt{576} + \sqrt{289})^2 =  (24 + 17)^2 = 1681,\] \[n_6 = ( \sqrt{2n_5} + \sqrt{n_5 + 1})^2 = ( 41\sqrt{2} + \sqrt{1682})^2 =  ((41 + 29)\sqrt{2})^2 = 9800,...\] Therefore, $t_{n_4} = t_{288} = \frac{288\cdot289}{2} = 41616 \implies 4+1+6+1+6=\boxed{18}$

Answer: D 18
Problem: Suppose $a$ is a real number such that the equation \[a\cdot(\sin{x}+\sin{(2x)}) = \sin{(3x)}\] has more than one solution in the interval $(0, \pi)$ . The set of all such $a$ that can be written
in the form \[(p,q) \cup (q,r),\] where $p, q,$ and $r$ are real numbers with $p < q< r$ . What is $p+q+r$
$\textbf{(A) } {-}4 \qquad \textbf{(B) } {-}1 \qquad \textbf{(C) } 0 \qquad \textbf{(D) } 1 \qquad \textbf{(E) } 4$

Solution: We are given that $a\cdot(\sin{x}+\sin{(2x)})=\sin{(3x)}$
Using the sine double angle formula combine with the fact that $\sin{(3x)} = \sin{x}\cdot(4\cos^2{x}-1)$ , which can be derived using sine angle addition with $\sin{(2x + x)}$ , we have \[a\cdot(\sin{x}+2\sin{x}\cos{x})=\sin{x}\cdot(4\cos^2{x}-1)\] Since $\sin{x} \ne 0$ as it is on the open interval $(0, \pi)$ , we can divide out $\sin{x}$ from both sides, leaving us with \[a\cdot(1+2\cos{x})=4\cos^2{x}-1\] Now, distributing $a$ and rearranging, we achieve the equation \[4\cos^2{x} - 2a\cos{x} - (1+a) = 0\] which is a quadratic in $\cos{x}$
Applying the quadratic formula to solve for $\cos{x}$ , we get \[\cos{x} =\frac{2a\pm\sqrt{4a^2+4(4)(1+a)}}{8}\] and expanding the terms under the radical, we get \[\cos{x} =\frac{2a\pm\sqrt{4a^2+16a+16}}{8}\] Factoring, since $4a^2+16a+16 = (2a+4)^2$ , we can simplify our expression even further to \[\cos{x} =\frac{a\pm(a+2)}{4}\]
Now, solving for our two solutions, $\cos{x} = -\frac{1}{2}$ and $\cos{x} = \frac{a+1}{2}$
Since $\cos{x} = -\frac{1}{2}$ yields a solution that is valid for all $a$ , that being $x = \frac{2\pi}{3}$ , we must now solve for the case where $\frac{a+1}{2}$ yields a valid value.
As $x\in (0, \pi)$ $\cos{x}\in (-1, 1)$ , and therefore $\frac{a+1}{2}\in (-1, 1)$ , and $a\in(-3,1)$
There is one more case we must consider inside this interval though, the case where $\frac{a+1}{2} = -\frac{1}{2}$ , as this would lead to a double root for $\cos{x}$ , yielding only one valid solution for $x$ . Solving for this case, $a \ne -2$
Therefore, combining this fact with our solution interval, $a\in(-3, -2) \cup (-2, 1)$ , so the answer is $-3-2+1 = \boxed{4}$

Answer: A 4
Problem: Suppose $a$ is a real number such that the equation \[a\cdot(\sin{x}+\sin{(2x)}) = \sin{(3x)}\] has more than one solution in the interval $(0, \pi)$ . The set of all such $a$ that can be written
in the form \[(p,q) \cup (q,r),\] where $p, q,$ and $r$ are real numbers with $p < q< r$ . What is $p+q+r$
$\textbf{(A) } {-}4 \qquad \textbf{(B) } {-}1 \qquad \textbf{(C) } 0 \qquad \textbf{(D) } 1 \qquad \textbf{(E) } 4$

Solution: We can optimize from the step from \[a\cdot(1+2\cos{x})=4\cos^2{x}-1\] in solution 1 by writing
\[a = \frac{4\cos^2{x}-1}{1+2\cos{x}} = 2\cos x - 1\]
and then get \[\cos x = \frac{a+1}{2}.\]
Now, solving for our two solutions, $\cos{x} = -\frac{1}{2}$ and $\cos{x} = \frac{a+1}{2}$
Since $\cos{x} = -\frac{1}{2}$ yields a solution that is valid for all $a$ , that being $x = \frac{2\pi}{3}$ , we must now solve for the case where $\frac{a+1}{2}$ yields a valid value.
As $x\in (0, \pi)$ $\cos{x}\in (-1, 1)$ , and therefore $\frac{a+1}{2}\in (-1, 1)$ , and $a\in(-3,1)$
There is one more case we must consider inside this interval though, the case where $\frac{a+1}{2} = -\frac{1}{2}$ , as this would lead to a double root for $\cos{x}$ , yielding only one valid solution for $x$ . Solving for this case, $a \ne -2$
Therefore, combining this fact with our solution interval, $a\in(-3, -2) \cup (-2, 1)$ , so the answer is $-3-2+1 = \boxed{4}$

Answer: A 4
Problem: Suppose $a$ is a real number such that the equation \[a\cdot(\sin{x}+\sin{(2x)}) = \sin{(3x)}\] has more than one solution in the interval $(0, \pi)$ . The set of all such $a$ that can be written
in the form \[(p,q) \cup (q,r),\] where $p, q,$ and $r$ are real numbers with $p < q< r$ . What is $p+q+r$
$\textbf{(A) } {-}4 \qquad \textbf{(B) } {-}1 \qquad \textbf{(C) } 0 \qquad \textbf{(D) } 1 \qquad \textbf{(E) } 4$

Solution: Use the sum to product formula to obtain $2a\cdot\sin{\frac{3x}{2}}\cos{\frac{x}{2}}=\sin{3x}$ . Use the double angle formula on the RHS to obtain $a\cdot\sin{\frac{3x}{2}}\cos{\frac{x}{2}}=\sin{\frac{3x}{2}}\cos{\frac{3x}{2}}$ . From here, it is obvious that $x=\frac{2\pi}{3}$ is always a solution, and thus we divide by $\sin{\frac{3x}{2}}$ to get \[a\cdot\cos{\frac{x}{2}}=\cos{\frac{3x}{2}}\] We wish to find all $a$ such that there is at least one more solution to this equation distinct from $x=\frac{2\pi}{3}$ . Letting $y=\cos{\frac{x}{2}}$ , and noting that $\cos{\frac{3x}{2}}=4y^3-3y$ , we can rearrange our equation to $4y^3=y(3+a)$ The smallest value $x$ where $y=0$ is $\pi$ , which is not in our domain so we divide by $y$ to obtain $4y^2=a+3$ . By the trivial inequality, $a+3\ge{0}$ . Furthermore, $y\neq{0}$ , so $a+3>0$ . Also, if $a=-2$ , then the solution to this equation would be shared with $x=\frac{2\pi}{3}$ , so there would only be one distinct solution. Finally, because $y\le{1}$ due to the restrictions of a sine wave, and that $y\neq{1}$ due to the restrictions on $x$ , we have $-3<a<1$ with $a\neq{-2}$ . Thus, $p=-3,q=-2, r=1$ , so our final answer is $-3+(-2)+1=\boxed{4}$

Answer: A 4
Problem: Let $c$ be a real number, and let $z_1$ and $z_2$ be the two complex numbers satisfying the equation $z^2 - cz + 10 = 0$ . Points $z_1$ $z_2$ $\frac{1}{z_1}$ , and $\frac{1}{z_2}$ are the vertices of (convex) quadrilateral $\mathcal{Q}$ in the complex plane. When the area of $\mathcal{Q}$ obtains its maximum possible value, $c$ is closest to which of the following?
$\textbf{(A) }4.5 \qquad\textbf{(B) }5 \qquad\textbf{(C) }5.5 \qquad\textbf{(D) }6\qquad\textbf{(E) }6.5$

Solution: Because $c$ is real, $z_2 = \bar z_1$ .
We have \begin{align*} 10 & = z_1 z_2 \\ & = z_1 \bar z_1 \\ & = |z_1|^2 , \end{align*} where the first equality follows from Vieta's formula.
Thus, $|z_1| = \sqrt{10}$
We have \begin{align*} c & = z_1 + z_2 \\ & = z_1 + \bar z_1 \\ & = 2 {\rm Re}(z_1), \end{align*} where the first equality follows from Vieta's formula.
Thus, ${\rm Re}(z_1) = \frac{c}{2}$
We have \begin{align*} \frac{1}{z_1} & = \frac{1}{10}\cdot\frac{10}{z_1} \\ & = \frac{1}{10}\cdot\frac{z_1 z_2}{z_1} \\ & = \frac{z_2}{10} \\ & = \frac{\bar z_1}{10}. \end{align*} where the second equality follows from Vieta's formula.
We have \begin{align*} \frac{1}{z_2} & = \frac{1}{10}\cdot\frac{10}{z_2} \\ & = \frac{1}{10}\cdot\frac{z_1 z_2}{z_2} \\ & = \frac{z_1}{10}. \end{align*} where the second equality follows from Vieta's formula.
Therefore, \begin{align*} {\rm Area} \ Q & = \frac{1}{2} \left| {\rm Re}(z_1) \right| \cdot 2 \left| {\rm Im}(z_1) \right| \cdot \left( 1 - \frac{1}{10^2} \right) \\ & = \frac{1}{2} |c| \sqrt{10 - \frac{c^2}{4}} \left( 1 - \frac{1}{10^2} \right) \\ & = \frac{1 - \frac{1}{10^2}}{4} \sqrt{c^2 \left( 40 - c^2 \right)} \\ & \leq \frac{1 - \frac{1}{10^2}}{4} \cdot \frac{c^2 + \left( 40 - c^2 \right)}{2} \\ & = \frac{1 - \frac{1}{10^2}}{4} \cdot 20 , \end{align*} where the inequality follows from the AM-GM inequality, and it is augmented to an equality if and only if $c^2 = 40 - c^2$ .
Thus, $|c| = 2 \sqrt{5} \approx \boxed{4.5}$

Answer: A 4.5
Problem: Let $c$ be a real number, and let $z_1$ and $z_2$ be the two complex numbers satisfying the equation $z^2 - cz + 10 = 0$ . Points $z_1$ $z_2$ $\frac{1}{z_1}$ , and $\frac{1}{z_2}$ are the vertices of (convex) quadrilateral $\mathcal{Q}$ in the complex plane. When the area of $\mathcal{Q}$ obtains its maximum possible value, $c$ is closest to which of the following?
$\textbf{(A) }4.5 \qquad\textbf{(B) }5 \qquad\textbf{(C) }5.5 \qquad\textbf{(D) }6\qquad\textbf{(E) }6.5$

Solution: Since $c$ , which is the sum of roots $z_1$ and $z_2$ , is real, $z_1=\overline{z_2}$
Let $z_1=a+bi$ . Then $z_2=a-bi$ . Note that the product of the roots is $10$ by Vieta's, so $z_1z_2=(a+bi)(a-bi)=a^2+b^2=10$
Thus, $\frac{1}{z_1}=\frac{1}{a-bi}\cdot\frac{a+bi}{a+bi}=\frac{a+bi}{a^2+b^2}=\frac{a+bi}{10}$ . With the same process, $\frac{1}{z_2}=\frac{a-bi}{10}$
So, our four points are $a+bi,\frac{a+bi}{10},a-bi,$ and $\frac{a-bi}{10}$ . WLOG let $a+bi$ be in the first quadrant and graph these four points on the complex plane. Notice how quadrilateral Q is a trapezoid with the real axis as its axis of symmetry. It has a short base with endpoints $\frac{a+bi}{10}$ and $\frac{a-bi}{10}$ , so its length is $\left|\frac{a+bi}{10} - \frac{a-bi}{10}\right| =\frac{b}{5}$ . Likewise, its long base has endpoints $a+bi$ and $a-bi$ , so its length is $|(a+bi)-(a-bi)|=2b$
The height, which is the distance between the two lines, is the difference between the real values of the two bases $\implies h= a-\frac{a}{10}=\frac{9a}{10}$
Plugging these into the area formula for a trapezoid, we are trying to maximize $\frac12\cdot\left(2b+\frac{b}{5}\right)\cdot\frac{9a}{10}=\frac{99ab}{100}$ . Thus, the only thing we need to maximize is $ab$
With the restriction that $a^2+b^2=(a-b)^2+2ab=10$ $ab$ is maximized when $a=b=\sqrt{5}$
Remember, $c$ is the sum of the roots, so $c=z_1+z_2=2a=2\sqrt5=\sqrt{20}\approx\boxed{4.5}$

Answer:  4.5
Problem: Let $c$ be a real number, and let $z_1$ and $z_2$ be the two complex numbers satisfying the equation $z^2 - cz + 10 = 0$ . Points $z_1$ $z_2$ $\frac{1}{z_1}$ , and $\frac{1}{z_2}$ are the vertices of (convex) quadrilateral $\mathcal{Q}$ in the complex plane. When the area of $\mathcal{Q}$ obtains its maximum possible value, $c$ is closest to which of the following?
$\textbf{(A) }4.5 \qquad\textbf{(B) }5 \qquad\textbf{(C) }5.5 \qquad\textbf{(D) }6\qquad\textbf{(E) }6.5$

Solution: Like the solutions above we can know that $|z_1| = |z_2| = \sqrt{10}$ and $z_1=\overline{z_2}$
Let $z_1=\sqrt{10}e^{i\theta}$ where $0<\theta<\pi$ , then $z_2=\sqrt{10}e^{-i\theta}$ $\frac{1}{z_1}=\frac{1}{\sqrt{10}}e^{-i\theta}$ $\frac{1}{z_2}= \frac{1}{\sqrt{10}}e^{i\theta}$
On the basis of symmetry, the area $A$ of $\mathcal{Q}$ is the difference between two isoceles triangles,so
$2A=10\sin2\theta-\frac{1}{10}\sin2\theta\leq10-\frac{1}{10}$ . The inequality holds when $2\theta=\frac{\pi}{2}$ , or $\theta=\frac{\pi}{4}$
Thus, $c= 2 {\rm Re} \ z_1 =2 \sqrt{10} \cos\frac{\pi}{4}=\sqrt{20} \approx \boxed{4.5}$

Answer: A 4.5
Problem: Let $c$ be a real number, and let $z_1$ and $z_2$ be the two complex numbers satisfying the equation $z^2 - cz + 10 = 0$ . Points $z_1$ $z_2$ $\frac{1}{z_1}$ , and $\frac{1}{z_2}$ are the vertices of (convex) quadrilateral $\mathcal{Q}$ in the complex plane. When the area of $\mathcal{Q}$ obtains its maximum possible value, $c$ is closest to which of the following?
$\textbf{(A) }4.5 \qquad\textbf{(B) }5 \qquad\textbf{(C) }5.5 \qquad\textbf{(D) }6\qquad\textbf{(E) }6.5$

Solution: Like in Solution 3, we find that $Q = \frac12\cdot\left(2b+\frac{b}{5}\right)\cdot\frac{9a}{10}= \frac{99}{100}ab$ , thus, $Q$ is maximized when $ab$ is maximized. $ab = a \cdot \sqrt{10 - a^2} = \sqrt{10a^2 - a^4}$ , let $f(a) =  \sqrt{10a^2 - a^4}$
By the Chain Rule and the Power Rule, $f'(a) = \frac12 \cdot (10a^2 - a^4)^{-\frac12} \cdot (10(2a)-4a^3)) = \frac{20a-4a^3}{ 2\sqrt{10a^2 - a^4} } = \frac{10a-2a^3}{ \sqrt{10a^2 - a^4} }$
$f'(a) = 0$ $10a-2a^3 = 0$ $a \neq 0$ $a^2 = 5$ $a = \sqrt{5}$
$\because f'(a) = 0$ when $a = \sqrt{5}$ $f'(a)$ is positive when $a < \sqrt{5}$ , and $f'(a)$ is negative when $a > \sqrt{5}$
$\therefore f(a)$ has a local maximum when $a = \sqrt{5}$
Notice that $ab = \frac{ (a+b)^2 - (a^2+b^2) }{2} = \frac{c^2 - 10}{2}$ $\frac{c^2 - 10}{2} = 5$ $c = \sqrt{2 \cdot 5 + 10} = \sqrt{20} \approx \boxed{4.5}$

Answer: A 4.5
Problem: Let $h_n$ and $k_n$ be the unique relatively prime positive integers such that \[\frac{1}{1}+\frac{1}{2}+\frac{1}{3}+\cdots+\frac{1}{n}=\frac{h_n}{k_n}.\] Let $L_n$ denote the least common multiple of the numbers $1, 2, 3, \ldots, n$ . For how many integers with $1\le{n}\le{22}$ is $k_n<L_n$
$\textbf{(A) }0 \qquad\textbf{(B) }3 \qquad\textbf{(C) }7 \qquad\textbf{(D) }8\qquad\textbf{(E) }10$

Solution: We are given that \[\sum_{i=1}^{n}\frac1i = \frac{1}{L_n}\sum_{i=1}^{n}\frac{L_n}{i} = \frac{h_n}{k_n}.\] Since $k_n < L_n,$ we need $\gcd\left(\sum_{i=1}^{n}\frac{L_n}{i}, L_n\right)>1.$
For all primes $p$ such that $p\leq n,$ let $v_p(L_n)=e\geq1$ be the largest power of $p$ that is a factor of $L_n.$
It is clear that $L_n\equiv0\pmod{p},$ so we test whether $\sum_{i=1}^{n}\frac{L_n}{i}\equiv0\pmod{p}.$ Note that \[\sum_{i=1}^{n}\frac{L_n}{i} \equiv \sum_{i=1}^{\left\lfloor\tfrac{n}{p^e}\right\rfloor}\frac{L_n}{ip^e} \ (\operatorname{mod} \ p^e) \equiv \sum_{i=1}^{\left\lfloor\tfrac{n}{p^e}\right\rfloor}\frac{L_n}{ip^e} \ (\operatorname{mod} \ p).\] We construct the following table for $v_p(L_n)=e:$ \[\begin{array}{c|c|l|c}  \textbf{Case of }\boldsymbol{(p,e)} & \textbf{Interval of }\boldsymbol{n} & \hspace{22.75mm}\textbf{Sum} & \boldsymbol{\stackrel{?}{\equiv}0\ (\operatorname{mod} \ p)} \\ [0.5ex] \hline\hline  & & & \\ [-2ex] (2,1) & [2,3] & L_n/2 &  \\      (2,2) & [4,7] & L_n/4 &  \\     (2,3) & [8,15] & L_n/8 & \\     (2,4) & [16,22] & L_n/16 &  \\ [0.5ex] \hline   & & & \\ [-2ex] (3,1) & [3,5] & L_n/3 & \\     & [6,8] & L_n/3 + L_n/6 & \checkmark \\    (3,2) & [9,17] & L_n/9 & \\  & [18,22] & L_n/9 + L_n/18 & \checkmark \\ [0.5ex] \hline   & & & \\ [-2ex] (5,1) & [5,9] & L_n/5 & \\     & [10,14] & L_n/5 + L_n/10 & \\ & [15,19] & L_n/5 + L_n/10 + L_n/15 & \\   & [20,22] & L_n/5 + L_n/10 + L_n/15 + L_n/20 & \checkmark \\ [0.5ex] \hline   & & & \\ [-2ex] (7,1) & [7,13] & L_n/7 & \\     & [14,20] & L_n/7 + L_n/14 & \\ & [21,22] & L_n/7 + L_n/14 + L_n/21 & \\ [0.5ex] \hline   & & & \\ [-2ex] (11,1) & [11,21] & L_n/11 &  \\ & \{22\} & L_n/11 + L_n/22 & \\ [0.5ex] \hline & & & \\ [-2ex] (13,1) & [13,22] & L_n/13 &  \\ [0.5ex] \hline & & & \\ [-2ex] (17,1) & [17,22] & L_n/17 &  \\ [0.5ex] \hline & & & \\ [-2ex] (19,1) & [19,22] & L_n/19 & \\ [0.5ex] \end{array}\] Note that:
Together, there are $\boxed{8}$ such integers $n,$ namely \[6,7,8,18,19,20,21,22.\] ~MRENTHUSIASM

Answer: D 8
Problem: A circle with integer radius $r$ is centered at $(r, r)$ . Distinct line segments of length $c_i$ connect points $(0, a_i)$ to $(b_i, 0)$ for $1 \le i \le 14$ and are tangent to the circle, where $a_i$ $b_i$ , and $c_i$ are all positive integers and $c_1 \le c_2 \le \cdots \le c_{14}$ . What is the ratio $\frac{c_{14}}{c_1}$ for the least possible value of $r$
$\textbf{(A)} ~\frac{21}{5} \qquad\textbf{(B)} ~\frac{85}{13} \qquad\textbf{(C)} ~7 \qquad\textbf{(D)} ~\frac{39}{5} \qquad\textbf{(E)} ~17$

Solution: Suppose that with a pair $(a_i,b_i)$ the circle is an excircle. Then notice that the hypotenuse must be $(r-x)+(r-y)$ , so it must be the case that \[a_i^2+b_i^2=(2r-a_i-b_i)^2.\] Similarly, if with a pair $(a_i,b_i)$ the circle is an incircle, the hypotenuse must be $(x-r)+(y-r)$ , leading to the same equation.
Notice that this equation can be simplified through SFFT to \[(a_i-2r)(b_i-2r)=2r^2.\] Thus, we want the smallest $r$ such that this equation has at least $14$ distinct pairs $(a_i,b_i)$ for which this holds. The obvious choice to check is $r=6$ . In this case, since $2r^2=2^3\cdot 3^2$ has $12$ positive factors, we get $12$ pairs, and we get another two if the factors are $-8,-9$ or vice versa. One can check that for smaller values of $r$ , we do not even get close to $14$ possible pairs.
When $r=6$ , the smallest possible $c$ -value is clearly when the factors are negative. When this occurs, $a_i=4, b_i=3$ (or vice versa), so the mimimal $c$ is $5$ . The largest possible $c$ -value occurs when the largest of $a_i$ and $b_i$ are maximized. This occurs when the factors are $72$ and $1$ , leading to $a_i=84, b_i=13$ (or vice-versa), leading to a maximal $c$ of $85$
Hence the answer is $\frac{85}5=\boxed{17}$

Answer:  17