Navvye/Olympiad-Problems-CSV huggingface dataset:
Problem: Let $\overline{MN}$ be a diameter of a circle with diameter 1. Let $A$ and $B$ be points on one of the semicircular arcs determined by $\overline{MN}$ such that $A$ is the midpoint of the semicircle and $MB=\frac{3}5$ . Point $C$ lies on the other semicircular arc. Let $d$ be the length of the line segment whose endpoints are the intersections of diameter $\overline{MN}$ with chords $\overline{AC}$ and $\overline{BC}$ . The largest possible value of $d$ can be written in the form $r-s\sqrt{t}$ , where $r, s$ and $t$ are positive integers and $t$ is not divisible by the square of any prime. Find $r+s+t$

Solution: By Pythagoras in $\triangle BMN,$ we get $BN=\dfrac{4}{5}.$
Since cross ratios are preserved upon projecting, note that $(M,Y;X,N)\stackrel{C}{=}(M,B;A,N).$ By definition of a cross ratio, this becomes \[\dfrac{XM}{NY}:\dfrac{NM}{NY}=\dfrac{AM}{AB}:\dfrac{MN}{NB}.\] Let $MY=a,YX=b,XN=c$ such that $a+b+c=1.$ We know that $XM=a+b,XY=b,NM=1,NY=b+c,$ so the LHS becomes $\dfrac{(a+b)(b+c)}{b}.$
In the RHS, we are given every value except for $AB.$ However, Ptolemy's Theorem on $MBAN$ gives $AB\cdot MN+AN\cdot BM=AM\cdot BN\implies AB+\dfrac{3}{5\sqrt{2}}=\dfrac{4}{5\sqrt{2}}\implies AB=\dfrac{1}{5\sqrt{2}}.$ Substituting, we get $\dfrac{(a+b)(b+c)}{b}=4\implies b(a+b+c)+ac=4b, b=\dfrac{ac}{3}$ where we use $a+b+c=1.$
Again using $a+b+c=1,$ we have $a+b+c=1\implies a+\dfrac{ac}{3}+c=1\implies a=3\dfrac{1-c}{c+3}.$ Then $b=\dfrac{ac}{3}=\dfrac{c-c^2}{c+3}.$ Since this is a function in $c,$ we differentiate WRT $c$ to find its maximum. By quotient rule, it suffices to solve \[(-2c+1)(c+3)-(c-c^2)=0 \implies c^2+6c-3,c=-3+2\sqrt{3}.\] Substituting back yields $b=7-4\sqrt{3},$ so $7+4+3=\boxed{014}$ is the answer.

Answer:  014
Problem: Of the students attending a school party, $60\%$ of the students are girls, and $40\%$ of the students like to dance. After these students are joined by $20$ more boy students, all of whom like to dance, the party is now $58\%$ girls. How many students now at the party like to dance?

Solution: Say that there were $3k$ girls and $2k$ boys at the party originally. $2k$ like to dance. Then, there are $3k$ girls and $2k + 20$ boys, and $2k + 20$ like to dance.
Thus, $\dfrac{3k}{5k + 20} = \dfrac{29}{50}$ , solving gives $k = 116$ . Thus, the number of people that like to dance is $2k + 20 = \boxed{252}$

Answer:  252
Problem: Of the students attending a school party, $60\%$ of the students are girls, and $40\%$ of the students like to dance. After these students are joined by $20$ more boy students, all of whom like to dance, the party is now $58\%$ girls. How many students now at the party like to dance?

Solution: Let the number of girls be $g$ . Let the number of total people originally be $t$
We know that $\frac{g}{t}=\frac{3}{5}$ from the problem.
We also know that $\frac{g}{t+20}=\frac{29}{50}$ from the problem.
We now have a system and we can solve.
The first equation becomes:
$3t=5g$
The second equation becomes:
$50g=29t+580$
Now we can sub in $30t=50g$ by multiplying the first equation by $10$ . We can plug this into our second equation.
$30t=29t+580$
$t=580$
We know that there were originally $580$ people. Of those, $\frac{2}{5}*580=232$ like to dance.
We also know that with these people, $20$ boys joined, all of whom like to dance. We just simply need to add $20$ to get $232+20=\boxed{252}$

Answer:  252
Problem: Of the students attending a school party, $60\%$ of the students are girls, and $40\%$ of the students like to dance. After these students are joined by $20$ more boy students, all of whom like to dance, the party is now $58\%$ girls. How many students now at the party like to dance?

Solution: Let $p$ denote the total number of people at the party. Then, because we know the proportions of boys to $p$ both before and after 20 boys arrived, we can create the following equation: \[0.4p+20 = 0.42(p+20)\] Solving for p gives us $p=580$ , so the solution is $0.4p+20 = \boxed{252}$

Answer:  252
Problem: Square $AIME$ has sides of length $10$ units.  Isosceles triangle $GEM$ has base $EM$ , and the area common to triangle $GEM$ and square $AIME$ is $80$ square units.  Find the length of the altitude to $EM$ in $\triangle GEM$

Solution: Note that if the altitude of the triangle is at most $10$ , then the maximum area of the intersection of the triangle and the square is $5\cdot10=50$ .
This implies that vertex G must be located outside of square $AIME$
Let $GE$ meet $AI$ at $X$ and let $GM$ meet $AI$ at $Y$ . Clearly, $XY=6$ since the area of trapezoid $XYME$ is $80$ . Also, $\triangle GXY \sim \triangle GEM$
Let the height of $GXY$ be $h$ . By the similarity, $\dfrac{h}{6} = \dfrac{h + 10}{10}$ , we get $h = 15$ . Thus, the height of $GEM$ is $h + 10 = \boxed{025}$

Answer:  025
Problem: Ed and Sue bike at equal and constant rates.  Similarly, they jog at equal and constant rates, and they swim at equal and constant rates.  Ed covers $74$ kilometers after biking for $2$ hours, jogging for $3$ hours, and swimming for $4$ hours, while Sue covers $91$ kilometers after jogging for $2$ hours, swimming for $3$ hours, and biking for $4$ hours.  Their biking, jogging, and swimming rates are all whole numbers of kilometers per hour.  Find the sum of the squares of Ed's biking, jogging, and swimming rates.

Solution: Let the biking rate be $b$ , swimming rate be $s$ , jogging rate be $j$ , all in km/h.
We have $2b + 3j + 4s = 74,2j + 3s + 4b = 91$ . Subtracting the second from twice the first gives $4j + 5s = 57$ . Mod 4, we need $s\equiv1\pmod{4}$ . Thus, $(j,s) = (13,1),(8,5),(3,9)$
$(13,1)$ and $(3,9)$ give non-integral $b$ , but $(8,5)$ gives $b = 15$ . Thus, our answer is $15^{2} + 8^{2} + 5^{2} = \boxed{314}$

Answer:  314
Problem: Ed and Sue bike at equal and constant rates.  Similarly, they jog at equal and constant rates, and they swim at equal and constant rates.  Ed covers $74$ kilometers after biking for $2$ hours, jogging for $3$ hours, and swimming for $4$ hours, while Sue covers $91$ kilometers after jogging for $2$ hours, swimming for $3$ hours, and biking for $4$ hours.  Their biking, jogging, and swimming rates are all whole numbers of kilometers per hour.  Find the sum of the squares of Ed's biking, jogging, and swimming rates.

Solution: Let $b$ $j$ , and $s$ be the biking, jogging, and swimming rates of the two people.  Hence, $2b + 3j + 4s = 74$ and $4b + 2j + 3s = 91$ .  Subtracting gives us that $2b - j - s = 17$ .  Adding three times this to the first equation gives that $8b + s = 125\implies b\le 15$ .  Adding four times the previous equation to the first given one gives us that $10b - j = 142\implies b > 14\implies b\ge 15$ .  This gives us that $b = 15$ , and then $j = 8$ and $s = 5$ .  Therefore, $b^2 + s^2 + j^2 = 225 + 64 + 25 = \boxed{314}$

Answer:  314
Problem: Ed and Sue bike at equal and constant rates.  Similarly, they jog at equal and constant rates, and they swim at equal and constant rates.  Ed covers $74$ kilometers after biking for $2$ hours, jogging for $3$ hours, and swimming for $4$ hours, while Sue covers $91$ kilometers after jogging for $2$ hours, swimming for $3$ hours, and biking for $4$ hours.  Their biking, jogging, and swimming rates are all whole numbers of kilometers per hour.  Find the sum of the squares of Ed's biking, jogging, and swimming rates.

Solution: Creating two systems, we get $2x+3y+4z=74$ , and $2y+3z+4x=91$ . Subtracting the two expressions we get $y+z-2x=-17$ . Note that $-17$ is odd, so one of $x,y,z$ is odd. We see from our second expression that $z$ must be odd, because $91$ is also odd and $2y$ and $4x$ are odd. Thus, with this information, we can test cases quickly:
When readdressing the first equation, we see that if $2x+3y$ will be a multiple of $6$ $4z \equiv 2 \pmod{6} = 5$ , we get that $x=15$ and $y=8$ , which works because of integer values. Therefore, $225+64+25=\boxed{314}$

Answer:  314
Problem: Ed and Sue bike at equal and constant rates.  Similarly, they jog at equal and constant rates, and they swim at equal and constant rates.  Ed covers $74$ kilometers after biking for $2$ hours, jogging for $3$ hours, and swimming for $4$ hours, while Sue covers $91$ kilometers after jogging for $2$ hours, swimming for $3$ hours, and biking for $4$ hours.  Their biking, jogging, and swimming rates are all whole numbers of kilometers per hour.  Find the sum of the squares of Ed's biking, jogging, and swimming rates.

Solution: Building on top of Solution 3, we can add $j+s-2b=17$ and $2b+3j+4s=74$ (sorry, I used different variables) to get $4j+5s=57$ . Logically speaking, most athletic people swim a lot faster than 1 km/h (0.62 mph), so we test out the next case that works, which is 5 km/h (3.1 mph). This seems much more logical, so we plug it into the equation to get $4j+25=57\implies j=8$ . This seems reasonable, as people usually jog at around 8 to 16 km/h. Plugging these values into $2b+3j+4s=74$ , we get $b=15$ . 15 km/h is a little slow (most people bike at 20 km/h), but is still reasonable. So, we get $5^2+8^2+15^2=\boxed{314}$

Answer:  314
Problem: There exist unique positive integers $x$ and $y$ that satisfy the equation $x^2 + 84x + 2008 = y^2$ . Find $x + y$

Solution: Completing the square $y^2 = x^2 + 84x + 2008 = (x+42)^2 + 244$ . Thus $244 = y^2 - (x+42)^2 = (y - x - 42)(y + x + 42)$ by difference of squares
Since $244$ is even, one of the factors is even. A parity check shows that if one of them is even, then both must be even. Since $244 = 2^2 \cdot 61$ , the factors must be $2$ and $122$ . Since $x,y > 0$ , we have $y - x - 42 = 2$ and $y + x + 42 = 122$ ; the latter equation implies that $x + y = \boxed{080}$

Answer:  080
Problem: There exist unique positive integers $x$ and $y$ that satisfy the equation $x^2 + 84x + 2008 = y^2$ . Find $x + y$

Solution: We complete the square like in the first solution: $y^2 = (x+42)^2 + 244$ . Since consecutive squares differ by the consecutive odd numbers, we note that $y$ and $x+42$ must differ by an even number. We can use casework with the even numbers, starting with $y-(x+42)=2$ \begin{align*}2(x+42)+1+2(x+42)+3&=244\\ \Leftrightarrow x&=18\end{align*}
Thus, $y=62$ and the answer is $\boxed{080}$

Answer:  080
Problem: There exist unique positive integers $x$ and $y$ that satisfy the equation $x^2 + 84x + 2008 = y^2$ . Find $x + y$

Solution: We see that $y^2 \equiv x^2 + 4 \pmod{6}$ . By quadratic residues , we find that either $x \equiv 0, 3 \pmod{6}$ . Also, $y^2 \equiv (x+42)^2 + 244 \equiv (x+2)^2 \pmod{4}$ , so $x \equiv 0, 2 \mod{4}$ . Combining, we see that $x \equiv 0 \mod{6}$
Testing $x = 6$ and other multiples of $6$ , we quickly find that $x = 18, y = 62$ is the solution. $18+62=\boxed{080}$

Answer:  080
Problem: There exist unique positive integers $x$ and $y$ that satisfy the equation $x^2 + 84x + 2008 = y^2$ . Find $x + y$

Solution: We solve for x: $x^2 + 84x + 2008-y^2 = 0$
$x=\dfrac{-84+\sqrt{84^2-4\cdot 2008+4y^2}}{2}=-42+\sqrt{y^2-244}$
So $y^2-244$ is a perfect square. Since 244 is even, the difference $\sqrt{y^2-244} -y^2$ is even, so we try $y^2-244=(y-2)^2$ $-244=-4y+4$ $y=62$
Plugging into our equation, we find that $x=18$ , and $(x,y)=(18,62)$ indeed satisfies the original equation. $x+y=\boxed{080}$

Answer:  080
Problem: There exist unique positive integers $x$ and $y$ that satisfy the equation $x^2 + 84x + 2008 = y^2$ . Find $x + y$

Solution: Let $y=x+d$ for some $d>0$ , substitute into the original equation to get $84x + 2008 = 2xd + d^2$
All terms except for the last one are even, hence $d^2$ must be even, hence let $d=2e$ . We obtain $21x + 502 = xe + e^2$ . Rearrange to $502-e^2 = x(e-21)$
Obviously for $0<e<21$ the right hand side is negative and the left hand side is positive. Hence $e\geq 21$ . Let $e=21+f$ , then $f\geq 0$
We have $502 - (21+f)^2 = xf$ . Left hand side simplifies to $61 - 42f + f^2$ . As $x$ must be an integer, $f$ must divide the left hand side. But $61$ is a prime, which only leaves two options: $f=1$ and $f=61$
Option $f=61$ gives us a negative $x$ . Option $f=1$ gives us $x=61/f - 42 + f = 18$ , and $y = x + d= x + 2e = x + 2(21+f) = 18 + 44 = 62$ , hence $x+y=\boxed{080}$

Answer:  080
Problem: There exist unique positive integers $x$ and $y$ that satisfy the equation $x^2 + 84x + 2008 = y^2$ . Find $x + y$

Solution: First complete the square to get $y^2 = (x+42)^2 + 244$ . Remember that squares are the sums of consecutive odd integers, so when the difference between the two squares is 244, the two squares must be an even number of odd integers apart. However, there is only one distinct solution, as the problem states, and very quickly you will realize that only two odd integers work. When there are four, then the numbers are not odd, and when it is any other even integer it does not divide. So we need two consecutive odd integers that sum to 244. Easily we find 121 and 123. 121 is the 61st odd integer and 123 is the 62nd odd integer, so $(x+42)^2$ is the sum of the first 60 odd integers, or $(60)^2$ , while $y^2$ is $62^2$ for the same reasons. That way we get $x=12$ $y=62$ , hence $x+y=\boxed{080}$

Answer:  080
Problem: A right circular cone has base radius $r$ and height $h$ .  The cone lies on its side on a flat table.  As the cone rolls on the surface of the table without slipping, the point where the cone's base meets the table traces a circular arc centered  at the point where the vertex touches the table.  The cone first returns to its original position on the table after making $17$ complete rotations.  The value of $h/r$ can be written in the form $m\sqrt {n}$ , where $m$ and $n$ are positive integers and $n$ is not divisible by the square of any prime.  Find $m + n$

Solution: The path is a circle with radius equal to the slant height of the cone, which is $\sqrt {r^{2} + h^{2}}$ . Thus, the length of the path is $2\pi\sqrt {r^{2} + h^{2}}$
Also, the length of the path is 17 times the circumference of the base, which is $34r\pi$ . Setting these equal gives $\sqrt {r^{2} + h^{2}} = 17r$ , or $h^{2} = 288r^{2}$ . Thus, $\dfrac{h^{2}}{r^{2}} = 288$ , and $\dfrac{h}{r} = 12\sqrt {2}$ , giving an answer of $12 + 2 = \boxed{014}$

Answer:  014
Problem: triangular array of numbers has a first row consisting of the odd integers $1,3,5,\ldots,99$ in increasing order.  Each row below the first has one fewer entry than the row above it, and the bottom row has a single entry.  Each entry in any row after the top row equals the sum of the two entries diagonally above it in the row immediately above it.  How many entries in the array are multiples of $67$

Solution: Let the $k$ th number in the $n$ th row be $a(n,k)$ . Writing out some numbers, we find that $a(n,k) = 2^{n-1}(n+2k-2)$
We wish to find all $(n,k)$ such that $67| a(n,k) = 2^{n-1} (n+2k-2)$ . Since $2^{n-1}$ and $67$ are relatively prime , it follows that $67|n+2k-2$ . Since every row has one less element than the previous row, $1 \le k \le 51-n$ (the first row has $50$ elements, the second $49$ , and so forth; so $k$ can range from $1$ to $50$ in the first row, and so forth). Hence
it follows that $67| n - 2k + 2$ implies that $n-2k+2 = 67$ itself.
Now, note that we need $n$ to be odd, and also that $n+2k-2 = 67 \le 100-n \Longrightarrow n \le 33$
We can check that all rows with odd $n$ satisfying $1 \le n \le 33$ indeed contains one entry that is a multiple of $67$ , and so the answer is $\frac{33+1}{2} = \boxed{017}$

Answer:  017
Problem: triangular array of numbers has a first row consisting of the odd integers $1,3,5,\ldots,99$ in increasing order.  Each row below the first has one fewer entry than the row above it, and the bottom row has a single entry.  Each entry in any row after the top row equals the sum of the two entries diagonally above it in the row immediately above it.  How many entries in the array are multiples of $67$

Solution: The result above is fairly intuitive if we write out several rows and then divide all numbers in row $r$ by $2^{r-1}$ (we can do this because dividing by a power of 2 doesn't affect divisibility by $67$ ). The second row will be $2, 4, 6, \cdots , 98$ , the third row will be $3, 5, \cdots, 97$ , and so forth. Clearly, only the odd-numbered rows can have a term divisible by $67$ . However, with each row the row will have one less element, and the $99-67+1 = 33$ rd row is the last time $67$ will appear. Therefore the number of multiples of 67 in the entire array is $\frac{33+1}{2} = \boxed{017}$

Answer:  017
Problem: Let $S_i$ be the set of all integers $n$ such that $100i\leq n < 100(i + 1)$ .  For example, $S_4$ is the set ${400,401,402,\ldots,499}$ .  How many of the sets $S_0, S_1, S_2, \ldots, S_{999}$ do not contain a perfect square?

Solution: The difference between consecutive squares is $(x + 1)^2 - x^2 = 2x + 1$ , which means that all squares above $50^2 = 2500$ are more than $100$ apart.
Then the first $26$ sets ( $S_0,\cdots S_{25}$ ) each have at least one perfect square because the differences between consecutive squares in them are all less than $100$ . Also, since $316$ is the largest $x$ such that $x^2 < 100000$ $100000$ is the upper bound which all numbers in $S_{999}$ must be less than), there are $316 - 50 = 266$ other sets after $S_{25}$ that have a perfect square.
There are $1000 - 266 - 26 = \boxed{708}$ sets without a perfect square.

Answer:  708
Problem: Find the positive integer $n$ such that
\[\arctan\frac {1}{3} + \arctan\frac {1}{4} + \arctan\frac {1}{5} + \arctan\frac {1}{n} = \frac {\pi}{4}.\]

Solution: Since we are dealing with acute angles, $\tan(\arctan{a}) = a$
Note that $\tan(\arctan{a} + \arctan{b}) = \dfrac{a + b}{1 - ab}$ , by tangent addition. Thus, $\arctan{a} + \arctan{b} = \arctan{\dfrac{a + b}{1 - ab}}$
Applying this to the first two terms, we get $\arctan{\dfrac{1}{3}} + \arctan{\dfrac{1}{4}} = \arctan{\dfrac{7}{11}}$
Now, $\arctan{\dfrac{7}{11}} + \arctan{\dfrac{1}{5}} = \arctan{\dfrac{23}{24}}$
We now have $\arctan{\dfrac{23}{24}} + \arctan{\dfrac{1}{n}} = \dfrac{\pi}{4} = \arctan{1}$ . Thus, $\dfrac{\dfrac{23}{24} + \dfrac{1}{n}}{1 - \dfrac{23}{24n}} = 1$ ; and simplifying, $23n + 24 = 24n - 23 \Longrightarrow n = \boxed{47}$

Answer:  47
Problem: Find the positive integer $n$ such that
\[\arctan\frac {1}{3} + \arctan\frac {1}{4} + \arctan\frac {1}{5} + \arctan\frac {1}{n} = \frac {\pi}{4}.\]

Solution: Adding a series of angles is the same as multiplying the complex numbers whose arguments they are. In general, $\arctan\frac{1}{n}$ , is the argument of $n+i$ . The sum of these angles is then just the argument of the product \[(3+i)(4+i)(5+i)(n+i)\] and expansion give us $(48n-46)+(48+46n)i$ . Since the argument of this complex number is $\frac{\pi}{4}$ , its real and imaginary parts must be equal; then, we can we set them equal to get \[48n - 46 = 48 + 46n.\] Therefore, $n=\boxed{47}$

Answer:  47
Problem: Ten identical crates each of dimensions $3\mathrm{ft}\times 4\mathrm{ft}\times 6\mathrm{ft}$ .  The first crate is placed flat on the floor.  Each of the remaining nine crates is placed, in turn, flat on top of the previous crate, and the orientation of each crate is chosen at random.  Let $\frac {m}{n}$ be the probability that the stack of crates is exactly $41\mathrm{ft}$ tall, where $m$ and $n$ are relatively prime positive integers.  Find $m$

Solution: Only the heights matter, and each crate is either 3, 4, or 6 feet tall with equal probability. We have the following:
\begin{align*}3a + 4b + 6c &= 41\\ a + b + c &= 10\end{align*}
Subtracting 3 times the second from the first gives $b + 3c = 11$ , or $(b,c) = (2,3),(5,2),(8,1),(11,0)$ . The last doesn't work, obviously. This gives the three solutions $(a,b,c) = (5,2,3),(3,5,2),(1,8,1)$ . In terms of choosing which goes where, the first two solutions are analogous.
For $(5,2,3),(3,5,2)$ , we see that there are $2\cdot\dfrac{10!}{5!2!3!} = 10\cdot9\cdot8\cdot7$ ways to stack the crates. For $(1,8,1)$ , there are $\dfrac{10!}{8!1!1!} = 90$ . Also, there are $3^{10}$ total ways to stack the crates to any height.
Thus, our probability is $\dfrac{10\cdot9\cdot8\cdot7 + 90}{3^{10}} = \dfrac{10\cdot8\cdot7 + 10}{3^{8}} = \dfrac{570}{3^8} = \dfrac{190}{3^{7}}$ . Our answer is the numerator, $\boxed{190}$

Answer:  190
Problem: Ten identical crates each of dimensions $3\mathrm{ft}\times 4\mathrm{ft}\times 6\mathrm{ft}$ .  The first crate is placed flat on the floor.  Each of the remaining nine crates is placed, in turn, flat on top of the previous crate, and the orientation of each crate is chosen at random.  Let $\frac {m}{n}$ be the probability that the stack of crates is exactly $41\mathrm{ft}$ tall, where $m$ and $n$ are relatively prime positive integers.  Find $m$

Solution: It would be helpful for this solution to be reformatted. To start with, let us observe the three numbers. Note that $3$ and $6$ are both divisible by $3$ , so the number of $4$ -crates must be congruent to $41\bmod{3}$ , which is also congruent to $2\bmod{3}$ . Our solutions for the number of $4$ -crates will repeat mod $3$ , so if $x$ is a solution, so is $x+3$ . By inspection, we have that $2$ is solution, and so are $5$ and $8$ . Each solution splits into its own case.We must solve the equation $41-4z=6x+3y$ , simultaneously with $x+y=10-z$ . Note that we already know the possible values of $z$ . Solving these (it's AIME $9$ , you should be able to do this and if anyone feels like they want to write a rundown of this please go ahead), we get the solution  sets $\{8,1,1\},\{5,2,3\},$ and $\{2,3,5\}$ . We can count the number of possible arrangements for each solution by taking $\dbinom{10}{z}$ and then multiplying by $\dbinom{10-z}{x}$ (the solution sets, for the sake of consistency, are in the form $z,x,y$ ). Summing the results for all the solutions gives us $5130$ . Finally, to calculate the probability we must determine our denominator. Since we have $3$ ways to arrange each block, our denominator is $3^{10}$ $\frac{5130}{3^{10}}=\frac{190}{3^7}$ . The answer is $m=\boxed{190}$

Answer:  190
Problem: Ten identical crates each of dimensions $3\mathrm{ft}\times 4\mathrm{ft}\times 6\mathrm{ft}$ .  The first crate is placed flat on the floor.  Each of the remaining nine crates is placed, in turn, flat on top of the previous crate, and the orientation of each crate is chosen at random.  Let $\frac {m}{n}$ be the probability that the stack of crates is exactly $41\mathrm{ft}$ tall, where $m$ and $n$ are relatively prime positive integers.  Find $m$

Solution: Let's make two observations. We are trying to find the number of ways we can add $3\text{s}, 4\text{s}$ , and $6\text{s}$ to get $41$ , and the total number of (non-distinct) sums possible is $3^{10}$ .
Then we just use casework to easily and directly solve for the number of ways to get $41$ . To begin, the minimum sum is produced with $10$ threes, so WLOG we can solve for the number of ways to get $11$ with $0\text{s}, 1\text{s}$ , and $3\text{s}$
Case I: $0$ zeroes, $0$ threes, $11$ ones
Impossible, because there are only ten available spots.
Case II: $1$ zero, $1$ three, $8$ ones
This is just $\frac{10!}{8!}$ , so there are $90$ possibilities.
Case III: $3$ zeroes, $2$ threes, $5$ ones
This is just $\frac{10!}{3!2!5!}$ . This gives $2520$ possibilities.
Case IV: 5 zeroes, 3 threes, and 2 ones.
This is the same as case $3$ , so also $2520$ possibilities.
$90+2520+2520=5130$
$5130$ has three powers of $3$ , so $5130$ divided by $27$ is $\boxed{190}$

Answer:  190