Navvye/Olympiad-Problems-CSV huggingface dataset:
Problem: Zou and Chou are practicing their $100$ -meter sprints by running $6$ races against each other. Zou wins the first race, and after that, the probability that one of them wins a race is $\frac23$ if they won the previous race but only $\frac13$ if they lost the previous race. The probability that Zou will win exactly $5$ of the $6$ races is $\frac mn$ , where $m$ and $n$ are relatively prime positive integers. Find $m+n.$

Solution: We have $5$ cases, depending on which race Zou lost. Let $\text{W}$ denote a won race, and $\text{L}$ denote a lost race for Zou. The possible cases are $\text{WWWWL, WWWLW, WWLWW, WLWWW, LWWWW}$ . The first case has probability $\left(\frac{2}{3} \right)^4 \cdot \frac{1}{3} = \frac{16}{3^5}$ . The second case has probability $\left( \frac{2}{3} \right)^3 \cdot \frac{1}{3} \cdot \frac{1}{3} = \frac{8}{3^5}$ . The third has probability $\left( \frac{2}{3} \right)^2 \cdot \frac{1}{3} \cdot \frac{1}{3} \cdot \frac{2}{3} = \frac{8}{3^5}$ . The fourth has probability $\frac{2}{3} \cdot \frac{1}{3} \cdot \frac{1}{3} \cdot \left( \frac{2}{3} \right)^2 = \frac{8}{3^5}$ . Lastly, the fifth has probability $\frac{1}{3} \cdot \frac{1}{3} \cdot \left( \frac{2}{3} \right)^3 = \frac{8}{3^5}$ . Adding these up, the total probability is $\frac{16 + 8 \cdot 4}{3^5} = \frac{16 \cdot 3}{3^5} = \frac{16}{81}$ , so $m+n = \boxed{097}$

Answer:  097
Problem: Zou and Chou are practicing their $100$ -meter sprints by running $6$ races against each other. Zou wins the first race, and after that, the probability that one of them wins a race is $\frac23$ if they won the previous race but only $\frac13$ if they lost the previous race. The probability that Zou will win exactly $5$ of the $6$ races is $\frac mn$ , where $m$ and $n$ are relatively prime positive integers. Find $m+n.$

Solution: Case 1: Zou loses the first race
In this case, Zou must win the rest of the races. Thus, our probability is $\frac{8}{243}$
Case 2: Zou loses the last race
There is only one possibility for this, so our probability is $\frac{16}{243}$
Case 3: Neither happens
There are three ways that this happens. Each has one loss that is not the last race. Therefore, the probability that one happens is $\frac{1}{3} \cdot \frac{1}{3} \cdot \frac{2}{3} \cdot \frac{2}{3} \cdot \frac{2}{3} = \frac{8}{243}$ . Thus, the total probability is $\frac{8}{243} \cdot 3 = \frac{24}{243}$
Adding these up, we get $\frac{48}{243} = \frac{16}{81}$ , so $16+81=\boxed{097}$

Answer:  097
Problem: Zou and Chou are practicing their $100$ -meter sprints by running $6$ races against each other. Zou wins the first race, and after that, the probability that one of them wins a race is $\frac23$ if they won the previous race but only $\frac13$ if they lost the previous race. The probability that Zou will win exactly $5$ of the $6$ races is $\frac mn$ , where $m$ and $n$ are relatively prime positive integers. Find $m+n.$

Solution: Note that Zou wins one race. The probability that he wins the last race is $\left(\frac{2}{3}\right)^4\left(\frac{1}{3}\right)=\frac{16}{243}.$ Now, if he doesn't win the last race, then there must be two races where the winner of the previous race loses. We can choose any $4$ of the middle races for Zou to win. So the probability for this case is $4\left(\frac{2}{3}\right)^3\left(\frac{1}{3}\right)^2=\frac{32}{243}.$ Thus, the answer is $\frac{16}{243}+\frac{32}{243}=\frac{16}{81}\implies\boxed{097}.$

Answer:  097
Problem: In the diagram below, $ABCD$ is a rectangle with side lengths $AB=3$ and $BC=11$ , and $AECF$ is a rectangle with side lengths $AF=7$ and $FC=9,$ as shown. The area of the shaded region common to the interiors of both rectangles is $\frac mn$ , where $m$ and $n$ are relatively prime positive integers. Find $m+n$ [asy] pair A, B, C, D, E, F; A=(0,3); B=(0,0); C=(11,0); D=(11,3); E=foot(C, A, (9/4,0)); F=foot(A, C, (35/4,3)); draw(A--B--C--D--cycle); draw(A--E--C--F--cycle); filldraw(A--(9/4,0)--C--(35/4,3)--cycle,gray*0.5+0.5*lightgray); dot(A^^B^^C^^D^^E^^F); label("$A$", A, W); label("$B$", B, W); label("$C$", C, (1,0)); label("$D$", D, (1,0)); label("$F$", F, N); label("$E$", E, S); [/asy]

Solution: Again, let the intersection of $AE$ and $BC$ be $G$ . By AA similarity, $\triangle AFG \sim \triangle CDG$ with a $\frac{7}{3}$ ratio. Define $x$ as $\frac{[CDG]}{9}$ . Because of similar triangles, $[AFG] = 49x$ . Using $ABCD$ , the area of the parallelogram is $33-18x$ . Using $AECF$ , the area of the parallelogram is $63-98x$ . These equations are equal, so we can solve for $x$ and obtain $x = \frac{3}{8}$ . Thus, $18x = \frac{27}{4}$ , so the area of the parallelogram is $33 - \frac{27}{4} = \frac{105}{4}$
Finally, the answer is $105+4=\boxed{109}$

Answer:  109
Problem: In the diagram below, $ABCD$ is a rectangle with side lengths $AB=3$ and $BC=11$ , and $AECF$ is a rectangle with side lengths $AF=7$ and $FC=9,$ as shown. The area of the shaded region common to the interiors of both rectangles is $\frac mn$ , where $m$ and $n$ are relatively prime positive integers. Find $m+n$ [asy] pair A, B, C, D, E, F; A=(0,3); B=(0,0); C=(11,0); D=(11,3); E=foot(C, A, (9/4,0)); F=foot(A, C, (35/4,3)); draw(A--B--C--D--cycle); draw(A--E--C--F--cycle); filldraw(A--(9/4,0)--C--(35/4,3)--cycle,gray*0.5+0.5*lightgray); dot(A^^B^^C^^D^^E^^F); label("$A$", A, W); label("$B$", B, W); label("$C$", C, (1,0)); label("$D$", D, (1,0)); label("$F$", F, N); label("$E$", E, S); [/asy]

Solution: Let $P = AD \cap FC$ , and $K = AE \cap BC$ . Also let $AP = x$
$CK$ also has to be $x$ by parallelogram properties. Then $PD$ and $BK$ must be $11-x$ because the sum of the segments has to be $11$
We can easily solve for $PC$ by the Pythagorean Theorem: \begin{align*} DC^2 + PD^2 &= PC^2\\ 9 + (11-x)^2 &= PC^2 \end{align*} It follows shortly that $PC = \sqrt{x^2-22x+30}$
Also, $FC = 9$ , and $FP + PC = 9$ . We can then say that $PC = \sqrt{x^2-22x+30}$ , so $FP = 9 - \sqrt{x^2-22x+30}$
Now we can apply the Pythagorean Theorem to $\triangle AFP$ \begin{align*} AF^2 + FP^2 = AP^2\\ 49 + \left(9 - \sqrt{x^2-22x+30}\right)^2 = x^2  \end{align*}
This simplifies (not-as-shortly) to $x = \dfrac{35}{4}$ . Now we have to solve for the area of $APCK$ . We know that the height is $3$ because the height of the parallelogram is the same as the height of the smaller rectangle.
From the area of a parallelogram (we know that the base is $\dfrac{35}{4}$ and the height is $3$ ), it is clear that the area is $\dfrac{105}{4}$ , giving an answer of $\boxed{109}$

Answer:  109
Problem: In the diagram below, $ABCD$ is a rectangle with side lengths $AB=3$ and $BC=11$ , and $AECF$ is a rectangle with side lengths $AF=7$ and $FC=9,$ as shown. The area of the shaded region common to the interiors of both rectangles is $\frac mn$ , where $m$ and $n$ are relatively prime positive integers. Find $m+n$ [asy] pair A, B, C, D, E, F; A=(0,3); B=(0,0); C=(11,0); D=(11,3); E=foot(C, A, (9/4,0)); F=foot(A, C, (35/4,3)); draw(A--B--C--D--cycle); draw(A--E--C--F--cycle); filldraw(A--(9/4,0)--C--(35/4,3)--cycle,gray*0.5+0.5*lightgray); dot(A^^B^^C^^D^^E^^F); label("$A$", A, W); label("$B$", B, W); label("$C$", C, (1,0)); label("$D$", D, (1,0)); label("$F$", F, N); label("$E$", E, S); [/asy]

Solution: Suppose $B=(0,0).$ It follows that $A=(0,3),C=(11,0),$ and $D=(11,3).$
Since $AECF$ is a rectangle, we have $AE=FC=9$ and $EC=AF=7.$ The equation of the circle with center $A$ and radius $\overline{AE}$ is $x^2+(y-3)^2=81,$ and the equation of the circle with center $C$ and radius $\overline{CE}$ is $(x-11)^2+y^2=49.$
We now have a system of two equations with two variables. Expanding and rearranging respectively give \begin{align*} x^2+y^2-6y&=72, &(1) \\ x^2+y^2-22x&=-72. &(2) \end{align*} Subtracting $(2)$ from $(1),$ we obtain $22x-6y=144.$ Simplifying and rearranging produce \[x=\frac{3y+72}{11}. \hspace{34.5mm} (*)\] Substituting $(*)$ into $(1)$ gives \[\left(\frac{3y+72}{11}\right)^2+y^2-6y=72,\] which is a quadratic of $y.$ We clear fractions by multiplying both sides by $11^2=121,$ then solve by factoring: \begin{align*} \left(3y+72\right)^2+121y^2-726y&=8712 \\ \left(9y^2+432y+5184\right)+121y^2-726y&=8712 \\ 130y^2-294y-3528&=0 \\ 2(5y+21)(13y-84)&=0 \\ y&=-\frac{21}{5},\frac{84}{13}. \end{align*} Since $E$ is in Quadrant IV, we have $E=\left(\frac{3\left(-\frac{21}{5}\right)+72}{11},-\frac{21}{5}\right)=\left(\frac{27}{5},-\frac{21}{5}\right).$ It follows that the equation of $\overleftrightarrow{AE}$ is $y=-\frac{4}{3}x+3.$
Let $G$ be the intersection of $\overline{AD}$ and $\overline{FC},$ and $H$ be the intersection of $\overline{AE}$ and $\overline{BC}.$ Since $H$ is the $x$ -intercept of $\overleftrightarrow{AE},$ we get $H=\left(\frac94,0\right).$
By symmetry, quadrilateral $AGCH$ is a parallelogram. Its area is $HC\cdot AB=\left(11-\frac94\right)\cdot3=\frac{105}{4},$ from which the requested sum is $105+4=\boxed{109}.$

Answer:  109
Problem: In the diagram below, $ABCD$ is a rectangle with side lengths $AB=3$ and $BC=11$ , and $AECF$ is a rectangle with side lengths $AF=7$ and $FC=9,$ as shown. The area of the shaded region common to the interiors of both rectangles is $\frac mn$ , where $m$ and $n$ are relatively prime positive integers. Find $m+n$ [asy] pair A, B, C, D, E, F; A=(0,3); B=(0,0); C=(11,0); D=(11,3); E=foot(C, A, (9/4,0)); F=foot(A, C, (35/4,3)); draw(A--B--C--D--cycle); draw(A--E--C--F--cycle); filldraw(A--(9/4,0)--C--(35/4,3)--cycle,gray*0.5+0.5*lightgray); dot(A^^B^^C^^D^^E^^F); label("$A$", A, W); label("$B$", B, W); label("$C$", C, (1,0)); label("$D$", D, (1,0)); label("$F$", F, N); label("$E$", E, S); [/asy]

Solution: Let the intersection of $AE$ and $BC$ be $G$ . It is useful to find $\tan(\angle DAE)$ , because $\tan(\angle DAE)=\frac{3}{BG}$ and $\frac{3}{\tan(\angle DAE)}=BG$ . From there, subtracting the areas of the two triangles from the larger rectangle, we get  Area = $33-3BG=33-\frac{9}{\tan(\angle DAE)}$
let $\angle CAD = \alpha$ . Let $\angle CAE = \beta$ . Note, $\alpha+\beta=\angle DAE$
$\alpha=\tan^{-1}\left(\frac{3}{11}\right)$
$\beta=\tan^{-1}\left(\frac{7}{9}\right)$
$\tan(\angle DAE) = \tan\left(\tan^{-1}\left(\frac{3}{11}\right)+\tan^{-1}\left(\frac{7}{9}\right)\right) = \frac{\frac{3}{11}+\frac{7}{9}}{1-\frac{3}{11}\cdot\frac{7}{9}} = \frac{\frac{104}{99}}{\frac{78}{99}} = \frac{4}{3}$
$\mathrm{Area}=33-\frac{9}{\frac{4}{3}} = 33-\frac{27}{4 } = \frac{105}{4}$ . The answer is $105+4=\boxed{109}$

Answer:  109
Problem: Find the number of positive integers less than $1000$ that can be expressed as the difference of two integral powers of $2.$

Solution: We want to find the number of positive integers $n<1000$ which can be written in the form $n = 2^a - 2^b$ for some non-negative integers $a > b \ge 0$ (note that if $a=b$ , then $2^a-2^b = 0$ ). We first observe $a$ must be at most 10; if $a \ge 11$ , then $2^a - 2^b \ge 2^{10} > 1000$ . As $2^{10} = 1024 \approx 1000$ , we can first choose two different numbers $a > b$ from the set $\{0,1,2,\ldots,10\}$ in $\binom{11}{2}=55$ ways. This includes $(a,b) = (10,0)$ $(10,1)$ $(10,2)$ $(10,3)$ $(10,4)$ which are invalid as $2^a - 2^b > 1000$ in this case. For all other choices $a$ and $b$ , the value of $2^a - 2^b$ is less than 1000.
We claim that for all other choices of $a$ and $b$ , the values of $2^a - 2^b$ are pairwise distinct. More specifically, if $(a_1,b_1) \neq (a_2,b_2)$ where $10 \ge a_1 > b_1 \ge 0$ and $10 \ge a_2 > b_2 \ge 0$ , we must show that $2^{a_1}-2^{b_1} \neq 2^{a_2} - 2^{b_2}$ . Suppose otherwise for sake of contradiction; rearranging yields $2^{a_1}+2^{b_2} = 2^{a_2}+2^{b_1}$ . We use the fact that every positive integer has a unique binary representation:
If $a_1 \neq b_2$ then $\{a_1,b_2\} = \{a_2,b_1\}$ ; from here we can deduce either $a_1=a_2$ and $b_1=b_2$ (contradicting the assumption that $(a_1,b_1) \neq (a_2,b_2)$ , or $a_1=b_1$ and $a_2=b_2$ (contradicting the assumption $a_1>b_1$ and $a_2>b_2$ ).
If $a_1 = b_2$ then $2^{a_1}+2^{b_2} = 2 \times 2^{a_1}$ , and it follows that $a_1=a_2=b_1=b_2$ , also contradicting the assumption $(a_1,b_1) \neq (a_2,b_2)$ . Hence we obtain contradiction.*
Then there are $\binom{11}{2}-5$ choices for $(a,b)$ for which $2^a - 2^b$ is a positive integer less than 1000; by the above claim, each choice of $(a,b)$ results in a different positive integer $n$ . Then there are $55-5 = \boxed{050}$ integers which can be expressed as a difference of two powers of 2.

Answer:  050
Problem: Find the number of positive integers less than $1000$ that can be expressed as the difference of two integral powers of $2.$

Solution: Case 1: When our answer is in the form $2^n-2^i$ , where $i$ is an integer such that $0\le i\le 4$
We start with the subcase where it is $2^n-2^0$ , for some integer $n$ where $n>0$ (this is because the case where $n=0$ yields $2^0-2^0=0$ , which doesn't work because it must be a positive integer.) 
Note that $2^{10}=1024$ , and $2^9=512$ . Our answer needs to be less than $1000$ , so the maximum possible result (in this case) is $2^9-2^0$ . Our lowest result is $2^1-2^0$ . All the positive powers of two less than $1024$ work, so we have $9$ possibilities for this subcase. For subcases $i=1, i=2, i=3,$ and $i=4$ , we have $8, 7, 6,$ and $5$ possibilities, respectively.
Case 2: When our answer is in the form of $2^n-2^j$ , where $j$ is an integer such that $5\le j\le 9$
We can start with the subcase where $j=5$ . We notice that $2^5=32$ , and $2^{10}-2^5=992$ which is less than $1000$ , so the greatest result in this subcase is actually $2^{10}-2^5$ , and the lowest is $2^6-2^5$ . Thus, we have $5$ possibilities. For the other four subcases, we have $4, 3, 2,$ and $1$ possibilities, respectively.
Answer: We note that these are our only cases, as numbers in the form of $2^n-2^{10}$ and beyond are greater than $1000$
Thus, our result is $9+8+7+6+5+5+4+3+2+1=(9+8+7+6+5+4+3+2+1)+5=\boxed{50}$ . ~jehu26

Answer:  50
Problem: Find the number of positive integers less than $1000$ that can be expressed as the difference of two integral powers of $2.$

Solution: We look for all positive integers of the form $2^a-2^b<1000,$ where $0\leq b<a.$ Performing casework on $a,$ we can enumerate all possibilities in the table below: \[\begin{array}{c|c} & \\ [-2.25ex] \boldsymbol{a} & \boldsymbol{b} \\  \hline  & \\ [-2ex] 1 & 0 \\   2 & 0,1 \\ 3 & 0,1,2 \\ 4 & 0,1,2,3 \\ 5 & 0,1,2,3,4 \\ 6 & 0,1,2,3,4,5 \\ 7 & 0,1,2,3,4,5,6 \\ 8 & 0,1,2,3,4,5,6,7 \\ 9 & 0,1,2,3,4,5,6,7,8 \\ 10 & \xcancel{0},\xcancel{1},\xcancel{2},\xcancel{3},\xcancel{4},5,6,7,8,9 \\ [0.5ex] \end{array}\] As indicated by the X-marks, the ordered pairs $(a,b)=(10,0),(10,1),(10,2),(10,3),(10,4)$ generate $2^a-2^b>1000,$ which are invalid.
Note that each of the remaining ordered pairs generates one unique desired positive integer.
We prove this statement as follows:
Together, we have justified our claim in bold. The answer is \[1+2+3+4+5+6+7+8+9+5=\boxed{050}.\]

Answer:  050
Problem: Find the number of positive integers less than $1000$ that can be expressed as the difference of two integral powers of $2.$

Solution: Because the difference is less than $1000$ , we can simply list out all numbers that satisfy $2^n < 1000$ . We get $0 \le n < 10$ , where n is an integer. Because the sequence $2^n$ is geometric, the difference of any two terms will be unique. $\binom{10}{2}$ will be the number of differences for $0\le n < 10$ . However, we also need to consider the case in which $n=10$ . With simple counting, we find that $5$ numbers: $(32, 64, 128, 256, 512)$ could be subtracted from $1024$ , which makes another 5 cases. There is no need to check for higher exponents since the lowest difference would be $2^{11} - 2^{10} = 1024$ , which exceeds $1000$
Thus, the final answer is $\binom{10}{2} + 5 = \boxed{050}.$

Answer:  050
Problem: Find the number of ways $66$ identical coins can be separated into three nonempty piles so that there are fewer coins in the first pile than in the second pile and fewer coins in the second pile than in the third pile.

Solution: Suppose we have $1$ coin in the first pile. Then $(1, 2, 63), (1, 3, 62), \ldots, (1, 32, 33)$ all work for a total of $31$ piles. Suppose we have $2$ coins in the first pile, then $(2, 3, 61), (2, 4, 60), \ldots, (2, 31, 33)$ all work, for a total of $29$ . Continuing this pattern until $21$ coins in the first pile, we have the sum \begin{align*} 31+29+28+26+25+\cdots+4+2+1 &= (31+28+25+22+\cdots+1)+(29+26+23+\cdots+2) \\ &= 176+155 \\ &= \boxed{331}

Answer:  331
Problem: Find the number of ways $66$ identical coins can be separated into three nonempty piles so that there are fewer coins in the first pile than in the second pile and fewer coins in the second pile than in the third pile.

Solution: We make an equation: $a+b+c=66,$ where $a<b<c.$ We don't have a clear solution, so we'll try complementary counting. First, let's find where $a\geq b\geq c.$ By stars and bars, we have $\dbinom{65}{2}=2080$ to assign positive integer solutions to $a + b + c = 66.$ Now we need to subtract off the cases where it doesn't satisfy the condition.
We start with $a = b.$ We can write that as $2b + c = 66.$ We can find there are 32 integer solutions to this equation. There are $32$ solutions for $b=c$ and $a = c$ by symmetry. We also need to add back $2$ because we subtracted $(a,b,c)=(22,22,22)$ $3$ times.
We then have to divide by $6$ because there are $3!=6$ ways to order $a, b,$ and $c.$ Therefore, we have $\dfrac{\dbinom{65}{2}-96+2}{6} = \dfrac{1986}{6} = \boxed{331}.$

Answer:  331
Problem: Find the number of ways $66$ identical coins can be separated into three nonempty piles so that there are fewer coins in the first pile than in the second pile and fewer coins in the second pile than in the third pile.

Solution: Let the piles have $a, b$ and $c$ coins, with $0 < a < b < c$ . Then, let $b = a + k_1$ , and $c = b + k_2$ , such that each $k_i \geq 1$ . The sum is then $a + a+k_1 + a+k_1+k_2 = 66 \implies 3a+2k_1 + k_2 = 66$ . This is simply the number of positive solutions to the equation $3x+2y+z = 66$ . Now, we take cases on $a$
If $a = 1$ , then $2k_1 + k_2 = 63 \implies 1 \leq k_1 \leq 31$ . Each value of $k_1$ corresponds to a unique value of $k_2$ , so there are $31$ solutions in this case. Similarly, if $a = 2$ , then $2k_1 + k_2 = 60 \implies 1 \leq k_1 \leq 29$ , for a total of $29$ solutions in this case. If $a = 3$ , then $2k_1 + k_2 = 57 \implies 1 \leq k_1 \leq 28$ , for a total of $28$ solutions. In general, the number of solutions is just all the numbers that aren't a multiple of $3$ , that are less than or equal to $31$
We then add our cases to get \begin{align*} 1 + 2 + 4 + 5 + \cdots + 31 &= 1 + 2 + 3 + \cdots + 31 - 3(1 + 2 + 3 + \cdots + 10) \\ &= \frac{31(32)}{2} - 3(55) \\ &= 31 \cdot 16 - 165 \\ &= 496 - 165 \\ &= \boxed{331} as our answer.

Answer:  331
Problem: Call a three-term strictly increasing arithmetic sequence of integers special if the sum of the squares of the three terms equals the product of the middle term and the square of the common difference. Find the sum of the third terms of all special sequences.

Solution: Let the terms be $a-b$ $a$ , and $a+b$ . Then we want $(a-b)^2+a^2+(a+b)^2=ab^2$ , or $3a^2+2b^2=ab^2$ . Rearranging, we get $b^2=\frac{3a^2}{a-2}$ . Simplifying further, $b^2=3a+6+\frac{12}{a-2}$ . Looking at this second equation, since the right side must be an integer, $a-2$ must equal $\pm1, 2, 3, 4, 6, 12$ . Looking at the first equation, we see $a>2$ since $b^2$ is positive. This means we must test $a=3, 4, 5, 6, 8, 14$ . After testing these, we see that only $a=5$ and $a=14$ work which give $b=5$ and $b=7$ respectively. Thus the answer is $10+21=\boxed{031}$ .
~JHawk0224

Answer:  031
Problem: Call a three-term strictly increasing arithmetic sequence of integers special if the sum of the squares of the three terms equals the product of the middle term and the square of the common difference. Find the sum of the third terms of all special sequences.

Solution: Let the common difference be $d$ and let the middle term be $x$ . Then, we have that the sequence is \[x-d,~x,~x+d.\] This means that the sum of the squares of the 3 terms of the sequence is \[(x-d)^2+x^2+(x+d)^2=x^2-2xd+d^2+x^2+x^2+2xd+d^2=3x^2+2d^2.\] We know that this must be equal to $xd^2,$ so we can write that \[3x^2+2d^2=xd^2,\] and it follows that \[3x^2-xd^2+2d^2=3x^2-\left(d^2\right)x+2d^2=0.\]
Now, we can treat $d$ as a constant and use the quadratic formula to get \[x=\frac{d^2\pm \sqrt{d^4-4(3)(2d^2)}}{6}.\] We can factor pull $d^2$ out of the square root to get \[x=\frac{d^2\pm d\sqrt{d^2-24}}{6}.\] Here, it is easy to figure out the values of $d$ . Let $\sqrt{d^2-24} = k$ , then $d^2-k^2=24$ which is $(d+k)(d-k)=24,$ note that $d$ $k$ are integers. Examining the parity, we find that $d+k$ and $d-k$ are of the same parity. Now, we solve by factoring. We can find that $d=5$ and $d=7$ are the only positive integer values of $d$ that make $\sqrt{d^2-24}$ a positive integer. $^{*}$ $d=5$ gives $x=5$ and $x=\frac{10}{3}$ , but we can ignore the latter. $d=7$ gives $x=14$ , as well as a fraction which we can ignore.
Since $d=5,~x=5$ and $d=7, x=14$ are the only two solutions and we want the sum of the third terms, our answer is $(5+5)+(7+14)=10+21=\boxed{031}$ . -BorealBear, minor edit by Kinglogic

Answer:  031
Problem: Call a three-term strictly increasing arithmetic sequence of integers special if the sum of the squares of the three terms equals the product of the middle term and the square of the common difference. Find the sum of the third terms of all special sequences.

Solution: As in Solution 1, write the three integers in the sequence as $a-d$ $a$ , and $a+d$
Then the sum of the squares of the three integers is $(a-d)^2+a^2+(a+d)^2 = 3a^2+2d^2$
Setting this equal to the middle term times the common difference squared, which is $ad^2$
and solving for $d^2$ we get:
$3a^2+2d^2 = ad^2 \implies ad^2-2d^2 = 3a^2 \implies d^2(a-2) = 3a^2 \implies d^2 = \frac{3a^2}{a-2}$
The numerator has to be positive, so the denominator has to be positive too for the sequence
to be strictly increasing; that is, $a>2$
For $\frac{3a^2}{a-2}$ to be a perfect square, $\frac{3}{a-2}$ must be a perfect square as well.
This means that $a-2$ is divisible by 3, and whatever left over is a perfect square.
We can express this as an equation: let the perfect square left over be $n^2$ . Then:
$3n^2 = a-2$ . Now when you divide the numerator and denominator by 3, you are left with
$d^2 = \frac{a^2}{n^2} \implies d = \frac{a}{n}$ . Because the sequence is of integers, d must also be an
integer, which means that $n$ must divide $a$
Taking the above equation we can solve for $a$ $3n^2 = a-2 \implies a = 3n^2+2$
This means that $3n^2+2$ is divisible by $n$ $3n^2$ is automatically divisible by $n$ , so
$2$ must be divisible by $n$ . Then $n$ must be either of $\{1,2\}$ . Plugging back into the 
equation,
$n = 1 \implies a = 5 \implies d = 5$ , so $a+d = 5+5 = 10$
$n = 2 \implies a = 14 \implies d = 7$ , so $a+d = 14+7 = 21$
Finally, $10+21 = \boxed{031}$

Answer:  031
Problem: Call a three-term strictly increasing arithmetic sequence of integers special if the sum of the squares of the three terms equals the product of the middle term and the square of the common difference. Find the sum of the third terms of all special sequences.

Solution: Following from previous solutions, we derive $3x^2+2a^2=xa^2.$ We divide both sides to get $3\left(\frac{x}{a}\right)^2+2=x.$ Since $x$ is an integer, $\frac{x}{a}$ must also be an integer, so we have $x=pa$ , for some factor $p$ . We then get $3p^2+2=pa.$ We then take this to modulo $p$ , getting $2\equiv 0 \pmod p.$ The only possibilities for $p$ are therefore 1 and 2. We plug these into $3p^2+2=pa$ , for $a=5$ and $x=5$ , giving us the sequence $0,5,10$ , or $2a=14$ and $x=14$ , for the sequence $7,14,21.$ $10+21 = \boxed{031}.$

Answer:  031