Navvye/Olympiad-Problems-CSV huggingface dataset:
Problem: Let $a, b, c,$ and $d$ be real numbers that satisfy the system of equations \begin{align*} a + b &= -3, \\ ab + bc + ca &= -4, \\ abc + bcd + cda + dab &= 14, \\ abcd &= 30. \end{align*} There exist relatively prime positive integers $m$ and $n$ such that \[a^2 + b^2 + c^2 + d^2 = \frac{m}{n}.\] Find $m + n$

Solution: For simplicity purposes, we number the given equations $(1),(2),(3),$ and $(4),$ in that order.
Rearranging $(2)$ and solving for $c,$ we have \begin{align*} ab+(a+b)c&=-4 \\ ab-3c&=-4 \\ c&=\frac{ab+4}{3}. \hspace{14mm} (5) \end{align*} Substituting $(5)$ into $(4)$ and solving for $d,$ we get \begin{align*} ab\left(\frac{ab+4}{3}\right)d&=30 \\ d&=\frac{90}{ab(ab+4)}. \hspace{5mm} (6)  \end{align*} Substituting $(5)$ and $(6)$ into $(3)$ and simplifying, we rewrite the left side of $(3)$ in terms of $a$ and $b$ only: \begin{align*} ab\left[\frac{ab+4}{3}\right] + b\left[\frac{ab+4}{3}\right]\left[\frac{90}{ab(ab+4)}\right] + \left[\frac{ab+4}{3}\right]\left[\frac{90}{ab(ab+4)}\right]a + \left[\frac{90}{ab(ab+4)}\right]ab &= 14 \\ ab\left[\frac{ab+4}{3}\right] + \underbrace{\frac{30}{a} + \frac{30}{b}}_{\text{Group them.}} + \frac{90}{ab+4} &= 14 \\ ab\left[\frac{ab+4}{3}\right] + \frac{30(a+b)}{ab} + \frac{90}{ab+4} &= 14 \\ ab\left[\frac{ab+4}{3}\right] + \underbrace{\frac{-90}{ab} + \frac{90}{ab+4}}_{\text{Group them.}} &= 14 \\ ab\left[\frac{ab+4}{3}\right] - \frac{360}{ab(ab+4)}&=14. \end{align*} Let $t=ab(ab+4),$ from which \[\frac{t}{3}-\frac{360}{t}=14.\] Multiplying both sides by $3t,$ rearranging, and factoring give $(t+18)(t-60)=0.$ Substituting back and completing the squares produce \begin{align*} \left[ab(ab+4)+18\right]\left[ab(ab+4)-60\right]&=0 \\ \left[(ab)^2+4ab+18\right]\left[(ab)^2+4ab-60\right]&=0 \\ \underbrace{\left[(ab+2)^2+14\right]}_{ab+2=\pm\sqrt{14}i\implies ab\not\in\mathbb R}\underbrace{\left[(ab+2)^2-64\right]}_{ab+2=\pm8}&=0 \\ ab&=6,-10. \end{align*} If $ab=6,$ then combining this with $(1),$ we know that $a$ and $b$ are the solutions of the quadratic $x^2+3x+6=0.$ Since the discriminant is negative, neither $a$ nor $b$ is a real number.
If $ab=-10,$ then combining this with $(1),$ we know that $a$ and $b$ are the solutions of the quadratic $x^2+3x-10=0,$ or $(x+5)(x-2)=0,$ from which $\{a,b\}=\{-5,2\}.$ Substituting $ab=-10$ into $(5)$ and $(6),$ we obtain $c=-2$ and $d=\frac32,$ respectively. Together, we have \[a^2+b^2+c^2+d^2=\frac{141}{4},\] so the answer is $141+4=\boxed{145}.$

Answer:  145
Problem: Let $a, b, c,$ and $d$ be real numbers that satisfy the system of equations \begin{align*} a + b &= -3, \\ ab + bc + ca &= -4, \\ abc + bcd + cda + dab &= 14, \\ abcd &= 30. \end{align*} There exist relatively prime positive integers $m$ and $n$ such that \[a^2 + b^2 + c^2 + d^2 = \frac{m}{n}.\] Find $m + n$

Solution: Let the four equations from top to bottom be listed 1 through 4 respectively. We factor equation 3 like so: \[abc+d(ab+bc+ca)=14\] Then we plug in equation 2 to receive $abc-4d=14$ . By equation 4 we get $abc=\frac{30}{d}$ . Plugging in, we get $\frac{30}{d}-4d=14$ . Multiply by $d$ on both sides to get the quadratic equation $4d^2+14d-30=0$ . Solving using the quadratic equation, we receive $d=\frac{3}{2},d=-5$ . So, we have to test which one is correct. We repeat a similar process as we did above for equations 1 and 2. We factor equation 2 to get \[ab+c(a+b)=-4\] After plugging in equation 1, we get $ab-3c=-4$ . Now we convert it into a quadratic to receive $3c^2-4c-abc=0$ . The value of $abc$ will depend on $d$ . So we obtain the discriminant $16+12abc$ .
Let d = -5.
Then $abc = \frac{30}{-5}$ , so $abc=-6$ , discriminant is $16-72$ , which makes this a dead end. Thus $d=\frac{3}{2}$ For $d=\frac{3}{2}$ , making $abc=20$ . This means the discriminant is just $256$ , so we obtain two values for $c$ as well. We get either $c=\frac{10}{3}$ or $c=-2$ . So, we must AGAIN test which one is correct.
We know $ab=3c-4$ , and $a+b=-3$ , so we use these values for testing.
Let $c=\frac{10}{3}$ .
Then $ab=6$ , so $a=\frac{6}{b}$ . We thus get $\frac{6}{b}+b=-3$ , which leads to the quadratic $b^2+3b+6$ . The discriminant for this is $9-24$ . That means this value of $c$ is wrong, so $c=-2$ . Thus we get polynomial $b^2+3b-10$ . The discriminant this time is $49$ , so we get two values for $b$ . Through simple inspection, you may see they are interchangeable, as if you take the value $b=2$ , you get $a=-5$ . If you take the value $b=-5$ , you get $a=2$ . So it doesn't matter. That means the sum of all their squares is \[\frac{9}{4}+4+4+25=\frac{141}{4}\] so the answer is $141+4=\boxed{145}.$ ~amcrunner

Answer:  145
Problem: Let $a, b, c,$ and $d$ be real numbers that satisfy the system of equations \begin{align*} a + b &= -3, \\ ab + bc + ca &= -4, \\ abc + bcd + cda + dab &= 14, \\ abcd &= 30. \end{align*} There exist relatively prime positive integers $m$ and $n$ such that \[a^2 + b^2 + c^2 + d^2 = \frac{m}{n}.\] Find $m + n$

Solution: Let the four equations from top to bottom be listed $(1)$ through $(4)$ respectively. Multiplying both sides of $(3)$ by $d$ and factoring some terms gives us $abcd + d^2(ab + ac + bc) = 14d$ . Substituting using equations $(4)$ and $(2)$ gives us $30 -4 d^2 = 14d$ , and solving gives us $d = -5$ or $d = \frac{3}{2}$ . Plugging this back into $(3)$ gives us $abc + d(ab + ac + bc) = abc + (-5)(-4) = abc + 20 = 14$ , or using the other solution for $d$ gives us $abc - 6 = 14$ . Solving both of these equations gives us $abc = -6$ when $d = -5$ and $abc = 14$ when $d = \frac{3}{2}$
Multiplying both sides of $(2)$ by $c$ and factoring some terms gives us $abc + c^2 (a + b) = abc -3c^2 = -4c$ . Testing $abc = -6$ will give us an imaginary solution for $c$ , so therefore $abc = 14$ and $d = \frac{3}{2}$ . This gets us $14 - 3c^2 = -4c$ . Solving for $c$ gives us $c = \frac{3}{10}$ or $c = -2$ . With a bit of testing, we can see that the correct value of $c$ is $c=-2$ . Now we know $a+b = -3$ and $ab + bc + ca = ab + c(a+b) = ab + 6 = -4$ $ab = -10$ , and it is obvious that $a = -5$ and $b = 2$ or the other way around, and therefore, $a^2 + b^2 + c^2 + d^2 = 25 + 4 + 4 + \frac{9}{4} = \frac{141}{4}$ , giving us the answer $141 + 4 = \boxed{145}$ ~hihitherethere

Answer:  145
Problem: An ant makes a sequence of moves on a cube where a move consists of walking from one vertex to an adjacent vertex along an edge of the cube. Initially the ant is at a vertex of the bottom face of the cube and chooses one of the three adjacent vertices to move to as its first move. For all moves after the first move, the ant does not return to its previous vertex, but chooses to move to one of the other two adjacent vertices. All choices are selected at random so that each of the possible moves is equally likely. The probability that after exactly $8$ moves that ant is at a vertex of the top face on the cube is $\frac{m}{n}$ , where $m$ and $n$ are relatively prime positive integers. Find $m + n.$

Solution: Let the state from bottom to top be $B2T,$ from top to top be $T2T,$ from top to bottom be $T2B,$ and from bottom to bottom be $B2B.$ We can draw the following State Transition Diagram with Markov Chain . The numbers on the transition arc are the transition probabilities.
The probabilities of being in a state after $n$ steps and after $n-1$ steps has the following relationships: \begin{align*} B2T(n) &= B2B(n-1) \cdot \frac12\\ T2T(n) &= B2T(n-1) + T2T(n-1) \cdot \frac12\\ T2B(n) &= T2T(n-1) \cdot \frac12\\ B2B(n) &= T2B(n-1) + B2B(n-1) \cdot \frac12 \end{align*} Those probabilities are calculated by Dynamic Programming in the following table: \[\begin{array}{c|cccc}  &  &  &  & \\ [-2ex] n & B2T(n) & T2T(n) & T2B(n) & B2B(n) \\ [1ex] \hline  &  &  &  & \\ [-1ex] 1 & \frac13 & 0 & 0 & \frac23\\  &  &  &  & \\ 2 & \frac23 \cdot \frac12 = \frac13 & \frac13 & 0 & \frac23 \cdot \frac12 = \frac13 \\  &  &  &  & \\ 3 & \frac13 \cdot \frac12 = \frac16 & \frac13 + \frac13 \cdot \frac12 = \frac12 & \frac13 \cdot \frac12 = \frac16 & \frac13 \cdot \frac12 = \frac16 \\  &  &  &  & \\ 4 & \frac16 \cdot \frac12 = \frac{1}{12} & \frac16 + \frac12 \cdot \frac12 = \frac{5}{12} & \frac12 \cdot \frac12 = \frac14 & \frac16 + \frac16 \cdot \frac12 = \frac14 \\  &  &  &  & \\ 5 & \frac14 \cdot \frac12 = \frac18 & \frac{1}{12} + \frac{5}{12} \cdot \frac{1}{2} = \frac{7}{24} & \frac{5}{12} \cdot \frac12 = \frac{5}{24} & \frac14 + \frac14 \cdot \frac12 = \frac38 \\  &  &  &  & \\ 6 & \frac38 \cdot \frac12 = \frac{3}{16} & \frac18 + \frac{7}{24} \cdot \frac12 = \frac{13}{48} & \frac{7}{24} \cdot \frac12 = \frac{7}{48} & \frac{5}{24} + \frac38 \cdot \frac12 = \frac{19}{48} \\  &  &  &  & \\ 7 & \frac{19}{48} \cdot \frac12 = \frac{19}{96}&  \frac{3}{16} + \frac{13}{48} \cdot \frac{1}{2} = \frac{31}{96}& \frac{13}{48} \cdot \frac{1}{2} = \frac{13}{96} & \frac{7}{48} + \frac{19}{48} \cdot \frac12 = \frac{11}{32}\\  &  &  &  & \\ 8 & \frac{11}{32} \cdot \frac{1}{2} = \frac{11}{64} & \frac{19}{96} + \frac{31}{96} \cdot \frac12 = \frac{23}{64} & \frac{31}{96} \cdot \frac12 = \frac{31}{192} & \frac{13}{96} + \frac{11}{32} \cdot \frac12 = \frac{59}{192} \\ [1ex] \end{array}\] Finally, the requested probability is $\frac{11}{64} + \frac{23}{64} = \frac{17}{32},$ from which the answer is $17 + 32 = \boxed{049}.$

Answer:  049
Problem: An ant makes a sequence of moves on a cube where a move consists of walking from one vertex to an adjacent vertex along an edge of the cube. Initially the ant is at a vertex of the bottom face of the cube and chooses one of the three adjacent vertices to move to as its first move. For all moves after the first move, the ant does not return to its previous vertex, but chooses to move to one of the other two adjacent vertices. All choices are selected at random so that each of the possible moves is equally likely. The probability that after exactly $8$ moves that ant is at a vertex of the top face on the cube is $\frac{m}{n}$ , where $m$ and $n$ are relatively prime positive integers. Find $m + n.$

Solution: Note that we don't care which exact vertex the ant is located at, just which level (either top face or bottom face).
Consider the ant to be on any of the two levels and having moved at least one move. Define $p_i$ to be the probability that after $i$ moves, the ant ends up on the level it started on.
On the first move, the ant can stay on the bottom level with $\frac{2}{3}$ chance and $7$ moves left. Or, it can move to the top level with $\frac{1}{3}$ chance and $6$ moves left (it has to spend another on the top as it can not return immediately). So the requested probability is $P = \frac{2}{3}(1 - p_7) + \frac{1}{3}p_6$
Consider when the ant has $i$ moves left (and it's not the ant's first move). It can either stay on its current level with $\frac{1}{2}$ chance and $i - 1$ moves left, or travel to the opposite level with $\frac{1}{2}$ chance, then move to another vertex on the opposite level, to have $i - 2$ moves left. Thus we obtain the recurrence \[p_i = \frac{1}{2}p_{i - 1} + \frac{1}{2}(1 - p_{i - 2})\]
Computing $p_i$ with the starting conditions $p_0 = 1$ and $p_1 = \frac{1}{2}$ , we obtain $p_6 = \frac{33}{64}$ and $p_7 = \frac{59}{128}$ . Hence $P = \frac{2}{3}(1 - p_7) + \frac{1}{3}p_6= \frac{17}{32}$ as desired; $\boxed{049}$

Answer:  049
Problem: An ant makes a sequence of moves on a cube where a move consists of walking from one vertex to an adjacent vertex along an edge of the cube. Initially the ant is at a vertex of the bottom face of the cube and chooses one of the three adjacent vertices to move to as its first move. For all moves after the first move, the ant does not return to its previous vertex, but chooses to move to one of the other two adjacent vertices. All choices are selected at random so that each of the possible moves is equally likely. The probability that after exactly $8$ moves that ant is at a vertex of the top face on the cube is $\frac{m}{n}$ , where $m$ and $n$ are relatively prime positive integers. Find $m + n.$

Solution: On each move, we can either stay on the level we previously were (stay on the bottom/top) or switch levels (go from top to bottom and vise versa). Since we start on the bottom, ending on the top means that we will have to switch an odd number of times; since we cannot switch twice in a row, over an eight-move period we can either make one or three switches. Furthermore, once we switch to a level we can choose one of two directions of traveling on that level: clockwise or counterclockwise (since we can't go back to our previous move, our first move on the level after switching determines our direction).
Our probability is then $\frac{176 + 28}{3 \cdot 2^7} = \frac{17}{32}$ , so the answer is $17+32=\boxed{049}$

Answer:  049
Problem: Find the number of ordered pairs $(m, n)$ such that $m$ and $n$ are positive integers in the set $\{1, 2, ..., 30\}$ and the greatest common divisor of $2^m + 1$ and $2^n - 1$ is not $1$

Solution: This solution refers to the Remarks section.
By the Euclidean Algorithm, we have \[\gcd\left(2^m+1,2^m-1\right)=\gcd\left(2,2^m-1\right)=1.\] We are given that $\gcd\left(2^m+1,2^n-1\right)>1.$ Multiplying both sides by $\gcd\left(2^m-1,2^n-1\right)$ gives \begin{align*} \gcd\left(2^m+1,2^n-1\right)\cdot\gcd\left(2^m-1,2^n-1\right)&>\gcd\left(2^m-1,2^n-1\right) \\ \gcd\left(\left(2^m+1\right)\left(2^m-1\right),2^n-1\right)&>\gcd\left(2^m-1,2^n-1\right) \hspace{12mm} &&\text{by }\textbf{Claim 1} \\ \gcd\left(2^{2m}-1,2^n-1\right)&>\gcd\left(2^m-1,2^n-1\right) \\ 2^{\gcd(2m,n)}-1&>2^{\gcd(m,n)}-1 &&\text{by }\textbf{Claim 2} \\ \gcd(2m,n)&>\gcd(m,n), \end{align*} which implies that $n$ must have more factors of $2$ than $m$ does.
We construct the following table for the first $30$ positive integers: \[\begin{array}{c|c|c} && \\ [-2.5ex] \boldsymbol{\#}\textbf{ of Factors of }\boldsymbol{2} & \textbf{Numbers} & \textbf{Count} \\ \hline && \\ [-2.25ex]  0 & 1,3,5,7,9,11,13,15,17,19,21,23,25,27,29 & 15 \\  && \\ [-2.25ex]  1 & 2,6,10,14,18,22,26,30 & 8 \\ && \\ [-2.25ex]  2 & 4,12,20,28 & 4 \\    && \\ [-2.25ex]  3 & 8,24 & 2 \\  && \\ [-2.25ex]  4 & 16 & 1 \\ \end{array}\] To count the ordered pairs $(m,n),$ we perform casework on the number of factors of $2$ that $m$ has:
Together, the answer is $225+56+12+2=\boxed{295}.$

Answer:  295
Problem: Two spheres with radii $36$ and one sphere with radius $13$ are each externally tangent to the other two spheres and to two different planes $\mathcal{P}$ and $\mathcal{Q}$ . The intersection of planes $\mathcal{P}$ and $\mathcal{Q}$ is the line $\ell$ . The distance from line $\ell$ to the point where the sphere with radius $13$ is tangent to plane $\mathcal{P}$ is $\tfrac{m}{n}$ , where $m$ and $n$ are relatively prime positive integers. Find $m + n$

Solution: The centers of the three spheres form a $49$ $49$ $72$ triangle. Consider the points at which the plane is tangent to the two bigger spheres; the line segment connecting these two points should be parallel to the $72$ side of this triangle. Take its midpoint $M$ , which is $36$ away from the midpoint $A$ of the $72$ side, and connect these two midpoints.
Now consider the point at which the plane is tangent to the small sphere, and connect $M$ with the small sphere's tangent point $B$ . Extend $\overline{MB}$ through $B$ until it hits the ray from $A$ through the center of the small sphere (convince yourself that these two intersect). Call this intersection $D$ , the center of the small sphere $C$ , we want to find $BD$
By Pythagoras, $AC=\sqrt{49^2-36^2}=\sqrt{1105}$ , and we know that $MA=36$ and $BC=13$ . We know that $\overline{MA}$ and $\overline{BC}$ must be parallel, using ratios we realize that $CD=\frac{13}{23}\sqrt{1105}$ . Apply the Pythagorean theorem to $\triangle BCD$ $BD=\frac{312}{23}$ , so $312 + 23 = \boxed{335}$

Answer:  335
Problem: Two spheres with radii $36$ and one sphere with radius $13$ are each externally tangent to the other two spheres and to two different planes $\mathcal{P}$ and $\mathcal{Q}$ . The intersection of planes $\mathcal{P}$ and $\mathcal{Q}$ is the line $\ell$ . The distance from line $\ell$ to the point where the sphere with radius $13$ is tangent to plane $\mathcal{P}$ is $\tfrac{m}{n}$ , where $m$ and $n$ are relatively prime positive integers. Find $m + n$

Solution: This solution refers to the Diagram section. [asy] /* Made by MRENTHUSIASM */ size(300); import graph3; import solids;  currentprojection=orthographic((10,-3,-40));  triple O1, O2, O3, T1, T2, T3, A, L1, L2, M; O1 = (0,-36,0); O2 = (0,36,0); O3 = (0,0,-sqrt(1105)); T1 = (864*sqrt(1105)/1105,-36,-828*sqrt(1105)/1105); T2 = (864*sqrt(1105)/1105,36,-828*sqrt(1105)/1105); T3 = (24*sqrt(1105)/85,0,-108*sqrt(1105)/85); A = (0,0,-36*sqrt(1105)/23); L1 = shift(0,-80,0)*A; L2 = shift(0,80,0)*A; M = midpoint(T1--T2);  draw(shift(O1)*rotate(-90,O1,O2)*scale3(36)*unithemisphere,yellow,light=White); draw(shift(O2)*rotate(-90,O1,O2)*scale3(36)*unithemisphere,yellow,light=White); draw(shift(O3)*rotate(-90,O1,O2)*scale3(13)*unithemisphere,red,light=White); draw(surface(L1--L2--(L2.x,L2.y,40)--(L1.x,L1.y,40)--cycle),gray); draw(shift(O1)*rotate(90,O1,O2)*scale3(36)*unithemisphere,yellow,light=White); draw(shift(O2)*rotate(90,O1,O2)*scale3(36)*unithemisphere,yellow,light=White); draw(shift(O3)*rotate(90,O1,O2)*scale3(13)*unithemisphere,red,light=White); draw(surface(T2--T1--T3--A--cycle),cyan); draw(surface(L1--L2--(L2.x,L2.y,L2.z-abs(T1.z))--(L1.x,L1.y,L1.z-abs(T2.z))--cycle),gray); draw(T1--T2--T3--cycle^^M--A--T2,blue);  dot("$O_1$",O1,(0,-1,1),linewidth(4.5)); dot("$O_2$",O2,(0,1,1),linewidth(4.5)); dot("$O$",O3,(0.5,-1,0),linewidth(4.5)); dot("$T_1$",T1,(0,-1,-1),blue+linewidth(4.5)); dot("$T_2$",T2,(0,1,-1),blue+linewidth(4.5)); dot("$T$",T3,(1,1,2),blue+linewidth(4.5)); dot("$M$",M,(0,0,5),blue+linewidth(4.5)); dot("$A$",A,(-0.5,-1.5,0),red+linewidth(4.5)); [/asy] The isosceles triangle of centers $O_1 O_2 O$ $O$ is the center of sphere of radii $13$ ) has sides $O_1 O = O_2 O = 36 + 13 = 49,$ and $O_1 O_2 = 36 + 36 = 72.$
Let $N$ be the midpoint $O_1 O_2$
The isosceles triangle of points of tangency $T_1 T_2 T$ has sides $T_1 T = T_2 T = 2 \sqrt{13 \cdot 36} = 12 \sqrt{13}$ and $T_1 T_2 = 72.$
Let $M$ be the midpoint $T_1 T_2.$
The height $TM$ is $\sqrt {12^2 \cdot 13 - 36^2} = 12 \sqrt {13-9} = 24.$
The tangents of the half-angle between the planes is $\frac {TO}{AT} = \frac {MN - TO}{TM},$ so $\frac {13}{AT} = \frac {36 - 13}{24},$ \[AT = \frac{24\cdot 13}{23} = \frac {312}{23} \implies  312 + 23 = \boxed{335}.\] vladimir.shelomovskii@gmail.com, vvsss

Answer:  335
Problem: A teacher was leading a class of four perfectly logical students. The teacher chose a set $S$ of four integers and gave a different number in $S$ to each student. Then the teacher announced to the class that the numbers in $S$ were four consecutive two-digit positive integers, that some number in $S$ was divisible by $6$ , and a different number in $S$ was divisible by $7$ . The teacher then asked if any of the students could deduce what $S$ is, but in unison, all of the students replied no.
However, upon hearing that all four students replied no, each student was able to determine the elements of $S$ . Find the sum of all possible values of the greatest element of $S$

Solution: We know right away that $42\not\in S$ and $84\not\in S$ as stated in Solution 1.
To get a feel for the problem, let’s write out some possible values of $S$ based on the teacher’s remarks. The first multiple of 7 that is two-digit is 14. The closest multiple of six from 14 is 12, and therefore there are two possible sets of four consecutive integers containing 12 and 14; $\{11,12,13,14\}$ and $\{12,13,14,15\}$ . Here we get our first crucial idea; that if the multiples of 6 and 7 differ by two, there will be 2 possible sets of $S$ without any student input. Similarly, if they differ by 3, there will be only 1 possible set, and if they differ by 1, 3 possible sets.
Now we read the student input. Each student says they can’t figure out what $S$ is just based on the teacher’s information, which means each student has to have a number that would be in 2 or 3 of the possible sets (This is based off of the first line of student input). However, now that each student knows that all of them have numbers that fit into more than one possible set, this means that S cannot have two possible sets because otherwise, when shifting from one set to the other, one of the end numbers would not be in the shifted set, but we know each number has to fall in two or more possible sets. For example, take $\{11,12,13,14\}$ and $\{12,13,14,15\}$ . The numbers at the end, 11 and 15, only fall in one set, but each number must fall in at least two sets. This means that there must be three possible sets of S, in which case the actual S would be the middle S.
Take for example $\{33,34,35,36\}$ $\{34,35,36,37\}$ , and $\{35,36,37,38\}$ . 37 and 34 fall in two sets while 35 and 36 fall in all three sets, so the condition is met. Now, this means that the multiple of 6 and 7 must differ by 1. 
Since 42 means the difference is 0, when you add/subtract 6 and 7, you will obtain the desired difference of 1, and similarly adding/subtracting 6 or 7 from 84 will also obtain the difference of 1. Thus there are four possible sets of $S$ $\{34,35,36,37\}$ $\{47,48,49,50\}$ $\{76,77,78,79\}$ and $\{89,90,91,92.\}$ . Therefore the sum of the greatest elements of the possible sets $S$ is $37+50+79+92=\boxed{258}$

Answer:  258
Problem: A teacher was leading a class of four perfectly logical students. The teacher chose a set $S$ of four integers and gave a different number in $S$ to each student. Then the teacher announced to the class that the numbers in $S$ were four consecutive two-digit positive integers, that some number in $S$ was divisible by $6$ , and a different number in $S$ was divisible by $7$ . The teacher then asked if any of the students could deduce what $S$ is, but in unison, all of the students replied no.
However, upon hearing that all four students replied no, each student was able to determine the elements of $S$ . Find the sum of all possible values of the greatest element of $S$

Solution: In a solution that satisfies these constraints, the multiple of 6 must be adjacent to multiple of 7. The other two numbers must be on either side.
WLOG assume the set is $\{a,6j,7k,b\}$ . The student with numbers $a$ $6j$ , and $7k$ can think the set is $\{a-1, a,6j,7k\}$ or $\{a,6j,7k,b\}$ , and the students with number $6j$ $7k$ , and $b$ can think the set is $\{a,6j,7k,b\}$ or $\{6j,7k,b, b+1\}$ . Therefore, none of the students know the set for sure.
Playing around with the arrangement of the multiple of 6 and multiple of 7 shows that this is the only configuration viewed as ambiguous to all the students. (Therefore when they hear nobody else knows either, they can find out it is this configuration)
Considering $S$ as $\{a,6j,7k,b\}$ ,b is 2 mod 6 and 1 mod 7, so $b$ is 8 mod 42. (since it is all 2-digit, the values are either 50 or 92).
Similarly, considering $S$ as $\{a,7j,6k,b\}$ $b$ is 1 mod 6 and 2 mod 7, so $b$ is 37 mod 42. The values that satisfy that are 37 and 79.
The total sum of all these values is therefore $50+92+37+89=\boxed{258}$

Answer:  258