Navvye/Olympiad-Problems-CSV huggingface dataset:
Problem: In $\triangle ABC, AB = AC = 10$ and $BC = 12$ . Point $D$ lies strictly between $A$ and $B$ on $\overline{AB}$ and point $E$ lies strictly between $A$ and $C$ on $\overline{AC}$ so that $AD = DE = EC$ . Then $AD$ can be expressed in the form $\dfrac{p}{q}$ , where $p$ and $q$ are relatively prime positive integers. Find $p+q$

Solution: Let's label $\angle A = \theta$ and $\angle ECD = \alpha$ . Using isosceles triangle properties and the triangle angle sum equation, we get \[180-(180-2\theta+\alpha) + \frac{180-\theta}{2} + \left(\frac{180-\theta}{2} - \alpha\right) = 180.\] Solving, we find $\theta = 2 \alpha$
Relabelling our triangle, we get $\angle ABC = 90 - \alpha$ . Dropping an altitude from $A$ to $BC$ and using the Pythagorean theorem, we find $[ABC] = 48$ . Using the sine area formula, we see $\frac12 \cdot 10 \cdot 12 \cdot \sin(90-\alpha) = 48$ . Plugging in our sine angle cofunction identity, $\sin(90-\alpha) = \cos(\alpha)$ , we get $\alpha = \cos{^{-1}}{\frac45}$
Now, using the Law of Sines on $\triangle ADE$ , we get \[\frac{\sin{2\alpha}}{\frac{p}{q}} = \frac{\sin{(180-4\alpha)}}{10-\frac{p}{q}}.\] After applying numerous trigonometric and algebraic tricks, identities, and simplifications, such as $\sin{(180-4\alpha)}=\sin{4\alpha}$ and $\sin{\left(\cos{^{-1}}{\frac45}\right)} = \frac35$ , we find $\frac{p}{q} = \frac{10\sin{2\alpha}}{\sin{4\alpha}+\sin{2\alpha}} = \frac{250}{39}$
Therefore, our answer is $250 + 39 = \boxed{289}$

Answer:  289
Problem: In $\triangle ABC, AB = AC = 10$ and $BC = 12$ . Point $D$ lies strictly between $A$ and $B$ on $\overline{AB}$ and point $E$ lies strictly between $A$ and $C$ on $\overline{AC}$ so that $AD = DE = EC$ . Then $AD$ can be expressed in the form $\dfrac{p}{q}$ , where $p$ and $q$ are relatively prime positive integers. Find $p+q$

Solution: We start by labelling a few angles (all of them in degrees). Let $\angle{BAC}=2\alpha = \angle{AED}, \angle{EDC}=\angle{ECD}=\alpha, \angle{DEC}=180-2\alpha, \angle{BDC}=3\alpha, \angle{DCB}=90-2\alpha, \angle{DBC}=90-\alpha$ . Also let $AD=a$ . By sine rule in $\triangle{ADE},$ we get $\frac{a}{\sin{2\alpha}}=\frac{10-a}{\sin{4\alpha}} \implies \cos{2\alpha}=\frac{5}{a}-\frac{1}{2}$ Using sine rule in $\triangle{ABC}$ , we get $\sin{\alpha}=\frac{3}{5}$ . Hence we get $\cos{2\alpha}=1-2\sin^2{\alpha}=1-\frac{18}{25}=\frac{7}{25}$ . Hence $\frac{5}{a}=\frac{1}{2}+\frac{7}{25}=\frac{39}{50} \implies a=\frac{250}{39}$ . Therefore, our answer is $\boxed{289}$

Answer:  289
Problem: For each ordered pair of real numbers $(x,y)$ satisfying \[\log_2(2x+y) = \log_4(x^2+xy+7y^2)\] there is a real number $K$ such that \[\log_3(3x+y) = \log_9(3x^2+4xy+Ky^2).\] Find the product of all possible values of $K$

Solution: Using the logarithmic property $\log_{a^n}b^n = \log_{a}b$ , we note that \[(2x+y)^2 = x^2+xy+7y^2\] That gives \[x^2+xy-2y^2=0\] upon simplification and division by $3$ . Factoring $x^2+xy-2y^2=0$ gives \[(x+2y)(x-y)=0\] Then, \[x=y \text{ or }x=-2y\] From the second equation, \[9x^2+6xy+y^2=3x^2+4xy+Ky^2\] If we take $x=y$ , we see that $K=9$ . If we take $x=-2y$ , we see that $K=21$ . The product is $\boxed{189}$

Answer:  189
Problem: For each ordered pair of real numbers $(x,y)$ satisfying \[\log_2(2x+y) = \log_4(x^2+xy+7y^2)\] there is a real number $K$ such that \[\log_3(3x+y) = \log_9(3x^2+4xy+Ky^2).\] Find the product of all possible values of $K$

Solution: Do as done in Solution 1 to get \[x^2+xy-2y^2=0\] \[\implies (\frac{x}{y})^2+\frac{x}{y}-2=0\] \[\implies \frac{x}{y}=\frac{-1\pm\sqrt{1+8}}{2}=1,-2\] Do as done in Solution 1 to get \[9x^2+6xy+y^2=3x^2+4xy+Ky^2\] \[\implies 6x^2+2xy+(1-K)y^2=0\] \[\implies 6(\frac{x}{y})^2+2\frac{x}{y}+(1-K)=0\] \[\implies \frac{x}{y}=\frac{-2\pm \sqrt{4-24(1-K)}}{12}\] \[\implies \frac{x}{y}=\frac{-2\pm 2\sqrt{6K-5}}{12}=\frac{-1\pm \sqrt{6K-5}}{6}\] If $\frac{x}{y}=1$ then \[1=\frac{-1\pm \sqrt{6K-5}}{6}\] \[\implies 6=-1\pm \sqrt{6K-5}\] \[\implies 7=\pm \sqrt{6K-5}\] \[\implies 49=6K-5\] \[\implies K=9\] If $\frac{x}{y}=-2$ then \[-2=\frac{-1\pm \sqrt{6K-5}}{6}\] \[\implies -12=-1\pm \sqrt{6K-5}\] \[\implies -11=\sqrt{6K-5}\] \[\implies 121=6K-5\] \[\implies 126=6K\] \[\implies K=21\] Hence our final answer is $21\cdot 9=\boxed{189}$ -vsamc $\newline$ -minor edit:einsteinstudent

Answer:  189
Problem: Let $N$ be the number of complex numbers $z$ with the properties that $|z|=1$ and $z^{6!}-z^{5!}$ is a real number. Find the remainder when $N$ is divided by $1000$

Solution: Let $a=z^{120}$ . This simplifies the problem constraint to $a^6-a \in \mathbb{R}$ . This is true if $\text{Im}(a^6)=\text{Im}(a)$ . Let $\theta$ be the angle $a$ makes with the positive x-axis. Note that there is exactly one $a$ for each angle $0\le\theta<2\pi$ . We are given $\sin\theta = \sin{6\theta}$ . Note that $\sin \theta = \sin (\pi - \theta)$ and $\sin \theta = \sin (\theta + 2\pi)$ . We can use these facts to create two types of solutions:
\[\sin \theta = \sin ((2m + 1)\pi - \theta)\]
which implies that $(2m+1)\pi-\theta = 6\theta$ and reduces to $\frac{(2m + 1)\pi}{7} = \theta$ . There are 7 solutions for this.
\[\sin \theta = \sin (2n\pi + \theta)\]
which implies that $2n\pi+\theta=6\theta$ and reduces to $\frac{2n\pi}{5} = \theta$ . There are 5 solutions for this, totaling 12 values of $a$
For each of these solutions for $a$ , there are necessarily $120$ solutions for $z$ . Thus, there are $12\cdot 120=1440$ solutions for $z$ , yielding an answer of $\boxed{440}$

Answer:  440
Problem: Let $N$ be the number of complex numbers $z$ with the properties that $|z|=1$ and $z^{6!}-z^{5!}$ is a real number. Find the remainder when $N$ is divided by $1000$

Solution: The constraint mentioned in the problem is equivalent to the requirement that the imaginary part is equal to $0$ . Since $|z|=1$ , let $z=\cos \theta + i\sin \theta$ , then we can write the imaginary part of $\Im(z^{6!}-z^{5!})=\Im(z^{720}-z^{120})=\sin\left(720\theta\right)-\sin\left(120\theta\right)=0$ . Using the sum-to-product formula, we get $\sin\left(720\theta\right)-\sin\left(120\theta\right)=2\cos\left(\frac{720\theta+120\theta}{2}\right)\sin\left(\frac{720\theta-120\theta}{2}\right)=2\cos\left(\frac{840\theta}{2}\right)\sin\left(\frac{600\theta}{2}\right)\implies \cos\left(\frac{840\theta}{2}\right)=0$ or $\sin\left(\frac{600\theta}{2}\right)=0$ . The former yields $840$ solutions, and the latter yields $600$ solutions, giving a total of $840+600=1440$ solution, so our answer is $\boxed{440}$

Answer:  440
Problem: Let $N$ be the number of complex numbers $z$ with the properties that $|z|=1$ and $z^{6!}-z^{5!}$ is a real number. Find the remainder when $N$ is divided by $1000$

Solution: As mentioned in solution one, for the difference of two complex numbers to be real, their imaginary parts must be equal. We use exponential form of complex numbers. Let $z = e^{i \theta}$ . We have two cases to consider. Either $z^{6!} = z^{5!}$ , or $z^{6!}$ and $z^{5!}$ are reflections across the imaginary axis.
If $z^{6!} = z^{5!}$ , then $e^{6! \theta i} = e^{5! \theta i}$ . Thus, $720 \theta = 120 \theta$ or $600\theta = 0$ , giving us 600 solutions. (Equalities are taken modulo $2 \pi$ )
For the second case, $e^{6! \theta i} = e^{(\pi - 5!\theta)i}$ . This means $840 \theta = \pi$ , giving us 840 solutions.
Our total count is thus $1440$ , yielding a final answer of $\boxed{440}$

Answer:  440
Problem: Let $N$ be the number of complex numbers $z$ with the properties that $|z|=1$ and $z^{6!}-z^{5!}$ is a real number. Find the remainder when $N$ is divided by $1000$

Solution: Because $|z| = 1,$ we know that $z\overline{z} = 1^2 = 1.$ Hence $\overline{z} = \frac 1 {z}.$ Because $z^{6!}-z^{5!}$ is real, it is equal to its complex conjugate. Hence $z^{6!}-z^{5!} = \overline{z^{6!}}-\overline{z^{5!}}.$ Substituting the expression we that we derived earlier, we get $z^{720}-z^{120} = \frac 1{z^{720}} - \frac 1{z^{120}}.$ This leaves us with a polynomial whose leading term is $z^{1440}.$ Hence our answer is $\boxed{440}$

Answer:  440
Problem: A right hexagonal prism has height $2$ . The bases are regular hexagons with side length $1$ . Any $3$ of the $12$ vertices determine a triangle. Find the number of these triangles that are isosceles (including equilateral triangles).

Solution: We can consider two cases: when the three vertices are on one base, and when the vertices are on two bases.
Case 1: vertices are on one base. Then we can call one of the vertices $A$ for distinction. Either the triangle can have sides $1, 1, \sqrt{3}$ with 6 cases or $\sqrt{3}, \sqrt{3}, \sqrt{3}$ with 2 cases. This can be repeated on the other base for $16$ cases.
Case 2: The vertices span two bases. WLOG call the only vertex on one of the bases $X$ . Call the closest vertex on the other base $B$ , and label clockwise $C, D, E, F, G$ . We will multiply the following scenarios by $12$ , because the top vertex can have $6$ positions and the top vertex can be on the other base. We can have $XCG, XDF$ , but we are not done! Don't forget that the problem statement implies that the longest diagonal in a base is $2$ and the height is $2$ , so $XBE$ is also correct! Those are the only three cases, so there are $12*3=36$ cases for this case.
In total there's $\boxed{052}$ cases.

Answer:  052
Problem: A right hexagonal prism has height $2$ . The bases are regular hexagons with side length $1$ . Any $3$ of the $12$ vertices determine a triangle. Find the number of these triangles that are isosceles (including equilateral triangles).

Solution: If there are two edges on a single diameter, there would be six diameters. There are four ways to put the third number, and four equilateral triangles. There are $4+ 2 \cdot 2\cdot 6 = 28$ ways. Then, if one length was $\sqrt{3}$ but no side on the diameter, there would be twelve was to put the $\sqrt{3}$ side, and two ways to put the other point. $2 \cdot 12 = 24$ for four ways to put the third point. Adding the number up, the final answer is $24+28 = \boxed{052}.$

Answer:  052
Problem: A right hexagonal prism has height $2$ . The bases are regular hexagons with side length $1$ . Any $3$ of the $12$ vertices determine a triangle. Find the number of these triangles that are isosceles (including equilateral triangles).

Solution: To start, let's find the distances from any vertex (call it A, doesn't matter since the prism is symmetrical) to all the other 11 vertices. Using Pythagorean relations, we find that the distances are $1$ $\sqrt{3}$ $2$ $\sqrt{3}$ $1$ $\sqrt{7}$ $\sqrt{5}$ $2$ $\sqrt{5}$ $\sqrt{7}$ , and $\sqrt{8}$
We can clearly form an isosceles triangle using any vertex and any two neighboring edges that are equal. There are 12 total vertices, all of which are symmetrical to the vertex A. Thus, for each vertex, we can form 5 isosceles triangles (with neighboring edges $1$ $1$ $2$ $2$ $\sqrt{3}$ $\sqrt{3}$ $\sqrt{5}$ $\sqrt{5}$ , and $\sqrt{7}$ $\sqrt{7}$ ) , so have so far $12 \times 5 = 60$ isosceles triangles.
To check for overcounting, note that isosceles triangles can be split into two major categories: equilateral isosceles, and non-equilateral isosceles. 
We only counted non-equilateral isosceles triangles once since there is only one vertex whose two neighboring edges are equal. 
But equilateral triangles were counted three times (namely once for each vertex).
By observation, there are only 4 equilateral triangles (1-1-1 side lengths), two on each hexagonal face. Since we counted each two more times than we should of (three instead of once), we will subtract ( $4$ eq. triangles) $\times$ $2$ overcounts per eq. triangle) = $8$ overcounts.
Finally, we have $60 - 8$ overcounts = $\boxed{052}$ isoceles triangles.

Answer:  052
Problem: A right hexagonal prism has height $2$ . The bases are regular hexagons with side length $1$ . Any $3$ of the $12$ vertices determine a triangle. Find the number of these triangles that are isosceles (including equilateral triangles).

Solution: Start by drawing the hexagonal prism. Start from one of the points, let's call it $A$ . From simple inspection, we can see that each point we choose creates $4$ isosceles triangles; $A$ and the two adjacent points, $A$ and the two points adjacent to those, $A$ and the adjacent point on the other face, and $A$ and the two points adjacent to those. Here, we have $12\cdot4\Rightarrow48$ isosceles triangles.
Next, notice that you can create an isosceles triangle with the bases of the hexagonal prism. Let's say that our base, starting from the very left-most point of the hexagon, has points $A, B, C, D, E, F$ , rotating counter-clockwise. We can create a triangle with either points $\bigtriangleup{BCE}$ or $\bigtriangleup{BCF}$ . WLOG, assuming we use $\bigtriangleup{BCE}$ , we can use the Law of Cosines and find that $\overline{CE}$ $\sqrt{3}$ . Thus, this would mean $\overline{CF}$ is $2$ , which is the exactly the same length as the height of the hexagonal prism. There are $4$ possibilities of this case, so our final solution is $48+4=\boxed{052}$

Answer:  052
Problem: Find the number of four-element subsets of $\{1,2,3,4,\dots, 20\}$ with the property that two distinct elements of a subset have a sum of $16$ , and two distinct elements of a subset have a sum of $24$ . For example, $\{3,5,13,19\}$ and $\{6,10,20,18\}$ are two such subsets.

Solution: This problem is tricky because it is the capital of a few "bashy" calculations. Nevertheless, the process is straightforward. Call the set $\{a, b, c, d\}$
Note that there are only two cases: 1 where $a + b = 16$ and $c + d = 24$ or 2 where $a + b = 16$ and $a + c = 24$ . Also note that there is no overlap between the two situations! This is because if they overlapped, adding the two equations of both cases and canceling out gives you $a=d$ , which cannot be true.
Case 1.
This is probably the simplest: just make a list of possible combinations for $\{a, b\}$ and $\{c, d\}$ . We get $\{1, 15\}\dots\{7, 9\}$ for the first and $\{4, 20\}\dots\{11, 13\}$ for the second. That appears to give us $7*8=56$ solutions, right? NO. Because elements can't repeat, take out the supposed sets \[\{1, 15, 9, 15\}, \{2, 14, 10, 14\}, \{3, 13, 11, 13\}, \{4, 12, 4, 20\}, \{5, 11, 5, 19\},\] \[\{5, 11, 11, 13\}, \{6, 10, 6, 18\}, \{6, 10, 10, 14\}, \{7, 9, 9, 15\}, \{7, 9, 7, 17\}\] That's ten cases gone. So $46$ for Case 1.
Case 2.
We can look for solutions by listing possible $a$ values and filling in the blanks. Start with $a=4$ , as that is the minimum. We find $\{4, 12, 20, ?\}$ , and likewise up to $a=15$ . But we can't have $a=8$ or $a=12$ because $a=b$ or $a=c$ , respectively! Now, it would seem like there are $10$ values for $a$ and $17$ unique values for each $?$ , giving a total of $170$ , but that is once again not true because there are some repeated values!
There are two cases of overcounting:
case 1) (5,11,13,19) & (5.11.19.13)
The same is for (6,10,14,18) and (7,9,15,17)
case 2) those that have the same b and c values
this case includes:
(1,15,9,7) and (7,9,15,1)
(2,14,10,6) and (6,10,14,2)
(3,13,11,5) and (5,11,13,3)
So we need to subtract 6 overcounts.
So, that's $164$ for Case 2.
Total gives $\boxed{210}$

Answer:  210
Problem: Find the number of four-element subsets of $\{1,2,3,4,\dots, 20\}$ with the property that two distinct elements of a subset have a sum of $16$ , and two distinct elements of a subset have a sum of $24$ . For example, $\{3,5,13,19\}$ and $\{6,10,20,18\}$ are two such subsets.

Solution: Let's say our four elements in our subset are $a,b,c,d$ . We have two cases. Note that the order of the elements / the element letters themselves don't matter since they are all on equal grounds at the start.
$\textrm{Case } 1 \textrm{:}$ $a+b = 16$ and $c+d = 24$
List out possibilities for $a+b$ $(\text{i.e. } 1+15, 2+14, 3+13 \text{ etc.})$ but don't list $8+8$ because those are the same elements and that is restricted.
Then list out the possibilities for $c+d \text{ }(\text{i.e. } 4+20, 5+19, 6+18, \text{ etc.})$ but don't list $12+12$ because they are the same elements.
This will give you $7 \cdot 8$ elements, which is $56$ . However, as stated above, we have overlap. Just count starting from $a+b$ $15,14,13,11,10,9,7,6,5,4$ all overlap once, which is $10$ , thus $56 - 10 = 46$ cases in this case. Note that $12$ wasn't included because again, if $c+d = 24$ $c$ and $d$ cannot be $12$
$\textrm{Case } 2 \textrm{:}$ $a+b = 16$ and $b+c = 24$
Here, $b$ is included in both equations. We can easily see that $a, b, c$ will never equal each other.
Furthermore, there are 17 choices for $d$ $20 - 3$ included elements) for each $b$ . Listing out the possible $b$ s, we go from $15,14,13,11,10,9,7,6,5,4$ . Do not include $8$ or $12$ because if they are included, then $a/c$ will be the same as $b$ , which is restricted.
There are $10$ options there, and thus $10 \cdot 17 = 170$ . But, note that if $d = b+8$ $a+d = a+b+8 = 24$ , and so we have a double-counted set. Starting with $b=15$ , we have $15, 14, 13, 11, 10, 9$ (where $d$ is $7, 6, 5, 3, 2, 1)$ . That means there are $6$ double-counted cases. Thus $170 - 6 = 164$ cases in this case.
Adding these up, we get $46+164 = \boxed{210}.$

Answer:  210
Problem: The wheel shown below consists of two circles and five spokes, with a label at each point where a spoke meets a circle. A bug walks along the wheel, starting at point $A$ . At every step of the process, the bug walks from one labeled point to an adjacent labeled point. Along the inner circle the bug only walks in a counterclockwise direction, and along the outer circle the bug only walks in a clockwise direction. For example, the bug could travel along the path $AJABCHCHIJA$ , which has $10$ steps. Let $n$ be the number of paths with $15$ steps that begin and end at point $A$ . Find the remainder when $n$ is divided by $1000.$

Solution: We divide this up into casework.  The "directions" the bug can go are $\text{Clockwise}$ $\text{Counter-Clockwise}$ , and $\text{Switching}$ .  Let an $I$ signal going clockwise (because it has to be in the inner circle), an $O$ signal going counter-clockwise, and an $S$ switching between inner and outer circles.  An example string of length fifteen that gets the bug back to $A$ would be $ISSIIISOOSISSII$ .
For the bug to end up back at $A$ , the difference between the number of $I$ 's and $O$ 's must be a multiple of $5$
So, the total number of ways is $1+2002+1001=3004$ which gives $\boxed{004}$ as the answer.

Answer:  004
Problem: The wheel shown below consists of two circles and five spokes, with a label at each point where a spoke meets a circle. A bug walks along the wheel, starting at point $A$ . At every step of the process, the bug walks from one labeled point to an adjacent labeled point. Along the inner circle the bug only walks in a counterclockwise direction, and along the outer circle the bug only walks in a clockwise direction. For example, the bug could travel along the path $AJABCHCHIJA$ , which has $10$ steps. Let $n$ be the number of paths with $15$ steps that begin and end at point $A$ . Find the remainder when $n$ is divided by $1000.$

Solution: Define $A_n$ to be the number of sequences of length $n$ that ends at $A$ and similarly for the other spokes. Also let \[S_n=A_n+B_n+C_n+D_n+E_n+F_n+G_n+H_n+I_n+J_n\] Apparently everytime the bug has $2$ choices for its next move, thus we have $S_n=2^n$ . Now we attempt to find a recursive formula for $A_n$ \begin{align*} A_n&=J_{n-1}+E_{n-1} \\ &=(I_{n-2}+A_{n-2})+(D_{n-2}+F_{n-2}) \\ &=(H_{n-3}+B_{n-3})+(J_{n-3}+E_{n-3})+(G_{n-3}+C_{n-3})+(J_{n-3}+E_{n-3}) \\ &=(B_{n-3}+C_{n-3}+E_{n-3}+G_{n-3}+H_{n-3}+J_{n-3})+(J_{n-3}+E_{n-3}) \\ &=(S_{n-3}-(I_{n-3}+A_{n-3}+D_{n-3}+F_{n-3}))+A_{n-2} \\ &=2^{n-3}-(J_{n-2}+E_{n-2})+A_{n-2} \\ &=2^{n-3}-A_{n-1}+A_{n-2} \\ \end{align*} Computing a few easy terms we have $A_0=0$ $A_1=0$ $A_2=1$ $A_3=0$ $A_4=3$ . Continuing the process yields $A_{15}=3\boxed{004}$

Answer:  004
Problem: Find the least positive integer $n$ such that when $3^n$ is written in base $143$ , its two right-most digits in base $143$ are $01$

Solution: Note that the given condition is equivalent to $3^n \equiv 1 \pmod{143^2}$ and $143=11\cdot 13$ . Because $\gcd(11^2, 13^2) = 1$ , the desired condition is equivalent to $3^n \equiv 1 \pmod{121}$ and $3^n \equiv 1 \pmod{169}$
If $3^n \equiv 1 \pmod{121}$ , one can see the sequence $1, 3, 9, 27, 81, 1, 3, 9...$ so $5|n$
Now if $3^n \equiv 1 \pmod{169}$ , it is harder. But we do observe that $3^3 \equiv 1 \pmod{13}$ , therefore $3^3 = 13a + 1$ for some integer $a$ . So our goal is to find the first number $p_1$ such that $(13a+1)^ {p_1} \equiv 1 \pmod{169}$ . Then, $p_1 \equiv 0 \pmod{13},$ which follows from the binomial theorem. It is not difficult to see that the smallest $p_1=13$ , so ultimately $3^{39} \equiv 1 \pmod{169}$ . Therefore, $39|n$
The first $n$ satisfying both criteria is thus $5\cdot 39=\boxed{195}$

Answer:  195
Problem: Find the least positive integer $n$ such that when $3^n$ is written in base $143$ , its two right-most digits in base $143$ are $01$

Solution: Note that Euler's Totient Theorem would not necessarily lead to the smallest $n$ and that in this case that $n$ is greater than $1000$
We wish to find the least $n$ such that $3^n \equiv 1 \pmod{143^2}$ . This factors as $143^2=11^{2}*13^{2}$ . Because $\gcd(121, 169) = 1$ , we can simply find the least $n$ such that $3^n \equiv 1 \pmod{121}$ and $3^n \equiv 1 \pmod{169}$
Quick inspection yields $3^5 \equiv 1 \pmod{121}$ and $3^3 \equiv 1 \pmod{13}$ . Now we must find the smallest $k$ such that $3^{3k} \equiv 1 \pmod{169}$ . Euler's gives $3^{156} \equiv 1 \pmod{169}$ . So $3k$ is a factor of $156$ . This gives $k=1,2, 4, 13, 26, 52$ . Some more inspection yields $k=13$ is the smallest valid $k$ . So $3^5 \equiv 1 \pmod{121}$ and $3^{39} \equiv 1 \pmod{169}$ . The least $n$ satisfying both is $lcm(5, 39)=\boxed{195}$ . (RegularHexagon)

Answer:  195
Problem: Find the least positive integer $n$ such that when $3^n$ is written in base $143$ , its two right-most digits in base $143$ are $01$

Solution: Listing out the powers of $3$ , modulo $169$ and modulo $121$ , we have: \[\begin{array}{c|c|c} n & 3^n\mod{169} & 3^n\mod{121}\\ \hline 0 & 1 & 1\\ 1 & 3 & 3\\ 2 & 9 & 9\\ 3 & 27 & 27\\ 4 & 81 & 81\\ 5 & 74 & 1\\ 6 & 53\\ 7 & 159\\ 8 & 139\\ 9 & 79\\ 10 & 68\\ 11 & 35\\ 12 & 105\\ 13 & 146\\ 14 & 100\\ 15 & 131\\ 16 & 55\\ 17 & 165\\ 18 & 157\\ 19 & 133\\ 20 & 61\\ 21 & 14\\ 22 & 42\\ 23 & 126\\ 24 & 40\\ 25 & 120\\ 26 & 22\\ 27 & 66\\ 28 & 29\\ 29 & 87\\ 30 & 92\\ 31 & 107\\ 32 & 152\\ 33 & 118\\ 34 & 16\\ 35 & 48\\ 36 & 144\\ 37 & 94\\ 38 & 113\\ 39 & 1\\ \end{array}\]
The powers of $3$ repeat in cycles of $5$ and $39$ in modulo $121$ and modulo $169$ , respectively. The answer is $\text{lcm}(5, 39) = \boxed{195}$

Answer:  195
Problem: Find the least positive integer $n$ such that when $3^n$ is written in base $143$ , its two right-most digits in base $143$ are $01$

Solution: We have that \[3^n \equiv 1 \pmod{143^2}.\] Now, $3^{110} \equiv 1 \pmod{11^2}$ so by the Fundamental Theorem of Orders, $\text{ord}_{11^2}(3)|110$ and with some bashing, we get that it is $5$ . Similarly, we get that $\text{ord}_{13^2}(3)=39$ . Now, $\text{lcm}(39,5)=\boxed{195}$ which is our desired solution.

Answer:  195
Problem: Find the least positive integer $n$ such that when $3^n$ is written in base $143$ , its two right-most digits in base $143$ are $01$

Solution: We wish to find the smallest $n$ such that $3^n\equiv 1\pmod{143^2}$ , so we want $n\equiv 1\pmod{121}$ and $n\equiv 1\pmod{169}$ . Note that $3^5\equiv 1\pmod{121}$ ,  so $3^n$ repeats $121$ with a period of $5$ , so $5|n$ . Now, in order for $n\equiv 1\pmod{169}$ , then $n\equiv 1\pmod{13}$ . Because $3^3\equiv 1\pmod{13}$ $3^n$ repeats with a period of $3$ , so $3|n$ . 
Hence, we have that for some positive integer $p$ $3^n\equiv (3^3)^p\equiv (26+1)^p\equiv \binom{p}{0}26^p+\binom{p}{1}26^{p-1}....+\binom{p}{p-2}26^2+\binom{p}{p-1}26+\binom{p}{p}\equiv 26p+1\equiv 1\pmod{169}$ , so $26p\equiv 0\pmod{169}$ and $p\equiv 0\pmod{13}$ . Thus, we have that $5|n$ $3|n$ , and $13|n$ , so the smallest possible value of $n$ is $3\times5\times13=\boxed{195}$ .
-Stormersyle

Answer:  195