Navvye/Olympiad-Problems-CSV huggingface dataset:
Problem: A beam of light strikes $\overline{BC}\,$ at point $C\,$ with angle of incidence $\alpha=19.94^\circ\,$ and reflects with an equal angle of reflection as shown.  The light beam continues its path, reflecting off line segments $\overline{AB}\,$ and $\overline{BC}\,$ according to the rule: angle of incidence equals angle of reflection.  Given that $\beta=\alpha/10=1.994^\circ\,$ and $AB=BC,\,$ determine the number of times the light beam will bounce off the two line segments.  Include the first reflection at $C\,$ in your count.
AIME 1994 Problem 14.png

Solution: At each point of reflection, we pretend instead that the light continues to travel straight. [asy] pathpen = linewidth(0.7); size(250);  real alpha = 28, beta = 36;  pair B = MP("B",(0,0),NW), C = MP("C",D((1,0))), A = MP("A",expi(alpha * pi/180),N); path r = C + .4 * expi(beta * pi/180) -- C - 2*expi(beta * pi/180);  D(A--B--(1.5,0));D(r);D(anglemark(C,B,D(A)));D(anglemark((1.5,0),C,C+.4*expi(beta*pi/180)));MP("\beta",B,(5,1.2),fontsize(9));MP("\alpha",C,(4,1.2),fontsize(9)); for(int i = 0; i < 180/alpha; ++i){  path l = B -- (1+i/2)*expi(-i * alpha * pi / 180);  D(l, linetype("4 4"));  D(IP(l,r)); } D(B); [/asy] Note that after $k$ reflections (excluding the first one at $C$ ) the extended line will form an angle $k \beta$ at point $B$ . For the $k$ th reflection to be just inside or at point $B$ , we must have $k\beta \le 180 - 2\alpha \Longrightarrow k \le \frac{180 - 2\alpha}{\beta} = 70.27$ . Thus, our answer is, including the first intersection, $\left\lfloor \frac{180 - 2\alpha}{\beta} \right\rfloor + 1 = \boxed{071}$

Answer:  071
Problem: How many even integers between 4000 and 7000 have four different digits?

Solution: The thousands digit is $\in \{4,5,6\}$
Case $1$ : Thousands digit is even
$4, 6$ , two possibilities, then there are only $\frac{10}{2} - 1 = 4$ possibilities for the units digit. This leaves $8$ possible digits for the hundreds and $7$ for the tens places, yielding a total of $2 \cdot 8 \cdot 7 \cdot 4 = 448$
Case $2$ : Thousands digit is odd
$5$ , one possibility, then there are $5$ choices for the units digit, with $8$ digits for the hundreds and $7$ for the tens place. This gives $1 \cdot 8 \cdot 7 \cdot 5= 280$ possibilities.
Together, the solution is $448 + 280 = \boxed{728}$

Answer:  728
Problem: How many even integers between 4000 and 7000 have four different digits?

Solution: Firstly, we notice that the thousands digit could be $4$ $5$ or $6$ .
Since the parity of the possibilities are different, we cannot cover all cases in one operation. Therefore, we must do casework.
Case $1$
Here we let thousands digit be $4$
4 _ _ _
We take care of restrictions first, and realize that there are 4 choices for the last digit, namely $2$ $6$ $8$ and $0$ . 
Now that there are no restrictions we proceed to find that there are $8$ choices for the hundreds digit and $7$ choices for the tens digit. Therefore we have $8 \cdot 7 \cdot 4 = 224$ numbers that satisfy the conditions posed by the problem.
Case $2$
Here we let thousands digit be $5$
5 _ _ _
Again, we take care of restrictions first. This time there are 5 choices for the last digit, which are all the even numbers because 5 is odd.
Now that there are no restrictions we proceed to find that there are $8$ choices for the hundreds digit and $7$ choices for the tens digit. Therefore we have $8 \cdot 7 \cdot 5 = 280$ numbers that satisfy the conditions posed by the problem.
Case $3$
Here we let thousands digit be $6$
6 _ _ _
Once again, we take care of restrictions first. Since 6 is even,  there are 4 choices for the last digit, namely $2$ $6$ $8$ and $0$
Now we proceed to find that there are $8$ choices for the hundreds digit and $7$ choices for the tens digit. Therefore we have $8 \cdot 7 \cdot 4 = 224$ numbers that satisfy the conditions posed by the problem.
Now that we have our answers for each possible thousands place digit, we add up our answers and get $224+280+224$ $\boxed{728}$ ~PEKKA

Answer:  728
Problem: During a recent campaign for office, a candidate made a tour of a country which we assume lies in a plane. On the first day of the tour he went east, on the second day he went north, on the third day west, on the fourth day south, on the fifth day east, etc. If the candidate went $n^{2}_{}/2$ miles on the $n^{\mbox{th}}_{}$ day of this tour, how many miles was he from his starting point at the end of the $40^{\mbox{th}}_{}$ day?

Solution: On the first day, the candidate moves $[4(0) + 1]^2/2\ \text{east},\, [4(0) + 2]^2/2\ \text{north},\, [4(0) + 3]^2/2\ \text{west},\, [4(0) + 4]^2/2\ \text{south}$ , and so on. The E/W displacement is thus $1^2 - 3^2 + 5^2 \ldots +37^2 - 39^2 = \left|\sum_{i=0}^9 \frac{(4i+1)^2}{2} - \sum_{i=0}^9 \frac{(4i+3)^2}{2}\right|$ . Applying difference of squares , we see that \begin{align*} \left|\sum_{i=0}^9 \frac{(4i+1)^2 - (4i+3)^2}{2}\right| &= \left|\sum_{i=0}^9 \frac{(4i+1+4i+3)(4i+1-(4i+3))}{2}\right|\\ &= \left|\sum_{i=0}^9 -(8i+4) \right|. \end{align*} The N/S displacement is \[\left|\sum_{i=0}^9 \frac{(4i+2)^2}{2} - \sum_{i=0}^9 \frac{(4i+4)^2}{2}\right| = \left|\sum_{i=0}^9 -(8i+6) \right|.\] Since $\sum_{i=0}^{9} i = \frac{9(10)}{2} = 45$ , the two distances evaluate to $8(45) + 10\cdot 4 = 400$ and $8(45) + 10\cdot 6 = 420$ . By the Pythagorean Theorem , the answer is $\sqrt{400^2 + 420^2} = 29 \cdot 20 = \boxed{580}$

Answer:  580
Problem: The table below displays some of the results of last summer's Frostbite Falls Fishing Festival, showing how many contestants caught $n\,$ fish for various values of $n\,$
In the newspaper story covering the event, it was reported that
What was the total number of fish caught during the festival?

Solution: Suppose that the number of fish is $x$ and the number of contestants is $y$ . The $y-(9+5+7)=y-21$ fishers that caught $3$ or more fish caught a total of $x - \left(0\cdot(9) + 1\cdot(5) + 2\cdot(7)\right) = x - 19$ fish. Since they averaged $6$ fish,
Similarily, those who caught $12$ or fewer fish averaged $5$ fish per person, so
Solving the two equation system, we find that $y = 175$ and $x = \boxed{943}$ , the answer.

Answer:  943
Problem: The table below displays some of the results of last summer's Frostbite Falls Fishing Festival, showing how many contestants caught $n\,$ fish for various values of $n\,$
In the newspaper story covering the event, it was reported that
What was the total number of fish caught during the festival?

Solution: Let $f$ be the total number of fish caught by the contestants who didn't catch $0, 1, 2, 3, 13, 14$ , or $15$ fish and let $a$ be the number of contestants who didn't catch $0, 1, 2, 3, 13, 14$ , or $15$ fish. From $\text{(b)}$ , we know that $\frac{69+f+65+28+15}{a+31}=6\implies f=6a+9$ . From $\text{(c)}$ we have $\frac{f+69+14+5}{a+44}=5\implies f=5a+132$ . Using these two equations gets us $a=123$ . Plug this back into the equation to get $f=747$ . Thus, the total number of fish caught is $5+14+69+f+65+28+15=\boxed{943}$ - Heavytoothpaste

Answer:  943
Problem: How many ordered four-tuples of integers $(a,b,c,d)\,$ with $0 < a < b < c < d < 500\,$ satisfy $a + d = b + c\,$ and $bc - ad = 93\,$

Solution: Let $k = a + d = b + c$ so $d = k-a, b=k-c$ . It follows that $(k-c)c - a(k-a) = (a-c)(a+c-k) = (c-a)(d-c) = 93$ . Hence $(c - a,d - c) = (1,93),(3,31),(31,3),(93,1)$
Solve them in terms of $c$ to get $(a,b,c,d) = (c - 93,c - 92,c,c + 1),$ $(c - 31,c - 28,c,c + 3),$ $(c - 1,c + 92,c,c + 93),$ $(c - 3,c + 28,c,c + 31)$ . The last two solutions don't follow $a < b < c < d$ , so we only need to consider the first two solutions.
The first solution gives us $c - 93\geq 1$ and $c + 1\leq 499$ $\implies 94\leq c\leq 498$ , and the second one gives us $32\leq c\leq 496$
So the total number of such quadruples is $405 + 465 = \boxed{870}$

Answer:  870
Problem: How many ordered four-tuples of integers $(a,b,c,d)\,$ with $0 < a < b < c < d < 500\,$ satisfy $a + d = b + c\,$ and $bc - ad = 93\,$

Solution: Let $b = a + m$ and $c = a + m + n$ . From $a + d = b + c$ $d = b + c - a = a + 2m + n$
Substituting $b = a + m$ $c = a + m + n$ , and $d = b + c - a = a + 2m + n$ into $bc - ad = 93$ \[bc - ad = (a + m)(a + m + n) - a(a + 2m + n) = m(m + n). = 93 = 3(31)\] Hence, $(m,n) = (1,92)$ or $(3,28)$
For $(m,n) = (1,92)$ , we know that $0 < a < a + 1 < a + 93 < a + 94 < 500$ , so there are $405$ four-tuples. For $(m,n) = (3,28)$ $0 < a < a + 3 < a + 31 < a + 34 < 500$ , and there are $465$ four-tuples. In total, we have $405 + 465 = \boxed{870}$ four-tuples.

Answer:  870
Problem: How many ordered four-tuples of integers $(a,b,c,d)\,$ with $0 < a < b < c < d < 500\,$ satisfy $a + d = b + c\,$ and $bc - ad = 93\,$

Solution: Square both sides of the first equation in order to get $bc$ and $ad$ terms, which we can plug $93$ in for. \begin{align*} (a+d)^2 = (b+c)^2 &\implies a^2 + 2ad + d^2 = b^2 + 2bc + c^2 \\ &\implies 2bc-2ad = a^2-b^2 + d^2-c^2 \\ &\implies 2(bc-ad) = (a-b)(a+b)+(d-c)(d+c) \end{align*} We can plug $93$ in for $bc - ad$ to get $186$ on the left side, and also observe that $a-b = c-d$ after rearranging the first equation. Plug in $c-d$ for $a-b$
$186 = (c-d)(a+b) + (d-c)(d+c) \implies 186 = -(d-c)(a+b) + (d-c)(d+c) \implies 186 = (d-c)(d+c-a-b)$
Now observe the possible factors of $186$ , which are $1 \cdot 186, 2\cdot 93, 3 \cdot 62, 6\cdot 31$ $(d-c)$ and $(d+c-a-b)$ must be factors of $186$ , and $(d+c-a-b)$ must be greater than $(d-c)$
$1 \cdot 186$ work, and yields $405$ possible solutions. $2 \cdot 93$ does not work, because if $c-d = 2$ , then $a+b$ must differ by 2 as well, but an odd number $93$ can only result from two numbers of different parity. $c-d$ will be even, and $a+b$ will be even, so $c+d - (a+b)$ must be even. $3 \cdot 62$ works, and yields $465$ possible solutions, while $6 \cdot 31$ fails for the same reasoning above.
Thus, the answer is $405 + 465 = \boxed{870}$

Answer:  870
Problem: How many ordered four-tuples of integers $(a,b,c,d)\,$ with $0 < a < b < c < d < 500\,$ satisfy $a + d = b + c\,$ and $bc - ad = 93\,$

Solution: Add the two conditions together to get $a+d+ad+93=b+c+bc$ . Rearranging and factorising with SFFT, $(a+1)(d+1)+93=(b+1)(c+1)$ . This implies that for every quadruple $(a,b,c,d)$ , we can replace $a\longrightarrow a+1$ $b\longrightarrow b+1$ , etc. and this will still produce a valid quadruple. This means, that we can fix $a=1$ , and then just repeatedly add $1$ to get the other quadruples.
Now, our conditions are $b+c=d+1$ and $bc=d+93$ . Replacing $d$ in the first equation, we get $bc-b-c=92$ . Factorising again with SFFT gives $(b-1)(c-1)=93$ . Since $b<c$ , we have two possible cases to consider.
Case 1: $b=2$ $c=94$ . This produces the quadruple $(1,2,94,95)$ , which indeed works.
Case 2: $b=4$ $c=32$ . This produces the quadruple $(1,4,32,35)$ , which indeed works.
Now, for case 1, we can add $1$ to each term exactly $404$ times (until we get the quadruple $(405,406,498,499)$ ), until we violate $d<500$ . This gives $405$ quadruples for case 1.
For case 2, we can add $1$ to each term exactly $464$ times (until we get the quadruple $(465,468,496,499)$ ). this gives $465$ quadruples for case 2.
In conclusion, having exhausted all cases, we can finish. There are hence $405+465=\boxed{870}$ possible quadruples.

Answer:  870
Problem: How many ordered four-tuples of integers $(a,b,c,d)\,$ with $0 < a < b < c < d < 500\,$ satisfy $a + d = b + c\,$ and $bc - ad = 93\,$

Solution: Let $r = d-c$ . From the equation $a+d = b+c$ , we have \[r = d-c = b-a ,\] so $b = a+r$ and $c = d-r$ . We then have \[93 = (a+r)(d-r) - ad = rd - ra - r^2 = r(d-a-r) .\] Since $c > b$ $d-r > a+r$ , or $d-a-r > r$ . Since the prime factorization of 93 is $3 \cdot 31$ , we must either have $r=1$ and $d-a-r = 93$ , or $r=3$ and $d-a-r = 31$ . We consider these cases separately.
If $r=1$ , then $d-a = 94$ $b= a+1$ , and $c = d-1$ . Thus $d$ can be any integer between 95 and 499, inclusive, and our choice of $d$ determines the four-tuple $(a,b,c,d)$ . We therefore have $499-95+1 = 405$ possibilities in this case.
If $r=3$ , then $d-a = 34$ $b = a+3$ , and $c=d-3$ . Thus $d$ can be any integer between 35 and 499, inclusive, and our choice of $d$ determines the four-tuple $(a,b,c,d)$ , as before. We therefore have $499-35+1 = 465$ possibilities in this case.
Since there are 405 possibilities in the first case and 465 possibilities in the second case, in total there are $405 + 465 = \boxed{870}$ four-tuples.

Answer:  870
Problem: How many ordered four-tuples of integers $(a,b,c,d)\,$ with $0 < a < b < c < d < 500\,$ satisfy $a + d = b + c\,$ and $bc - ad = 93\,$

Solution: Assume $d = x+m, a = x-m, c = x+n$ , and $b = x-n$ . This clearly satisfies the condition that $a+d = b+c$ since ( $2x = 2x$ ) . Now plug this into $bc-ad = 93$ . You get $(x+n)(x-n) - (x+m)(x-m) = 93 \Rightarrow m^2 - n^2 = 93 \Rightarrow (m-n)(m+n) = 93$
Since $m>n$ (as given by the condition that $a<b<c<d$ ), $m+n>m-n$ and $m$ and $n$ are integers, there are two cases we have to consider since $93 = 3\cdot 31$ . We first have to consider $m-n = 1, m+n = 93$ , and then consider $m-n=3, m+n = 31$
In the first case, we get $m=47, n=46$ and in the second case we get $m=17, n=14$ . Now plug these values (in separate cases) back into $a,b,c,d$ . Since the only restriction is that all numbers have to be greater than $0$ or less than $500$ , we can write two inequalities. Either $x+47 < 500, x-47 > 0$ , or $x+17 < 500, x-17 > 0$ (using the inequalities given by $d$ and $a$ , and since $b$ and $c$ are squeezed in between $d$ and $a$ , we only have to consider these two inequalities).
This gives us either $47 < x < 453$ or $17 < x < 483$ , and using simple counting, there are $405$ values for $x$ in the first case and $465$ values for $x$ in the second case, and hence our answer is $405+465 = \boxed{870}$

Answer:  870
Problem: Let $P_0(x) = x^3 + 313x^2 - 77x - 8\,$ .  For integers $n \ge 1\,$ , define $P_n(x) = P_{n - 1}(x - n)\,$ .  What is the coefficient of $x\,$ in $P_{20}(x)\,$

Solution: Notice that \begin{align*}P_{20}(x) &= P_{19}(x - 20)\\ &= P_{18}((x - 20) - 19)\\ &= P_{17}(((x - 20) - 19) - 18)\\ &= \cdots\\ &= P_0(x - (20 + 19 + 18 + \ldots + 2 + 1)).\end{align*}
Using the formula for the sum of the first $n$ numbers, $1 + 2 + \cdots + 20 = \frac{20(20+1)}{2} = 210$ . Therefore, \[P_{20}(x) = P_0(x - 210).\]
Substituting $x - 210$ into the function definition, we get $P_0(x-210) = (x - 210)^3 + 313(x - 210)^2 - 77(x - 210) - 8$ . We only need the coefficients of the linear terms, which we can find by the binomial theorem
Adding up the coefficients, we get $630 \cdot 210 - 626 \cdot 210 - 77 = \boxed{763}$

Answer:  763
Problem: Let $P_0(x) = x^3 + 313x^2 - 77x - 8\,$ .  For integers $n \ge 1\,$ , define $P_n(x) = P_{n - 1}(x - n)\,$ .  What is the coefficient of $x\,$ in $P_{20}(x)\,$

Solution: Notice the transformation of $P_{n-1}(x)\to P_n(x)$ adds $n$ to the roots. Thus, all these transformations will take the roots and add $1+2+\cdots+20=210$ to them. (Indeed, this is very easy to check in general.)
Let the roots be $r_1,r_2,r_3.$ Then $P_{20}(x)=(x-r_1-210)(x-r_2-210)(x-r_3-210).$ By Vieta's/expanding/common sense, you see the coefficient of $x$ is $(r_1+210)(r_2+210)+(r_2+210)(r_3+210)+(r_3+210)(r_1+210).$ Expanding yields $r_1r_2+r_2r_3+r_3r_1+210\cdot 2(r_1+r_2+r_3)+3\cdot 210^2.$ Using Vieta's (again) and plugging stuff in yields $-77+210\cdot 2\cdot -313+3\cdot 210^2=\boxed{763}.$

Answer:  763
Problem: What is the smallest positive integer that can be expressed as the sum of nine consecutive integers, the sum of ten consecutive integers, and the sum of eleven consecutive integers?

Solution: Let the desired integer be $n$ . From the information given, it can be determined that, for positive integers $a, \ b, \ c$
$n = 9a + 36 = 10b + 45 = 11c + 55$
This can be rewritten as the following congruences:
$n \equiv 0 \pmod{9}$
$n \equiv 5 \pmod{10}$
$n \equiv 0 \pmod{11}$
Since 9 and 11 are relatively prime, n is a multiple of 99. It can then easily be determined that the smallest multiple of 99 with a units digit 5 (this can be interpreted from the 2nd congruence) is $\boxed{495}$

Answer:  495
Problem: What is the smallest positive integer that can be expressed as the sum of nine consecutive integers, the sum of ten consecutive integers, and the sum of eleven consecutive integers?

Solution: Let $n$ be the desired integer. From the given information, we have \begin{align*}9x &= a \\ 11y &= a \\ 10z + 5 &= a, \end{align*} here, $x,$ and $y$ are the middle terms of the sequence of 9 and 11 numbers, respectively. Similarly, we have $z$ as the 4th term of the sequence. Since, $a$ is a multiple of $9$ and $11,$ it is also a multiple of $\text{lcm}[9,11]=99.$ Hence, $a=99m,$ for some $m.$ So, we have $10z + 5 = 99m.$ It follows that $99(5) = \boxed{495}$ is the smallest integer that can be represented in such a way.

Answer:  495
Problem: What is the smallest positive integer that can be expressed as the sum of nine consecutive integers, the sum of ten consecutive integers, and the sum of eleven consecutive integers?

Solution: By the method in Solution 1, we find that the number $n$ can be written as $9a+36=10b+45=11c+55$ for some integers $a,b,c$ . From this, we can see that $n$ must be divisible by 9, 5, and 11. This means $n$ must be divisible by 495. The only multiples of 495 that are small enough to be AIME answers are 495 and 990. From the second of the three expressions above, we can see that $n$ cannot be divisible by 10, so $n$ must equal $\boxed{495}$ . Solution by Zeroman.

Answer:  495
Problem: What is the smallest positive integer that can be expressed as the sum of nine consecutive integers, the sum of ten consecutive integers, and the sum of eleven consecutive integers?

Solution: First note that the integer clearly must be divisible by $9$ and $11$ since we can use the "let the middle number be x" trick. Let the number be $99k$ for some integer $k.$ Now let the $10$ numbers be $x,x+1, \cdots x+9.$ We have $10x+45 = 99k.$ Taking mod $5$ yields $k \equiv 0 \pmod{5}.$ Since $k$ is positive, we take $k=5$ thus obtaining $99 \cdot 5 = \boxed{495}$ as our answer.

Answer:  495
Problem: Three numbers, $a_1, a_2, a_3$ , are drawn randomly and without replacement from the set $\{1, 2, 3,\ldots, 1000\}$ . Three other numbers, $b_1, b_2, b_3$ , are then drawn randomly and without replacement from the remaining set of $997$ numbers. Let $p$ be the probability that, after suitable rotation, a brick of dimensions $a_1 \times a_2 \times a_3$ can be enclosed in a box of dimension $b_1 \times b_2 \times b_3$ , with the sides of the brick parallel to the sides of the box. If $p$ is written as a fraction in lowest terms, what is the sum of the numerator and denominator?

Solution: There is a total of $P(1000,6)$ possible ordered $6$ -tuples $(a_1,a_2,a_3,b_1,b_2,b_3).$
There are $C(1000,6)$ possible sets $\{a_1,a_2,a_3,b_1,b_2,b_3\}.$ We have five valid cases for the increasing order of these six elements:
Note that the $a$ 's are different from each other, as there are $3!=6$ ways to permute them as $a_1,a_2,$ and $a_3.$ Similarly, the $b$ 's are different from each other, as there are $3!=6$ ways to permute them as $b_1,b_2,$ and $b_3.$
So, there is a total of $C(1000,6)\cdot5\cdot6^2$ valid ordered $6$ -tuples. The requested probability is \[p=\frac{C(1000,6)\cdot5\cdot6^2}{P(1000,6)}=\frac{C(1000,6)\cdot5\cdot6^2}{C(1000,6)\cdot6!}=\frac14,\] from which the answer is $1+4=\boxed{005}.$

Answer:  005
Problem: Three numbers, $a_1, a_2, a_3$ , are drawn randomly and without replacement from the set $\{1, 2, 3,\ldots, 1000\}$ . Three other numbers, $b_1, b_2, b_3$ , are then drawn randomly and without replacement from the remaining set of $997$ numbers. Let $p$ be the probability that, after suitable rotation, a brick of dimensions $a_1 \times a_2 \times a_3$ can be enclosed in a box of dimension $b_1 \times b_2 \times b_3$ , with the sides of the brick parallel to the sides of the box. If $p$ is written as a fraction in lowest terms, what is the sum of the numerator and denominator?

Solution: Call the six numbers selected $x_1 > x_2 > x_3 > x_4 > x_5 > x_6$ . Clearly, $x_1$ must be a dimension of the box, and $x_6$ must be a dimension of the brick.
The total number of arrangements is ${6\choose3} = 20$ ; therefore, $p = \frac{3 + 2}{20} = \frac{1}{4}$ , and the answer is $1 + 4 = \boxed{005}$

Answer:  005
Problem: Three numbers, $a_1, a_2, a_3$ , are drawn randomly and without replacement from the set $\{1, 2, 3,\ldots, 1000\}$ . Three other numbers, $b_1, b_2, b_3$ , are then drawn randomly and without replacement from the remaining set of $997$ numbers. Let $p$ be the probability that, after suitable rotation, a brick of dimensions $a_1 \times a_2 \times a_3$ can be enclosed in a box of dimension $b_1 \times b_2 \times b_3$ , with the sides of the brick parallel to the sides of the box. If $p$ is written as a fraction in lowest terms, what is the sum of the numerator and denominator?

Solution: As in Solutions 2 and 3, we let $x_1>x_2>x_3>x_4>x_5>x_6$ where each $x_i$ is a number selected.  It is clear that when choosing whether each number must be in the set with larger dimensions (the box) or the set with smaller dimensions (the brick) there must always be at least as many numbers in the former set as the latter.  We realize that this resembles Catalan numbers, where the indices of the numbers in the first set can be replaced with rising sections of a mountain, and the other indices representing falling sections of a mountain.  The formula for the $n$ th Catalan number (where $n$ is the number of pairs of rising and falling sections) is \[\frac{\binom{2n}{n}}{n+1}\] Thus, there are $\frac{\binom{6}{3}}{4}$ ways to pick which of $x_1,x_2,x_3,x_4,x_5,$ and $x_6$ are the dimensions of the box, and which are the dimensions of the brick, such that the condition is fulfilled.  There are $\binom{6}{3}$ total ways to choose which numbers make up the brick and box, so the probability of the condition being fulfilled is $\frac{\binom{6}{3}/4}{\binom{6}{3}}=\frac14\Longrightarrow \boxed{005}$

Answer:  005
Problem: Let $S\,$ be a set with six elements. In how many different ways can one select two not necessarily distinct subsets of $S\,$ so that the union of the two subsets is $S\,$ ?  The order of selection does not matter; for example, the pair of subsets $\{a, c\},\{b, c, d, e, f\}$ represents the same selection as the pair $\{b, c, d, e, f\},\{a, c\}.$

Solution: Call the two subsets $m$ and $n.$ For each of the elements in $S,$ we can assign it to either $m,n,$ or both. This gives us $3^6$ possible methods of selection. However, because the order of the subsets does not matter, each possible selection is double counted, except the case where both $m$ and $n$ contain all $6$ elements of $S.$ So our final answer is then $\frac {3^6 - 1}{2} + 1 = \boxed{365}.$

Answer:  365
Problem: Let $S\,$ be a set with six elements. In how many different ways can one select two not necessarily distinct subsets of $S\,$ so that the union of the two subsets is $S\,$ ?  The order of selection does not matter; for example, the pair of subsets $\{a, c\},\{b, c, d, e, f\}$ represents the same selection as the pair $\{b, c, d, e, f\},\{a, c\}.$

Solution: Given one of ${6 \choose k}$ subsets with $k$ elements, the other also has $2^k$ possibilities; this is because it must contain all of the "missing" $n - k$ elements and thus has a choice over the remaining $k.$ We want $\sum_{k = 0}^6 {6 \choose k}2^k = (2 + 1)^6 = 729$ by the Binomial Theorem. But the order of the sets doesn't matter, so we get $\dfrac{729 - 1}{2} + 1 = \boxed{365}.$

Answer:  365
Problem: Let $S\,$ be a set with six elements. In how many different ways can one select two not necessarily distinct subsets of $S\,$ so that the union of the two subsets is $S\,$ ?  The order of selection does not matter; for example, the pair of subsets $\{a, c\},\{b, c, d, e, f\}$ represents the same selection as the pair $\{b, c, d, e, f\},\{a, c\}.$

Solution: We evaluate $f(6)$ recursively: \begin{alignat*}{6} f(0)&=1, \\ f(1)&=3f(0)-1&&=2, \\ f(2)&=3f(1)-1&&=5, \\ f(3)&=3f(2)-1&&=14, \\ f(4)&=3f(3)-1&&=41, \\ f(5)&=3f(4)-1&&=122, \\ f(6)&=3f(5)-1&&=\boxed{365} ~MRENTHUSIASM

Answer:  365