Navvye/Olympiad-Problems-CSV huggingface dataset:
Problem: A five-legged Martian has a drawer full of socks, each of which is red, white or blue, and there are at least five socks of each color. The Martian pulls out one sock at a time without looking. How many socks must the Martian remove from the drawer to be certain there will be 5 socks of the same color?
$\textbf{(A)}\ 6\qquad\textbf{(B)}\ 9\qquad\textbf{(C)}\ 12\qquad\textbf{(D)}\ 13\qquad\textbf{(E)}\ 15$

Solution: The Martian can pull out $12$ socks, $4$ of each color, without having $5$ of the same kind yet. However, the next one he pulls out must be the fifth of one of the colors so he must remove $\boxed{13}$ socks.

Answer: D 13
Problem: How many three-digit numbers are divisible by 13?
$\textbf{(A)}\ 7\qquad\textbf{(B)}\ 67\qquad\textbf{(C)}\ 69\qquad\textbf{(D)}\ 76\qquad\textbf{(E)}\ 77$

Solution: Let $k$ be any positive integer so that $13k$ is a multiple of $13$ . For the smallest three-digit number, $13k>100$ and $k>\frac{100}{13} \approx 7.7$ . For the greatest three-digit number, $13k<999$ and $k<\frac{999}{13} \approx 76.8$ . The number $k$ can range from $8$ to $76$ so there are $\boxed{69}$ three-digit numbers.

Answer: C 69
Problem: What is the perimeter of trapezoid $ABCD$
[asy]size(3inch, 1.5inch); pair a=(0,0), b=(18,24), c=(68,24), d=(75,0), f=(68,0), e=(18,0); draw(a--b--c--d--cycle); draw(b--e); draw(shift(0,2)*e--shift(2,2)*e--shift(2,0)*e); label("30", (9,12), W); label("50", (43,24), N); label("25", (71.5, 12), E); label("24", (18, 12), E); label("$A$", a, SW); label("$B$", b, N); label("$C$", c, N); label("$D$", d, SE); label("$E$", e, S);[/asy]
$\textbf{(A)}\ 180\qquad\textbf{(B)}\ 188\qquad\textbf{(C)}\ 196\qquad\textbf{(D)}\ 200\qquad\textbf{(E)}\ 204$

Solution: Draw altitudes from $B$ and $C$ to base $AD$ to create a rectangle and two right triangles. The side opposite $BC$ is equal to $50$ . The bases of the right triangles can be found using Pythagorean or special triangles to be $18$ and $7$ . Add it together to get $AD=18+50+7=75$ . The perimeter is $75+30+50+25=\boxed{180}$

Answer: A 180
Problem: Alice and Bob play a game involving a circle whose circumference is divided by 12 equally-spaced points. The points are numbered clockwise, from 1 to 12. Both start on point 12. Alice moves clockwise and Bob, counterclockwise.
In a turn of the game, Alice moves 5 points clockwise and Bob moves 9 points counterclockwise. The game ends when they stop on the same point. How many turns will this take?
$\textbf{(A)}\ 6\qquad\textbf{(B)}\ 8\qquad\textbf{(C)}\ 12\qquad\textbf{(D)}\ 14\qquad\textbf{(E)}\ 24$

Solution: Alice moves $5k$ steps and Bob moves $9k$ steps, where $k$ is the turn they are on. Alice and Bob coincide when the number of steps they move collectively, $14k$ , is a multiple of $12$ . Since this number must be a multiple of $12$ , as stated in the previous sentence, $14$ has a factor $2$ $k$ must have a factor of $6$ . The smallest number of turns that is a multiple of $6$ is $\boxed{6}$

Answer: A 6
Problem: How many distinct triangles can be drawn using three of the dots below as vertices?
[asy]dot(origin^^(1,0)^^(2,0)^^(0,1)^^(1,1)^^(2,1));[/asy]
$\textbf{(A)}\ 9\qquad\textbf{(B)}\ 12\qquad\textbf{(C)}\ 18\qquad\textbf{(D)}\ 20\qquad\textbf{(E)}\ 24$

Solution: The number of ways to choose three points to make a triangle is $\binom 63 = 20$ . However, two* of these are a straight line so we subtract $2$ to get $\boxed{18}$

Answer: C 18
Problem: How many distinct triangles can be drawn using three of the dots below as vertices?
[asy]dot(origin^^(1,0)^^(2,0)^^(0,1)^^(1,1)^^(2,1));[/asy]
$\textbf{(A)}\ 9\qquad\textbf{(B)}\ 12\qquad\textbf{(C)}\ 18\qquad\textbf{(D)}\ 20\qquad\textbf{(E)}\ 24$

Solution: Case 1: One vertex is on the top 3 points
Here, there is $\binom 31 =3$ ways to choose the vertex on the top and $\binom 32 =3$ ways the choose the $2$ on the bottom, so there is $3 \cdot 3=9$ triangles.
Case 2: One vertex is on the bottom 3 points
By symmetry, there is $9$ triangles.
Otherwise, the triangle is degenerate.
Our final answer is $9+9=\boxed{18}$ Ddk001

Answer: C 18
Problem: Isosceles right triangle $ABC$ encloses a semicircle of area $2\pi$ . The circle has its center $O$ on hypotenuse $\overline{AB}$ and is tangent to sides $\overline{AC}$ and $\overline{BC}$ . What is the area of triangle $ABC$
[asy]pair a=(4,4), b=(0,0), c=(0,4), d=(4,0), o=(2,2); draw(circle(o, 2)); clip(a--b--c--cycle); draw(a--b--c--cycle); dot(o); label("$C$", c, NW); label("$A$", a, NE); label("$B$", b, SW);[/asy]
$\textbf{(A)}\ 6\qquad\textbf{(B)}\ 8\qquad\textbf{(C)}\ 3\pi\qquad\textbf{(D)}\ 10\qquad\textbf{(E)}\ 4\pi$

Solution: First, we notice half a square so first let's create a square. Once we have a square, we will have a full circle. This circle has a diameter of 4 which will be the side of the square. The area would be $4\cdot 4 = 16.$ Divide 16 by 2 to get the original shape and you get $\boxed{8}$

Answer:  8
Problem: A certain calculator has only two keys [+1] and [x2]. When you press one of the keys, the calculator automatically  displays the result. For instance, if the calculator originally displayed "9" and you pressed [+1], it would display "10." If you then pressed [x2], it would display "20." Starting with the display "1," what is the fewest number of keystrokes you would need to reach "200"?
$\textbf{(A)}\ 8\qquad\textbf{(B)}\ 9\qquad\textbf{(C)}\ 10\qquad\textbf{(D)}\ 11\qquad\textbf{(E)}\ 12$

Solution: We can start at $200$ and work our way down to $1$ . We want to press the button that multiplies by $2$ the most, but since we are going down instead of up, we divide by $2$ instead. If we come across an odd number, then we will subtract that number by $1$ . Notice
Since we've reached $1$ , it's clear that the answer should be $\boxed{9}$ . Because we only subtracted $1$ when we had to, this is optimal. ~Roy2020

Answer: B 9
Problem: A square with side length 2 and a circle share the same center. The total area of the regions that are inside the circle and outside the square is equal to the total area of the regions that are outside the circle and inside the square. What is the radius of the circle?
[asy] pair a=(4,4), b=(0,0), c=(0,4), d=(4,0), o=(2,2); draw(a--d--b--c--cycle); draw(circle(o, 2.5)); [/asy]
$\textbf{(A)}\ \frac{2}{\sqrt{\pi}} \qquad \textbf{(B)}\ \frac{1+\sqrt{2}}{2} \qquad \textbf{(C)}\ \frac{3}{2} \qquad \textbf{(D)}\ \sqrt{3} \qquad \textbf{(E)}\ \sqrt{\pi}$

Solution: Let the region within the circle and square be $a$ . In other words, it is the area inside the circle $\textbf{and}$ the square.  Let $r$ be the radius. We know that the area of the circle minus $a$ is equal to the area of the square, minus $a$
We get:
$\pi r^2 -a=4-a$
$r^2=\frac{4}{\pi}$
$r=\frac{2}{\sqrt{\pi}}$
So the answer is $\boxed{2}$

Answer: A 2
Problem: A square with side length 2 and a circle share the same center. The total area of the regions that are inside the circle and outside the square is equal to the total area of the regions that are outside the circle and inside the square. What is the radius of the circle?
[asy] pair a=(4,4), b=(0,0), c=(0,4), d=(4,0), o=(2,2); draw(a--d--b--c--cycle); draw(circle(o, 2.5)); [/asy]
$\textbf{(A)}\ \frac{2}{\sqrt{\pi}} \qquad \textbf{(B)}\ \frac{1+\sqrt{2}}{2} \qquad \textbf{(C)}\ \frac{3}{2} \qquad \textbf{(D)}\ \sqrt{3} \qquad \textbf{(E)}\ \sqrt{\pi}$

Solution: We realize that since the areas of the regions outside of the circle and the square are equal to each other, the area of the circle must be equal to the area of the square.
$\pi r^2=4$
$r^2=\frac{4}{\pi}$
$r=\frac{2}{\sqrt{\pi}}$
So the answer is $\boxed{2}$

Answer: A 2
Problem: On a map, a $12$ -centimeter length represents $72$ kilometers. How many kilometers does a $17$ -centimeter length represent?
$\textbf{(A)}\ 6\qquad\textbf{(B)}\ 102\qquad\textbf{(C)}\ 204\qquad\textbf{(D)}\ 864\qquad\textbf{(E)}\ 1224$

Solution: We set up the proportion $\frac{12  \text{cm}}{72  \text{km}}=\frac{17 \text{cm}}{x \text{km}}$ . Thus $x=102 \Rightarrow \boxed{102}$

Answer: B 102
Problem: How many different four-digit numbers can be formed by rearranging the four digits in $2004$
$\textbf{(A)}\ 4\qquad\textbf{(B)}\ 6\qquad\textbf{(C)}\ 16\qquad\textbf{(D)}\ 24\qquad\textbf{(E)}\ 81$

Solution: We can solve this problem easily, just by calculating how many choices there are for each of the four digits.
First off, we know there are only $2$ choices for the first digit, because $0$ isn't a valid choice, or the number would a 3-digit number, which is not what we want. 
We have $3$ choices for the second digit, since we already used up one of the digits, and $2$ choices for the third, and finally just $1$ choices for the fourth and final one. $2*3*2*1$ is 12, but there are 2 zeros that have been counted as different numbers, so divide by 2 to get $\boxed{6}$

Answer: B 6
Problem: How many different four-digit numbers can be formed by rearranging the four digits in $2004$
$\textbf{(A)}\ 4\qquad\textbf{(B)}\ 6\qquad\textbf{(C)}\ 16\qquad\textbf{(D)}\ 24\qquad\textbf{(E)}\ 81$

Solution: Note that the four-digit number must start with either a $2$ or a $4$ . The four-digit numbers that start with $2$ are $2400, 2040$ , and $2004$ . The four-digit numbers that start with $4$ are $4200, 4020$ , and $4002$ which gives us a total of $\boxed{6}$

Answer: B 6
Problem: Twelve friends met for dinner at Oscar's Overstuffed Oyster House, and each ordered one meal. The portions were so large, there was enough food for $18$ people. If they shared, how many meals should they have ordered to have just enough food for the $12$ of them?
$\textbf{(A)}\ 8\qquad\textbf{(B)}\ 9\qquad\textbf{(C)}\ 10\qquad\textbf{(D)}\ 15\qquad\textbf{(E)}\ 18$

Solution: Set up the proportion $\frac{12\ \text{meals}}{18\ \text{people}}=\frac{x\ \text{meals}}{12\ \text{people}}$ . Solving for $x$ gives us $x= \boxed{8}$

Answer: A 8
Problem: Ms. Hamilton’s eighth-grade class wants to participate in the annual three-person-team basketball tournament.
Lance, Sally, Joy, and Fred are chosen for the team. In how many ways can the three starters be chosen?
$\textbf{(A)}\ 2\qquad\textbf{(B)}\ 4\qquad\textbf{(C)}\ 6\qquad\textbf{(D)}\ 8\qquad\textbf{(E)}\ 10$

Solution: There are $\binom{4}{3}$ ways to choose three starters. Thus the answer is $\boxed{4}$

Answer: B 4
Problem: Ms. Hamilton’s eighth-grade class wants to participate in the annual three-person-team basketball tournament.
Lance, Sally, Joy, and Fred are chosen for the team. In how many ways can the three starters be chosen?
$\textbf{(A)}\ 2\qquad\textbf{(B)}\ 4\qquad\textbf{(C)}\ 6\qquad\textbf{(D)}\ 8\qquad\textbf{(E)}\ 10$

Solution: We can choose $3$ people by eliminating one from a set of $4$ one at a time and the other three get selected. There are $4$ ways to remove a person from a group of four (without considering order), so there are $\boxed{4}$ ways to choose three people, where order doesn't matter.

Answer: B 4
Problem: Ms. Hamilton's eighth-grade class wants to participate in the annual three-person-team basketball tournament. The losing team of each game is eliminated from the tournament. If sixteen teams compete, how many games will be played to determine the winner?
$\textbf{(A)}\ 4\qquad\textbf{(B)}\ 7\qquad\textbf{(C)}\ 8\qquad\textbf{(D)}\ 15\qquad\textbf{(E)}\ 16$

Solution: The remaining team will be the only undefeated one. The other $\boxed{15}$ teams must have lost a game before getting out, thus fifteen games yielding fifteen losers.

Answer: D 15
Problem: Ms. Hamilton's eighth-grade class wants to participate in the annual three-person-team basketball tournament. The losing team of each game is eliminated from the tournament. If sixteen teams compete, how many games will be played to determine the winner?
$\textbf{(A)}\ 4\qquad\textbf{(B)}\ 7\qquad\textbf{(C)}\ 8\qquad\textbf{(D)}\ 15\qquad\textbf{(E)}\ 16$

Solution: There will be $8$ games the first round, $4$ games the second round, $2$ games the third round, and $1$ game in the final round, giving us a total of $8+4+2+1=15$ games. $\boxed{15}$

Answer: D 15
Problem: After Sally takes $20$ shots, she has made $55\%$ of her shots. After she takes $5$ more shots, she raises her percentage to $56\%$ . How many of the last $5$ shots did she make?
$\textbf{(A)}\ 1\qquad\textbf{(B)}\ 2\qquad\textbf{(C)}\ 3\qquad\textbf{(D)}\ 4\qquad\textbf{(E)}\ 5$

Solution: Sally made $0.55*20=11$ shots originally. Letting $x$ be the number of shots she made, we have $\frac{11+x}{25}=0.56$ . Solving for $x$ gives us $x=\boxed{3}$

Answer: C 3
Problem: An athlete's target heart rate, in beats per minute, is $80\%$ of the theoretical maximum heart rate. The maximum heart rate is found by subtracting the athlete's age, in years, from $220$ . To the nearest whole number, what is the target heart rate of an athlete who is $26$ years old?
$\textbf{(A)}\ 134\qquad\textbf{(B)}\ 155\qquad\textbf{(C)}\ 176\qquad\textbf{(D)}\ 194\qquad\textbf{(E)}\ 243$

Solution: The maximum heart rate is $220-26=194$ beats per minute. The target heart rate is then $0.8*194 \approx \boxed{155}$ beats per minute.

Answer: B 155
Problem: Find the number of two-digit positive integers whose digits total $7$
$\textbf{(A)}\ 6\qquad\textbf{(B)}\ 7\qquad\textbf{(C)}\ 8\qquad\textbf{(D)}\ 9\qquad\textbf{(E)}\ 10$

Solution: The numbers are $16, 25, 34, 43, 52, 61, 70$ which gives us a total of $\boxed{7}$

Answer: B 7
Problem: The average of the five numbers in a list is $54$ . The average of the first two
numbers is $48$ . What is the average of the last three numbers?
$\textbf{(A)}\ 55 \qquad \textbf{(B)}\ 56 \qquad \textbf{(C)}\ 57 \qquad \textbf{(D)}\ 58 \qquad \textbf{(E)}\ 59$

Solution: Let the $5$ numbers be $a, b, c, d$ , and $e$ . Thus $\frac{a+b+c+d+e}{5}=54$ and $a+b+c+d+e=270$ . Since $\frac{a+b}{2}=48$ $a+b=96$ . Substituting back into our original equation, we have $96+c+d+e=270$ and $c+d+e=174$ . Dividing by $3$ gives the average of $\boxed{58}$

Answer: D 58
Problem: Handy Aaron helped a neighbor $1 \frac14$ hours on Monday, $50$ minutes on Tuesday, from 8:20 to 10:45 on Wednesday morning, and a half-hour on Friday. He is paid $\textdollar 3$ per hour. How much did he earn for the week?
$\textbf{(A)}\ \textdollar 8 \qquad \textbf{(B)}\ \textdollar 9 \qquad \textbf{(C)}\ \textdollar 10 \qquad \textbf{(D)}\ \textdollar 12 \qquad \textbf{(E)}\ \textdollar 15$

Solution: Let's convert everything to minutes and add them together. On Monday he worked for $\frac54 \cdot 60 = 75$ minutes. On Tuesday he worked $50$ minutes. On Wednesday he worked for $2$ hours $25$ minutes, or $2(60)+25=145$ minutes. On Friday he worked $\frac{60}{2}=30$ minutes. This adds up to $75+50+145+30=300$ minutes, or $300/60=5$ hours and $5\cdot 3 = \boxed{15}$

Answer: E 15
Problem: The numbers $-2, 4, 6, 9$ and $12$ are rearranged according to these rules:
What is the average of the first and last numbers?
$\textbf{(A)}\ 3.5 \qquad \textbf{(B)}\ 5 \qquad \textbf{(C)}\ 6.5 \qquad \textbf{(D)}\ 7.5 \qquad \textbf{(E)}\ 8$

Solution: From rule 1, the largest number, $12$ , can be second or third. From rule 2, because there are five places, the smallest number $-2$ can either be third or fourth. The median, $6$ can be second, third, or fourth. Because we know the middle three numbers, the first and last numbers are $4$ and $9$ , disregarding their order. Their average is $(4+9)/2 = \boxed{6.5}$

Answer: C 6.5
Problem: Niki usually leaves her cell phone on. If her cell phone is on but
she is not actually using it, the battery will last for $24$ hours. If
she is using it constantly, the battery will last for only $3$ hours.
Since the last recharge, her phone has been on $9$ hours, and during
that time she has used it for $60$ minutes. If she doesn’t use it any
more but leaves the phone on, how many more hours will the battery last?
$\textbf{(A)}\ 7 \qquad \textbf{(B)}\ 8 \qquad \textbf{(C)}\ 11 \qquad \textbf{(D)}\ 14 \qquad \textbf{(E)}\ 15$

Solution: When not being used, the cell phone uses up $\frac{1}{24}$ of its battery per hour. When being used, the cell phone uses up $\frac{1}{3}$ of its battery per hour. Since Niki's phone has been on for $9$ hours, of those $8$ simply on and $1$ being used to talk, $8(\frac{1}{24}) + 1(\frac{1}{3}) = \frac{2}{3}$ of its battery has been used up. To drain the remaining $\frac{1}{3}$ the phone can last for $\frac{\frac{1}{3}}{\frac{1}{24}}=\boxed{8}$ more hours without being used.

Answer: B 8
Problem: Thirteen black and six white hexagonal tiles were used to create the figure below. If a new figure is created by attaching a border of white tiles with the same size and shape as the others, what will be the difference between the total number of white tiles and the total number of black tiles in the new figure?
[asy] defaultpen(linewidth(1)); real x=sqrt(3)/2; path p=rotate(30)*polygon(6); filldraw(p^^shift(0,3)*p^^shift(4x,0)*p^^shift(3x,1.5)*p^^shift(2x,3)*p^^shift(-4x,0)*p^^shift(-3x,1.5)*p^^shift(-2x,3)*p^^shift(3x,-1.5)*p^^shift(-3x,-1.5)*p^^shift(2x,-3)*p^^shift(-2x,-3)*p^^shift(0,-3)*p, black, black); draw(shift(2x,0)*p^^shift(-2x,0)*p^^shift(x,1.5)*p^^shift(-x,1.5)*p^^shift(x,-1.5)*p^^shift(-x,-1.5)*p); [/asy]
$\textbf{(A)}\ 5\qquad \textbf{(B)}\ 7\qquad \textbf{(C)}\ 11\qquad \textbf{(D)}\ 12 \qquad \textbf{(E)}\ 18$

Solution: The first ring around the middle tile has $6$ tiles, and the second has $12$ . From this pattern, the third ring has $18$ tiles. Of these, $6+18=24$ are white and $1+12=13$ are black, with a difference of $24-13 = \boxed{11}$

Answer: C 11
Problem: Three friends have a total of $6$ identical pencils, and each one has at least one pencil. In how many ways can this happen?
$\textbf{(A)}\ 1\qquad \textbf{(B)}\ 3\qquad \textbf{(C)}\ 6\qquad \textbf{(D)}\ 10 \qquad \textbf{(E)}\ 12$

Solution: For each person to have at least one pencil, assign one pencil to each of the three friends so that you have $3$ left. In partitioning the remaining $3$ pencils into $3$ distinct groups, use Ball-and-urn to find the number of possibilities is $\binom{3+3-1}{3-1} = \binom{5}{2} = \boxed{10}$

Answer: D 10
Problem: Three friends have a total of $6$ identical pencils, and each one has at least one pencil. In how many ways can this happen?
$\textbf{(A)}\ 1\qquad \textbf{(B)}\ 3\qquad \textbf{(C)}\ 6\qquad \textbf{(D)}\ 10 \qquad \textbf{(E)}\ 12$

Solution: Like in solution 1, for each person to have at least one pencil, assign one of the pencils to each of the three friends so that you have $3$ left. In partitioning the remaining $3$ pencils into $3$ distinct groups, use number of non-negative integral solutions. 
Let the three friends be $a, b, c$ respectively.
$a + b + c = 3$ The total being 3 and 2 plus signs, which implies $\binom{3+2}{2} = \binom{5}{2} = \boxed{10}$

Answer: D 10
Problem: Three friends have a total of $6$ identical pencils, and each one has at least one pencil. In how many ways can this happen?
$\textbf{(A)}\ 1\qquad \textbf{(B)}\ 3\qquad \textbf{(C)}\ 6\qquad \textbf{(D)}\ 10 \qquad \textbf{(E)}\ 12$

Solution: For each of the 3 People to have at least one pencil when distributing 6 pencil amongst them, we can use another formula from the Ball-and-urn counting technique, shown below:
for n = number of items, and s = slots:
Now we can plug in our values,
number of items = 6, and slots = 3:
$\binom{6-1}{3-1}     =     \binom{5}{2}     =    \boxed{10}$

Answer: D 10
Problem: Three friends have a total of $6$ identical pencils, and each one has at least one pencil. In how many ways can this happen?
$\textbf{(A)}\ 1\qquad \textbf{(B)}\ 3\qquad \textbf{(C)}\ 6\qquad \textbf{(D)}\ 10 \qquad \textbf{(E)}\ 12$

Solution: Like in solution 1 and solution 2, assign one pencil to each of the three friends so that you have $3$ left. In partitioning the remaining $3$ pencils into $3$ distinct groups use casework. 
Let the three friends be $a$ $b$ $c$ repectively.
$a + b + c = 3$
Case $1:a=0$
$b + c = 3$
$b = 0,1,2,3$
$c = 3,2,1,0$
$\boxed{4}$ solutions.

Answer:  4
Problem: Two-thirds of the people in a room are seated in three-fourths of the chairs. The rest of the people are standing. If there are $6$ empty chairs, how many people are in the room?
$\textbf{(A)}\ 12\qquad \textbf{(B)}\ 18\qquad \textbf{(C)}\ 24\qquad \textbf{(D)}\ 27\qquad \textbf{(E)}\ 36$

Solution: Working backwards, if $3/4$ of the chairs are taken and $6$ are empty, then there are three times as many taken chairs as empty chairs, or $3 \cdot 6 = 18$ . If $x$ is the number of people in the room and $2/3$ are seated, then $\frac23 x = 18$ and $x = \boxed{27}$

Answer: D 27
Problem: Spinners $A$ and $B$ are spun. On each spinner, the arrow is equally likely to land on each number. What is the probability that the product of the two spinners' numbers is even?
[asy] pair A=(0,0); pair B=(3,0); draw(Circle(A,1)); draw(Circle(B,1));  draw((-1,0)--(1,0)); draw((0,1)--(0,-1)); draw((3,0)--(3,1)); draw((3+sqrt(3)/2,-.5)--(3,0)); draw((3,0)--(3-sqrt(3)/2,-.5));   label("$A$",(-1,1)); label("$B$",(2,1));  label("$1$",(-.4,.4)); label("$2$",(.4,.4)); label("$3$",(.4,-.4)); label("$4$",(-.4,-.4)); label("$1$",(2.6,.4)); label("$2$",(3.4,.4)); label("$3$",(3,-.5));  [/asy]
$\textbf{(A)}\ \frac14\qquad \textbf{(B)}\ \frac13\qquad \textbf{(C)}\ \frac12\qquad \textbf{(D)}\ \frac23\qquad \textbf{(E)}\ \frac34$

Solution: An even number comes from multiplying an even and even, even and odd, or odd and even. Since an odd number only comes from multiplying an odd and odd, there are less cases and it would be easier to find the probability of spinning two odd numbers from $1$ . Multiply the independent probabilities of each spinner getting an odd number together and subtract it from $1$
\[1-\frac24 \cdot \frac23 = 1- \frac13 = \boxed{23}\]

Answer: D 23
Problem: Spinners $A$ and $B$ are spun. On each spinner, the arrow is equally likely to land on each number. What is the probability that the product of the two spinners' numbers is even?
[asy] pair A=(0,0); pair B=(3,0); draw(Circle(A,1)); draw(Circle(B,1));  draw((-1,0)--(1,0)); draw((0,1)--(0,-1)); draw((3,0)--(3,1)); draw((3+sqrt(3)/2,-.5)--(3,0)); draw((3,0)--(3-sqrt(3)/2,-.5));   label("$A$",(-1,1)); label("$B$",(2,1));  label("$1$",(-.4,.4)); label("$2$",(.4,.4)); label("$3$",(.4,-.4)); label("$4$",(-.4,-.4)); label("$1$",(2.6,.4)); label("$2$",(3.4,.4)); label("$3$",(3,-.5));  [/asy]
$\textbf{(A)}\ \frac14\qquad \textbf{(B)}\ \frac13\qquad \textbf{(C)}\ \frac12\qquad \textbf{(D)}\ \frac23\qquad \textbf{(E)}\ \frac34$

Solution: We can make a chart and the we see that the 12 possibilities: 1, 2, 3, 2, 4, 6, 3, 6, 9, 4, 8, and 12. Out of these only 8 work; thus the probability is \[\boxed{23}\]

Answer: D 23
Problem: Spinners $A$ and $B$ are spun. On each spinner, the arrow is equally likely to land on each number. What is the probability that the product of the two spinners' numbers is even?
[asy] pair A=(0,0); pair B=(3,0); draw(Circle(A,1)); draw(Circle(B,1));  draw((-1,0)--(1,0)); draw((0,1)--(0,-1)); draw((3,0)--(3,1)); draw((3+sqrt(3)/2,-.5)--(3,0)); draw((3,0)--(3-sqrt(3)/2,-.5));   label("$A$",(-1,1)); label("$B$",(2,1));  label("$1$",(-.4,.4)); label("$2$",(.4,.4)); label("$3$",(.4,-.4)); label("$4$",(-.4,-.4)); label("$1$",(2.6,.4)); label("$2$",(3.4,.4)); label("$3$",(3,-.5));  [/asy]
$\textbf{(A)}\ \frac14\qquad \textbf{(B)}\ \frac13\qquad \textbf{(C)}\ \frac12\qquad \textbf{(D)}\ \frac23\qquad \textbf{(E)}\ \frac34$

Solution: We do a little bit of casework. In order to get a product that's even, we need at least one even number. First, we consider the probability of getting an even number on the first spinner, and then multiply by 1 because the second spinner can be anything. \[\frac12 \cdot 1 = \frac12\] Next, we look at the chance that we don't get an even on the first spinner, but get an even on the second spinner (we don't do the probability of even on the second spinner multiplied by one because we would be double counting both spinners are even). \[\frac12 \cdot \frac13 = \frac16\] Add these two together to get the total probability, and we get \[\frac12 + \frac16 = \frac46 = \boxed{23}\]

Answer: D 23