Navvye/Olympiad-Problems-CSV huggingface dataset:
Problem: A group of $12$ pirates agree to divide a treasure chest of gold coins among themselves as follows.  The $k^{\text{th}}$ pirate to take a share takes $\frac{k}{12}$ of the coins that remain in the chest.  The number of coins initially in the chest is the smallest number for which this arrangement will allow each pirate to receive a positive whole number of coins.  How many coins does the $12^{\text{th}}$ pirate receive?
$\textbf{(A)}\ 720\qquad\textbf{(B)}\ 1296\qquad\textbf{(C)}\ 1728\qquad\textbf{(D)}\ 1925\qquad\textbf{(E)}\ 3850$

Solution: Solution $1$ mentioned the expression $x \cdot \frac{11}{12} \cdot \frac{10}{12} \cdot ... \cdot \frac{1}{12}$ . Note that this is equivalent to $\frac{x \cdot 11!}{12^{11}}$
We can compute the amount of factors of $2$ $3$ $5$ , etc. but this is not necessary. To minimize this expression, we must take out factors of $2$ and $3$ , since $12^{11}=2^{22} \cdot 3^{11}$ $11!$ has neither $22$ factors of $2$ , nor $11$ factors of $3$ . This means that if $11!$ contains $a$ factors of $2$ , then $x$ will contain $22-a$ factors of $2$ . This also holds for factors of $3$
Thus, once simplified, the expression will have no factors of $2$ . It will also have no factors of $3$
Looking at the answer choices, there is only one answer which is not even, which is $\boxed{1925}$

Answer: D 1925
Problem: A group of $12$ pirates agree to divide a treasure chest of gold coins among themselves as follows.  The $k^{\text{th}}$ pirate to take a share takes $\frac{k}{12}$ of the coins that remain in the chest.  The number of coins initially in the chest is the smallest number for which this arrangement will allow each pirate to receive a positive whole number of coins.  How many coins does the $12^{\text{th}}$ pirate receive?
$\textbf{(A)}\ 720\qquad\textbf{(B)}\ 1296\qquad\textbf{(C)}\ 1728\qquad\textbf{(D)}\ 1925\qquad\textbf{(E)}\ 3850$

Solution: We know that the 11th pirate takes $\frac{11}{12}$ of what is left from the 10th pirate, so we have the 12th pirate taking \[\frac{12}{12}\cdot(1-\frac{11}{12})=1\cdot\frac{1}{12}\] of what is left from the 10th pirate. Similarly, since the 10th pirate takes $\frac{10}{12}$ of what is left from the 9th pirate, we have the 11th pirate taking \[\frac{11}{12}\cdot(1-\frac{10}{12})=\frac{11}{12}\cdot\frac{2}{12}\] of what is left from the 9th pirate. Thus, the 12th pirate takes \[1\cdot\frac{1}{12}\cdot\frac{2}{12}\] of what is left from the 9th pirate. Repeating the method, we can find that the 12th pirate takes \[1\cdot\frac{1}{12}\cdot\frac{2}{12}\cdot\frac{3}{12}\cdot...\cdot\frac{10}{12}\cdot\frac{11}{12}\] of what is left from the 1st pirate, or \[1\cdot\frac{1}{12}\cdot\frac{2}{12}\cdot\frac{3}{12}\cdot...\cdot\frac{10}{12}\cdot\frac{11}{12}\cdot\frac{12}{12}\] of the total amount of coins.
Now canceling out the denominator with the numerator as possible, we are left with $\frac{1\cdot5\cdot7\cdot5\cdot11}{...}=\frac{1925}{...}$ with some factors of 12 in the denominator. For this fraction to be an integer, the smallest possible number of coins is the same as the denominator, so the numerator is the number of coins taken by the 12th pirate, or $1925 \,\boxed{1925}$

Answer: D 1925
Problem: In $\triangle ABC$ $AB = 86$ , and $AC=97$ .  A circle with center $A$ and radius $AB$ intersects $\overline{BC}$ at points $B$ and $X$ .  Moreover $\overline{BX}$ and $\overline{CX}$ have integer lengths.  What is $BC$
$\textbf{(A)}\ 11\qquad\textbf{(B)}\ 28\qquad\textbf{(C)}\ 33\qquad\textbf{(D)}\ 61\qquad\textbf{(E)}\ 72$

Solution: Let $BX = q$ $CX = p$ , and $AC$ meets the circle at $Y$ and $Z$ , with $Y$ on $AC$ .  Then $AZ = AY = 86$ .  Using the Power of a Point (Secant-Secant Power Theorem), we get that $p(p+q) = 11(183) = 11 * 3 * 61$ .  We know that $p+q>p$ , so $p$ is either $3$ $11$ , or $33$ .  We also know that $p>11$ by the triangle inequality on $\triangle ACX$ .  Thus, $p$ is $33$ so we get that $BC = p+q = \boxed{61}$

Answer: D 61
Problem: In $\triangle ABC$ $AB = 86$ , and $AC=97$ .  A circle with center $A$ and radius $AB$ intersects $\overline{BC}$ at points $B$ and $X$ .  Moreover $\overline{BX}$ and $\overline{CX}$ have integer lengths.  What is $BC$
$\textbf{(A)}\ 11\qquad\textbf{(B)}\ 28\qquad\textbf{(C)}\ 33\qquad\textbf{(D)}\ 61\qquad\textbf{(E)}\ 72$

Solution: Let $x$ represent $CX$ , and let $y$ represent $BX$ . Since the circle goes through $B$ and $X$ $AB = AX = 86$ .
Then by Stewart's Theorem,
$xy(x+y) + 86^2 (x+y) = 97^2 y + 86^2 x.$
$x^2 y + xy^2 + 86^2 x + 86^2 y = 97^2 y + 86^2 x$
$x^2 y + xy^2 + 86^2 y = 97^2 y$
$x^2 + xy + 86^2 = 97^2$
(Since $y$ cannot be equal to $0$ , dividing both sides of the equation by $y$ is allowed.)
$x(x+y) = (97+86)(97-86)$
$x(x+y) = 2013$
The prime factors of $2013$ are $3$ $11$ , and $61$ . Obviously, $x < x+y$ . In addition, by the Triangle Inequality, $BC < AB + AC$ , so $x+y < 183$ . Therefore, $x$ must equal $33$ , and $x+y$ must equal $\boxed{61}$

Answer: D 61
Problem: In $\triangle ABC$ $AB = 86$ , and $AC=97$ .  A circle with center $A$ and radius $AB$ intersects $\overline{BC}$ at points $B$ and $X$ .  Moreover $\overline{BX}$ and $\overline{CX}$ have integer lengths.  What is $BC$
$\textbf{(A)}\ 11\qquad\textbf{(B)}\ 28\qquad\textbf{(C)}\ 33\qquad\textbf{(D)}\ 61\qquad\textbf{(E)}\ 72$

Solution: Let $CX=x, BX=y$ . Let the circle intersect $AC$ at $D$ and the diameter including $AD$ intersect the circle again at $E$ .
Use power of a point on point C to the circle centered at A.
So $CX \cdot CB=CD \cdot CE \Rightarrow x(x+y)=(97-86)(97+86) \Rightarrow x(x+y)=3*11*61$
Obviously $x+y>x$ so we have three solution pairs for $(x,x+y)=(1,2013),(3,671),(11,183),(33,61)$ .
By the Triangle Inequality, only $x+y=61$ yields a possible length of $BX+CX=BC$
Therefore, the answer is $\boxed{61}$

Answer: D 61
Problem: In $\triangle ABC$ $AB = 86$ , and $AC=97$ .  A circle with center $A$ and radius $AB$ intersects $\overline{BC}$ at points $B$ and $X$ .  Moreover $\overline{BX}$ and $\overline{CX}$ have integer lengths.  What is $BC$
$\textbf{(A)}\ 11\qquad\textbf{(B)}\ 28\qquad\textbf{(C)}\ 33\qquad\textbf{(D)}\ 61\qquad\textbf{(E)}\ 72$

Solution: [asy] unitsize(2); import olympiad; import graph;  pair A,B,C,D,E; A = (0,0); B = (70,51); C = (97,0); D = (82,29); E = (76,40);  draw(Circle((0,0),86.609)); draw(A--B--C--A); draw(A--B--E--A); draw(A--D); dot(A); dot(B,blue); dot(C); dot(D,blue); dot(E); label("A",A,S); label("B",B,NE); label("C",C,S); label("D",D,NE); label("E",E,NE); label("86",(A+B)/2,NW); label("86",(A+D)/2,SE); label("97",(A+C)/2,S); label("h",(A+E)/2,N); label("k",(E+D)/2,NE); label("k",(B+E)/2,NE); label("m",(C+D)/2,NE);   fill(anglemark(A,E,D,100),black); label("$90^\circ$",anglemark(A,E,D),3*S); [/asy]
We first draw the height of isosceles triangle $ABD$ and get two equations by the Pythagorean Theorem .
First, $h^2 + k^2 = 86^2$ . Second, $h^2 + (k + m)^2 = 97^2$ .
Subtracting these two equations, we get $2km + m^2 = 97^2 - 86^2 = (97 - 86)(97 + 86) = 2013$ .
We then add $k^2$ to both sides to get $k^2 + 2km + m^2 = 2013 + k^2$ .
We then complete the square to get $(k + m)^2 = 2013 + k^2$ . Because $k$ and $m$ are both integers, we get that $2013 + k^2$ is a square number. Simple guess and check reveals that $k = 14$ .
Because $k$ equals $14$ , therefore $m = 33$ . We want $\overline{BC} = 2k + m$ , so we get that $\overline{BC} = \boxed{61}$

Answer: D 61
Problem: In $\triangle ABC$ $AB = 86$ , and $AC=97$ .  A circle with center $A$ and radius $AB$ intersects $\overline{BC}$ at points $B$ and $X$ .  Moreover $\overline{BX}$ and $\overline{CX}$ have integer lengths.  What is $BC$
$\textbf{(A)}\ 11\qquad\textbf{(B)}\ 28\qquad\textbf{(C)}\ 33\qquad\textbf{(D)}\ 61\qquad\textbf{(E)}\ 72$

Solution: Let $E$ be the foot of the altitude from $A$ to $BX.$ Since $\triangle ABX$ is isosceles $AX=AB=86,EB=EX,$ and the answer is $EC+EB=EC+EX.$ $(EC+EX)(EC-EX)=EC^2-EX^2=(97^2-AE^2)-(86^2-AE^2)=97^2-86^2=2013$ by the Pythagorean Theorem. Only $EC+EX=\boxed{61}.$

Answer: D 61
Problem: Central High School is competing against Northern High School in a backgammon match. Each school has three players, and the contest rules require that each player play two games against each of the other school's players. The match takes place in six rounds, with three games played simultaneously in each round. In how many different ways can the match be scheduled?
$\textbf{(A)}\ 540\qquad\textbf{(B)}\ 600\qquad\textbf{(C)}\ 720\qquad\textbf{(D)}\ 810\qquad\textbf{(E)}\ 900$

Solution: Let us label the players of the first team $A$ $B$ , and $C$ , and those of the second team, $X$ $Y$ , and $Z$
$\textbf{1}$ . One way of scheduling all six distinct rounds could be:
Round 1: $AX$ $BY$ $CZ$
Round 2: $AX$ $BZ$ $CY$
Round 3: $AY$ $BX$ $CZ$
Round 4: $AY$ $BZ$ $CX$
Round 5: $AZ$ $BX$ $CY$
Round 6: $AZ$ $BY$ $CX$
The above mentioned schedule ensures that each player of one team plays twice with each player from another team. Now you can generate a completely new schedule by permutating those $6$ rounds and that can be done in $6!=720$ ways.
$\textbf{2}$ . One can also make the schedule in such a way that two rounds are repeated.
(a)
Round 1: $AX$ $BZ$ $CY$
Round 2: $AX$ $BZ$ $CY$
Round 3: $AY$ $BX$ $CZ$
Round 4: $AY$ $BX$ $CZ$
Round 5: $AZ$ $BY$ $CX$
Round 6: $AZ$ $BY$ $CX$
(b)
Round 1: $AX$ $BY$ $CZ$
Round 2: $AX$ $BY$ $CZ$
Round 3: $AY$ $BZ$ $CX$
Round 4: $AY$ $BZ$ $CX$
Round 5: $AZ$ $BX$ $CY$
Round 6: $AZ$ $BX$ $CY$
As mentioned earlier any permutation of (a) and (b) will also give us a new schedule. For both (a) and (b) the number of permutations are $\frac{6!}{2!2!2!}$ $90$
So the total number of schedules is $720+90+90$ $\boxed{900}$

Answer: E 900
Problem: Central High School is competing against Northern High School in a backgammon match. Each school has three players, and the contest rules require that each player play two games against each of the other school's players. The match takes place in six rounds, with three games played simultaneously in each round. In how many different ways can the match be scheduled?
$\textbf{(A)}\ 540\qquad\textbf{(B)}\ 600\qquad\textbf{(C)}\ 720\qquad\textbf{(D)}\ 810\qquad\textbf{(E)}\ 900$

Solution: Label the players of the first team $A$ $B$ , and $C$ , and those of the second team, $X$ $Y$ , and $Z$ . We can start by assigning an opponent to person $A$ for all $6$ games. Since $A$ has to play each of $X$ $Y$ , and $Z$ twice, there are $\frac{6!}{2!2!2!} = 90$ ways to do this. We can assume that the opponents for $A$ in the $6$ rounds are $X$ $X$ $Y$ $Y$ $Z$ $Z$ and multiply by $90$ afterwards.
Notice that for every valid assignment of the opponents of $A$ and $B$ , there is only $1$ valid assignment of opponents for $C$ . More specifically, the opponents for $C$ are the leftover opponents after the opponents for $A$ and $B$ are chosen in each round. Therefore, all we have to do is assign the opponents for $B$ . This is the same as finding the number of permutations of $X$ $Y$ , and $Z$ that do not have a $X$ in the first two spots, an $Y$ in the next two spots, and a $Z$ in the final two spots.
We can use casework to find this by using the fact that after we put down the $X$ 's and $Y$ 's first there is $1$ way to put down the $Z$ 's (the two remaining spots).
If $X$ 's are put in the middle $2$ spots, then there is $1$ way to assign spots to $Y$ , namely the last $2$ spots. (If one of the last two spots are left empty, there will have to be a $Z$ there, which which not valid).
If $X$ 's are put in the last $2$ spots, then there is $1$ way to assign spots to $Y$
Finally, if one $X$ was put in on of the middle two spots and one $X$ was put in one of the last two spots, there are $2\cdot2$ ways to assign spots to $X$ and $2\cdot1$ ways to assign spots to $Y$ (one of the first two spots and the remaining spot in the last $2$ ).
There are $1+1+2\cdot2\cdot2\cdot1 = 10$ ways to assign opponents to $B$ and $90\cdot10 = 900$ ways to order the games. $\boxed{900}$

Answer: E 900
Problem: All $20$ diagonals are drawn in a regular octagon. At how many distinct points in the interior
of the octagon (not on the boundary) do two or more diagonals intersect?
$\textbf{(A)}\ 49\qquad\textbf{(B)}\ 65\qquad\textbf{(C)}\ 70\qquad\textbf{(D)}\ 96\qquad\textbf{(E)}\ 128$

Solution: If you draw a clear diagram like the one below, it is easy to see that there are $\boxed{49}$ points.

Answer: A 49
Problem: All $20$ diagonals are drawn in a regular octagon. At how many distinct points in the interior
of the octagon (not on the boundary) do two or more diagonals intersect?
$\textbf{(A)}\ 49\qquad\textbf{(B)}\ 65\qquad\textbf{(C)}\ 70\qquad\textbf{(D)}\ 96\qquad\textbf{(E)}\ 128$

Solution: Let the number of intersections be $x$ .  We know that $x\le \dbinom{8}{4} = 70$ , as every $4$ vertices on the octagon forms a quadrilateral with intersecting diagonals which is an intersection point.  However, four diagonals intersect in the center, so we need to subtract $\dbinom{4}{2} -1 = 5$ from this count, $70-5 = 65$ .  Note that diagonals like $\overline{AD}$ $\overline{CG}$ , and $\overline{BE}$ all intersect at the same point.  There are $8$ of this type with three diagonals intersecting at the same point, so we need to subtract $2$ of the $\dbinom{3}{2}$ (one is kept as the actual intersection). In the end, we obtain $65 - 16 = \boxed{49}$

Answer: A 49
Problem: All $20$ diagonals are drawn in a regular octagon. At how many distinct points in the interior
of the octagon (not on the boundary) do two or more diagonals intersect?
$\textbf{(A)}\ 49\qquad\textbf{(B)}\ 65\qquad\textbf{(C)}\ 70\qquad\textbf{(D)}\ 96\qquad\textbf{(E)}\ 128$

Solution: We know that the amount of intersection points is at most $\dbinom{8}{4} = 70$ , as in solution $2$ . There's probably going to be more than $5$ intersections counted multiple times (to get $\textbf{(B) }65$ ), leading us to the only reasonable answer, $\boxed{49}$ .
-Lcz

Answer: A 49
Problem: All $20$ diagonals are drawn in a regular octagon. At how many distinct points in the interior
of the octagon (not on the boundary) do two or more diagonals intersect?
$\textbf{(A)}\ 49\qquad\textbf{(B)}\ 65\qquad\textbf{(C)}\ 70\qquad\textbf{(D)}\ 96\qquad\textbf{(E)}\ 128$

Solution: Like solution one, we may draw. Except note that the octagon has eight regions, and each region has an equal number of points, so drawing only one of the eight regions and the intersection points suffices. One of the eight regions contains $8$ points (not including the octagon center). However each adjacent region share one side in common and that side contains $2$ intersection points, so in actuality there are $8 - 2 = 6$ points per region. We multiply this by $8$ to get $6\cdot 8 = 48$ and add the one center point to get $48 + 1 = \boxed{49}$

Answer: A 49
Problem: All $20$ diagonals are drawn in a regular octagon. At how many distinct points in the interior
of the octagon (not on the boundary) do two or more diagonals intersect?
$\textbf{(A)}\ 49\qquad\textbf{(B)}\ 65\qquad\textbf{(C)}\ 70\qquad\textbf{(D)}\ 96\qquad\textbf{(E)}\ 128$

Solution: [asy] size(8cm); pathpen = black; // draw the circle pair[] A; for (int i=0; i<8; ++i) { A[i] = dir(45*i); } D(CR((0,0), 1)); // draw the octagon and diagonals // choose pen colors pen[] colors; colors[1] = yellow; colors[2] = purple; colors[3] = green; colors[4] = orange; for (int d=1; d<=4; ++d) { pathpen = colors[d]; for (int j=0; j<8; ++j) { D(A[j]--A[(j+d) % 8]); } } // draw the 3 or more line intersections pointpen = red + 5; D(IP(A[0]--A[4], A[2]--A[6])); // center of the circle for (int x1=0; x1<8; ++x1) { for (int x2=0; x2<8; ++x2) { int y1 = (x1 + 4)%8; int y2 = (x2 + 3)%8; if (x1 != x2 && y1 != y2 && x1 != y1 && x1 != y2 && x2 != y1 && x2 != y2) D(IP(A[x1]--A[y1], A[x2]--A[y2])); } }  // draw the 2 line intersections pointpen = blue + 4; for (int x1 = 0; x1 < 8; ++x1) { int x2 = (x1 + 1)%8; D(IP(A[x1]--A[(x1+2)%8], A[x2]--A[(x2+2)%8])); D(IP(A[x1]--A[(x1+2)%8], A[x2]--A[(x2+3)%8])); D(IP(A[x1]--A[(x1+2)%8], A[x2]--A[(x2+4)%8])); D(IP(A[x1]--A[(x1+2)%8], A[x2]--A[(x2+5)%8])); D(IP(A[x1]--A[(x1+3)%8], A[x2]--A[(x2+5)%8])); } [/asy]
This problem is a counting problem of combinatoric geometry. There are 2 cases for the above diagram:
Case 1: Red Dots
The red dots are the intersection of 3 or more lines. It consists of 8 dots that make up an octagon and 1 dot in the center. Hence, there are $9$ red dots.
Case 2: Blue Dots
The blue dots are the intersection of 2 lines. Each vertex of the octagon has 2 purple lines, 2 green lines, and 1 orange line coming out of it. There are 5 dots of intersection on a purple line, 6 dots on a green line, and 5 dots on an orange line. There are $2 \cdot 5+2 \cdot 6+5=27$ dots that come out of 1 vertex, which includes 7 red dots already counted. So there are $27-7 = 20$ blue dots coming out of 1 vertex. There are 8 vertices, but each blue dot is the intersection of 2 lines, corresponding to $2 \cdot 2 = 4$ vertices. So there are $\frac{20 \cdot 8}{4} = 40$ blue dots.
The number of intersection dots are the sum of the number of red and blue dots. Hence, the answer is $40 + 9 = \boxed{49}$

Answer: A 49
Problem: Mr. Green measures his rectangular garden by walking two of the sides and finds that it is $15$ steps by $20$ steps. Each of Mr. Green's steps is $2$ feet long. Mr. Green expects a half a pound of potatoes per square foot from his garden. How many pounds of potatoes does Mr. Green expect from his garden?
$\textbf{(A)}\ 600 \qquad \textbf{(B)}\ 800 \qquad \textbf{(C)}\ 1000 \qquad \textbf{(D)}\ 1200 \qquad \textbf{(E)}\ 1400$

Solution: Since each step is $2$ feet, his garden is $30$ by $40$ feet.  Thus, the area of $30(40) = 1200$ square feet.  Since he is expecting $\frac{1}{2}$ of a pound per square foot, the total amount of potatoes expected is $1200 \times \frac{1}{2} =  \boxed{600}$

Answer: A 600
Problem: On a particular January day, the high temperature in Lincoln, Nebraska, was $16$ degrees higher than the low temperature, and the average of the high and low temperatures was $3$ . In degrees, what was the low temperature in Lincoln that day?
$\textbf{(A)}\ -13 \qquad \textbf{(B)}\ -8 \qquad \textbf{(C)}\ -5 \qquad \textbf{(D)}\ -3 \qquad \textbf{(E)}\ 11$

Solution: Let $L$ be the low temperature. The high temperature is $L+16$ . The average is $\frac{L+(L+16)}{2}=3$ . Solving for $L$ , we get $L=\boxed{5}$

Answer: C 5
Problem: When counting from $3$ to $201$ $53$ is the $51^{st}$ number counted. When counting backwards from $201$ to $3$ $53$ is the $n^{th}$ number counted. What is $n$
$\textbf{(A)}\ 146 \qquad \textbf{(B)}\ 147 \qquad \textbf{(C)}\ 148 \qquad \textbf{(D)}\ 149 \qquad \textbf{(E)}\ 150$

Solution: Note that $n$ is equal to the number of integers between $53$ and $201$ , inclusive. Thus, $n=201-53+1=\boxed{149}$

Answer: D 149
Problem: Positive integers $a$ and $b$ are each less than $6$ . What is the smallest possible value for $2 \cdot a - a \cdot b$
$\textbf{(A)}\ -20\qquad\textbf{{(B)}}\ -15\qquad\textbf{{(C)}}\ -10\qquad\textbf{{(D)}}\ 0\qquad\textbf{{(E)}}\ 2$

Solution: Factoring the equation gives $a(2 - b)$ . From this we can see that to obtain the least possible value, $2 - b$ should be negative, and should be as small as possible. To do so, $b$ should be maximized. Because $2 - b$ is negative, we should maximize the positive value of $a$ as well. The maximum values of both $a$ and $b$ are $5$ , so the answer is $5(2 - 5) = \boxed{15}$

Answer: B 15
Problem: The average age of $33$ fifth-graders is $11$ . The average age of $55$ of their parents is $33$ . What is the average age of all of these parents and fifth-graders?
$\textbf{(A)}\ 22 \qquad \textbf{(B)}\ 23.25 \qquad \textbf{(C)}\ 24.75 \qquad \textbf{(D)}\ 26.25 \qquad \textbf{(E)}\ 28$

Solution: The sum of the ages of the fifth graders is $33 * 11$ , while the sum of the ages of the parents is $55 * 33$ . Therefore, the total sum of all their ages must be $2178$ , and given $33 + 55 = 88$ people in total, their average age is $\frac{2178}{88} = \frac{99}{4} = \boxed{24.75}$

Answer: C 24.75
Problem: Ray's car averages $40$ miles per gallon of gasoline, and Tom's car averages $10$ miles per gallon of gasoline. Ray and Tom each drive the same number of miles. What is the cars' combined rate of miles per gallon of gasoline? $\textbf{(A)}\ 10 \qquad \textbf{(B)}\ 16 \qquad \textbf{(C)}\ 25 \qquad \textbf{(D)}\ 30 \qquad \textbf{(E)}\ 40$

Solution: Let Ray and Tom drive 40 miles. Ray's car would require $\frac{40}{40}=1$ gallon of gas and Tom's car would require $\frac{40}{10}=4$ gallons of gas. They would have driven a total of $40+40=80$ miles, on $1+4=5$ gallons of gas, for a combined rate of $\frac{80}{5}=$ $\boxed{16}$

Answer: B 16
Problem: Ray's car averages $40$ miles per gallon of gasoline, and Tom's car averages $10$ miles per gallon of gasoline. Ray and Tom each drive the same number of miles. What is the cars' combined rate of miles per gallon of gasoline? $\textbf{(A)}\ 10 \qquad \textbf{(B)}\ 16 \qquad \textbf{(C)}\ 25 \qquad \textbf{(D)}\ 30 \qquad \textbf{(E)}\ 40$

Solution: Taking the harmonic mean of the two rates, we get \[\left(\frac{40^{-1} + 10^{-1}}{2}\right)^{-1} = \frac{2}{\frac{1}{40}+\frac{1}{10}} = \frac{2}{\frac{5}{40}} = \frac{2}{\frac{1}{8}} = \boxed{16}.\]

Answer: B 16
Problem: Ray's car averages $40$ miles per gallon of gasoline, and Tom's car averages $10$ miles per gallon of gasoline. Ray and Tom each drive the same number of miles. What is the cars' combined rate of miles per gallon of gasoline? $\textbf{(A)}\ 10 \qquad \textbf{(B)}\ 16 \qquad \textbf{(C)}\ 25 \qquad \textbf{(D)}\ 30 \qquad \textbf{(E)}\ 40$

Solution: Let the number of miles that Ray and Tom each drive be denoted by $m$ . Thus the number of gallons Ray's car uses can be represented by $\frac{m}{40}$ and the number of gallons that Tom's car uses can likewise be expressed as $\frac{m}{10}$ . Thus the total amount of gallons used by both cars can be expressed as $\frac{m}{40} + \frac{m}{10} = \frac{m}{8}$ . The total distance that both drive is equal to $2m$ so the total miles per gallon can be expressed as $\frac{2m}{\frac{m}{8}} = \boxed{16}$

Answer: B 16
Problem: Three positive integers are each greater than $1$ , have a product of $27000$ , and are pairwise relatively prime. What is their sum?
$\textbf{(A)}\ 100\qquad\textbf{(B)}\ 137\qquad\textbf{(C)}\ 156\qquad\textbf{(D)}\ 160\qquad\textbf{(E)}\ 165$

Solution: The prime factorization of $27000$ is $2^3*3^3*5^3$ . These three factors are pairwise relatively prime, so the sum is $2^3+3^3+5^3=8+27+125=$ $\boxed{160}$

Answer: D 160
Problem: A basketball team's players were successful on 50% of their two-point shots and 40% of their three-point shots, which resulted in 54 points. They attempted 50% more two-point shots than three-point shots. How many three-point shots did they attempt?
$\textbf{(A) }10\qquad\textbf{(B) }15\qquad\textbf{(C) }20\qquad\textbf{(D) }25\qquad\textbf{(E) }30$

Solution: Let $x$ be the number of two point shots attempted and $y$ the number of three point shots attempted. Because each two point shot is worth two points and the team made 50% and each three point shot is worth 3 points and the team made 40%, $0.5(2x)+0.4(3y)=54$ or $x+1.2y=54$ . Because the team attempted 50% more two point shots then threes, $x=1.5y$ . Substituting $1.5y$ for $x$ in the first equation gives $1.5y+1.2y=54$ , which equals $2.7y=54$ so $y=$ $\boxed{20}$

Answer: C 20
Problem: Real numbers $x$ and $y$ satisfy the equation $x^2+y^2=10x-6y-34$ . What is $x+y$
$\text{(A)}\ 1 \qquad \text{(B)}\ 2 \qquad \text{(C)}\ 3 \qquad \text{(D)}\ 6 \qquad \text{(E)}\ 8$

Solution: If we move every term dependent on $x$ or $y$ to the LHS, we get $x^2 - 10x + y^2 + 6y = -34$ . Adding $34$ to both sides, we have $x^2 - 10x + y^2 + 6y + 34 = 0$ . We can split the $34$ into $25$ and $9$ to get $(x - 5)^2 + (y + 3)^2 = 0$ . Notice this is a circle with radius $0$ , which only contains one point. So, the only point is $(5, -3)$ , so the sum is $5 + (-3) = 2 \implies \boxed{2}$ . ~ asdf334

Answer: B 2