Navvye/Olympiad-Problems-CSV huggingface dataset:
Problem: In a circle parallel chords of lengths 2, 3, and 4 determine central angles of $\alpha$ $\beta$ , and $\alpha + \beta$ radians , respectively, where $\alpha + \beta < \pi$ . If $\cos \alpha$ , which is a positive rational number , is expressed as a fraction in lowest terms, what is the sum of its numerator and denominator?

Solution: It’s easy to see in triangle which lengths 2, 3, and 4, that the angle opposite the side 2 is $\frac{\alpha}{2}$ , and using the Law of Cosines , we get: \[2^2 = 3^2 + 4^2 - 2\cdot3\cdot4\cos\frac{\alpha}{2}\] Which, rearranges to: \[21 = 24\cos\frac{\alpha}{2}\] And, that gets us: \[\cos\frac{\alpha}{2} = 7/8\] Using $\cos 2\theta = 2\cos^2 \theta - 1$ , we get that: \[\cos\alpha = 17/32\] Which gives an answer of $\boxed{049}$

Answer:  049
Problem: In a circle parallel chords of lengths 2, 3, and 4 determine central angles of $\alpha$ $\beta$ , and $\alpha + \beta$ radians , respectively, where $\alpha + \beta < \pi$ . If $\cos \alpha$ , which is a positive rational number , is expressed as a fraction in lowest terms, what is the sum of its numerator and denominator?

Solution: Using the first diagram above, \[\sin \frac{\alpha}{2} = \frac{1}{r}\] \[\sin \frac{\beta}{2} = \frac{1.5}{r}\] \[\sin(\frac{\alpha}{2}+\frac{\beta}{2})=\frac{2}{r}\] by the Pythagorean trig identities, \[\cos\frac{\alpha}{2}=\sqrt{1-\frac{1}{r^2}}\] \[\cos\frac{\beta}{2}=\sqrt{1-\frac{2.25}{r^2}}\] so by the composite sine identity \[\frac{2}{r}=\frac{1}{r}\sqrt{1-\frac{2.25}{r^2}}+\frac{1.5}{r}\sqrt{1-\frac{1}{r^2}}\] multiply both sides by $2r$ , then subtract $\sqrt{4-\frac{9}{r^2}}$ from both sides
squaring both sides, we get \[16 - 8\sqrt{4-\frac{9}{r^2}} + 4 - \frac{9}{r^2}=9 - \frac{9}{r^2}\] \[\Longrightarrow 16+4=9+8\sqrt{4-\frac{9}{r^2}}\Longrightarrow\frac{11}{8}=\sqrt{4-\frac{9}{r^2}}\Longrightarrow\frac{121}{64}=4-\frac{9}{r^2}\] \[\Longrightarrow\frac{(256-121)r^2}{64}=9\Longrightarrow r^2= \frac{64}{15}\] plugging this back in, \[\cos^2(\frac{\alpha}{2})=1-\frac{15}{64}=\frac{49}{64}\] so \[\cos(\alpha)=2(\frac{49}{64})-1=\frac{34}{64}=\frac{17}{32}\] and the answer is $17+32=\boxed{049}$

Answer:  049
Problem: How many of the first 1000 positive integers can be expressed in the form
$\lfloor 2x \rfloor + \lfloor 4x \rfloor + \lfloor 6x \rfloor + \lfloor 8x \rfloor$
where $x$ is a real number , and $\lfloor z \rfloor$ denotes the greatest integer less than or equal to $z$

Solution: Noting that all of the numbers are even, we can reduce this to any real number $x$ between $0$ to $\frac 12$ , as this will be equivalent to $\frac n2$ to $\frac {n+1}2$ for any integer $n$ (same reasoning as above). So now we only need to test every 10 numbers; and our answer will be 100 times the number of integers we can reach between 1 and 10.
We can now approach this by directly searching for the integers (this solution) or brute forcing all of the cases (next solution):
We can match up the greatest integer functions with one of the partitions of the integer. If we let $x = \frac 12$ then we get the solution $10$ ; now consider when $x < \frac 12$ $\lfloor 2x \rfloor = 0$ $\lfloor 4x \rfloor \le 1$ $\lfloor 6x \rfloor \le 2$ $\lfloor 8x \rfloor \le 3$ . But according to this the maximum we can get is $1+2+3 = 6$ , so we only need to try the first 6 numbers.
Out of these 6 cases, only 3 fails. So between 1 and 10 we can reach only the integers $1,2,4,5,6,10$ ; hence our solution is $6 \cdot 100 = \boxed{600}$

Answer:  600
Problem: How many of the first 1000 positive integers can be expressed in the form
$\lfloor 2x \rfloor + \lfloor 4x \rfloor + \lfloor 6x \rfloor + \lfloor 8x \rfloor$
where $x$ is a real number , and $\lfloor z \rfloor$ denotes the greatest integer less than or equal to $z$

Solution: As we change the value of $x$ , the value of our expression changes only when $x$ crosses rational number of the form $\frac{m}{n}$ , where $n$ is divisible by 2, 4, 6 or 8.  Thus, we need only see what happens at the numbers of the form $\frac{m}{\textrm{lcm}(2, 4, 6, 8)} = \frac{m}{24}$ .  This gives us 24 calculations to make; we summarize the results here:
$\frac{1}{24}, \frac{2}{24} \to 0$
$\frac{3}{24} \to 1$
$\frac{4}{24}, \frac{5}{24} \to 2$
$\frac{6}{24}, \frac{7}{24} \to 4$
$\frac{8}{24} \to 5$
$\frac{9}{24}, \frac{10}{24}, \frac{11}{24} \to 6$
$\frac{12}{24}, \frac{13}{24}, \frac{14}{24} \to 10$
$\frac{15}{24} \to 11$
$\frac{16}{24},\frac{17}{24} \to 12$
$\frac{18}{24}, \frac{19}{24} \to 14$
$\frac{20}{24}\to 15$
$\frac{21}{24}, \frac{22}{24}, \frac{23}{24} \to16$
$\frac{24}{24} \to 20$
Thus, we hit 12 of the first 20 integers and so we hit $50 \cdot 12 = \boxed{600}$ of the first $1000$

Answer:  600
Problem: How many of the first 1000 positive integers can be expressed in the form
$\lfloor 2x \rfloor + \lfloor 4x \rfloor + \lfloor 6x \rfloor + \lfloor 8x \rfloor$
where $x$ is a real number , and $\lfloor z \rfloor$ denotes the greatest integer less than or equal to $z$

Solution: Because $2,4,6,8$ are all multiples of $2$ , we can speed things up. We only need to check up to $\frac{12}{24}$ , and the rest should repeat. As shown before, we hit 6 integers ( $1,2,4,5,6,10$ ) from $\frac{1}{24}$ to $\frac{12}{24}$ . Similarly, this should repeat 100 times, for $\boxed{600}$

Answer:  600
Problem: How many of the first 1000 positive integers can be expressed in the form
$\lfloor 2x \rfloor + \lfloor 4x \rfloor + \lfloor 6x \rfloor + \lfloor 8x \rfloor$
where $x$ is a real number , and $\lfloor z \rfloor$ denotes the greatest integer less than or equal to $z$

Solution: We only need to check the numbers where it increments, namely $\frac{1}{8}, \frac{1}{6}, \frac{1}{4}, \frac{1}{3}, \frac{3}{8}, \frac{1}{2}$ . As shown before, we hit 6 integers ( $1,2,4,5,6,10$ ) from $\frac{1}{24}$ to $\frac{1}{2}$ . Similarly, this should repeat 100 times, for $\boxed{600}$

Answer:  600
Problem: How many of the first 1000 positive integers can be expressed in the form
$\lfloor 2x \rfloor + \lfloor 4x \rfloor + \lfloor 6x \rfloor + \lfloor 8x \rfloor$
where $x$ is a real number , and $\lfloor z \rfloor$ denotes the greatest integer less than or equal to $z$

Solution: Recall from Hermite's Identity that $\sum_{k = 0}^{n - 1}\left\lfloor x + \frac kn\right\rfloor = \lfloor nx\rfloor$ . Then we can rewrite $\lfloor 2x \rfloor + \lfloor 4x \rfloor + \lfloor 6x \rfloor + \lfloor 8x \rfloor = 4\lfloor x\rfloor + \left\lfloor x + \frac18\right\rfloor + \left\lfloor x + \frac16\right\rfloor + 2\left\lfloor x + \frac14\right\rfloor + \left\lfloor x + \frac13\right\rfloor$ $+ \left\lfloor x + \frac38\right\rfloor + 4\left\lfloor x + \frac12\right\rfloor + \left\lfloor x + \frac58\right\rfloor + \left\lfloor x + \frac23\right\rfloor + 2\left\lfloor x + \frac34\right\rfloor + \left\lfloor x + \frac56\right\rfloor + \left\lfloor x + \frac78\right\rfloor$ . There are $12$ terms here (we don't actually have to write all of it out; we can just see where there will be duplicates and subtract accordingly from $20$ ). Starting from every integer $x$ , we can keep adding to achieve one higher value for each of these terms, but after raising the last term, we will have raised the whole sum by $20$ while only achieving $12$ of those $20$ values. We can conveniently shift the $1000$ (since it can be achieved) to the position of the $0$ so that there are only complete cycles of $20$ , and the answer is $\frac {12}{20}\cdot1000 = \boxed{600}$

Answer:  600
Problem: How many of the first 1000 positive integers can be expressed in the form
$\lfloor 2x \rfloor + \lfloor 4x \rfloor + \lfloor 6x \rfloor + \lfloor 8x \rfloor$
where $x$ is a real number , and $\lfloor z \rfloor$ denotes the greatest integer less than or equal to $z$

Solution: Let $x=\lfloor x\rfloor+\{x\}$ then \begin{align*} \lfloor 2x\rfloor+\lfloor 4x\rfloor+\lfloor 6x\rfloor+\lfloor 8x\rfloor&=\lfloor 2(\lfloor x\rfloor+\{x\})\rfloor+\lfloor 4(\lfloor x\rfloor+\{x\})\rfloor+\lfloor 6(\lfloor x\rfloor+\{x\})\rfloor+\lfloor 8(\lfloor x\rfloor+\{x\})\rfloor\\ &=2\lfloor x\rfloor+4\lfloor x\rfloor+6\lfloor x\rfloor+8\lfloor x\rfloor+\lfloor 2\{x\}\rfloor+\lfloor 4\{x\}\rfloor+\lfloor 6\{x\}\rfloor+\lfloor 8\{x\}\rfloor\\ &=20\lfloor x\rfloor+(\lfloor 2\{x\}\rfloor+\lfloor 4\{x\}\rfloor+\lfloor 6\{x\}\rfloor+\lfloor 8\{x\}\rfloor) \end{align*} Similar to the previous solutions, the value of $\lfloor 2\{x\}\rfloor+\lfloor 4\{x\}\rfloor+\lfloor 6\{x\}\rfloor+\lfloor 8\{x\}\rfloor$ changes when $\{x\}=\frac{m}{n}$ , where $m\in\{1,2,3,...,n-1\}$ $n\in\{2,4,6,8\}$ . Using Euler's Totient Function \[\sum\limits_{k=0}^4 \phi(2k)\] to obtain $12$ different values for $\{x\}=\frac{m}{n}$ . (note that here Euler's Totient Function counts the number of $\{x\}=\frac{m}{n}$ where $m$ $n$ are relatively prime so that the values of $\{x\}$ won't overlap.).
Thus if $k$ can be expressed as $\lfloor 2x\rfloor+\lfloor 4x\rfloor+\lfloor 6x\rfloor+\lfloor 8x\rfloor$ , then $k=20a+b$ for some non-negative integers $a$ $b$ , where there are $12$ values for $b$
Exclusively, there are $49$ values for $a$ in the range $0<k<1000$ , or $49\cdot12=588$ ordered pairs $(a,b)$
If $a=0$ $b\neq0$ , which includes $11$ ordered pairs.
If $a=50$ $b=0$ , which includes $1$ ordered pair.
In total, there are $588+11+1=\boxed{600}$ values for $k$

Answer:  600
Problem: How many of the first 1000 positive integers can be expressed in the form
$\lfloor 2x \rfloor + \lfloor 4x \rfloor + \lfloor 6x \rfloor + \lfloor 8x \rfloor$
where $x$ is a real number , and $\lfloor z \rfloor$ denotes the greatest integer less than or equal to $z$

Solution: To simplify the question, let $y = 2x$ . Then, the expression in the question becomes $\lfloor y \rfloor + \lfloor 2y \rfloor + \lfloor 3y \rfloor + \lfloor 4y \rfloor$
Let $\{x\}$ represent the non-integer part of $x$ (For example, $\{2.8\} = 0.8$ ). Then,
\begin{align*} \lfloor y \rfloor + \lfloor 2y \rfloor + \lfloor 3y \rfloor + \lfloor 4y \rfloor &= y - \{y\} + 2y - \{2y\} + 3y - \{3y\} + 4y - \{4y\} \\ &= 10y - (\{y\} + \{2y\} + \{3y\} + \{4y\}) \\ &= 10(\lfloor y \rfloor + \{y\}) - (\{y\} + \{2y\} + \{3y\} + \{4y\}) \\ &= 10\lfloor y \rfloor + 10\{y\} - (\{y\} + \{2y\} + \{3y\} + \{4y\}) \\ &= 10\lfloor y \rfloor + 9\{y\} - (\{2y\} + \{3y\} + \{4y\}) \\ \end{align*}
Since $\lfloor y \rfloor$ is always an integer, $10\lfloor y \rfloor$ will be a multiple of 10. Thus, we look for the range of the other part of the expression. We will be able to reach the same numbers when $y$ ranges from $0$ to $1$ , because the curly brackets ( $\{\}$ ) gets rid of any integer part. Let the combined integer part of $2y$ $3y$ , and $4y$ be $k$ (In other words, $k = \lfloor 2y \rfloor + \lfloor 3y \rfloor + \lfloor 4y \rfloor$ ). Then,
\begin{align*} 9\{y\} - (\{2y\} + \{3y\} + \{4y\}) &= 9\{y\} - (2\{y\} + 3\{y\} + 4\{y\} - k) \\ &= 9\{y\} - (9\{y\} - k) \\ &= k \end{align*}
The maximum value of $k$ will be when $y$ is slightly less than $1$ , which means $k = 1 + 2 + 3 = 6$ . As $y$ increases from $0$ to $1$ $k$ will increase whenever $2y$ $3y$ , or $4y$ is an integer, which happens when $y$ hits one of the numbers in the set $\left\{\dfrac14, \dfrac13, \dfrac12, \dfrac23, \dfrac34 \right\}$ . When $y$ reaches $\dfrac12$ , both $2y$ and $4y$ will be an integer, so $k$ will increase by $2$ . For all the other numbers in the set, $k$ increases by $1$ since only $1$ number in the set will be an integer. Thus, the possible values of $k$ are $\{0, 1, 2, 4, 5, 6\}$
Finally, looking back at the original expression, we plug in $k$ to get a multiple of $10$ plus any number in $\{0, 1, 2, 4, 5, 6\}$ . Thus, we hit all numbers ending in $\{0, 1, 2, 4, 5, 6\}$ , of which there are $\boxed{600}$

Answer:  600
Problem: An ellipse has foci at $(9,20)$ and $(49,55)$ in the $xy$ -plane and is tangent to the $x$ -axis. What is the length of its major axis

Solution: An ellipse is defined to be the locus of points $P$ such that the sum of the distances between $P$ and the two foci is constant.  Let $F_1 = (9, 20)$ $F_2 = (49, 55)$ and $X = (x, 0)$ be the point of tangency of the ellipse with the $x$ -axis.  Then $X$ must be the point on the axis such that the sum $F_1X + F_2X$ is minimal.  (The last claim begs justification:  Let $F'_2$ be the reflection of $F_2$ across the $x$ -axis. Let $Y$ be where the line through $F_1$ and $F’_2$ intersects the ellipse.  We will show that $X=Y$ . Note that $X F_2 = X F’_2$ since $X$ is on the $x$ -axis.  Also, since the entire ellipse is on or above the $x$ -axis and the line through $F_2$ and $F’_2$ is perpendicular to the $x$ -axis, we must have $F_2 Y \leq F’_2 Y$ with equality if and only if $Y$ is on the $x$ -axis.  Now, we have \[F_1 X + F'_2 X = F_1 X + F_2 X = F_1 Y + F_2 Y \leq F_1 Y + F’_2 Y\] But the right most sum is the straight-line distance from $F_1$ to $F’_2$ and the left is the distance of some path from $F_1$ to $F_2$ ., so this is only possible if we have equality and thus $X = Y$ ). Finding the optimal location for $X$ is a classic problem: for any path from $F_1$ to $X$ and then back to $F_2$ , we can reflect (as above) the second leg of this path (from $X$ to $F_2$ ) across the $x$ -axis.  Then our path connects $F_1$ to the reflection $F_2'$ of $F_2$ via some point on the $x$ -axis, and this path will have shortest length exactly when our original path has shortest length.  This occurs exactly when we have a straight-line path, and by the above argument, this path passes through the point of tangency of the ellipse with the $x-$ axis.
The sum of the two distances $F_1 X$ and $F_2X$ is therefore equal to the length of the segment $F_1F_2'$ , which by the distance formula is just $d = \sqrt{(9 - 49)^2 + (20 + 55)^2} = \sqrt{40^2 + 75^2} = 5\sqrt{8^2 + 15^2} = 5\cdot 17 = 85$
Finally, let $A$ and $B$ be the two endpoints of the major axis of the ellipse.  Then by symmetry $AF_1 = F_2B$ so $AB = AF_1 + F_1B = F_2B + F_1B = d$ (because $B$ is on the ellipse), so the answer is $\boxed{085}$

Answer:  085
Problem: An ellipse has foci at $(9,20)$ and $(49,55)$ in the $xy$ -plane and is tangent to the $x$ -axis. What is the length of its major axis

Solution: An ellipse is defined as the set of points where the sum of the distances from the foci to the point is fixed. The length of major axis is equal to the sum of these distances $(2a)$ . Thus if we find the sum of the distances, we get the answer. Let k be this fixed sum; then we get, by the distance formula:
$k = \sqrt{(x - 9)^2 + 20^2} + \sqrt{(x - 49)^2 + 55^2}$
This is the equation of the ellipse expressed in terms of $x$ . The line tangent to the ellipse at the given point $P(x, 0)$ will thus have slope $0$ . Taking the derivative gives us the slope of this line. To simplify, let $f(x) = (x - 9)^2 + 20^2$ and $g(x) = (x - 49)^2 + 55^2$ . Then we get:
$0 = \frac{f^\prime(x)}{2\sqrt{f(x)}} + \frac{g^\prime(x)}{2\sqrt{g(x)}}$
Next, we multiply by the conjugate to remove square roots. We next move the resulting $a^2 - b^2$ form expression into form $a^2 = b^2$
$\frac{(f^\prime(x))^2}{4\cdot f(x)} = \frac{(g^\prime(x))^2}{4\cdot g(x)}$
We know $f^\prime(x) = 2x - 18$ and $g^\prime(x) = 2x - 98$ . Simplifying yields:
$\frac{(x - 9)^2}{(x - 9)^2 + 20^2} = \frac{(x - 49)^2}{(x - 49)^2 + 55^2}$
To further simplify, let $a = (x - 9)^2$ and $b = (x - 49)^2$ . This means $\frac{a}{a + 400} = \frac{b}{b + 3025}$ . Solving yields that $16b = 121a$ .
Substituting back $a$ and $b$ yields:
$16 \cdot (x - 49)^2 = 121 \cdot (x - 9)^2$
Solving for $x$ yields $x = \frac{59}{3}$ . Substituting back into our original distance formula, solving for $k$ yields $\boxed{085}$

Answer:  085
Problem: Let $A$ $B$ $C$ and $D$ be the vertices of a regular tetrahedron, each of whose edges measures $1$ meter. A bug, starting from vertex $A$ , observes the following rule: at each vertex it chooses one of the three edges meeting at that vertex, each edge being equally likely to be chosen, and crawls along that edge to the vertex at its opposite end. Let $p = \frac{n}{729}$ be the probability that the bug is at vertex $A$ when it has crawled exactly $7$ meters. Find the value of $n$

Solution: We evaluate $P(7)$ recursively: \begin{alignat*}{6} P(0)&=1, \\ P(1)&=\frac13(1-P(0))&&=0, \\ P(2)&=\frac13(1-P(1))&&=\frac13, \\ P(3)&=\frac13(1-P(2))&&=\frac29, \\ P(4)&=\frac13(1-P(3))&&=\frac{7}{27}, \\ P(5)&=\frac13(1-P(4))&&=\frac{20}{81}, \\ P(6)&=\frac13(1-P(5))&&=\frac{61}{243},\\ P(7)&=\frac13(1-P(6))&&=\frac{182}{729}. \end{alignat*} Therefore, the answer is $n=\boxed{182}.$

Answer:  182
Problem: Let $A$ $B$ $C$ and $D$ be the vertices of a regular tetrahedron, each of whose edges measures $1$ meter. A bug, starting from vertex $A$ , observes the following rule: at each vertex it chooses one of the three edges meeting at that vertex, each edge being equally likely to be chosen, and crawls along that edge to the vertex at its opposite end. Let $p = \frac{n}{729}$ be the probability that the bug is at vertex $A$ when it has crawled exactly $7$ meters. Find the value of $n$

Solution: Let $P(k)=Q(k)+c$ for some function $Q(k)$ and constant $c.$ For all $k\geq1,$ the recursive formula for $P(k)$ becomes \[Q(k)+c=\frac13(1-(Q(k-1)+c))=\frac13-\frac13Q(k-1)-\frac13c.\] Solving for $Q(k),$ we get \[Q(k)=\frac13-\frac13Q(k-1)-\frac43c.\] For simplicity purposes, we set $c=\frac14,$ which gives \[Q(k)=-\frac13Q(k-1).\] Clearly, $Q(0),Q(1),Q(2),\ldots$ is a geometric sequence with the common ratio $\frac{Q(k)}{Q(k-1)}=-\frac13.$ Substituting $P(0)=1$ and $c=\frac14$ into $P(0)=Q(0)+c$ produces $Q(0)=\frac34,$ the first term of the geometric sequence.
For all nonnegative integers $k,$ the explicit formula for $Q(k)$ is \[Q(k)=\frac34\left(-\frac13\right)^k,\] and the explicit formula for $P(k)$ is \[P(k)=\frac34\left(-\frac13\right)^k+\frac14.\] Finally, the requested probability is $p=P(7)=\frac{182}{729},$ from which $n=\boxed{182}.$

Answer:  182
Problem: Let $A$ $B$ $C$ and $D$ be the vertices of a regular tetrahedron, each of whose edges measures $1$ meter. A bug, starting from vertex $A$ , observes the following rule: at each vertex it chooses one of the three edges meeting at that vertex, each edge being equally likely to be chosen, and crawls along that edge to the vertex at its opposite end. Let $p = \frac{n}{729}$ be the probability that the bug is at vertex $A$ when it has crawled exactly $7$ meters. Find the value of $n$

Solution: Denominator
There are $3^7$ ways for the bug to make $7$ independent crawls without restrictions.
Numerator
Let $V_k$ denote the number of ways for the bug to crawl exactly $k$ meters starting from vertex $V$ and ending at vertex $A,$ where $V\in\{A,B,C,D\}$ and $k$ is a positive integer. We wish to find $A_7.$
Since the bug must crawl to vertex $B,C,$ or $D$ on the first move, we have \begin{align*} A_7&=B_6+C_6+D_6 \\ &=(A_5+C_5+D_5)+(A_5+B_5+D_5)+(A_5+B_5+C_5) \\ &=A_5+2(A_5+B_5+C_5+D_5) \\ &=A_5+2S_5, \end{align*} where $S_k=A_k+B_k+C_k+D_k.$
More generally, we get \[A_{k+2}=A_k+2S_k. \qquad\qquad (\spadesuit)\] For $S_k,$ note that
Clearly, $S_1,S_2,S_3,\ldots$ is a geometric sequence with the first term $S_1=3$ and the common ratio $\frac{S_{k+1}}{S_k}=3.$ Therefore, its explicit formula is \[S_k=3^k. \qquad\qquad (\clubsuit)\] Recall that $A_1=0.$ By $(\spadesuit)$ and $(\clubsuit),$ we rewrite $A_7$ recursively: \begin{align*} A_7 &= A_5+2S_5 \\ &=\left(A_3+2S_3\right)+2S_5 \\ &=\left(\left(A_1+2S_1\right)+2S_3\right)+2S_5 \\ &=2S_1+2S_3+2S_5 \\ &=2(3)+2\left(3^3\right)+2\left(3^5\right). \end{align*} Answer
The requested probability is \[p=\frac{A_7}{3^7}=\frac{2(3)+2\left(3^3\right)+2\left(3^5\right)}{3^7}=\frac{2(1)+2\left(3^2\right)+2\left(3^4\right)}{3^6}=\frac{182}{729},\] from which $n=\boxed{182}.$

Answer:  182
Problem: Let $A$ $B$ $C$ and $D$ be the vertices of a regular tetrahedron, each of whose edges measures $1$ meter. A bug, starting from vertex $A$ , observes the following rule: at each vertex it chooses one of the three edges meeting at that vertex, each edge being equally likely to be chosen, and crawls along that edge to the vertex at its opposite end. Let $p = \frac{n}{729}$ be the probability that the bug is at vertex $A$ when it has crawled exactly $7$ meters. Find the value of $n$

Solution: Define notation $V_k$ as Solution 2 does.
In fact, we can generalize the following relationships for all nonnegative integers $k:$ \begin{align*} A_0&=1, \\ B_0&=0, \\ C_0&=0, \\ D_0&=0, \\ A_{k+1}&=B_k+C_k+D_k, \\ B_{k+1}&=A_k+C_k+D_k, \\ C_{k+1}&=A_k+B_k+D_k, \\ D_{k+1}&=A_k+B_k+C_k. \\ \end{align*} Using these equations, we recursively fill out the table below: \[\begin{array}{c||c|c|c|c|c|c|c|c}    \hspace{7mm}&\hspace{6mm}&\hspace{6mm}&\hspace{6mm}&\hspace{6mm}&\hspace{6mm}&\hspace{6mm}&& \\ [-2.5ex] \boldsymbol{k} & \boldsymbol{0} & \boldsymbol{1} & \boldsymbol{2} & \boldsymbol{3} & \boldsymbol{4} & \boldsymbol{5} & \boldsymbol{6} & \boldsymbol{7} \\  \hline \hline &&&&&&&& \\ [-2.25ex] \boldsymbol{A_k} &1&0&3&6&21&60&183&546 \\ \hline   &&&&&&&& \\ [-2.25ex] \boldsymbol{B_k} &0&1&2&7&20&61&182&547 \\ \hline  &&&&&&&& \\ [-2.25ex] \boldsymbol{C_k} &0&1&2&7&20&61&182&547 \\ \hline    &&&&&&&& \\ [-2.25ex] \boldsymbol{D_k} &0&1&2&7&20&61&182&547 \\ \end{array}\] Note that the paths from $V$ to $A$ and the paths from $A$ to $V$ have one-to-one correspondence. So, we must get \[A_k+B_k+C_k+D_k=3^k\] for all values of $k.$
The requested probability is \[p=\frac{A_7}{3^7}=\frac{546}{2187}=\frac{182}{729},\] from which $n=\boxed{182}.$

Answer:  182
Problem: Let $A$ $B$ $C$ and $D$ be the vertices of a regular tetrahedron, each of whose edges measures $1$ meter. A bug, starting from vertex $A$ , observes the following rule: at each vertex it chooses one of the three edges meeting at that vertex, each edge being equally likely to be chosen, and crawls along that edge to the vertex at its opposite end. Let $p = \frac{n}{729}$ be the probability that the bug is at vertex $A$ when it has crawled exactly $7$ meters. Find the value of $n$

Solution: Let $a_n$ denotes the number of ways that the bug arrives at $A$ after crawling $n$ meters, then we have $a_1=0$
Notice that there is respectively $1$ way to arrive at $A$ for each of the different routes after the previous $n-1$ crawls, excluding the possibility that the bug ends up at $A$ after the $(n-1)$ th crawl (as it will be forced to move somewhere else.). Thus, we get the recurrence relation \[a_n=3^{n-1}-a_{n-1}.\] Quick calculations yield \begin{align*} a_7&=3^6-a_6\\ &=3^6-\left(3^5-3^4+3^3-3^2+3-a_1\right)\\ &=546. \end{align*} Thus, $p=\frac{546}{3^7}=\frac{182}{729}\Longrightarrow n=\boxed{182}$

Answer:  182