Navvye/Olympiad-Problems-CSV huggingface dataset:
Problem: A pyramid has a triangular base with side lengths $20$ $20$ , and $24$ . The three edges of the pyramid from the three corners of the base to the fourth vertex of the pyramid all have length $25$ . The volume of the pyramid is $m\sqrt{n}$ , where $m$ and $n$ are positive integers, and $n$ is not divisible by the square of any prime. Find $m+n$

Solution: Let the triangular base be $\triangle ABC$ , with $\overline {AB} = 24$ . We find that the altitude to side $\overline {AB}$ is $16$ , so the area of $\triangle ABC$ is $(24*16)/2 = 192$
Let the fourth vertex of the tetrahedron be $P$ , and let the midpoint of $\overline {AB}$ be $M$ . Since $P$ is equidistant from $A$ $B$ , and $C$ , the line through $P$ perpendicular to the plane of $\triangle ABC$ will pass through the circumcenter of $\triangle ABC$ , which we will call $O$ . Note that $O$ is equidistant from each of $A$ $B$ , and $C$ . Then,
\[\overline {OM} + \overline {OC} = \overline {CM} = 16\]
Let $\overline {OM} = d$ . Then $OC=OA=\sqrt{d^2+12^2}.$ Equation $(1)$ \[d + \sqrt {d^2 + 144} = 16\]
Squaring both sides, we have
\[d^2 + 144 + 2d\sqrt {d^2+144} + d^2 = 256\]
\[2d^2 + 2d\sqrt {d^2+144} = 112\]
\[2d(d + \sqrt {d^2+144}) = 112\]
Substituting with equation $(1)$
\[2d(16) = 112\]
\[d = 7/2\]
We now find that $\sqrt{d^2 + 144} = 25/2$
Let the distance $\overline {OP} = h$ . Using the Pythagorean Theorem on triangle $AOP$ $BOP$ , or $COP$ (all three are congruent by SSS):
\[25^2 = h^2 + (25/2)^2\]
\[625 = h^2 + 625/4\]
\[1875/4 = h^2\]
\[25\sqrt {3} / 2 = h\]
Finally, by the formula for volume of a pyramid,
\[V = Bh/3\]
\[V = (192)(25\sqrt{3}/2)/3\] This simplifies to $V = 800\sqrt {3}$ , so $m+n = \boxed{803}$

Answer:  803
Problem: A pyramid has a triangular base with side lengths $20$ $20$ , and $24$ . The three edges of the pyramid from the three corners of the base to the fourth vertex of the pyramid all have length $25$ . The volume of the pyramid is $m\sqrt{n}$ , where $m$ and $n$ are positive integers, and $n$ is not divisible by the square of any prime. Find $m+n$

Solution: We can place a three dimensional coordinate system on this pyramid. WLOG assume the vertex across from the line that has length $24$ is at the origin, or $(0, 0, 0)$ . Then, the two other vertices can be $(-12, -16, 0)$ and $(12, -16, 0)$ . Let the fourth vertex have coordinates of $(x, y, z)$ . We have the following $3$ equations from the distance formula.
\[x^2+y^2+z^2=625\]
\[(x+12)^2+(y+16)^2+z^2=625\]
\[(x-12)^2+(y+16)^2+z^2=625\]
Adding the last two equations and substituting in the first equation, we get that $y=-\frac{25}{2}$ . If you drew a good diagram, it should be obvious that $x=0$ . Now, solving for $z$ , we get that $z=\frac{25\sqrt{3}}{2}$ . So, the height of the pyramid is $\frac{25\sqrt{3}}{2}$ . The base is equal to the area of the triangle, which is $\frac{1}{2} \cdot 24 \cdot 16 = 192$ . The volume is $\frac{1}{3} \cdot 192 \cdot \frac{25\sqrt{3}}{2} = 800\sqrt{3}$ . Thus, the answer is $800+3 = \boxed{803}$

Answer:  803
Problem: A pyramid has a triangular base with side lengths $20$ $20$ , and $24$ . The three edges of the pyramid from the three corners of the base to the fourth vertex of the pyramid all have length $25$ . The volume of the pyramid is $m\sqrt{n}$ , where $m$ and $n$ are positive integers, and $n$ is not divisible by the square of any prime. Find $m+n$

Solution: Label the four vertices of the tetrahedron and the midpoint of $\overline {AB}$ , and notice that the area of the base of the tetrahedron, $\triangle ABC$ , equals $192$ , according to Solution 1.
Notice that the altitude of $\triangle CPM$ from $\overline {CM}$ to point $P$ is the height of the tetrahedron. Side $\overline {PM}$ is can be found using the Pythagorean Theorem on $\triangle APM$ , giving us $\overline {PM}=\sqrt{481}.$
Using Heron's Formula, the area of $\triangle CPM$ can be written as \[\sqrt{\frac{41+\sqrt{481}}{2}(\frac{41+\sqrt{481}}{2}-16)(\frac{41+\sqrt{481}}{2}-25)(\frac{41+\sqrt{481}}{2}-\sqrt{481})}\] \[=\frac{\sqrt{(41+\sqrt{481})(9+\sqrt{481})(-9+\sqrt{481})(41-\sqrt{481})}}{4}\]
Notice that both $(41+\sqrt{481})(41-\sqrt{481})$ and $(9+\sqrt{481})(-9+\sqrt{481})$ can be rewritten as differences of squares; thus, the expression can be written as \[\frac{\sqrt{(41^2-481)(481-9^2)}}{4}=\frac{\sqrt{480000}}{4}=100\sqrt{3}.\]
From this, we can determine the height of both $\triangle CPM$ and tetrahedron $ABCP$ to be $\frac{100\sqrt{3}}{8}$ ; therefore, the volume of the tetrahedron equals $\frac{100\sqrt{3}}{8} \cdot 192=800\sqrt{3}$ ; thus, $m+n=800+3=\boxed{803}.$

Answer:  803
Problem: A pyramid has a triangular base with side lengths $20$ $20$ , and $24$ . The three edges of the pyramid from the three corners of the base to the fourth vertex of the pyramid all have length $25$ . The volume of the pyramid is $m\sqrt{n}$ , where $m$ and $n$ are positive integers, and $n$ is not divisible by the square of any prime. Find $m+n$

Solution: Notation is shown on diagram. \[AM = MB = c = 12, AC = BC = b = 20,\] \[DA = DB = DC = a = 25.\] \[CM = x + y = \sqrt{b^2-c^2} = 16,\] \[x^2 - y^2 = CD^2 – DM^2 = CD^2 – (BD^2 – BM^2) = c^2 = 144,\] \[x – y = \frac{x^2 – y^2}{x+y} = \frac {c^2} {16} = 9,\] \[x = \frac {16 + 9}{2} = \frac {a}{2},\] \[h = \sqrt{a^2 -\frac{ a^2}{4}} = a \frac {\sqrt{3}}{2},\] \[V = \frac{h\cdot CM \cdot c}{3}= \frac{16\cdot 25 \sqrt{3} \cdot 12}{3} = 800 \sqrt{3} \implies \boxed{803}.\] vladimir.shelomovskii@gmail.com, vvsss

Answer:  803
Problem: A rational number written in base eight is $\underline{ab} . \underline{cd}$ , where all digits are nonzero. The same number in base twelve is $\underline{bb} . \underline{ba}$ . Find the base-ten number $\underline{abc}$

Solution: First, note that the first two digits will always be a positive number. We will start with base twelve because of its repetition. List all the positive numbers in base eight that have equal tens and ones digits in base 12.
$11_{12}=15_8$
$22_{12}=32_8$
$33_{12}=47_8$
$44_{12}=64_8$
$55_{12}=101_8$
We stop because we only can have two-digit numbers in base 8 and 101 is not a 2 digit number.
Compare the ones places to check if they are equal. We find that they are equal if $b=2$ or $b=4$ .
Evaluating the places to the right side of the decimal point gives us $22.23_{12}$ or $44.46_{12}$ .
When the numbers are converted into base 8, we get $32.14_8$ and $64.30_8$ . Since $d\neq0$ , the first value is correct. Compiling the necessary digits leaves us a final answer of $\boxed{321}$

Answer:  321
Problem: A rational number written in base eight is $\underline{ab} . \underline{cd}$ , where all digits are nonzero. The same number in base twelve is $\underline{bb} . \underline{ba}$ . Find the base-ten number $\underline{abc}$

Solution: The parts before and after the decimal points must be equal. Therefore $8a + b = 12b + b$ and $c/8 + d/64 = b/12 + a/144$ .  Simplifying the first equation gives $a = (3/2)b$ .  Plugging this into the second equation gives $3b/32 = c/8 + d/64$ . Multiplying both sides by 64 gives $6b = 8c + d$ $a$ and $b$ are both digits between 1 and 7 (they must be a single non-zero digit in base eight) so using $a = 3/2b$ $(a,b) = (3,2)$ or $(6,4)$ .  Testing these gives that $(6,4)$ doesn't work, and $(3,2)$ gives $a = 3, b = 2, c = 1$ , and $d = 4$ .  Therefore $abc = \boxed{321}$

Answer:  321
Problem: A rational number written in base eight is $\underline{ab} . \underline{cd}$ , where all digits are nonzero. The same number in base twelve is $\underline{bb} . \underline{ba}$ . Find the base-ten number $\underline{abc}$

Solution: Converting to base $10$ we get
$4604a+72c+9d=6960b$
Since $72c$ and $9d$ are much smaller than the other two terms, dividing by $100$ and approximating we get
$46a=70b$
Writing out the first few values of $a$ and $b$ , the first possible tuple is
$a=3, b=2, c=1, d=4$
and the second possible tuple is
$a=6, b=4, c=3, d=0$
Note that $d$ can not be $0$ , therefore the answer is $\boxed{321}$

Answer:  321
Problem: A rational number written in base eight is $\underline{ab} . \underline{cd}$ , where all digits are nonzero. The same number in base twelve is $\underline{bb} . \underline{ba}$ . Find the base-ten number $\underline{abc}$

Solution: In the problem, we are given that \[8a+b+\dfrac c8+\dfrac d{64}=12b+b+\dfrac b{12}+\dfrac a{144}.\] We multiply by the LCM of the denominators, which is $576$ to get \[4608a+576b+72c+9d=6912b+576b+48b+4a.\] We then group like terms and factor to get \[4604a+72c+9d=48b(144+1)=145\cdot48b=24\cdot290b.\]
Observe that the coefficients of $a$ $c$ , and $b$ are all divisible by $4$ . Therefore, we know that $d$ must also be divisible by $4$ to compensate. (To observe this, one could rearrange the terms to see $9d=24\cdot290b-4604a-72c=4(6\cdot290b-1151a-18c)$ . At this point, it is obvious that $9d$ must be divisible by $4$ , so $d$ must be divisible by $4$ .) Thus, we let $d=4y$ to see $4604a+72c+9\cdot4y=24\cdot290b$ . We divide by $4$ in the equation to get \[1151a+18c+9y=6\cdot290b.\]
Observe that the coefficients of $c$ $y$ , and $b$ are all divisible by $3$ . By similar reasoning, $a$ must be divisible by $3$ , so we let $a=3x$ . Substituting and dividing by $3$ , we get \[1151x+6c+3y=2\cdot290b.\] We observe that this can no longer be reduced by similar means.
We know that $0<a,b,c<8\implies0<3x,b,4y<8$ . We examine $0<4y<8$ ; this becomes $0<y<2$ . It is apparent that $y=1\implies d=4$ . However, the problem does not even ask for $d$ , so it may appear that this find is meaningless. However, we substitute our value of $y$ in to get \[1151x+6c+3=580b.\]
We know that $x$ is either $1$ or $2$ , since $0<3x<8$ . We take our equation modulo $6$ to find \[5x+3\equiv4b\pmod{6}\implies3\equiv4b+x\pmod{6}.\] If $x=2$ , then $1\equiv4b\pmod{6}$ - this is equivalent to saying that $1+6n=4b\implies 1=4b-6n$ . However, $4b-6n$ is always even, and $1$ is odd. Thus, this case is not possible, so $x=1\implies a=3$ . We know that $x=1$ ; thus, $3\equiv4b+1\pmod{6}\implies2\equiv4b\pmod{6}\implies1\equiv2b\pmod{3}$ . Obviously, $b=2$ and $b=5$ work; however, if $b=5$ , then the RHS of our original equation ( $2\cdot290b$ ) is much too large to be equal to the LHS (the maximum possible value of the LHS is $1151\cdot1+6\cdot7+3\cdot1$ , which is less than $1200$ while the RHS would become $2\cdot290\cdot5=2900$ ), so we have $b=2$
We recall our equation of $1151x+6c+3=580b$ . Plugging in what we know, we have $1151\cdot1+6c+3=580\cdot2\implies6c=6\implies c=1$
Therefore, $\overline{abc}=\boxed{321}$

Answer:  321
Problem: A circle is circumscribed around an isosceles triangle whose two congruent angles have degree measure $x$ . Two points are chosen independently and uniformly at random on the circle, and a chord is drawn between them. The probability that the chord intersects the triangle is $\frac{14}{25}$ . Find the difference between the largest and smallest possible values of $x$

Solution: The probability that the chord doesn't intersect the triangle is $\frac{11}{25}$ . The only way this can happen is if the two points are chosen on the same arc between two of the triangle vertices. The probability that a point is chosen on one of the arcs opposite one of the base angles is $\frac{2x}{360}=\frac{x}{180}$ (this comes from the Central Angle Theorem, which states that the central angle from two points on a circle is always twice the inscribed angle from those two points), and the probability that a point is chosen on the arc between the two base angles is $\frac{180-2x}{180}$ . Therefore, we can write \[2\left(\frac{x}{180}\right)^2+\left(\frac{180-2x}{180}\right)^2=\frac{11}{25}\] This simplifies to \[x^2-120x+3024=0\]
(Note that the simplification is quite tedious)
Which factors as \[(x-84)(x-36)=0\] So $x=84, 36$ . The difference between these is $\boxed{048}$

Answer:  048
Problem: A circle is circumscribed around an isosceles triangle whose two congruent angles have degree measure $x$ . Two points are chosen independently and uniformly at random on the circle, and a chord is drawn between them. The probability that the chord intersects the triangle is $\frac{14}{25}$ . Find the difference between the largest and smallest possible values of $x$

Solution: Because we know that we have an isosceles triangle with angles of $x$ (and we know that x is an inscribed angle), that means that the arc that is intercepted by this angle is $2x$ . We form this same conclusion for the other angle $x$ , and $180-2x$ . Therefore we get $3$ arcs, namely, $2x$ $2x$ , and $360-4x$ . To have the chords intersect the triangle, we need the two points selected (to make a chord) to be on completely different arcs. An important idea to understand is that order matters in this case, so we have the equation $2$ $\frac{2x}{360}$ $\frac{2x}{360}$ $2$ $2$ $\frac{2x}{360}$ $\frac{360-4x}{360}$ $\frac{14}{25}$ which using trivial algebra gives you $x^2-120x+3024$ and factoring gives you $(x-84)(x-36)$ and so your answer is $\boxed{048}$ . 
~jske25

Answer:  048
Problem: A circle is circumscribed around an isosceles triangle whose two congruent angles have degree measure $x$ . Two points are chosen independently and uniformly at random on the circle, and a chord is drawn between them. The probability that the chord intersects the triangle is $\frac{14}{25}$ . Find the difference between the largest and smallest possible values of $x$

Solution: After constructing the circumscribed circle, realize that the only time when the chord does not intersect the circle is when our $2$ points fall on only one arc formed by the sides of the triangle. Thus, lets call our isosceles triangle $ABC$ , where $AB=BC$ . Thus, the arcs formed by $BC$ and $AB$ can be called $a$ , and the arc formed by $AC$ is called $b$ . So, we can create the following system
\[2a+b=1\] \[2a^2+4ab=\frac{14}{25}\] \[2a^2+b^2=\frac{11}{25}\]
Notice we are denoting $a$ and $b$ as our probabilities, which we will be converting to degrees later. The 2 remaining systems can be calculated by using our rule about intersecting arcs and chords. So, after some hairy algebra we get:
\[a=\frac{1}{5}\] if \[b=\frac{3}{5}\] \[a=\frac{7}{15}\] if \[b=\frac{1}{15}\]
From here we find the absolute difference by doing $\frac{7}{15}-\frac{1}{5} = \frac{4}{15}$ . Converting to degrees, since the angles of a triangle add up to $180^o$ , we find that $\frac{4}{15} \cdot 180 =\boxed{048}$ , which is our answer.

Answer:  048
Problem: For nonnegative integers $a$ and $b$ with $a + b \leq 6$ , let $T(a, b) = \binom{6}{a} \binom{6}{b} \binom{6}{a + b}$ . Let $S$ denote the sum of all $T(a, b)$ , where $a$ and $b$ are nonnegative integers with $a + b \leq 6$ . Find the remainder when $S$ is divided by $1000$

Solution: Let $c=6-(a+b)$ , and note that $\binom{6}{a + b}=\binom{6}{c}$ . The problem thus asks for the sum $\binom{6}{a} \binom{6}{b} \binom{6}{c}$ over all $a,b,c$ such that $a+b+c=6$ . Consider an array of 18 dots, with 3 columns of 6 dots each. The desired expression counts the total number of ways to select 6 dots by considering each column separately, which is equal to $\binom{18}{6}=18564$ . Therefore, the answer is $\boxed{564}$

Answer:  564
Problem: For nonnegative integers $a$ and $b$ with $a + b \leq 6$ , let $T(a, b) = \binom{6}{a} \binom{6}{b} \binom{6}{a + b}$ . Let $S$ denote the sum of all $T(a, b)$ , where $a$ and $b$ are nonnegative integers with $a + b \leq 6$ . Find the remainder when $S$ is divided by $1000$

Solution: Alternatively, one can note that we can consider groups where $a+b$ is constant, say $c$ . Fix any value of $c$ . Then the sum of all of the values of $T(a,b)$ such that $a+b=c$ is $\binom{6}{a+b} \sum_{a+b=c} \binom{6}{a}\binom{6}{b}$ which by Vandermonde's is $\binom{6}{a+b}\binom{12}{a+b}$ . Remember, that expression is the resulting sum for a fixed $a+b$ . So, for $a+b\le 6$ , we want $\sum_{c=0}^{6} \binom{6}{c}\binom{12}{c}$ . This is (by Vandermonde's or committee forming) $\binom{18}{6} = 18564 \implies \boxed{564}$ ~ firebolt360

Answer:  564
Problem: For nonnegative integers $a$ and $b$ with $a + b \leq 6$ , let $T(a, b) = \binom{6}{a} \binom{6}{b} \binom{6}{a + b}$ . Let $S$ denote the sum of all $T(a, b)$ , where $a$ and $b$ are nonnegative integers with $a + b \leq 6$ . Find the remainder when $S$ is divided by $1000$

Solution: Treating $a+b$ as $n$ , this problem asks for \[\sum_{n=0}^{6} \left[\binom{6}{n}\sum_{m=0}^{n}\left[\binom{6}{m}\binom{6}{n-m}\right]\right].\] But \[\sum_{m=0}^{n} \left[\binom{6}{m} \binom{6}{n-m}\right]\] can be computed through the following combinatorial argument.  Choosing $n$ elements from a set of size $12$ is the same as splitting the set into two sets of size $6$ and choosing $m$ elements from one, $n-m$ from the other where $0\leq m\leq n$ .  The number of ways to perform such a procedure is simply $\binom{12}{n}$ . Therefore, the requested sum is \[\sum_{n=0}^{6} \left[\binom{6}{n} \binom{12}{n}\right] = 18564.\] As such, our answer is $\boxed{564}$

Answer:  564
Problem: For nonnegative integers $a$ and $b$ with $a + b \leq 6$ , let $T(a, b) = \binom{6}{a} \binom{6}{b} \binom{6}{a + b}$ . Let $S$ denote the sum of all $T(a, b)$ , where $a$ and $b$ are nonnegative integers with $a + b \leq 6$ . Find the remainder when $S$ is divided by $1000$

Solution: Case 1: $a<b$
Subcase 1: $a=0$ \[\binom{6}{0}\binom{6}{1}\binom{6}{1}=36\] \[\binom{6}{0}\binom{6}{2}\binom{6}{2}=225\] \[\binom{6}{0}\binom{6}{3}\binom{6}{3}=400\] \[\binom{6}{0}\binom{6}{4}\binom{6}{4}=225\] \[\binom{6}{0}\binom{6}{5}\binom{6}{5}=36\] \[\binom{6}{0}\binom{6}{6}\binom{6}{6}=1\] \[36+225+400+225+36+1=923\] Subcase 2: $a=1$ \[\binom{6}{1}\binom{6}{2}\binom{6}{3}=1800 \equiv 800 \pmod {1000}\] \[\binom{6}{1}\binom{6}{3}\binom{6}{4}=1800 \equiv 800 \pmod {1000}\] \[\binom{6}{1}\binom{6}{4}\binom{6}{5}=540\] \[\binom{6}{1}\binom{6}{5}\binom{6}{6}=36\] \[800+800+540+36=2176 \equiv 176 \pmod {1000}\] Subcase 3: $a=2$ \[\binom{6}{2}\binom{6}{3}\binom{6}{5}=1800\equiv800\pmod{1000}\] \[\binom{6}{2}\binom{6}{4}\binom{6}{6}=225\] \[800+225=1025\equiv25\pmod{1000}\]
\[923+176+25=1124\equiv124\pmod{1000}\]
Case 2: $b<a$
By just switching $a$ and $b$ in all of the above cases, we will get all of the cases such that $b>a$ is true. Therefore, this case is also $124\pmod{1000}$
Case 3: $a=b$ \[\binom{6}{0}\binom{6}{0}\binom{6}{0}=1\] \[\binom{6}{1}\binom{6}{1}\binom{6}{2}=540\] \[\binom{6}{2}\binom{6}{2}\binom{6}{4}=3375\equiv375\pmod{1000}\] \[\binom{6}{3}\binom{6}{3}\binom{6}{6}=400\] \[1+540+375+400=1316\equiv316\pmod{1000}\]
\[316+124+124=\boxed{564}\]

Answer:  564
Problem: For nonnegative integers $a$ and $b$ with $a + b \leq 6$ , let $T(a, b) = \binom{6}{a} \binom{6}{b} \binom{6}{a + b}$ . Let $S$ denote the sum of all $T(a, b)$ , where $a$ and $b$ are nonnegative integers with $a + b \leq 6$ . Find the remainder when $S$ is divided by $1000$

Solution: We begin as in solution 1 to rewrite the sum as $\binom{6}{a} \binom{6}{b} \binom{6}{c}$ over all $a,b,c$ such that $a+b+c=6$ .
Consider the polynomial $P(x)=\left(\binom{6}{0} + \binom{6}{1}x + \binom{6}{2}x^2 + \binom{6}{3}x^3 + \cdot \cdot \cdot + \binom{6}{6}x^6\right)^3$ .
We can see the sum we wish to compute is just the coefficient of the $x^6$ term. However $P(x)=((1+x)^6)^3=(1+x)^{18}$ . Therefore, the coefficient of the $x^6$ term is just $\binom{18}{6} = 18564$ so the answer is $\boxed{564}$

Answer:  564
Problem: For nonnegative integers $a$ and $b$ with $a + b \leq 6$ , let $T(a, b) = \binom{6}{a} \binom{6}{b} \binom{6}{a + b}$ . Let $S$ denote the sum of all $T(a, b)$ , where $a$ and $b$ are nonnegative integers with $a + b \leq 6$ . Find the remainder when $S$ is divided by $1000$

Solution: Let $c=6-(a+b)$ . Then $\binom{6}{a+b}=\binom{6}{c}$ , and $a+b+c=6$ . The problem thus asks for \[\sum_{a+b+c=6}\binom{6}{a}\binom{6}{b}\binom{6}{c} \pmod {1000}.\] Suppose we have $6$ red balls, $6$ green balls, and $6$ blue balls lined up in a row, and we want to choose $6$ balls from this set of $18$ balls by considering each color separately. Over all possible selections of $6$ balls from this set, there are always a nonnegative number of balls in each color group. The answer is $\binom{18}{6} \pmod {1000}=18\boxed{564}$

Answer:  564
Problem: For nonnegative integers $a$ and $b$ with $a + b \leq 6$ , let $T(a, b) = \binom{6}{a} \binom{6}{b} \binom{6}{a + b}$ . Let $S$ denote the sum of all $T(a, b)$ , where $a$ and $b$ are nonnegative integers with $a + b \leq 6$ . Find the remainder when $S$ is divided by $1000$

Solution: Since $\binom{6}{n}=\binom{6}{6-n}$ , we can rewrite $T(a,b)$ as $\binom{6}{a}\binom{6}{b}\binom{6}{6-(a+b)}$ . Consider the number of ways to choose a committee of 6 people from a group of 6 democrats, 6 republicans, and 6 independents. We can first pick $a$ democrats, then pick $b$ republicans, provided that $a+b \leq 6$ . Then we can pick the remaining $6-(a+b)$ people from the independents. But this is just $T(a,b)$ , so the sum of all $T(a,b)$ is equal to the number of ways to choose this committee.
On the other hand, we can simply pick any 6 people from the $6+6+6=18$ total politicians in the group. Clearly, there are $\binom{18}{6}$ ways to do this. So the desired quantity is equal to $\binom{18}{6}$ . We can then compute (routinely) the last 3 digits of $\binom{18}{6}$ as $\boxed{564}$

Answer:  564
Problem: Two real numbers $a$ and $b$ are chosen independently and uniformly at random from the interval $(0, 75)$ . Let $O$ and $P$ be two points on the plane with $OP = 200$ . Let $Q$ and $R$ be on the same side of line $OP$ such that the degree measures of $\angle POQ$ and $\angle POR$ are $a$ and $b$ respectively, and $\angle OQP$ and $\angle ORP$ are both right angles. The probability that $QR \leq 100$ is equal to $\frac{m}{n}$ , where $m$ and $n$ are relatively prime positive integers. Find $m + n$

Solution: Noting that $\angle OQP$ and $\angle ORP$ are right angles, we realize that we can draw a semicircle with diameter $\overline{OP}$ and points $Q$ and $R$ on the semicircle. Since the radius of the semicircle is $100$ , if $\overline{QR} \leq 100$ , then $\overarc{QR}$ must be less than or equal to $60^{\circ}$
This simplifies the problem greatly. Since the degree measure of an angle on a circle is simply half the degree measure of its subtended arc, the problem is simply asking:
Given $a, b$ such that $0<a, b<75$ , what is the probability that $|a-b| \leq 30$ ? 
Through simple geometric probability, we get that $P = \frac{16}{25}$
The answer is $16+25=\boxed{041}$

Answer:  041