Navvye/Olympiad-Problems-CSV huggingface dataset:
Problem: Misha rolls a standard, fair six-sided die until she rolls $1-2-3$ in that order on three consecutive rolls. The probability that she will roll the die an odd number of times is $\dfrac{m}{n}$ where $m$ and $n$ are relatively prime positive integers. Find $m+n$

Solution: Let $P_n$ be the probability of getting consecutive $1,2,3$ rolls in $n$ rolls and not rolling $1,2,3$ prior to the nth roll.
Let $x = P_3+P_5+...=1-(P_4+P_6+..)$ . Following Solution 2, one can see that \[P_{n+1}=P_{n}-\frac{P_{n-2}}{6^3}\] for all positive integers $n \ge 5$ . Summing for $n=5,7,...$ gives \[(1-x)-\frac{1}{6^3}=x-\frac{1}{6^3}-\frac{x}{6^3}\] \[\implies x = \frac{m}{n} = \frac{216}{431} \implies m+n=216+431= \boxed{647}\]

Answer:  647
Problem: Misha rolls a standard, fair six-sided die until she rolls $1-2-3$ in that order on three consecutive rolls. The probability that she will roll the die an odd number of times is $\dfrac{m}{n}$ where $m$ and $n$ are relatively prime positive integers. Find $m+n$

Solution: Let $P_{odd}=\frac{m}{n}$ , with the subscript indicating an odd number of rolls. Then $P_{even}=1-\frac{m}{n}$
The ratio of $\frac{P_{odd}}{P_{even}}$ is just $\frac{P_{odd}}{1-P_{odd}}$
We see that $P_{odd}$ is the sum of $P_3$ $P_5$ $P_7$ ,... , while $P_{even}$ is the sum of $P_4$ $P_6$ $P_8$ ,... .
$P_3$ , the probability of getting rolls of 1-2-3 in exactly 3 rolls, is obviously $\frac{1}{216}$
We set this probability of $P_3$ aside, meaning we totally remove the chance of getting 1-2-3 in 3 rolls. Now the ratio of $P_4+P_6+P_8+...$ to $P_5+P_7+P_9+...$ should be equal to the ratio of $\frac{P_{odd}}{P_{even}}$ , because in this case the 1st roll no longer matters, so we can disregard the very existence of it in counting how many times of rolls, and thus, 4 rolls, 6 rolls, 8 rolls... would become an odd number of rolls (while 5 rolls, 7 rolls, 9 rolls... would become even number of rolls).
Notice $P_4+P_6+P_8+...=P_{even}$ , and also $P_5+P_7+P_9+...=P_{odd}-P_3=P_{odd}-\frac{1}{216}$
So we have $\frac{P_{even}}{P_{odd}-\frac{1}{216}}=\frac{P_{odd}}{P_{even}}$
Finally, we get $P_{odd}=\frac{m}{n}=\frac{216}{431}$ . 
Therefore, $m+n = \boxed{647}$

Answer:  647
Problem: Misha rolls a standard, fair six-sided die until she rolls $1-2-3$ in that order on three consecutive rolls. The probability that she will roll the die an odd number of times is $\dfrac{m}{n}$ where $m$ and $n$ are relatively prime positive integers. Find $m+n$

Solution: Call the probability you win on a certain toss $f_n$ , where $n$ is the toss number.
Obviously, since the sequence has length 3, $f_1=0$ and $f_2=0$ .
Additionally, $f_3=\left(\frac{1}{6}\right)^3$ . We can call this value $x$ , to keep our further equations looking clean.
We can now write our general form for $f$ as $f_n=x\left(1-\sum_{i=1}^{n-3}f_i\right)$ . This factors the probability of the last 3 rolls being 1-2-3, and the important probability that the sequence has not been rolled in the past (because then the game would already be over).
Note that $\sum_{i=1}^{\infty}f_i=1$ since you'll win at some point.
An intermediate step here is figuring out $f_n-f_{n+1}$ . This is equal to $x\left(1-\sum_{i=1}^{n-3}f_i\right)-x\left(1-\sum_{i=1}^{n-2}f_i\right)=x\left(\sum_{i=1}^{n-2}f_i-\sum_{i=1}^{n-3}f_i\right)=xf_{n-2}$ .
Adding up all the differences, i.e. $\sum_{i=2}^{\infty}(f_{2n-1}-f_{2n})$ will give us the amount by which the odds probability exceeds the even probability. Since they sum to 1, that means the odds probability will be half of the difference above one-half. Subbing in our earlier result from the intermediate step, the odd probability is equal to $\frac{1}{2}+\frac{1}{2}x\sum_{i=2}^{\infty}f_{2n-3}$ .
Another way to find the odd probability is simply summing it up, which turns out to be $\sum_{i=1}^{\infty}f_{2n-1}$ . Note the infinite sums in both expressions are equal; let's call it $P$ . Equating them gives $\frac{1}{2}+\frac{1}{2}xP=P$ , or $P=\frac{1}{2-x}$ .
Finally, substituting $x=\frac{1}{216}$ , we find that $P=\frac{216}{431}$ , giving us a final answer of $216 + 431 = \boxed{647}$

Answer:  647
Problem: Misha rolls a standard, fair six-sided die until she rolls $1-2-3$ in that order on three consecutive rolls. The probability that she will roll the die an odd number of times is $\dfrac{m}{n}$ where $m$ and $n$ are relatively prime positive integers. Find $m+n$

Solution: Let $S(n)$ be the number of strings of length $n$ containing the digits $1$ through $6$ that do not contain the string $123$ . Then we have $S(n) = 6 \cdot S(n-1) - S(n-3)$ because we can add any digit to end of a string with length $n-1$ but we have to subtract all the strings that end in $123$ . We rewrite this as \begin{align*} S(n) &= 6 \cdot S(n-1) - S(n-3) \\ &= 6 \cdot (6 \cdot S(n-2) - S(n-4)) - (6 \cdot (S(n-4) - S(n-6)) \\ &= 36 \cdot S(n-2) - 12 \cdot S(n-4) + S(n-6) \end{align*} We wish to compute $P=\sum_{n=0}^\infty \frac{S(2n)}{6^{2n+3}}$ since the last three rolls are $123$ for the game to end. Summing over the recursion, we obtain \[\sum_{n=0}^\infty \frac{S(2n)}{6^{2n+3}} =36 \cdot \sum_{n=0}^\infty \frac{S(2n-2)}{6^{2n+3}} - 12 \cdot  \sum_{n=0}^\infty \frac{S(2n-4)}{6^{2n+3}}+ \sum_{n=0}^\infty \frac{S(2n-6)}{6^{2n+3}}\] Now shift the indices so that the inside term is the same: \begin{align*} \sum_{n=3}^\infty  \frac{S(2n)}{6^{2n+3}} &= \frac{36}{6^2} \cdot \sum_{n=2}^\infty \frac{S(2n)}{6^{2n+3}} - \frac{12}{6^4} \cdot \sum_{n=1}^\infty \frac{S(2n)}{6^{2n+3}} + \frac{1}{6^6} \cdot \sum_{n=0}^\infty \frac{S(2n)}{6^{2n+3}}  \\  \left(P - \frac{S(0)}{6^3} -  \frac{S(2)}{6^5} -\frac{S(4)}{6^7}  \right) &= \frac{36}{6^2} \cdot \left( P -   \frac{S(0)}{6^3} -  \frac{S(2)}{6^5}\right) - \frac{12}{6^4} \cdot \left(  P - \frac{S(0)}{6^3} \right) + \frac{1}{6^6} \cdot P  \end{align*} Note that $S(0) = 1, S(2) = 36$ and $S(4) = 6^4 - 2 \cdot 6 = 1284$ . Therefore, \begin{align*} \left(P - \frac{1}{6^3} -  \frac{36}{6^5} -\frac{1284}{6^7}  \right) = \frac{36}{6^2} \cdot \left( P -   \frac{1}{6^3} -  \frac{36}{6^5}\right) - \frac{12}{6^4} \cdot \left(  P - \frac{1}{6^3} \right) + \frac{1}{6^6} \cdot P   \end{align*} Solving for $P$ , we obtain $P = \frac{216}{431} \implies m+n = \boxed{647}$

Answer:  647
Problem: Misha rolls a standard, fair six-sided die until she rolls $1-2-3$ in that order on three consecutive rolls. The probability that she will roll the die an odd number of times is $\dfrac{m}{n}$ where $m$ and $n$ are relatively prime positive integers. Find $m+n$

Solution: Let $A=\frac{1}{6} \begin{bmatrix} 5 & 1 & 0 & 0 \\ 4 & 1 & 1 & 0 \\ 4 & 1 & 0 & 1 \\ 0 & 0 & 0 & 0 \\ \end{bmatrix}$ $A$ is a transition matrix for the prefix of 1-2-3 matched so far. The state corresponding to a complete match has no outgoing probability mass. The probability that we roll the dice exactly $k$ times is $(A^k)_{1,4}$ . Thus the probability that we roll the dice an odd number of times is $1-\left(\sum_{k=0}^\infty A^{2k}\right)_{1,4} = 1-\left((I - A^2)^{-1}\right)_{1,4} = \frac{216}{431}$ . Thus the answer is $216+431=\boxed{647}$

Answer:  647
Problem: Misha rolls a standard, fair six-sided die until she rolls $1-2-3$ in that order on three consecutive rolls. The probability that she will roll the die an odd number of times is $\dfrac{m}{n}$ where $m$ and $n$ are relatively prime positive integers. Find $m+n$

Solution: Consider it as a contest of Odd and Even. Let $P_o$ and $P_e$ be probability that Odd and Even wins, respectively. If we consider every 3 rolls as an atomic action, then we can have a simple solution. If the rolls is 1-2-3, Odd wins; otherwise, Odd and Even switch the odds of winning. Therefore, we have \[P_o = \frac{1}{216} + \frac{215}{216}P_e\] Plug in $P_e = 1 - P_o$ and we can easily solve for $Po=\frac{216}{431}$
$\boxed{647}$

Answer:  647
Problem: Misha rolls a standard, fair six-sided die until she rolls $1-2-3$ in that order on three consecutive rolls. The probability that she will roll the die an odd number of times is $\dfrac{m}{n}$ where $m$ and $n$ are relatively prime positive integers. Find $m+n$

Solution: Let $P_o$ $P_e$ be the winning probabilities respectively. We call Odd "in position" when a new sequence of 1-2-3 starts at odd position, and likewise, call Even is "in position" when a new sequence starts at even position.
Now, consider the situation when the first roll is $1.$ The conditional probability for Odd or Even to eventually win out depends on whose is in position. So let's denote by $P_o(1), P_e(1)$ the probabilities of Odd and Even winning out, respectively, both when Odd is in position. Remember that the probabilities simply switch if Even is in position. Similarly, after 1-2 is rolled, we denote by $P_o(2), P_e(2)$ the conditional probabilities of Odd and Even winning out, when Odd is in position.
Consider the first roll. If it's not a 1, the sequence restarts, but Even is now in position; if it's a 1, then Odd's winning probability becomes $P_o(1)$ . So, \[P_o = \frac{1}{6}P_o(1) + \frac{5}{6}P_e\] In the next roll, there are 3 outcomes. If the roll is 2, then Odd's winning probability becomes $P_o(2)$ ; if the roll is 1, then we stay in the sequence, but Even is now in position, so the probability of Odd winning now becomes $P_e(1)$ ; if the rolls is any other number, then the sequence restarts, and Odd is still in position. So, \[P_o(1) = \frac{1}{6}P_o(2) + \frac{1}{6}P_e(1) + \frac{4}{6}P_o\] In the next roll after a 1-2 sequence, there are 3 outcomes. If the roll is a 3, Odd wins; if it's a 1, we go back to the state when 1 is just rolled, and Odd is in position; if it's any other number, then the sequence restarts, and Even is in position. So, \[P_o(2) = \frac{1}{6} + \frac{1}{6}P_o(1) + \frac{4}{6}P_e\]
Plug in $P_e = 1-P_o$ and $P_e(1) = 1 - P_o(1)$ , we have a 3-equation linear system which is not hard to solve. The final answer is $Po=\frac{216}{431}$ $\boxed{647}$ . (We want P_o because if it starts odd, it will also end odd)

Answer:  647
Problem: Misha rolls a standard, fair six-sided die until she rolls $1-2-3$ in that order on three consecutive rolls. The probability that she will roll the die an odd number of times is $\dfrac{m}{n}$ where $m$ and $n$ are relatively prime positive integers. Find $m+n$

Solution: Take a block of any three possible rolls. You have $6^3$ different possibilities for a block and $6^3 - 1$ different possibilities to have a block without 1-2-3. This gives two probabilities of $\frac{6^3-1}{6^3}$ and $\frac{1}{6^3}$ , respectively. To get an odd number of total rolls, we need an even number of blocks without a 1-2-3. Then, we can sum up the probabilities up to infinity to see the total probability of getting an even number of rolls. \[\frac{1}{6^3}\sum_{k=0}^{\infty}\left(\frac{6^3-1}{6^3}\right)^{2k}\] \[=\frac{\frac{1}{6^3}}{1-\left(\frac{6^3-1}{6^3}\right)^2}\] Using difference of squares on the bottom half we can reduce it to $\left(1-\frac{6^3-1}{6^3}\right)\left(1+\frac{6^3-1}{6^3}\right)=\frac{2\times 6^3 -1}{(6^3)^2}$ . Plugging this back in to the equation we get- \[\frac{\frac{1}{6^3}}{\frac{2\times 6^3 -1}{(6^3)^2}} = \frac{6^3}{2\times 6^3 - 1} = \frac{216}{431}\] Then, our answer is $\boxed{647}$ -Mathiscool109

Answer:  647
Problem: The incircle $\omega$ of triangle $ABC$ is tangent to $\overline{BC}$ at $X$ . Let $Y \neq X$ be the other intersection of $\overline{AX}$ with $\omega$ . Points $P$ and $Q$ lie on $\overline{AB}$ and $\overline{AC}$ , respectively, so that $\overline{PQ}$ is tangent to $\omega$ at $Y$ . Assume that $AP = 3$ $PB = 4$ $AC = 8$ , and $AQ = \dfrac{m}{n}$ , where $m$ and $n$ are relatively prime positive integers. Find $m+n$

Solution: Let the sides $\overline{AB}$ and $\overline{AC}$ be tangent to $\omega$ at $Z$ and $W$ , respectively. Let $\alpha = \angle BAX$ and $\beta = \angle AXC$ . Because $\overline{PQ}$ and $\overline{BC}$ are both tangent to $\omega$ and $\angle YXC$ and $\angle QYX$ subtend the same arc of $\omega$ , it follows that $\angle AYP = \angle QYX = \angle YXC = \beta$ . By equal tangents, $PZ = PY$ . Applying the Law of Sines to $\triangle APY$ yields \[\frac{AZ}{AP} = 1 + \frac{ZP}{AP} = 1 + \frac{PY}{AP} = 1 + \frac{\sin\alpha}{\sin\beta}.\] Similarly, applying the Law of Sines to $\triangle ABX$ gives \[\frac{AZ}{AB} = 1 - \frac{BZ}{AB} = 1 - \frac{BX}{AB} = 1 - \frac{\sin\alpha}{\sin\beta}.\] It follows that \[2 = \frac{AZ}{AP} + \frac{AZ}{AB} = \frac{AZ}3 + \frac{AZ}7,\] implying $AZ = \tfrac{21}5$ . Applying the same argument to $\triangle AQY$ yields \[2 = \frac{AW}{AQ} + \frac{AW}{AC} = \frac{AZ}{AQ} + \frac{AZ}{AC} = \frac{21}5\left(\frac{1}{AQ} + \frac 18\right),\] from which $AQ = \tfrac{168}{59}$ . The requested sum is $168 + 59 = \boxed{227}$

Answer:  227
Problem: The incircle $\omega$ of triangle $ABC$ is tangent to $\overline{BC}$ at $X$ . Let $Y \neq X$ be the other intersection of $\overline{AX}$ with $\omega$ . Points $P$ and $Q$ lie on $\overline{AB}$ and $\overline{AC}$ , respectively, so that $\overline{PQ}$ is tangent to $\omega$ at $Y$ . Assume that $AP = 3$ $PB = 4$ $AC = 8$ , and $AQ = \dfrac{m}{n}$ , where $m$ and $n$ are relatively prime positive integers. Find $m+n$

Solution: Let the incircle of $ABC$ be tangent to $AB$ and $AC$ at $Z$ and $W$ . By Brianchon's theorem on tangential hexagons $QWCBZP$ and $PYQCXB$ , we know that $ZW,CP,BQ$ and $XY$ are concurrent at a point $O$ . Let $PQ \cap BC = M$ . Then by La Hire's $A$ lies on the polar of $M$ so $M$ lies on the polar of $A$ . Therefore, $ZW$ also passes through $M$ . Then projecting through $M$ , we have \[-1 = (A,O;Y,X) \stackrel{M}{=} (A,Z;P,B) \stackrel{M}{=} (A,W;Q,C).\] Therefore, $\frac{AP \cdot ZB}{MP \cdot AB} = 1 \implies \frac{3 \cdot ZB}{ZP \cdot 7} = 1$ . Since $ZB+ZP=4$ we know that $ZP = \frac{6}{5}$ and $ZB = \frac{14}{5}$ . Therefore, $AW = AZ = \frac{21}{5}$ and $WC = 8 - \frac{21}{5} = \frac{19}{5}$ . Since $(A,W;Q,C) = -1$ , we also have $\frac{AQ \cdot WC}{NQ \cdot AC} = 1 \implies \frac{AQ \cdot \tfrac{19}{5}}{(\tfrac{21}{5} - AQ) \cdot 8} = 1$ . Solving for $AQ$ , we obtain $AQ = \frac{168}{59} \implies m+n = \boxed{227}$ .
😃
-Vfire

Answer:  227
Problem: The incircle $\omega$ of triangle $ABC$ is tangent to $\overline{BC}$ at $X$ . Let $Y \neq X$ be the other intersection of $\overline{AX}$ with $\omega$ . Points $P$ and $Q$ lie on $\overline{AB}$ and $\overline{AC}$ , respectively, so that $\overline{PQ}$ is tangent to $\omega$ at $Y$ . Assume that $AP = 3$ $PB = 4$ $AC = 8$ , and $AQ = \dfrac{m}{n}$ , where $m$ and $n$ are relatively prime positive integers. Find $m+n$

Solution: Let the center of the incircle of $\triangle ABC$ be $O$ . Link $OY$ and $OX$ . Then we have $\angle OYP=\angle OXB=90^{\circ}$
$\because$ $OY=OX$
$\therefore$ $\angle OYX=\angle OXY$
$\therefore$ $\angle PYX=\angle YXB$
$\therefore$ $\sin \angle PYX=\sin \angle YXB=\sin \angle YXC=\sin \angle PYA$
Let the incircle of $ABC$ be tangent to $AB$ and $AC$ at $M$ and $N$ , let $MP=YP=x$ and $NQ=YQ=y$
Use Law of Sine in $\triangle APY$ and $\triangle AXB$ , we have
$\frac{\sin \angle PAY}{PY}=\frac{\sin \angle PYA}{PA}$
$\frac{\sin \angle BAX}{BX}=\frac{\sin \angle AXB}{AB}$
therefore we have
$\frac{3}{x}=\frac{7}{4-x}$
Solve this equation, we have $x=\frac{6}{5}$
As a result, $MB=4-x=\frac{14}{5}=BX$ $AM=x+3=\frac{21}{5}=AN$ $NC=8-AN=\frac{19}{5}=XC$ $AQ=\frac{21}{5}-y$ $PQ=\frac{6}{5}+y$
So, $BC=\frac{14}{5}+\frac{19}{5}=\frac{33}{5}$
Use Law of Cosine in $\triangle BAC$ and $\triangle PAQ$ , we have
$\cos \angle BAC=\frac{AB^2+AC^2-BC^2}{2\cdot AB\cdot AC}=\frac{7^2+8^2-{(\frac{33}{5})}^2}{2\cdot 7\cdot 8}$
$\cos \angle PAQ=\frac{AP^2+AQ^2-PQ^2}{2\cdot AP\cdot AQ}=\frac{3^2+{(\frac{21}{5}-y)}^2-{(\frac{6}{5}+y)}^2}{2\cdot {(\frac{21}{5}-y)}\cdot 3}$
And we have
$\cos \angle BAC=\cos \angle PAQ$
So
$\frac{7^2+8^2-{(\frac{33}{5})}^2}{2\cdot 7\cdot 8}=\frac{3^2+{(\frac{21}{5}-y)}^2-{(\frac{6}{5}+y)}^2}{2\cdot {(\frac{21}{5}-y)}\cdot 3}$
Solve this equation, we have $y=\frac{399}{295}=QN$
As a result, $AQ=AN-QN=\frac{21}{5}-\frac{399}{295}=\frac{168}{59}$
So, the final answer of this question is $168+59=\boxed{227}$

Answer:  227
Problem: Find the number of functions $f$ from $\{0, 1, 2, 3, 4, 5, 6\}$ to the integers such that $f(0) = 0$ $f(6) = 12$ , and \[|x - y| \leq |f(x) - f(y)| \leq 3|x - y|\] for all $x$ and $y$ in $\{0, 1, 2, 3, 4, 5, 6\}$

Solution: Because $f(n)$ and $f(n+1)$ can differ by at most 3 for $n=0,1,2,3,4,5$ $f$ can decrease at most once. This is because if $f$ decreases $2$ times (or more) then we get a contradiction: $12 = f(6) \le 3\cdot 4 + (-1)\cdot 2 = 10$
If $f$ never decreases, then $f(n+1)-f(n)\in \{1,2,3\}$ for all $n$ . Let $a, b$ , and $c$ denote the number of times this difference is $1, 2$ , and $3$ , respectively. Then $a+b+c=6$ and $a+2b+3c = 12$ . Subtracting the first equation from the second yields $b+2c=6$ , so $(b,c)=(6,0), (4,1), (2,2)$ , or $(0,3)$ . These yield $a=0,1,2$ , or $3$ , respectively, so the number of possibilities in this case is \[\binom{6}{0,6,0}+\binom{6}{1,4,1}+\binom{6}{2,2,2}+\binom{6}{3,0,3}=141.\] If $f$ decreases from $f(0)$ to $f(1)$ or from $f(5)$ to $f(6)$ , then $f(2)$ or $f(4)$ , respectively, is determined. The only solutions to $a+b+c=4$ and $a+2b+3c=10$ are $(a,b,c)=(1,0,3)$ and $(a,b,c)=(0,2,2)$ , so the number of functions is \[2\left[\binom{4}{1,0,3}+\binom{4}{0,2,2}\right]=20.\] Finally, suppose that $f(n+1)<f(n)$ for some $n=1,2,3,4$ . Note that the condition $|(n+1)-(n-1)|\le |f(n+1)-f(n-1)|$ implies that $f(n+1)-f(n-1)\ge 2$ , so it must be that $f(n)-f(n+1)=1$ and \[f(n+2)-f(n+1)=f(n)-f(n-1)=3.\] This means that $f(n-1)$ and $f(n+2)$ are uniquely determined by the value of $f(n)$ , and, in particular, that $f(n+2)-f(n-1)=5$ . As a result, there are three more values of $f$ to determine, and they must provide a total increase of $7$ . The only ways to do this are either to have two differences of $3$ and one difference of $1$ , which can be arranged in $3$ ways, or to have one difference of $3$ and two differences of $2$ , which can be arranged in $3$ ways. Thus for each of the $4$ possibilities for $n$ , there are $6$ ways to arrange the increases, giving a total of $24$ ways.
The total number of functions is $141+20+24=\boxed{185}$

Answer:  185
Problem: Fifteen distinct points are designated on $\triangle ABC$ : the 3 vertices $A$ $B$ , and $C$ $3$ other points on side $\overline{AB}$ $4$ other points on side $\overline{BC}$ ; and $5$ other points on side $\overline{CA}$ . Find the number of triangles with positive area whose vertices are among these $15$ points.

Solution: Every triangle is uniquely determined by 3 points. There are $\binom{15}{3}=455$ ways to choose 3 points, but that counts the degenerate triangles formed by choosing three points on a line. There are $\binom{5}{3}$ invalid cases on segment $AB$ $\binom{6}{3}$ invalid cases on segment $BC$ , and $\binom{7}{3}$ invalid cases on segment $CA$ for a total of $65$ invalid cases. The answer is thus $455-65=\boxed{390}$

Answer:  390
Problem: When each of $702$ $787$ , and $855$ is divided by the positive integer $m$ , the remainder is always the positive integer $r$ . When each of $412$ $722$ , and $815$ is divided by the positive integer $n$ , the remainder is always the positive integer $s \neq r$ . Find $m+n+r+s$

Solution: Let's work on both parts of the problem separately. First, \[855 \equiv 787 \equiv 702 \equiv r \pmod{m}.\] We take the difference of $855$ and $787$ , and also of $787$ and $702$ . We find that they are $85$ and $68$ , respectively. Since the greatest common divisor of the two differences is $17$ (and the only one besides one), it's safe to assume that $m = 17$
Then, we divide $855$ by $17$ , and it's easy to see that $r = 5$ . Dividing $787$ and $702$ by $17$ also yields remainders of $5$ , which means our work up to here is correct.
Doing the same thing with $815$ $722$ , and $412$ , the differences between $815$ and $722$ and $412$ are $310$ and $93$ , respectively. Since the only common divisor (besides $1$ , of course) is $31$ $n = 31$ . Dividing all $3$ numbers by $31$ yields a remainder of $9$ for each, so $s = 9$ . Thus, $m + n + r + s = 17 + 5 + 31 + 9 = \boxed{062}$

Answer:  062
Problem: When each of $702$ $787$ , and $855$ is divided by the positive integer $m$ , the remainder is always the positive integer $r$ . When each of $412$ $722$ , and $815$ is divided by the positive integer $n$ , the remainder is always the positive integer $s \neq r$ . Find $m+n+r+s$

Solution: We know that $702 = am + r, 787 = bm + r,$ and $855 = cm+r$ where $a-c$ are integers.
Subtracting the first two, the first and third, and the last two, we get $85 = (b-a)m, 153=(c-a)m,$ and $68=(c-b)m.$
We know that $b-a, c-a$ and $c-b$ must be integers, so all the numbers are divisible by $m.$
Factorizing the numbers, we get $85 = 5 \cdot 17, 153 = 3^2 \cdot 17,$ and $68 = 2^2 \cdot 17.$ We see that all these have a factor of 17, so $m=17.$
Finding the remainder when $702$ is divided by $17,$ we get $n=5.$
Doing the same thing for the next three numbers, we get $17 + 5 + 31 + 9 = \boxed{062}$

Answer:  062
Problem: When each of $702$ $787$ , and $855$ is divided by the positive integer $m$ , the remainder is always the positive integer $r$ . When each of $412$ $722$ , and $815$ is divided by the positive integer $n$ , the remainder is always the positive integer $s \neq r$ . Find $m+n+r+s$

Solution: As in Solution 1, we are given $855\equiv787\equiv702\equiv r\pmod{m}$ and $815\equiv722\equiv412\equiv s\pmod{n}.$ Tackling the first equation, we can simply look at $855\equiv787\equiv702\pmod{m}$ . We subtract $702$ from each component of the congruency to get $153\equiv85\equiv0\pmod{m}$ . Thus, we know that $153$ and $85$ must both be divisible by $m$ . The only possible $m$ , in this case, become $17$ and $1$ ; obviously, $m\neq1$ , so we know $m=17$ . We go back to the original equation, plug in $m$ , and we find that $r=5$
Similarly, we can subtract out the smallest value in the second congruency, $412$ . We end up with $403\equiv310\equiv0\pmod{n}$ . Again, we find that $n=31$ or $1$ , so $n=31$ . We also find that $s=9$
Thus, our answer is $\boxed{062}$

Answer:  062
Problem: For a positive integer $n$ , let $d_n$ be the units digit of $1 + 2 + \dots + n$ . Find the remainder when \[\sum_{n=1}^{2017} d_n\] is divided by $1000$

Solution: We see that $d_n$ appears in cycles of $20$ and the cycles are \[1,3,6,0,5,1,8,6,5,5,6,8,1,5,0,6,3,1,0,0,\] adding a total of $70$ each cycle.
Since $\left\lfloor\frac{2017}{20}\right\rfloor=100$ , we know that by $2017$ , there have been $100$ cycles and $7000$ has been added. This can be discarded as we're just looking for the last three digits.
Adding up the first $17$ of the cycle of $20$ , we can see that the answer is $\boxed{069}$

Answer:  069