Navvye/Olympiad-Problems-CSV huggingface dataset:
Problem: Let $A = (0,0)$ and $B = (b,2)$ be points on the coordinate plane. Let $ABCDEF$ be a convex equilateral hexagon such that $\angle FAB = 120^\circ,$ $\overline{AB}\parallel \overline{DE},$ $\overline{BC}\parallel \overline{EF,}$ $\overline{CD}\parallel \overline{FA},$ and the y-coordinates of its vertices are distinct elements of the set $\{0,2,4,6,8,10\}.$ The area of the hexagon 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: [asy] size(200); draw((0,0)--(10/sqrt(3),2)--(18/sqrt(3),6)--(10/sqrt(3),10)--(0,8)--(-8/sqrt(3),4)--cycle); dot((0,0));dot((10/sqrt(3),2));dot((18/sqrt(3),6));dot((10/sqrt(3),10));dot((0,8));dot((-8/sqrt(3),4)); label("$A (0,0)$",(0,0),S);label("$B (b,2)$",(10/sqrt(3),2),SE);label("$C$",(18/sqrt(3),6),E);label("$D$",(10/sqrt(3),10),N);label("$E$",(0,8),NW);label("$F$",(-8/sqrt(3),4),W); [/asy] From this image, we can see that the y-coordinate of F is 4, and from this, we can gather that the coordinates of E, D, and C, respectively, are 8, 10, and 6.
[asy] size(200); draw((0,0)--(10/sqrt(3),2)--(18/sqrt(3),6)--(10/sqrt(3),10)--(0,8)--(-8/sqrt(3),4)--cycle); dot((0,0));dot((10/sqrt(3),2));dot((18/sqrt(3),6));dot((10/sqrt(3),10));dot((0,8));dot((-8/sqrt(3),4)); label("$A (0,0)$",(0,0),SE);label("$B (b,2)$",(10/sqrt(3),2),SE);label("$C$",(18/sqrt(3),6),E);label("$D$",(10/sqrt(3),10),N);label("$E$",(0,8),NW);label("$F$",(-8/sqrt(3),4),W); xaxis("$x$");yaxis("$y$"); pair b=foot((10/sqrt(3),2),(0,0),(10,0)); pair f=foot((-8/sqrt(3),4),(0,0),(-10,0)); draw(b--(10/sqrt(3),2),dotted); draw(f--(-8/sqrt(3),4),dotted); label("$\theta$",(0,0),7*dir((0,0)--(10/sqrt(3),2)+(4*sqrt(21)/3,0))); [/asy]
Let the angle between the $x$ -axis and segment $AB$ be $\theta$ , as shown above. Thus, as $\angle FAB=120^\circ$ , the angle between the $x$ -axis and segment $AF$ is $60-\theta$ , so $\sin{(60-\theta)}=2\sin{\theta}$ . Expanding, we have
Isolating $\sin{\theta}$ we see that $\frac{\sqrt{3}\cos{\theta}}{2}=\frac{5\sin{\theta}}{2}$ , or $\cos{\theta}=\frac{5}{\sqrt{3}}\sin{\theta}$ . Using the fact that $\sin^2{\theta}+\cos^2{\theta}=1$ , we have $\frac{28}{3}\sin^2{\theta}=1$ , or $\sin{\theta}=\sqrt{\frac{3}{28}}$ . Letting the side length of the hexagon be $y$ , we have $\frac{2}{y}=\sqrt{\frac{3}{28}}$ . After simplification we find that that $y=\frac{4\sqrt{21}}{3}$
In particular, note that by the Pythagorean theorem, $b^2+2^2=y^2$ , hence $b=10\sqrt{3}/3$ . Also, if $C=(c,6)$ , then $y^2=BC^2=4^2+(c-b)^2$ , hence $c-b=8\sqrt{3}/3,$ and thus $c=18\sqrt{3}/3$ . Using similar methods (or symmetry), we determine that $D=(10\sqrt{3}/3,10)$ $E=(0,8)$ , and $F=(-8\sqrt{3}/3,4)$ . By the Shoelace theorem, \[[ABCDEF]=\frac12\left|\begin{array}{cc} 0&0\\ 10\sqrt{3}/3&2\\ 18\sqrt{3}/3&6\\ 10\sqrt{3}/3&10\\ 0&8\\ -8\sqrt{3}/3&4\\ 0&0\\ \end{array}\right|=\frac12|60+180+80-36-60-(-64)|\sqrt{3}/3=48\sqrt{3}.\]
Hence the answer is $\boxed{51}$

Answer:  51
Problem: Let $A = (0,0)$ and $B = (b,2)$ be points on the coordinate plane. Let $ABCDEF$ be a convex equilateral hexagon such that $\angle FAB = 120^\circ,$ $\overline{AB}\parallel \overline{DE},$ $\overline{BC}\parallel \overline{EF,}$ $\overline{CD}\parallel \overline{FA},$ and the y-coordinates of its vertices are distinct elements of the set $\{0,2,4,6,8,10\}.$ The area of the hexagon 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: This is similar to solution 2 but faster and easier.
First off we see that the y coordinate of F must be 4, the y coordinate of E must be 8, the y coordinate of D must be 10, and the y coordinate of C must be 6 (from the parallel sides of the hexagon).
We then use the sine sum angle formula to find the x coordinate of B (lets call it $x$ ): $2\cdot\cos(120)+x\sin(120)=\frac{x\sqrt3}{2}-1=4\rightarrow x=\frac{10\sqrt3}{3}$ .
Now that we know $x$ we can find the x coordinate of F in multiple ways, including using the cosine sum angle formula or using the fact that triangle AFE is isosceles and AE is on the y axis.  Either way, we find that the x coordinate of F is $-\frac{8\sqrt3}{3}$ .
Now, divide ABCDEF into two congruent triangles and a parallelogram:  AFE, BCD, and ABDE.  The areas of AFE and BCD are each $\frac12\cdot\frac{8\sqrt3}{3}\cdot8=\frac{32\sqrt3}{3}$ .  The area of ABDE is $\frac12\cdot8\cdot\frac{10\sqrt3}{3}=\frac{80\sqrt3}3$ .
The total area of the hexagon is $2\cdot\frac{32\sqrt3}3+\frac{80\sqrt3}3=\frac{144\sqrt3}{3}=48\sqrt3\rightarrow48+3=\boxed{051}$

Answer:  051
Problem: Let $A = (0,0)$ and $B = (b,2)$ be points on the coordinate plane. Let $ABCDEF$ be a convex equilateral hexagon such that $\angle FAB = 120^\circ,$ $\overline{AB}\parallel \overline{DE},$ $\overline{BC}\parallel \overline{EF,}$ $\overline{CD}\parallel \overline{FA},$ and the y-coordinates of its vertices are distinct elements of the set $\{0,2,4,6,8,10\}.$ The area of the hexagon 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: [asy] size(200); draw((0,0)--(10/sqrt(3),2)--(18/sqrt(3),6)--(10/sqrt(3),10)--(0,8)--(-8/sqrt(3),4)--cycle); dot((0,0));dot((10/sqrt(3),2));dot((18/sqrt(3),6));dot((10/sqrt(3),10));dot((0,8));dot((-8/sqrt(3),4)); label("$A (0,0)$",(0,0),SE);label("$B (b,2)$",(10/sqrt(3),2),SE);label("$C$",(18/sqrt(3),6),E);label("$D$",(10/sqrt(3),10),N);label("$E$",(0,8),NW);label("$F$",(-8/sqrt(3),4),W); xaxis("$x$");yaxis("$y$"); pair b=foot((10/sqrt(3),2),(0,0),(10,0)); pair f=foot((-8/sqrt(3),4),(0,0),(-10,0)); draw(b--(10/sqrt(3),2),dotted); draw(f--(-8/sqrt(3),4),dotted); label("$\theta$",(0,0),7*dir((0,0)--(10/sqrt(3),2)+(4*sqrt(21)/3,0))); [/asy]
First, we see that the y-coordinates of F, E, D, and C must be 4, 8, and 10, and 6, respectively, as in the previous solutions. We can draw a rectangle around the hexagon ABCDEF and use negative space to find the area of the hexagon. If we call the distance from the foot of the perpendiculars of B and F to A $x$ and $z$ , respectively, and the distance from the bottom left vertex of the rectangle to the foot of the perpendicular from B $y$ . This tells us that the area of the entire rectangle is $10(x+y+z)$ , since the opposite sides are parallel and thus the length of the rectangle is $4+4+2=10$ . Then, 
if we find the area of the extra triangles and subtract, we find that the area of hexagon ABCDEF as $6x+8z+2y$ . However, noticing that $x=y$ , the area of ABCDEF can also be expressed as $8(x+z)$ . Now we just need to find $x+z$ . Since $AB=AF$ and $\angle BAF = 120$ degrees, $BF=AB\sqrt{3}$ . However, we can find AB by using the Pythagorean Theorem on either of the right triangles formed by dropping perpendiculars from B and F to the x-axis (let's call them ABX and AFY).
From triangle ABX we have that $AB=\sqrt{4+x^2}$ , so $BF=\sqrt{3x^2+12}$ . Since AB=AF, we can also form the equation $4+x^2=16+z^2$ .
We can also find BF by dropping a perpendicular from B to line FY and using the Pythagorean Theorem on the right triangle formed. This gives us $BF=\sqrt{4+(x+z)^2}$ . Setting our two values of BF equal and substituting $x^2$ as $12+z^2$ and simplifying, we get the equation $3z^4-16z^2-1024=0$ . Now we can use the quadratic formula to get that $z^2=\frac{64}{3}$ or $-18$ , so $z^2=\frac{64}{3}$ . Plugging this value back into the equation $x^2=12+z^2$ , we get that $x^2=\frac{100}{3}$ . Now we get that $x+z$ is $6\sqrt{3}$ , so the area of the hexagon is $8 \cdot 6\sqrt{3}=48\sqrt{3}$ , so the answer is $48+3=\boxed{051}$

Answer:  051
Problem: Let \[P(x) = 24x^{24} + \sum_{j = 1}^{23}(24 - j)(x^{24 - j} + x^{24 + j}).\] Let $z_{1},z_{2},\ldots,z_{r}$ be the distinct zeros of $P(x),$ and let $z_{k}^{2} = a_{k} + b_{k}i$ for $k = 1,2,\ldots,r,$ where $a_{k}$ and $b_{k}$ are real numbers. Let
where $m, n,$ and $p$ are integers and $p$ is not divisible by the square of any prime. Find $m + n + p.$

Solution: This can be factored as:
\[P(x) = x\left( x^{23} + x^{22} + \cdots + x^2 + x + 1 \right)^2\]
Note that $\left( x^{23} + x^{22} + \cdots + x^2 + x + 1 \right) \cdot (x-1) = x^{24} - 1$ .
So the roots of $x^{23} + x^{22} + \cdots + x^2 + x + 1$ are exactly all $24$ -th complex roots of $1$ , except for the root $x=1$
Let $\omega=\cos \frac{360^\circ}{24} + i\sin \frac{360^\circ}{24}$ . Then the distinct zeros of $P$ are $0,\omega,\omega^2,\dots,\omega^{23}$
We can clearly ignore the root $x=0$ as it does not contribute to the value that we need to compute.
The squares of the other roots are $\omega^2,~\omega^4,~\dots,~\omega^{24}=1,~\omega^{26}=\omega^2,~\dots,~\omega^{46}=\omega^{22}$
Hence we need to compute the following sum:
\[R = \sum_{k = 1}^{23} \left|\, \sin \left( k\cdot \frac{360^\circ}{12} \right) \right|\]
Using basic properties of the sine function, we can simplify this to
\[R = 4 \cdot \sum_{k = 1}^{5} \sin \left( k\cdot \frac{360^\circ}{12} \right)\]
The five-element sum is just $\sin 30^\circ + \sin 60^\circ + \sin 90^\circ + \sin 120^\circ + \sin 150^\circ$ . 
We know that $\sin 30^\circ = \sin 150^\circ = \frac 12$ $\sin 60^\circ = \sin 120^\circ = \frac{\sqrt 3}2$ , and $\sin 90^\circ = 1$ .
Hence our sum evaluates to:
\[R = 4 \cdot \left( 2\cdot \frac 12 + 2\cdot \frac{\sqrt 3}2 + 1 \right) = 8 + 4\sqrt 3\]
Therefore the answer is $8+4+3 = \boxed{015}$

Answer:  015
Problem: Let \[P(x) = 24x^{24} + \sum_{j = 1}^{23}(24 - j)(x^{24 - j} + x^{24 + j}).\] Let $z_{1},z_{2},\ldots,z_{r}$ be the distinct zeros of $P(x),$ and let $z_{k}^{2} = a_{k} + b_{k}i$ for $k = 1,2,\ldots,r,$ where $a_{k}$ and $b_{k}$ are real numbers. Let
where $m, n,$ and $p$ are integers and $p$ is not divisible by the square of any prime. Find $m + n + p.$

Solution: As in Solution 1, we find that the roots of $P(x)$ we care about are the 24th roots of unity except $1$ . Therefore, the squares of these roots are the 12th roots of unity. In particular, every 12th root of unity is counted twice, except for $1$ , which is only counted once.
The possible imaginary parts of the 12th roots of unity are $0$ $\pm\frac{1}{2}$ $\pm\frac{\sqrt{3}}{2}$ , and $\pm 1$ . We can disregard $0$ because it doesn't affect the sum.
$8$ squares of roots have an imaginary part of $\pm\frac{1}{2}$ $8$ squares of roots have an imaginary part of $\pm\frac{\sqrt{3}}{2}$ , and $4$ squares of roots have an imaginary part of $\pm 1$ . Therefore, the sum equals $8\left(\frac{1}{2}\right) + 8\left(\frac{\sqrt{3}}{2}\right) + 4(1) = 8 + 4\sqrt{3}$
The answer is $8+4+3=\boxed{015}$

Answer:  015
Problem: Many states use a sequence of three letters followed by a sequence of three digits as their standard license-plate pattern. Given that each three-letter three-digit arrangement is equally likely,  the probability that such a license plate will contain at least one palindrome (a three-letter arrangement or a three-digit arrangement that reads the same left-to-right as it does right-to-left) is $\dfrac{m}{n}$ , where $m$ and $n$ are relatively prime positive integers. Find $m+n.$

Solution: Using complementary counting, we count all of the license plates that do not have the desired property.  To not be a palindrome, the first and third characters of each string must be different.  Therefore, there are $10\cdot 10\cdot 9$ three-digit non-palindromes, and there are $26\cdot 26\cdot 25$ three-letter non-palindromes.  As there are $10^3\cdot 26^3$ total three-letter three-digit arrangements, the probability that a license plate does not have the desired property is $\frac{10\cdot 10\cdot 9\cdot 26\cdot 26\cdot 25}{10^3\cdot 26^3}=\frac{45}{52}$ .  We subtract this from 1 to get $1-\frac{45}{52}=\frac{7}{52}$ as our probability.  Therefore, our answer is $7+52=\boxed{059}$

Answer:  059
Problem: Many states use a sequence of three letters followed by a sequence of three digits as their standard license-plate pattern. Given that each three-letter three-digit arrangement is equally likely,  the probability that such a license plate will contain at least one palindrome (a three-letter arrangement or a three-digit arrangement that reads the same left-to-right as it does right-to-left) is $\dfrac{m}{n}$ , where $m$ and $n$ are relatively prime positive integers. Find $m+n.$

Solution: Note that we can pick the first and second letters/numbers freely with one choice left for the last letter/number for there to be a palindrome. Thus, the probability of no palindrome is \[\frac{25}{26}\cdot \frac{9}{10}=\frac{45}{52}\] thus we have $1-\frac{45}{52}=\frac{7}{52}$ so our answer is $7+52 = \boxed{059}.$

Answer:  059
Problem: The diagram shows twenty congruent circles arranged in three rows and enclosed in a rectangle. The circles are tangent to one another and to the sides of the rectangle as shown in the diagram. The ratio of the longer dimension of the rectangle to the shorter dimension can be written as $\dfrac{1}{2}(\sqrt{p}-q)$ where $p$ and $q$ are positive integers. Find $p+q$

Solution: Let the radius of the circles be $r$ . The longer dimension of the rectangle can be written as $14r$ , and by the Pythagorean Theorem , we find that the shorter dimension is $2r\left(\sqrt{3}+1\right)$
Therefore, $\frac{14r}{2r\left(\sqrt{3}+1\right)}= \frac{7}{\sqrt{3} + 1} \cdot \left[\frac{\sqrt{3}-1}{\sqrt{3}-1}\right] = \frac{1}{2}\left(7\sqrt{3} - 7\right) = \frac{1}{2}\left(\sqrt{p}-q\right)$ . Thus we have $p=147$ and $q=7$ , so $p+q=\boxed{154}$

Answer:  154
Problem: The diagram shows twenty congruent circles arranged in three rows and enclosed in a rectangle. The circles are tangent to one another and to the sides of the rectangle as shown in the diagram. The ratio of the longer dimension of the rectangle to the shorter dimension can be written as $\dfrac{1}{2}(\sqrt{p}-q)$ where $p$ and $q$ are positive integers. Find $p+q$

Solution: Since we only care about the ratio between the longer side and shorter side, we can set the longer side to $14$ . So, this means that each of the radii is $1$ . Now, we connect the radii of three circles such that they form an equilateral triangle with side length 4. Obviously, the height this triangle is $2\sqrt{3}$ , and the shorter side if the triangle is therefore $2\sqrt{3}+2$ and we use simplification similar to as showed above, and we reach the result $\frac{1}{2} \cdot (\sqrt{147}-7)$ and the final answer is $147+7 = \boxed{154}$

Answer:  154
Problem: Jane is 25 years old. Dick is older than Jane. In $n$ years, where $n$ is a positive integer, Dick's age and Jane's age will both be two-digit numbers and will have the property that Jane's age is obtained by interchanging the digits of Dick's age. Let $d$ be Dick's present age. How many ordered pairs of positive integers $(d,n)$ are possible?

Solution: Let Jane's age $n$ years from now be $10a+b$ , and let Dick's age be $10b+a$ . If $10b+a>10a+b$ , then $b>a$ . The possible pairs of $a,b$ are:
That makes 36. But $10a+b>25$ , so we subtract all the extraneous pairs: $(1,2), (1,3), (2,3), (1,4), (2,4), (1,5), (2,5), (1,6), (1,7), (1,8),$ and $(1,9)$ $36-11=\boxed{025}$

Answer:  025
Problem: Jane is 25 years old. Dick is older than Jane. In $n$ years, where $n$ is a positive integer, Dick's age and Jane's age will both be two-digit numbers and will have the property that Jane's age is obtained by interchanging the digits of Dick's age. Let $d$ be Dick's present age. How many ordered pairs of positive integers $(d,n)$ are possible?

Solution: Start by assuming that $n < 5$ (essentially, Jane is in the 20s when their ages are 'reverses' of each other). Then we get the pairs \[(61,1),(70,2),(79,3),(88,4).\] Repeating this for the 30s gives \[(34,9),(43,10),(52,11),(61,12),(70,13),(79,14).\] From here, it's pretty clear that every decade we go up we get $(d,n+11)$ as a pair. Since both ages must always be two-digit numbers, we can show that each decade after the 30s, we get 1 fewer option. Therefore, our answer is $4+6+5+\dots+2+1=4+21=\boxed{025}.$

Answer:  025
Problem: Let $A_1,A_2,A_3,\cdots,A_{12}$ be the vertices of a regular dodecagon. How many distinct squares in the plane of the dodecagon have at least two vertices in the set $\{A_1,A_2,A_3,\cdots,A_{12}\} ?$

Solution: Proceed as above to initially get 198 squares (with overcounting). Then note that any square with all four vertices on the dodecagon has to have three sides "between" each vertex, giving us a total of three squares. However, we counted these squares with all four of their sides plus both of their diagonals, meaning we counted them 6 times. Therefore, our answer is $198-3(6-1)=198-15=\boxed{183}.$

Answer:  183
Problem: The solutions to the system of equations
are $(x_1,y_1)$ and $(x_2,y_2)$ . Find $\log_{30}\left(x_1y_1x_2y_2\right)$

Solution: Let $A=\log_{225}x$ and let $B=\log_{64}y$
From the first equation: $A+B=4 \Rightarrow B = 4-A$
Plugging this into the second equation yields $\frac{1}{A}-\frac{1}{B}=\frac{1}{A}-\frac{1}{4-A}=1 \Rightarrow A = 3\pm\sqrt{5}$ and thus, $B=1\pm\sqrt{5}$
So, $\log_{225}(x_1x_2)=\log_{225}(x_1)+\log_{225}(x_2)=(3+\sqrt{5})+(3-\sqrt{5})=6$ $\Rightarrow x_1x_2=225^6=15^{12}$
And $\log_{64}(y_1y_2)=\log_{64}(y_1)+\log_{64}(y_2)=(1+\sqrt{5})+(1-\sqrt{5})=2$ $\Rightarrow y_1y_2=64^2=2^{12}$
Thus, $\log_{30}\left(x_1y_1x_2y_2\right) = \log_{30}\left(15^{12}\cdot2^{12} \right) = \log_{30}\left(30^{12} \right) = \boxed{012}$

Answer:  012
Problem: The Binomial Expansion is valid for exponents that are not integers. That is, for all real numbers $x,y$ and $r$ with $|x|>|y|$
\[(x+y)^r=x^r+rx^{r-1}y+\dfrac{r(r-1)}{2}x^{r-2}y^2+\dfrac{r(r-1)(r-2)}{3!}x^{r-3}y^3 \cdots\]
What are the first three digits to the right of the decimal point in the decimal representation of $(10^{2002}+1)^{\frac{10}{7}}$

Solution: $1^n$ will always be 1, so we can ignore those terms, and using the definition ( $2002 / 7 = 286$ ):
\[(10^{2002} + 1)^{\frac {10}7} = 10^{2860}+\dfrac{10}{7}10^{858}+\dfrac{15}{49}10^{-1144}+\cdots\]
Since the exponent of the $10$ goes down extremely fast, it suffices to consider the first few terms. Also, the $10^{2860}$ term will not affect the digits after the decimal, so we need to find the first three digits after the decimal in
\[\dfrac{10}{7}10^{858}\]
(The remainder after this term is positive by the Remainder Estimation Theorem ). Since the repeating decimal of $\dfrac{10}{7}$ repeats every 6 digits, we can cut out a lot of 6's from $858$ to reduce the problem to finding the first three digits after the decimal of
$\dfrac{10}{7}$
That is the same as $1+\dfrac{3}{7}$ , and the first three digits after $\dfrac{3}{7}$ are $\boxed{428}$

Answer:  428
Problem: Find the smallest integer $k$ for which the conditions
(1) $a_1,a_2,a_3\cdots$ is a nondecreasing sequence of positive integers
(2) $a_n=a_{n-1}+a_{n-2}$ for all $n>2$
(3) $a_9=k$
are satisfied by more than one sequence.

Solution: From $(2)$ $a_9=$ $a_8+a_7=2a_7+a_6=3a_6+2a_5=5a_5+3a_4=8a_4+5a_3=13a_3+8a_2=21a_2+13a_1$ $=k$
Suppose that $a_1=x_0$ is the smallest possible value for $a_1$ that yields a good sequence, and $a_2=y_0$ in this sequence. So, $13x_0+21y_0=k$
Since $\gcd(13,21)=1$ , the next smallest possible value for $a_1$ that yields a good sequence is $a_1=x_0+21$ . Then, $a_2=y_0-13$
By $(1)$ $a_2 \ge a_1 \Rightarrow y_0-13 \ge x_0+21 \Rightarrow y_0 \ge x_0+34 \ge 35$ . So the smallest value of $k$ is attained when $(x_0,y_0)=(1,35)$ which yields $(a_1,a_2)=(1,35)$ or $(22,22)$
Thus, $k=13(1)+21(35)=\boxed{748}$ is the smallest possible value of $k$

Answer:  748
Problem: Harold, Tanya, and Ulysses paint a very long picket fence.
Call the positive integer $100h+10t+u$ paintable when the triple $(h,t,u)$ of positive integers results in every picket being painted exactly once. Find the sum of all the paintable integers.

Solution: Note that it is impossible for any of $h,t,u$ to be $1$ , since then each picket will have been painted one time, and then some will be painted more than once.
$h$ cannot be $2$ , or that will result in painting the third picket twice. If $h=3$ , then $t$ may not equal anything not divisible by $3$ , and the same for $u$ . Now for fourth and fifth pickets to be painted, $t$ and $u$ must be $3$ as well. This configuration works, so $333$ is paintable.
If $h$ is $4$ , then $t$ must be even. The same for $u$ , except that it can't be $2 \mod 4$ . Thus $u$ is $0 \mod 4$ and $t$ is $2 \mod 4$ . Since this is all $\mod 4$ $t$ must be $2$ and $u$ must be $4$ , in order for $5,6$ to be paint-able. Thus $424$ is paintable.
$h$ cannot be greater than $4$ , since if that were the case then the answer would be greater than $999$ , which would be impossible for the AIME.
Thus the sum of all paintable numbers is $\boxed{757}$

Answer:  757
Problem: Harold, Tanya, and Ulysses paint a very long picket fence.
Call the positive integer $100h+10t+u$ paintable when the triple $(h,t,u)$ of positive integers results in every picket being painted exactly once. Find the sum of all the paintable integers.

Solution: Again, note that $h,t,u \neq 1$ . The three conditions state that no picket number $n$ may satisfy any two of the conditions: $n \equiv 1 \pmod{h},\ n \equiv 2 \pmod{t},\ n \equiv 3 \pmod{u}$ . By the Chinese Remainder Theorem , the greatest common divisor of any pair of the three numbers $\{h,t,u\}$ cannot be $1$ (since otherwise without loss of generality consider $\text{gcd}\,(h,t) = 1$ ; then there will be a common solution $\pmod{h \times t}$ ).
Now for $4$ to be paint-able, we require either $h = 3$ or $t=2$ , but not both.
Thus the answer is $333+424 = \boxed{757}$

Answer:  757
Problem: Harold, Tanya, and Ulysses paint a very long picket fence.
Call the positive integer $100h+10t+u$ paintable when the triple $(h,t,u)$ of positive integers results in every picket being painted exactly once. Find the sum of all the paintable integers.

Solution: The three conditions state that no picket number $n$ may satisfy any two of the conditions: $n \equiv 1 \pmod{h},\ n \equiv 2 \pmod{t},\ n \equiv 3 \pmod{u}$ . Note that the smallest number, $min \{ h,t,u \},$ divides the other $2$ , and the next smallest divide the largest number, otherwise there is a common solution by the Chinese Remainder Theorem . It is also a necessary condition so that it paints exactly once. Note that the smallest number can't be at least $5$ , otherwise not all picket will be painted. We are left with few cases (we could also exclude $1$ as the possibility) which could be done quickly. Thus the answer is $333+424 = \boxed{757}$

Answer:  757
Problem: In the diagram below, angle $ABC$ is a right angle. Point $D$ is on $\overline{BC}$ , and $\overline{AD}$ bisects angle $CAB$ . Points $E$ and $F$ are on $\overline{AB}$ and $\overline{AC}$ , respectively, so that $AE=3$ and $AF=10$ . Given that $EB=9$ and $FC=27$ , find the integer closest to the area of quadrilateral $DCFG$

Solution: By the Pythagorean Theorem, $BC=35$ . Letting $BD=x$ we can use the Angle Bisector Theorem on triangle $ABC$ to get $x/12=(35-x)/37$ , and solving gives $BD=60/7$ and $DC=185/7$
The area of triangle $AGF$ is $10/3$ that of triangle $AEG$ , since they share a common side and angle, so the area of triangle $AGF$ is $10/13$ the area of triangle $AEF$
Since the area of a triangle is $\frac{ab\sin{C}}2$ , the area of $AEF$ is $525/37$ and the area of $AGF$ is $5250/481$
The area of triangle $ABD$ is $360/7$ , and the area of the entire triangle $ABC$ is $210$ . Subtracting the areas of $ABD$ and $AGF$ from $210$ and finding the closest integer gives $\boxed{148}$ as the answer.

Answer:  148