Problem number one states, what is the value of 20 times 19 plus 20 plus 19? In order to solve this problem, we have to keep in mind the order of operations, which is commonly abbreviated as PEMDAS. First, we have to do the P, which is parentheses. There are no parentheses in the problem, so we can continue. The E stands for exponents, and there are no exponents in the expression, so we can continue. The M stands for multiplication, and there is multiplication. 20 times 19 is going to be 20 times 20 minus 20, so 380. The D stands for division, and again, there is no division, so we can continue. The A stands for addition, and 380 plus 20 is 400, and 400 plus 19 is 419. At this point, we've reduced our expression down to one number, so we can just stop here. So the question asked us, what is the value of this expression? The answer is 419, letter D. Problem number 2 states, a model train takes exactly 1 minute and 11 seconds for each round on a course. How long does it take for 6 rounds? So we are told that the train takes 1 minute and 11 seconds per round. If we multiply this rate by 6 rounds, we will get the answer. In this case, the rounds unit cancels out, and dividing a number by 1 is useless because that just gives you that number again, so we can cancel that out. And we're left with 1 minute and 11 seconds times 6. We distribute the 6 over the minutes and the seconds to get 6 minutes and 66 seconds. There are more seconds in the seconds space than there are in a minute, so the seconds would overflow, causing an increase in the minute and a reduction of 60 in the seconds. So you get 7 minutes and 6 seconds. The question asked us, how long does it take for 6 rounds? The answer is 7 minutes, 6 seconds, letter B. Problem number 3 states, a barber wants to write the word shave on a board in such a way that a client looking in the mirror reads the word correctly. How could the barber write it on the board? When a mirror flips an image, it flips it over the x-axis. Let's draw a line of symmetry through the letter A. The letter A is symmetrical with respect to the x-axis, so when it is flipped over, nothing changes. The H and the V are also symmetrical with respect to the x-axis, so when they are flipped, only their positions flip. Now for the S and the E. The S and the E are not symmetrical with respect to the x-axis, so when they are flipped over the red line, the S will look backwards and the E will look backwards. In addition to this, they will also swap spaces. So in the end, we have a backwards E, a V, an A, an H, and a backwards S. So the question asked us, how should the barber write it on the board? Well, the answer is letter E. Problem number four states, how many different sums of dots can you get by rolling three standard dice simultaneously? A standard dice has six sides. Each side is labeled with one through six dots. In order to get the maximum possible roll with three dice, you would have to roll three sixes. This would result in three times six dots, or 18 dots. Again, this is the maximum possible number of dots you can get in one roll of three six-sided dice. The minimum you can get is if you roll three ones. Then you would get three times one dots, or just three dice. This is again the minimum number of dots you can get by rolling three dice. It is possible to get any number of dots in between three and 18. The number of possible rolls between three and 18 inclusive is 16. The question asked us, how many different sums of dots can you get by rolling three standard dice simultaneously? The answer is 16, letter C. Problem number five states, five identical glasses are filled with water. Four of them contain the same amount of water. Which one contains a different amount? First, let's take a look at glass A. We need to find the volume of this region. This is impossible since we do not know the dimensions of the bottom of the glass. Luckily, we know that all the glasses cover at least this portion, so we can just not count it. The equation for the volume of the glass above the red line is pi r-squared h. This glass is five units tall, so the volume would be five pi r-squared. Now let's move on to glass B. Again, let's disregard the bottom region, since all the glasses have it filled up to at least that point, and let's bring up the volume equation. Now we've reached a problem, because h is not a constant, it varies. It would be useful if we found the average height of the water, which would allow us to use our equation. In order to find the average height, we'll take the height of both sides of the glass. We'll take the height of both sides and divide it by two. The height on the left side is eight units, and the height on the right side is three units. Again, we count only above the red line. Three plus eight over two is five and a half, so the volume of the water is five and a half pi r-squared. It is good to know that A and B are different. It means that we only have to do C to see which one is not like the others. So let's do C. Again, let's disregard the bottom region, and let's bring up the equations we previously used. The height of the water on the left side is three, and on the right side it is seven. Three plus seven over two is five, so the volume is five pi r-squared. Now we know that glass B has a different amount of water to the other glasses. The question asked us, which one contains a different amount? The answer is glass B, letter B. Problem number six states, a park has five gates. Monica wants to enter through one gate and exit through a different one. In how many ways can she enter and exit the park? At the start, Monica has five gates to choose from. She can enter through any of them. So far, there have been five possibilities. Then, Monica has four options. She can exit through any one of the remaining four gates. This gives us four additional options per gate, which gives us five times four options in total, giving us 20 options. The question asked us, in how many ways can she enter and exit the park? The answer is 20 ways, letter B. Problem number seven states, the weights of three kangaroos are different whole numbers. Their total weight is 97 kilograms. How much at most can the lightest one of them weigh? So we have a restriction, and we know that A plus B plus C must be equal to 97. A plus B plus C are the weights of the kangaroos. And our goal is going to be to maximize A, B, and C. So let's try to distribute the weight amongst them. First, let's distribute the 90 kilograms by giving each kangaroo 30 kilograms. We have 7 kilograms left, so let's distribute 6 of those kilograms evenly to give each kangaroo 32 kilograms with 1 kilogram spare. Let's give the remaining kilogram to kangaroo A. The question specifically said that the three kangaroos weigh different whole numbers, which means that this cannot be the weights of them, because B and C are the same number. Therefore, we have to change the weight of one of the kangaroos that weighs 32 kilograms, and add it to the one of the kangaroos that weighs 1 kilogram. Let's change the weight of one of the kangaroos that weighs 32 kilograms, and add it to the one that weighs 33 kilograms. Now all the kangaroos weigh different whole numbers. The smallest weight is 31 kilograms. The question asked us, how much at most can the lightest one of them weigh? The answer is 31 kilograms, letter C. Problem number 8 states, which of the following statements is true for the marked angles in the given figure made of 9 identical squares? First, let's mark some key points of interest. It is easy to see that angle A, B, D, and A, C, E are equal to 90 degrees, because they are the corners of a square, and by definition the corners of a square are 90 degrees. Length A, C, and length A, B are equal to one another. We get this by counting squares. Each are 3 squares long. The same is true for B, D, and C, E. They are both 2 squares long. Using side-angle-side, we can see that triangle A, C, E, and triangle A, B, D are congruent. Since the two triangles are congruent, this means that angle E, A, C, and D, A, B are the same angle. Since we know that angle B, A, C is 90 degrees, we know that 2 alpha plus beta equals 90 degrees. This is one of the statements the problem gives us. So the question asked us, which of the following statements is true for the marked angles in the given figure made of 9 identical squares? The answer is 2 alpha plus beta equals 90 degrees. Letter B. Problem number 9 states, inside each unit square, a certain part has been shaded. In which square is the total shaded area the largest? Let's first start off with the most simple square. Let's say it has width A. Since it is a square, it also has height A. The formula for the area of a triangle is 1 half base times height. In this case, the base and the height are both A, so the area is A squared over 2, or 1 half A squared. The next square is more complicated. It contains 7 smaller triangles. So in order to find the area of the shaded region, we have to add up all the individual triangles. The base of each of these triangles is labeled B1, B2, B3, and so on, all the way until B7. In order to simplify the area equation, let's factor out A over 2 from each of these terms. Now we have to notice that B1 plus B2 plus B3 plus B4 plus B5 plus B6 plus B7 is the length of A. The bases of all the triangles make up the side length, so we can substitute that sum for A. Therefore, the shaded area of this square is also 1 half A squared. A similar process occurs with this square. Let's use our factored triangle area equation. Here it is more clear to see that B1 plus B2 plus B3 equal the side length A, so the area is again equal to 1 half A squared. In this next one, there are only two triangles, and it is again easy to see that the base of the first triangle plus the base of the second triangle is equal to the side length, so the area is again equal to 1 half A squared. For this one, the shaded area is going to be the sum of the areas of the triangles plus the area of the rectangle. Again, the triangles have bases B1, B2, and the rectangle in this case is going to have base BR. BR times A is the base times the height of the rectangle, which is the equation for the area of the rectangle. Now let's factor out a 1 half from the two left terms, and let's factor out an A from every single term to get the following equation. Now it is important to note that B1 plus B2 plus BR are equal to A. If we subtract BR from both sides, we get that B1 plus B2 is equal to A minus BR, which we substitute in for B1 plus B2. Now we factor the 1 half to the A and to the base of the rectangle, and then combine like terms. Lastly, we distribute the A to the remaining terms to get 1 half A squared plus 1 half the base of the rectangle times A. This, of course, is greater than 1 half A squared. So the question asked us, in which square is the total shaded area the largest? The answer is square A, letter A. Problem number 10 states, a five digit number is written on each of three pieces of paper as shown. Three of the digits are covered. The sum of the three digits is 57,263. What are the covered digits? In order to solve this problem, let's just do standard long addition. Let's start out in the rightmost column. 8 plus 4 plus 1 is 13, which makes sense since there's a 3 at the end, and we would carry a 1. 1 plus 2 plus 0 plus 3 is going to be 6, which makes sense because there's a 6 at the end. So that all checks out. Then we have 7 plus A plus 3 has to have something that ends in a 2. So it would be 10 plus A. So if we replace A with 2, we would get a sum of 12, which would result in an ending 2, which would make sense, and then we would just carry a 1. For this one, we have 1 plus 5 plus 2, which would be 8. 8 plus a number has to end in a 7, so if we replace the B with a 9, we would get a total of 17, and we would carry a 1 and there would be a 7 on the end, which would make sense with the sum. So now finally, we have 1 plus 1 plus 2 plus C equals 5, so C must be equal to 1. So the numbers hidden by the piece of paper were 1, 2, and 9. So the question asked us, what are the cover digits? The answer is 1, 2, and 9, letter B. Problem number 11 states, a square has vertices A, B, C, D labeled clockwise. An equilateral triangle is constructed with labels AEC labeled clockwise. What is the measure of angle CBE in degrees? So first let's create a line that runs through D, B, and E. Now we know that angle DCB is 90 degrees, and we know that line AC cuts that angle in half. So therefore we know that angle ACB is 45 degrees. We also know that triangle ACE is an equilateral triangle, which means that each of its angles has an angle of 60 degrees. So angle ACE is 60 degrees. By subtracting these two angles, we can see that angle BCE is 15 degrees. Since ACE is an equilateral triangle, AEC is 60 degrees. And we know that the line segment that we constructed cuts this angle in half. So we know that angle BEC is 30 degrees. We know that the sum of the angles of triangle CBE add up to 180 degrees. And we already know the value of two of the angles, angle ECB and BEC. This allows us to solve for angle CBE, and we see that it equals 135 degrees. So the question asked us, what is the measure of angle CBE in degrees? The answer is 135. Letter C. Problem number 12 states, the numbers ABCD are distinct positive integers chosen from 1 to 10. What is the least possible value A over B plus C over D could have? So we know that A, B, C, and D have to be between 1 and 10 and cannot be the same number. We know that as we increase the denominator, the number gets smaller, and we also know that as we decrease the numerator, the number also gets smaller. So we want the lowest numerator and the largest denominator. So in our example of A over B plus C over D, we want A and C to be as small as possible, and B and D to be as large as possible. So we want A and C to be 1 and 2, and B and D to be 9 and 10. We have four possibilities. Since addition is commutative, half of these possibilities are the same, so we can just not check them. First let's do A equals 1, B equals 9, C equals 2, and D equals 10. This gets us 1 over 9 plus 2 over 10, which is 28 90ths. Now let's try the other way. A equals 2, C equals 1. 2 9ths plus 1 10th. This equals 29 90ths. 28 90ths was less, so we'll use this one. 28 and 90 are both divisible by 2, so we can simplify by dividing both by 2 to get 14 45ths. So the question asked us, what is the least possible value A over B plus C over D could have? The answer is 14 45ths. Letter C. Problem number 13 states, the flag of Kangaroo area is a rectangle with side lengths in the ratio 3 to 5. The flag is divided into four rectangles of equal area as shown. What is the ratio of the side lengths of the white rectangle? So the question told us that the ratio of the sides of the flag is 3 to 5, so we can assume that the height of the flag is 3 and the width of the flag is 5 units. With this knowledge, and the knowledge that the area of a rectangle is its base times its height, we can determine that the area of the flag is 3 times 5, or 15 units squared. Since the question tells us that all the stripes are of the same area, we know that if we divide the area of the flag by 4, we will get the area of each individual stripe. In this case, the area is 15 4ths. Since there are three stripes, and they make up together a length of 3 units, each must be 1 unit tall. We know that the area of a rectangle is its width times its height, and we know the area of the stripes. We can now determine what the width of the rectangle is, 15 4ths. With this knowledge, we can find the ratio of the height to the width of the rectangle, which is 4 to 15. The question asked us, what is the ratio of the side lengths of the white rectangle? The answer is 4 to 15, letter E. Problem number 14 states, a 3 by 2 rectangle can be completely covered by two of the L-shaped figures in two different ways, as shown below. In how many different ways can the figure below be covered by the L-shaped figures? So first, let's start off by trying to fill in the top right corner of the shape. If we fill it in like this, it will be impossible to reach this section, because the L-shaped figures do not have the length to reach it. If we do it like this, the top right square cannot be covered. If we do it like this, the top left square can't be covered. And doing it like this looks promising, so let's continue. We can only continue from this point in one way, by putting an L-shaped figure right here. In this case, these two squares are impossible to fill, so this cannot be a possibility. Let's try to fill it in this way. If we put in an L-shaped piece here, we'll have the same problem as last time, so instead, let's put it here. Now we have a 3 by 2 rectangle. And as you recall, the problem told us that a 3 by 2 rectangle can be completely covered by two of the L-shaped figures in two different ways. And they even showed us an example. They showed it can be done in this way, and in this way. So the shape can be filled by the L-shaped figure in two different ways. So the question asked us, in how many different ways can the figure below be covered by the L-shaped figures? The answer is in two ways, letter B. Problem number 15 states, a triathlon consists of swimming, running, and biking. The biking is three-quarters of the total distance, the running is one-fifth, and the swimming is two kilometers. What is the total distance of this triathlon in kilometers? So we know that the length of the total triathlon is going to be equal to the length of the biking section plus the running length plus the swimming length. The question told us that the biking section would be three-fourths the length of the total race, and that the running portion would be one-fifth the portion of the total length. They also told us the swimming portion would be exactly two kilometers. Let's substitute these values in in our equation. Now let's combine like terms. Three-fourths plus one-fifth is going to be equal to nineteen-twentieths. Let's subtract nineteen-twentieths L from both sides to get one-twentieth L equals two, and finally let's multiply both sides by twenty. We get that the length of the total race is forty kilometers. So the question asked us, what is the total distance of this triathlon in kilometers? The answer is forty, letter D. Problem number sixteen states, Some diluted juice is to be made out of concentrate and water in the ratio one to seven by volume. The juice concentrate is in a one liter flask, and the flask is half full. What fraction of this concentrate should be used to produce two liters of diluted juice? So first let's take the ratio component of the concentrate and divide it by the total sum of the ratio components, and then let's multiply this by the amount of juice we need. This will get us the amount of concentrate we need to use to make that amount of juice. One-eighth times two will be one-fourth liters of concentrate. Next, let's figure out how much concentrate we have. We're told that we have a one liter flask of concentrate, but that it's only half full. So that means that we only have one-half liters of concentrate in the flask. So the question asked us what portion of this concentrate in the flask do we need to use to get the amount of concentrate we need to make the juice. So one-half times what is going to be equal to one-fourth? One-half times one-half is going to be equal to one-fourth, which means we only have to use half of what's in the flask. So the question asked us, What fraction of this concentrate should be used to produce two liters of diluted juice? The answer is one-half, letter B. Problem number 17 states, What is the perimeter of the shape? In order to simplify this problem, we have to notice that the shape is both symmetrical in the x and the y axis, which means we only have to find the perimeter of one-fourth of the shape, and then we can just multiply it by four. First, let's notice that this length right here is r. This length is r because it is the radius of the middle circle. This length is also r because it is the radius of the leftmost circle. This length is also r because it is the radius of the middlemost circle. As we can see, we formed a triangle where all three sides have length r, meaning the triangle is equilateral. All the angles on this triangle have a value of 60 degrees. The value of the orange angle plus the value of the green angle have to equal 180 degrees. This means the value of the green angle has to equal 120 degrees. The length of the arc in blue is going to be equal to r theta, where r is the radius of the circle and theta is the angle swept by the arc. We cannot use degrees, we have to use radians in this case, so let's convert 120 degrees into radians. That would be 2 pi over 3 radians. So the length of the arc is r times 2 pi over 3. We know that the orange angle plus the green angle in this case have to equal 90 degrees. This means that the green angle has to be equal to 30 degrees. Let's use the same arc formula for the other arc. This time the angle swept by the arc is going to be 30 degrees, so in radians that would be pi over 6 radians. 2 pi over 3 plus pi over 6 is going to be equal to 5 pi over 6. That means that the length of one quarter of the shape is going to be r 5 pi over 6. In order to find the total perimeter of the shape, let's multiply this length by 4. We get 20 pi over 6. This simplifies down to 10 pi over 3, so the perimeter of the shape is 10 pi r over 3. So the question asked us, what is the perimeter of the shape? The answer is 10 pi r over 3, letter A. Problem number 18 states, the 7 digits of the phone number AAABBBB add up to the 2 digit number AB. What is the sum of A plus B? So first let's figure out how to represent the sum AB mathematically. A is in the tens place, so AB could be represented as 10A plus B, as long as A and B are less than or equal to 9 and greater than or equal to 0. Next let's figure out how to represent the sum mathematically. We know that the sum of AAABBBB has to be equal to the 2 digit number AB. So let's just say that 3A plus 4B is going to be equal to the 2 digit sum AB. Let's substitute in 3A plus 4B for AB in our top equation. Now let's combine like terms. First let's subtract 3A from both sides to get 4B equals 7A plus B, and then let's subtract B from both sides to get 3B equals 7A. Now we have to figure out a whole number between 0 and 9 inclusive, such that 3B equals 7A, where B is not equal to A. Since multiplication is commutative, we can just say that B is 7 and A is 3, and we know that 3 times 7 equals 7 times 3, and they both equal 21, so this works. So we'll say that B equals 7 and A equals 3. Of course the sum A plus B is 7 plus 3, or 10. So the question asked us, what is the sum A plus B? The answer is 10, letter C. Problem number 19 states, 60 apples and 60 pears are packed into boxes so that each box contains the same number of apples, and so that no two boxes contain the same number of pears. What is the largest possible number of boxes that can be packed in this way? So first, let's start off with the apples. We have 60 apples, and if we divide the number of apples by the number of boxes that we use, we need to get a whole number. The possibilities for the number of boxes is known as the divisors of 60. 60 is 12 divisors, so these are all the possible number of boxes that we can have. Now let's move on to the pears. For the pears, let's just try to distribute them as efficiently as possible. So for the first box, let's put in 0 pears, and for the second box, let's put in 1 pear. For the third box, let's put in 2 pears, and for the fourth box, let's put in 3 pears, and so on and so on. After 11 boxes, we have 5 pears left. This means that we cannot have more than 11 boxes. So if we go back to our list of possible box numbers, we can cross off any number that's greater than 11. The largest possible number is 10, so let's try to make a combination of pears so that we have 10 boxes. Now there are many combinations of 10 different whole numbers whose sum is 60. One is shown here, and as you can see, the sum is in fact equal to 60. So it is possible to have 10 boxes with different amounts of pears in them who add up to 60. So the question asked us, what is the largest possible number of boxes that can be packed in this way? The answer is 10 boxes, letter D. Problem number 20 states, The diagram shows a net of an octahedron. When this net is folded to form the octahedron, which of the labeled line segments will coincide with the line segment marked with the X? Before we start the problem, I'm going to show you an animation of how an octahedron net folds into an octahedron. So as you can see in the animation, the two sides of the octahedron net folded into pyramids. This one side marked in red folded into one pyramid, and the other side not marked in red folded into another pyramid. In this pyramid, the two lines marked in yellow were connected together. When the two pyramids came together, we know that this edge was connected. But what was this edge connected to? Well this vertex over here swung down connected to that vertex, so this edge connected with the other edge. We work our way around the net figuring out which edges stitched together. Orange to orange, green to green. Now let's put back up the numbers again. It's clear to see that the X lines up with the 5. So the question asked us, which of the labeled line segments will coincide with the line segment marked with the X? The answer is 5, letter E. Problem number 21 states, a square has two of its vertices on a semicircle and the other two on the diameter of the semicircle as shown. The radius of the circle is 1 centimeter. What is the area of the square? So we know that this length is 1 centimeter. Let's call the side length of the square H. We know then that this length is H over 2. We know this length is H over 2 because the square hits the circle at the same height twice. Therefore, the center of the square is at the center of the circle. Now we will use Pythagoras' theorem, which states that the sum of the square of the edge lengths equals the square of the hypotenuse. We will take the square of the fraction and get H squared over 4 plus H squared equals 1. Then we will combine like terms to get 5H squared over 4 equals 1. We will then multiply both sides by 4 fifths. At this point there is no point continuing because we know that one edge of the square is H, which means that the area of the square is H squared. And that's what we're after, the area of the square. So we know now that it is 4 fifths. So the question asked us, what is the area of the square? The answer is 4 fifths centimeters squared. Letter A. Problem number 22 states, Two points are marked on a disk that is rotating around its center. One of them is 3 centimeters further than the other from the center of the disk and moves at a constant speed that is 2.5 times as fast as the other. What is the distance from the center of the disk to this far point? So let's call the point closer to the center point A and the one further away point B. The question tells us that the distance between them is 3 centimeters. We also know that the speed of B is 2.5 times that of the speed of A. Let's define the distance between A and the center of the disk as x. The distance point A travels is going to be 2 pi x. We get this from the equation for the perimeter of a circle, 2 pi r. In this case, the radius is x. The distance B travels per revolution is going to be 2 pi x plus 3. The ratio of the distances traveled is the ratio of their speeds, because point A and B both travel at a constant speed. We will turn this ratio into a fraction. Now we cross out 2 pi from the numerator and denominator on the left side, and then we cross-multiply. Then we subtract x from both sides, and divide both sides by 1.5 to get x equals 2. This means that the distance from the center of the disk to point B is 5 centimeters. So the question asked us, what is the distance from the center of the disk to this far point? The answer is 5 centimeters, letter E. Problem number 23 states, the integers from 1 to 99 are written in ascending order without gaps. The sequence of digits is then divided into triplets of digits. Which of the following is not one of the triplets? Since the majority of this problem deals with double digit integers, we'll just cut out the first triples, because they only deal with single digit integers. Now let's expand the amount of triplets we have. So now we have 10, 11, 12, 13, 14, 15, 16, 17, and 18. And if you notice, the 10 is at the beginning, and then the 11 is cut in half, and the 12 is at the end. And then the cycle begins again. The 13 is at the beginning, the 14 is cut in half, and the 15 is at the end. So let's split these up into pairs of triplets. The first pair has 10, 11, 12, then the second one has 13, 14, 15, then the third one has 16, 17, 18, and so on. As you notice, the last number in each of these pairs is divisible by 3. 12 is divisible by 3, 15 is divisible by 3, and 18 is divisible by 3. And the first number is divisible by 3 if you subtract 1 from it. So 10 minus 1 is 9, which is divisible by 3. 13 minus 1 is 12, which is divisible by 3. And 16 minus 1 is 15, which is divisible by 3. So now we know how these triplets are formed, and we know what properties they have. So let's go through each of the triplets in the problem, and see if they have a possibility of appearing. So let's start out with 2, 2, 2. We'll start out with doing 22, 23, 24. And 24 is divisible by 3, and 22 minus 1 is 21, which is also divisible by 3. So A works. B is 4, 4, 4, so let's first do 44, 45, 46. 44 minus 1 is 43, which is prime, so it is not divisible by 3, and 46 is also not divisible by 3. So let's do it in the other direction. 42, 43, 44, 45. 24 minus 1 is 23, which is not divisible by 3, and 44 is not divisible by 3, which means that 4, 4, 4 cannot appear. For C, let's do 46, 47, 48. 46 minus 1 is 45, which is divisible by 3. 48 is also divisible by 3, so C works. For D, 64, 65, 66. 66 is divisible by 3, and 64 minus 1 is 63, which is divisible by 3, so it also works. For E, let's do 88, 89, 90. 90 is divisible by 3, and 88 minus 1 is 87, which is also divisible by 3, so it also works. So the only one that doesn't work is 4, 4, 4. So the question asked us, which of the following is not one of the triplets? The answer is 4, 4, 4, letter B. Problem number 24 states, how many planes pass through exactly three vertices of a given cube? So here we have a cube. The vertices of the cube are the corners of the cube. Let's mark the vertices A through H. First, let's try to find a plane that runs through only three vertices. We'll choose the plane AFH. As you can see, it only runs through the vertices AFH and does not run through any other vertices. In order to find another plane like this, we'll just rotate this plane 90 degrees to get plane DEG. We'll do this again to get plane CFH, and once more to get plane BEG. If we rotate it again, we'll get back to plane AFH, so let's not do that. In order to get another plane combination, we can flip this whole plane vertically and get plane ACF. Now we can continue to rotate this one 90 degrees as we did the other one to get planes BED, ACH, and BDG. There are no more planes that run through exactly three vertices of a cube. We have eight possible planes. So the question asked us, how many planes pass through exactly three vertices of a given cube? The answer is eight planes, letter D. Problem number 25 states, a graph consists of 16 vertices and some edges that connect them, as shown in the picture. An ant is now at vertex labeled A. In each move, it can walk from one vertex to any neighboring vertex, crawling along a connecting edge. At which of the vertices labeled PQRST can the ant be after 2019 moves? So let's now color code all of our vertices, such that every red vertex has only neighboring blue vertices, and blue vertices only have neighboring red vertices. Now let's put our ant at vertex A and start a simulation. The ant can only move to a blue space, so in one move, the ant will be on a blue vertex. Now the ant can only move to a red vertex, so after two moves, it will be on a red vertex. After three moves, the ant will be on a blue vertex, and after four moves, the ant will be on a red vertex. You may see a pattern here. When the number of moves is even, the ant is on a red vertex, but when the number of moves is odd, the ant is on a blue vertex. This means that vertices P, S, T, and R are impossible to reach in 2019 moves, because 2019 is odd, and P, S, T, R are all red. So now let's just make sure that it's actually possible to get to position Q. So the ant will make its way towards position Q as fast and as efficient as possible. In five moves, it is at position Q. In order to use up its remaining 2014 moves, it will simply oscillate between vertex Q and vertex S, and after 2019 moves, it will end on vertex Q. So the question asked us, at which of the vertices labeled P, Q, R, S, T can the ant be after 2019 moves? The answer is only Q, letter C. Problem number 26 states that positive integers A, B, and C have three digits each, and for each integer, the first digit is the same as the last digit. Also, B equals 2A plus 1, and C equals 2B plus 1. How many possibilities are there for the integer A? So we know that A, B, and C have to both be between 100 and 999 inclusive, since they have to be three digits each. We can restrict the minimums of each of them a little bit further, since we know the first digit has to be equal to the last digit. We know that the minimum is actually 101, not 100. The problem gave us these two equations, and we're going to get to know these equations very well, because we are going to use them a lot throughout this problem. Since A is an integer, 2 times A will give an even number, plus 1 will give an odd number. Therefore, B has to be an odd number. The same is true for C. Since B is an integer, 2 times B is an even number, plus 1 will give an odd number. Now let's substitute one equation in for the other to get an equation for C with respect to A. Using this equation, let's restrict the maximum value of A by plugging the maximum value of C in for C. We know that 4A plus 3 has to be less than or equal to 999. Solving for A, we get that A has to be less than or equal to 249. Let's plug that in as our maximum value for A. Using this, we can also restrict the maximum for B. We're given the equation that B has to equal 2A plus 1. Let's plug in the maximum value for A, and then solve for B. We get that B is going to be equal to 499. So B is restricted between 101 and 499. But this is not true, because 499's first digit and last digit are not the same. We cannot go down to 494, because B has to be odd, so we have to go all the way down to 393. Because we had such a jump in the maximum value of B, we have to re-restrict the maximum value of A. So let's bring up our equation again, and plug in 393 for B. As we solve for A, we see that it is equal to 196. So A cannot be greater than 196. The first digit and last digit of A are not the same, so the actual maximum is 191. Now let's restrict the bounds of B even more. First, let's start out by restricting the minimum of B. Using our equation again, we plug in the minimum A value, and get that the minimum B value is 203. The first digit is not the same as the last digit, so 203 is not the proper minimum. 303 is. Since we had such a jump in A-max, let's re-restrict B-max. Let's again use 2A plus 1 equals B as our equation, and plug in 191 for A. We get that B is equal to 383. So the maximum value of B is 383. Now let's restrict the bounds of C with the values of B. So the equation for C was C equals 2B plus 1. We plug in the minimum of B in for B to get 2 times 303 plus 1, which is C equals 607. The first and last digit are not the same, and 6 is an even number, so we have to restrict it all the way up to 707. Now let's restrict C-max. Let's use our equation again, and plug in 383 for B. We get that the maximum value that C can have is 767, which fits both of our rules so we don't have to change it. If you recall, the minimum value of c jumped, so we have to re-account for that by recalculating the minimum of b and then recalculating the minimum of a. So using our equation again, we'll plug in 707 for c and then solve for b. The minimum value of b is now 353. Now let's recalculate the value of a using the other equation. Let's plug in the minimum of b for b to get 2a plus 1 equals 353, and then solve for a. We get that the minimum value of a is 176. But again, the first digit is not equal to the last digit, so we have to change it to 181. At this point, there are only two possible values for a, 181 or 191. Let's plug both of those in to our equations to see if they work. Let's start out with a equals 181. Plugging it into our first equation, we get that b equals 363. And plugging it into our second equation, we get that c equals 727. 363 is in the valid range of b, and 727 is in the valid range for c. For 191, we have a similar story. 383 is within the acceptable range of b, and 767 is in the acceptable range of c. Therefore, there are only two sets of numbers that work, one where a is 181 and the other when a is 191. So the question asked us, how many possibilities are there for the integer a? The answer is two, letter c. Problem number 27 states, At each vertex of a square, one positive integer is placed. For any two numbers joined by an edge of the square, one is a multiple of the other. However, for any two diagonally opposite numbers, neither is a multiple of the other. What is the smallest possible sum of the four numbers? So first let's label the vertices of our square a, b, c, d. Now we want to use the smallest possible numbers, so let's use 1 and 2. However, this is problematic, because every number is a multiple of 1, so we cannot use 1. Instead, let's use 2 and 3. 2 is not a multiple of 3, so we can put them diagonally from each other. Now we need to fill in b and d, such that they're both multiples of 2 and 3, but are not multiples of each other. The lowest multiples of 2 and 3 are 6 and 12. If we put them diagonally from each other, it does not work, because 6 is a multiple of 12. Let's use the next highest numbers, 12 and 18. If we put them diagonally from each other, it works. Let's check everything to make sure. 2 is a multiple of 18 and 12. 3 is a multiple of 18 and 12. 12 is not a multiple of 18, and 2 is not a multiple of 3, so everything works. The sum of these four numbers is 35. So the question asked us, what is the smallest possible sum of the four numbers? The answer is 35, letter d. Problem number 28 states, What is the least number of elements we have to delete from the set 10, 20, 30, 40, 50, 60, 70, 80, 90, so that the product of the elements remaining in the set is a perfect square? So first, let's review, what is a square? So as I'm sure you know, the number 49 is a square. It is 7 squared. The number 36 is also a square. It is 6 squared. However, we can break up the 6 in the parentheses to be 3 times 2, and then we can distribute the 2 over to be that 36 is equal to 3 squared times 2 squared. Here we see that as long as the prime factorization of the number has even exponents, we know the number is a square. For example, the number 2 to the 12th times 19 to the 6th is a square, because both of the exponents of its prime factorization are even. So let's go back to our numbers. Let's get their prime factorizations. And now let's multiply all the numbers together by simply adding the exponents of all the prime factorizations. As you can see, it's almost a square. The only problem is the 7. In order to get rid of the 7, let's take 70 out of the set. Now we have 2 to the 15th times 3 to the 4th times 5 to the 9th. Now the 2 and the 5 are the problem. We have to take away a 2 and a 5, and 10 is the perfect candidate, because its prime factorization is 2 times 5. So let's take away that. Now we get 2 to the 14th times 3 to the 4th times 5 to the 8th. Perfect. All the exponents are even. We did this by only removing two numbers from the set. So the question asked us, what is the least number of elements we have to delete from the set so that the product of the elements remaining in the set is a perfect square? The answer is two elements, letter B. Problem number 29 states, given triangle ABC with the area S, let D be the midpoint of BC. Take points PQR on the lines AB, AD, AC respectively as shown in the picture, and such that AP equals 2 times AB, and AQ equals 3 times AD, and AR equals 4 times AC. What is the area of triangle PQR? Before we go on to find the area of triangle PQR, let me show you a cool trick to find the area of two triangles with a similar vertical angle. So angle NXO and ZXY are the same value because they are vertical angles. We'll say this angle is theta. Now the area of triangle XYZ, the script A just means area, is going to be equal to 1 half AB sine theta, where A is one side length and B is the other side length. So in this case, the first side length would be XZ, the next side length would be XY, and the angle is theta. Now the area of the other triangle, triangle XNO, is going to be the proportion of XO, the length of XO, over the length of XZ, times the length of XN, over the length of XY, times the area of XYZ. So now let's go on to the actual problem. First, let's look at what the problem told us. It told us the length of AP is twice that of AB, the length of AQ is three times that of AD, and the length of AR is four times that of AC. We're trying to find the area of triangle PQR. So first, let's find the area of triangle PAQ. To do this, we know that the area of triangle APQ is going to be equal to the proportion of the length of side AP to the length of AB, times the length of AQ over AD, times the area of triangle ADB. If we plug in what we know, we get that the area, we get that the area of triangle APQ is equal to six times that of triangle ADB. So if we call the area of triangle ADB alpha, we know that the area of triangle APQ is six alpha. Now let's find the area of triangle ARQ. We know that the area of triangle ARQ is going to be AQ over AD times AR over AC times the area of triangle ACD. Again, this is the formula that I showed you before. So we get that the area of triangle ARQ is equal to twelve times that of the area of triangle ACD. So if we call the area of triangle ACD beta, we know that the area of triangle ARQ is going to be twelve beta. Now finally, let's do this one more time for triangle ARP and triangle ACB. We get that the area of triangle ARP is eight times that of triangle ACB. The question told us that area ACB is called S, but we've been using Greek letters, so we'll just use gamma. So if ACB is gamma, we know that ARP is eight gamma. So we know that D splits C and B in half, which means that triangle ACD and triangle ADB are equal in area. This means that alpha is equal to beta. We also know that alpha and beta are the two components of gamma, so we know that alpha plus beta equals gamma. By using substitution, we can see that two alpha is going to be equal to gamma. Now let's look at our other areas. We know that six alpha is just another way of saying three gamma now, and twelve beta is just another way of saying six gamma. So we know that the total area of quadrilateral PQRA is nine gamma. We also know that the area of triangle PRA is eight gamma. So we now know that the area of triangle PQR is gamma. Again, at the beginning of the problem, they told us that triangle ACB has an area of S, but we called that gamma, so triangle PQR actually has an area of S. So the question asked us, what is the area of triangle PQR? The answer is S, letter A. Problem number 30 states, if any digit of a given four-digit number is deleted, the resulting three-digit number is a divisor of the original number. How many four-digit numbers have this property? So let's say that N is a natural number, where N is greater than or equal to ten, and that L is a digit, and then we'll define M as being equal to ten times N plus L. So for example, let's say that N is 135 and that L is six. We get that M is equal to 1,356. So now let's look at some properties of these numbers. The first one is that L is always the last digit of M. The second one is that M without the last digit is always N. The third one is that M is always greater than or equal to ten times N, and is always less than eleven times N. And the fourth one is that if M is a multiple of N, then L equals zero, and M equals ten times N. Let me show you an example of each of these properties. Again, we'll use our example with N equals 135, L equals six, and M equals 1,356. So in this case, L is the last digit of M, because L is six and the last digit in M is six, because M is 1,356. For two, if we take out the last digit, M becomes N, and this is true. If we take out the six at the end of the M, we just get 135, which is N. Now I could show you that number three is correct, but I think it's better for me to prove it for you. So M is going to be sandwiched between ten times N plus the minimum digit we can have, and ten times N plus the maximum digit we can have. The minimum digit is plus zero, the maximum digit is plus nine. Since we know that N has to be greater than or equal to ten, if we replace the plus nine with N, we get that M has to be less than ten times N plus N, or 11N. This is how we derive property three. So now for number four, we have to change our example a little bit. If we want M to be a multiple of N, then L has to be equal to zero, and we can see this. M is a multiple of N, M is 1350, and N is 135, and M ends in a zero. So now let's get on to the actual problem. Let's say we have a four-digit number. The digits are A, B, C, D. We want to make sure that if we divide ABCD by either ABC, ABD, ACD, or BCD, we will get an integer. Integers are represented with J. So it is quite easy to see that by utilizing property four, that L has to be equal to zero. This makes sure that the first possibility is always true. However, we have to make sure the remaining three possibilities are also true. So let's continue. Now let's say that M is only equal to ABC, and N is equal to AB, and L is equal to C. The properties 1, 2, 3, 4 still work for this. So again, we want L to be equal to zero, which makes M AB zero. Going back to our four-digit example, we can see that the second option is also guaranteed now. Only the remaining two options have to be guaranteed. That AB over A has to be an integer, and that AB over B has to be an integer. So AB represented mathematically is 10 times A plus B. So if we take 10 times A plus B over A, that has to be equal to an integer, as well as 10 times A plus B over B also has to be an integer. Let's split up the fractions and simplify, and get that it actually equals 10 plus B over A, and 10 A over B plus 1. In order for the bottom two expressions to be integers, and for rule number three to hold, there are only a couple of possibilities we can have. One of them is that B equals A. In this case, on the left expression, it would just cancel out and it would be 1, and on the right expression, it would also cancel out and just be 10 plus 1. Another possibility is if B equals 2A. In this case, on the left side, we would always just have 10 plus 2, and on the right side, we would have 10A over 2A, which would be 5. So both would be integers. And finally, the last possibility is that B equals 5A. On the left side, we would get 10 plus 5A over A, which would be 5, which would be an integer. And on the right side, we would get 10A over 5A, which would be 2, which would also be an integer, so it would work. So these are the only possible numbers, and there are 14 of them. So the question asked us, how many four-digit numbers have this property? The answer is 14, letter C.

Order of Operations

PEMDAS

Geometric Properties

Arithmetic

Side-length Ratios

Divisibility

Integer Properties

Time Calculations

Mirror Symmetry

Probability

Weight Distribution