Problem number one states, the flag of Kangerland is a rectangle which is divided into three equal smaller rectangles as shown. What is the ratio of the side lengths of the white rectangle? So first, let's call the height of the white rectangle h. Since all the rectangles are even, we know all of these values are also h. Let's call the width of the rectangle of the white rectangle w. And therefore, we know that all of these lengths are also w. So if you see here, we can see that w is equal to 2h. So now if we take the ratio of the sides, we'll take h to w. We know that w is equal to 2h, so we can substitute it in to have h to 2h. And we can divide out the h to get 1 to 2. So the question asks us, what is the ratio of the side lengths of the white rectangle? The answer is 1 to 2, letter a. Problem number two states, the numbers 1, 2, 3, and 4 are each written in different cells of a 2x2 table. After that, the sum of the numbers in each row and column is calculated. Two of these sums are 4 and 5. What are the other two sums? So we know that one of the sums of the rows or columns has to be 4. And the only way to make 4 is to add 1 and 3. So we know that these two have to be 1 and 3. We need to make another sum of 5, and we cannot make it in the next row because we only have a 2 and a 4 remaining, which would make a sum of 6. So we have to do it in a column. If we replace d with 2, we get 3 plus 2, which is equal to 5. This means that c would have to equal 4. The remaining sums are 1 plus 4 equals 5, and 4 plus 2 equals 6. Those are the ones that were not given to us, 5 and 6. So the question asked us, what are the other two sums? The answer is 5 and 6, letter E. Problem number three states, a rectangle has been shaded in five different ways as shown. In which diagram does the shaded part have the largest area? So first, let's call the width of our rectangle W and the height of our rectangle H. In this case, the shaded area is just going to be the area of a triangle, which is 1 half base times height. The base of the triangle we call W, and the height is H. So the shaded area is 1 half WH. For the second half, it's going to be the sum of the two triangles. For the first one, it's going to be 1 half B1W, where B1 is the base of the first triangle, plus 1 half B2W, where B2 is the base of the second triangle. Let's factor out a 1 half W from both terms to get 1 half W times B1 plus B2. Now it is important to see that B1 plus B2 makes up the side of H, so we can substitute B1 and B2 for H. Again, we get that the area is 1 half WH. For this one, it's going to be a similar story to the previous one, only this time we're dealing with a different side. So the area is going to be the sum of all the triangles, each of the triangles is going to have a base, B1, B2, B3, B4, B5, and they all have the same height, H. So we'll factor out a 1 half H like we did last time, and again we can notice that B1 plus B2 plus B3 plus B4 plus B5 is going to be the length of the width of the rectangle. So we can make the substitution, and again get that the area is going to be equal to 1 half WH. For this one, the area of the shaded region is going to be the sum of the three triangles, plus the area of the rectangle. The areas of the triangles we already know, 1 half B1H plus 1 half B2H plus 1 half B3H, and the area of the rectangle is going to be BRH, where BR is the base of the rectangle. So now let's factor out a 1 half from the right side of the equation, and factor an H from the equation. We can see that the width of the rectangle is equal to the base of the rectangle plus the base of triangle 1 plus the base of triangle 2 plus the base of triangle 3. If we subtract the base of the rectangle from both sides, we get the following. Using this, we can substitute it in for B1 plus B2 plus B3 to get the following equation. Now let's finally distribute the 1 half, and combine like terms, and finally distribute the H. We can see that the area of this shaded region is 1 half HW plus 1 half the height of the rectangle times the base of the rectangle. For this one, the bases of the triangle don't cover the whole side, so we know that the area of the shaded region is going to be less than 1 half WH, so there's no point in calculating the exact area of this one since we know it's going to be less. So the question asked us, in which diagram does the shaded part have the largest area? The answer is diagram E, letter E. Diagram number 14 states, three triangles are linked as shown. Which of the following pictures shows these three triangles linked in the same way? So first, let's notice that the gray triangle and the black triangle are not linked directly. In the first figure and in the last figure, the gray triangle and the black triangle are linked directly, so they cannot be the answer. Next, let's notice that the gray triangle and the white triangle are linked together. In the second and third figures, we can see that the white and the gray triangle are not linked together, so therefore the answer cannot be the second or the third figure. This just leaves us with the fourth figure, which has to be the answer. So the question asked us, which of the following pictures shows these three triangles linked in the same way? The answer is picture D, letter D. Problem number 5 states, a given pyramid has 23 triangular faces. How many edges does the pyramid have? So if we know the pyramid has 23 triangular faces, this means that the base of the pyramid is a 23-sided polygon. So let's take one L-segment of the pyramid. This will be an L-segment. And we know that if we copy and paste and rotate it around, we'll cover up the whole pyramid once we do it 23 times. Each of these L-figures has two edges, so if we multiply 23 by 2, we get the number of edges, 46. So the question asked us, how many edges does the pyramid have? The answer is 46, letter C. Problem number 6 states, three 4-digit numbers are written on three pieces of paper as shown. The sum of the three numbers is 11,126. Three of the digits are covered. What are the covered digits? So in order to solve this problem, let's do long addition. Let's start out on the right. 3 plus 7 plus 6 is going to be 16, which makes sense since the bottom number is a 6, and then we would carry a 1. Now we need to find a number such that 1 plus 4 plus 2 plus that number ends in a 2. We're too high for it to be 2, so we have to make it be 12. If we say that our number is 5, we will get 1 plus 4 plus 5 plus 2 is 12, which is good. We carry the 1. Now we need to find a number such that 1 plus 2 plus 1 plus our number is going to end in a 1. Again, we're way past the 1, so let's make it 11. If we make our number a 7, we get a sum of 11, which works, and again, we carry the 1. Now 1 plus 7 plus 2 plus some number has to be equal to 11. We know this since our last two numbers are 1 and 1. So if we replace our number with a 1, we get a sum of 11. So the covered numbers were 1, 5, and 7. So the question asked us, what are the covered digits? The answer is 1, 5, and 7, letter B. Problem number 7 states, what is the first, leftmost, digit of the smallest positive integer whose digits add up to 2019? So we want the sum of all the digits of a number to be equal to 2019, and we want this number to be as small as possible. Therefore, we want the least number of digits in this number. In order to get the least number of digits, we want to make sure that each digit maximizes the sum. So we want a lot of 9s at the end of our number. It's okay if the first digit isn't a 9. The first digit doesn't have to be a 9, because it has to be whatever it has to be for the sum to be equal to 2019, not some multiple of 9. So now we have an expression for our number. A, the first digit, plus 9 times n, the number of 9s in the digit, has to equal 2019. Let's figure out what n is by dividing 2019 by 9. We get 224. 9 times 224 is 2016. This means that A, the first digit, has to be a 3. So the question asked us, what is the first leftmost digit of the smallest positive integer whose digits add up to 2019? The answer is 3, letter B. Problem number 8 states, each of the faces of a special die is marked with 1, 2, or 3 dots in such a way that the probability of rolling a 1 is 1 in 2, the probability of rolling a 2 is 1 in 3, and the probability of rolling a 3 is 1 in 6. Which of the following cannot be a view of this dice? So we know that 1 has a 1 in 2 chance, 2 has a 1 in 3 chance, and 3 has a 1 in 6 chance. We know this die has 6 sides. So if we multiply the probability of getting each of these numbers by the number of sides the dice has, we will have the number of sides each number of dots should appear on. So if we multiply 1 half by 6, we get 3 sides, 1 third by 6, we get 2 sides, and 1 sixth by 6, we get 1 side. This means that 1 should only appear on 3 sides, 2 should appear on 2 sides, and 3 should only appear on 1 side. Let's bring back our dice. As you can see, in the third dice there are 2 sides with 3 dots, and at maximum there can be only 1 side with 3 dots. So this dice is impossible. So the question asked us, which of the following cannot be a view of this dice? The answer is view number 3, letter C. Problem number 9 states, Michael invented a new diamond operation for real numbers, defined as x diamond y equals y minus x. If a, b, and c satisfy a diamond b diamond c equals a diamond b diamond c, which of the following statements is necessarily true? So let's start out by looking at what this diamond operation really does. What the diamond operation really does is flip the 2 values on either side of it, and then the diamond gets replaced by a minus sign. So now let's look at the equation that the problem gave us. First let's work in the parentheses. Let's flip both of the numbers in the parentheses, the a and the b, and the b and the c, and then let's replace the diamond with a minus sign. Now let's do it for the other one. Flip both of the terms, and then replace the diamond with a minus sign. Now for the left let's distribute the negative and drop the parentheses, and on the right we don't have to distribute a negative so we can just drop the parentheses. Now let's subtract c from both sides, and add b to both sides. We get that a equals negative a. The only way that a can equal negative a is if a equals 0. So the question asked us, which of the following statements is necessarily true? The answer is a equals 0, letter d. Problem number 10 states, how many of the numbers from 2 to the 10th to 2 to the 13th inclusive are divisible by 2 to the 10th? So we know we have the range of numbers from 2 to the 10th to 2 to the 13th. We can break up 2 to the 13th by taking out a 2 to the 3rd, to make it 2 to the 10th times 2 to the 3rd. 2 to the 3rd is 8, so we're actually looking for all numbers divisible between 2 to the 10th and 2 to the 10th times 8. In order for a number to be divisible by 2 to the 10th, it has to be a multiple of 2 to the 10th. So we have to ask ourselves, how many numbers are there that are multiples between these two? And obviously the answer is 8. In this case, a can be any number, 1, 2, 3, 4, 5, 6, 7, or 8, and still fit within the range inclusive. So the question asked us, how many of the numbers from 2 to the 10th to 2 to the 13th inclusive are divisible by 2 to the 10th? The answer is 8, letter D. Problem number 11 states, what is the highest power of 3, dividing the number 7 factorial plus 8 factorial plus 9 factorial? So in order to do this, we need to break up this number into its prime factorization. But before we can do that, we need to turn this number into a product. So let's try to factor out a 7 factorial. So let's break up the 8 factorial into 7 factorial times 8, and the 9 factorial into 7 factorial times 8 times 9. Now we can factor out a 7 factorial. Now let's add up all the numbers in the parentheses to get 7 factorial times 81. Now that it is down to one product, it is easier to find the prime factorization. So first let's break down the 7 factorial. 7 factorial is going to be 1 times 2 times 3 times 4 times 5 times 6 times 7, and the 81 is just going to be the 81. So now the 4 turns into 2 squared, the 6 turns into 2 times 3, and the 81 turns into 3 to the 4th. And we get that our number is 2 to the 4th times 3 to the 6th times 5 times 7. So we can see that the largest power of 3 that divides into this number is 3 to the 6th. So the question asked us, what is the highest power of 3 dividing the number 7 factorial plus 8 factorial plus 9 factorial? The answer is 3 to the 6th, letter D. Problem number 12 states, this year the number of boys in my class increased by 20% and the number of girls decreased by 20%. We now have one student more than before. Which of the following could be the number of students in my class now? So first let's say that B was the number of boys in the class last year, and G was the number of girls in the class last year. We know that now there's going to be 1.2B plus 0.8G number of students this year. We know this because the problem said that there are 20% more boys and 20% less girls this year. We also know that there's going to be B plus G plus 1 this year as well, because the problem said that there's one more student this year than there was last year. Now if we subtract B from both sides and subtract 0.8G from both sides, we get 0.2B equals 0.2G plus 1. And now we'll multiply both sides by 5 to get rid of the ugly decimals to get B equals G plus 5. So now it seems that we're kind of at a standstill, but we missed some information earlier. They told us that 0.8g is the number of girls next year. This means that 0.8g has to be a whole number. With this information, we know that g is equal to 5 times p, where p is a natural number. Now if we substitute this in for g in our previous equation, we get that b equals 5 times p plus 1. Using both of our equations now, we can plug it into our third equation to get the total number of kids in our class. Then we simplify it, and we get that the total number of students in our class is 5 times 2p plus 1 plus 1. So now let's look at the possibilities of the number of students in the class. If we just plug in numbers for p, we get that when p equals 1, we would have 16 students, and when p equals 2, we would have 26 students, which means that 26 is a possibility. So the question asked us, which of the following could be the number of students in my class now? The answer is 26, letter B. Problem number 13 states, A glass container in the shape of a rectangular prism is partially filled with 120 meters cubed of water. The depth of the water is 2 meters, 3 meters, or 5 meters, depending on which side of the prism lies on the floor as shown, not to scale. What is the volume of the container? So we know that each side of the prism is on the bottom of the container for at least one of all the views. So we know that the base of one of them is x by y, another one is x by z, and the last one is y by z. This means that the volume of water in the first one is going to be 2 times x y equals 120, in the next one 3 x z equals 120, and in the last one 5 y z equals 120. Let's combine all these equations by multiplying them all together. Now let's simplify by dividing both sides by 30. We get x squared y squared z squared equals 4 times 120 times 120. We are trying to find x times y times z, the volume of the container. So in order to do this, let's take the square root of both sides. The left side is quite easy, we just get x y z, and on the right side, the square root of 4 is 2, and then 120 times 120, the square root of that is just 120. So we get x times y times z equals 2 times 120, or 240. So the question asked us, what is the volume of the container? The answer is 240 meters cubed, letter E. Problem number 14 states, three kangaroos, Alex, Bob, and Carl, go for a walk every day. If Alex doesn't wear a hat, then Bob wears a hat. If Bob doesn't wear a hat, then Carl wears a hat. Today, Carl is not wearing a hat. Who is certainly wearing a hat today? So first, let's take a look at all the possible combinations of wearing a hat and not wearing a hat, where Carl is not wearing a hat. We know Carl is not wearing a hat, because the problem told us Carl is not wearing a hat. So now let's take a look at the statements the problem told us. The first one was that if Alex is not wearing a hat, then Bob is wearing a hat. In the upper left situation, Alex is not wearing a hat, and Bob is also not wearing a hat. Therefore, this possibility is impossible. Next, let's look at the second statement. If Bob is not wearing a hat, then Carl is wearing a hat. In the upper right possibility, Bob is not wearing a hat, and Carl is not wearing a hat. So therefore, that situation is impossible. We have two situations that remain. In both, Bob is certainly wearing a hat, but Alex can either be wearing a hat or not wearing a hat. So the question asked us, who is certainly wearing a hat today? The answer is only Bob, letter E. Problem number 15 states, The system shown consists of three pulleys with vertical sections of rope between them. The endpoint P is moved down 24 centimeters. How many centimeters does the point Q move up? So we know that the point P is going to be going down. And if we see the first pulley attached, we can see that all that pulley does is change the direction of the rope. So we can just remove it, and we can just move point P and say that point P is going up. Now we know point P is going to go up 24 centimeters. That distance is going to be absorbed by these two ropes evenly. So each of them will take 12 centimeters, moving that whole pulley 12 centimeters up. Now that we know this pulley is moving up, we can analyze the second system. Again, the distances would be split evenly across the ropes, so the middle section would move up only by 6 centimeters. So point Q would only move up 6 centimeters. So the question asked us, how many centimeters does point Q move up? The answer is 6, letter D. Problem number 16 states, a positive integer n is called good if its largest divisor, excluding n, is equal to n minus 6. How many good positive integers are there? So we know that the n over n minus 6 has to be equal to an integer. So let's find out how many good positive integers are there. So we know that the n over n minus 6 has to be equal to an integer. So let's manipulate this a little bit. Let's add 0 to the numerator by subtracting 6 and then adding 6. And then let's bring out an n minus 6 over n minus 6. This of course is equal to 1. So we get that our original n over n minus 6 is equal to 1 plus 6 over n minus 6. This of course means that the value of n minus 6 must be a divisor of 6. Luckily, there are only a few divisors of 6. There's negative 6, negative 3, negative 2, negative 1, 1, 2, 3, and 6. The first thing that jumps out is that negative 6 cannot be the answer. Because if n minus 6 equals negative 6, we get that n equals 0. And the problem told us that n has to be a positive integer. So now let's make a table with all the remaining values of n minus 6. Let's compute the values of n and the largest divisor of n that is less than n. The problem told us that when the value of n minus 6 is equal to the largest divisor, then we have a good number. In our case, there are three such numbers, 7, 9, and 12. So the question asked us, how many good positive integers are there? The answer is 3, letter C. Problem number 17 states, A box contains four chocolates and one fruit chew. John and Mary take turns drawing a treat out of the box without replacement. Whoever draws the fruit chew wins. John draws first. What is the probability that Mary wins? So let's represent all the candy that's drawn in a line like this. In this case, the first chocolate would be drawn first, the second chocolate would be drawn second, the third chocolate would be drawn third, and the fourth chocolate would be drawn fourth, and then the fruit chew would be drawn last. In total, there are five factorial different orders that these candies can be drawn in. Mary would win if the fruit chew was either in position 2 or in position 4. When the fruit chew is in position 2, there are four factorial ways of arranging the additional chocolates. If the fruit chew is in position 4, there is an additional four factorial ways of arranging the additional chocolates, which means the chance that Mary is going to win is 4 factorial plus 4 factorial over 5 factorial, the number of possibilities where she wins over the total number of possibilities. When we turn these into actual numbers, we get 48 over 120, which simplifies down to 2 fifths. So the question asked us, what is the probability that Mary wins? The answer is 2 fifths, letter A. Problem number 18 states, two adjacent squares with side lengths a and b where a is less than b are shown. What is the area of the shaded triangle? Since we know one side length of both of the squares, we know what these lengths also are because all the sides of a square have the same length. In order to solve this problem, let's split up the shaded triangle into two separate triangles, a blue triangle and a red triangle. In order to find the area of both of these triangles, we just have to do one half base times height. In this case, the base will be the length of a minus the section shaded in yellow. It's quite easy to find the section shaded in yellow by using similar triangles. We get that the height of the little triangle is ab over a plus b. Now that we know that length, let's find the yellow length, which will be the base of both of the triangles. In order to do this, it's just a minus ab over a plus b. If we clean it up, we get that the base is equal to a squared over a plus b. Now let's find the areas of the triangles. First, let's start out with the blue triangle. The area is, of course, one half base times the height. We already know the base, and the height is just going to be a. If we simplify it down, we get one half a cubed over a plus b. For the red triangle, it's the same process, one half base times height. We know the base, and we get one half ba squared over a plus b. So now, let's add up the two areas to find the area of the shaded triangle. Since we have a common denominator, we can add the numerators, and then factor out a squared, and then cancel out a plus b over b plus a, to get an area of one half a squared. So the question asked us, what is the area of the shaded triangle? The answer is one half a squared, letter b. Problem number 19 states, what is the integer part of the square root of 20, So first, let's notice the square root of 20 is between 4 and 5. It's greater than the square root of 16, yet less than the square root of 25. Let's add 20 to all sides, to get 24 is less than 20, plus the square root of 20, which is less than 25, and then let's take the square root of all sides. On the left, we get the square root of 24, which is greater than 4, and on the right, we get the square root of 25, which is equal to 5. Therefore, we know that our middle section is again sandwiched between 4 and 5. Let's continue this process until we get to the expression that the question stated. So let's add 20 again to both sides, and then let's take the square root, and again we see that that expression is sandwiched between 4 and 5. In the problem, there were 5 20s, so we gotta keep on going on. Add 20 again, take the square root of all sides, again sandwiched between 4 and 5, and then just one more time, add 20, and take the square root, and finally, we see that the expression is sandwiched between 4 and 5. This means that our value will be 4 point something something, which means the integer part of the value is 4. So the question asked us, what is the integer part of this expression? The answer is 4, letter A. Problem number 20 states, Sarah needs to calculate the result of A plus B over C. On a calculator, she types A plus B divides C equals, and the result is 11. A, B, and C are positive integers. She then types B plus A divides C equals, and she is surprised to see that the result is 14. She realizes that the calculator is designed to calculate division before addition. What is the correct result of A plus B over C? So we were told that A plus B over C is equal to 11, and B plus A over C is equal to 14. Let's subtract the lone variable from both sides of both equations, and then let's combine them by adding them together. Now let's collect like terms, and then finally let's add the A plus B to both sides. Now let's factor out the A plus B. Now let's multiply the 1 by C over C, and then add it to 1 over C, and now let's move the divide C over to the left side, so we have the desired A plus B over C. Now it is clear that A plus B over C has to be an integer, and that C plus 1 also has to be an integer. This means, of course, that C plus 1 has to be 5. Therefore, A plus B over C also has to equal 5. So the question asked us, what is the correct result of A plus B over C? The answer is 5, letter E. Problem number 21 states, let A be the sum of all positive divisors of 1024, and B the product of all divisors of 1024. So first, let's find all the divisors of 1024. We know 1024 is 2 to the 10th, and if we have a number where it's 2 to some power, all of its divisors are 2 to the 0, 2 to the 1st, 2 to the 2nd, and so on, all the way up to 2 to the 10th. The sum of all of these divisors is going to be 2047, which is going to be 2 to the 11th minus 1. So that's our value for A. For B, we need to find the product of all the divisors, so we can just add the exponents of all of the 2 to the power of in the divisors to get 2 to the 55th. So now we know that B is 2 to the 55th, and A is 2 to the 11th minus 1. Let's bring back up our options that we had from the problem, and we can notice that if we plug in this one, it works, since the plus 1 cancels out the minus 1, and then 2 to the 11th to the 5th power would be 2 to the 55th, which is the same value as B. So the question asked us, let A be the sum of all positive divisors of 1024, and B the product of all positive divisors of 1024. Then, A plus 1 to the 5th equals B. Letter B. Problem number 22 states, What is the set of all values of the parameter A, for which there are two solutions for the equation 2 minus the absolute value of x equals Ax. So let's solve this problem graphically. First, let's graph 2 minus the absolute value of x. Next, let's graph Ax. Now of course, A can be whatever value it wants to be, so the blue line can shift like this, or like this, or in any orientation it wants. Let's try to find the maximum value of A, such that the blue line crosses the red line at two points. As we approach a slope of 1, we can see that the blue line becomes parallel to the left red line. This means that only one solution exists. So our upper bound is 1. Now let's try to find the lowest bound. As A approaches negative 1, the blue line gets closer and closer to being parallel to the red right line. This means that it will only cross at one point, which is what we want. So we know the minimum value of A is slightly greater than negative 1. So the question asked us, what is the set of all values of the parameter a for which there are two solutions for the equation 2 minus the absolute value of x equals ax? The answer is the open interval between negative 1 and 1, letter b. Problem number 23 states, the vertices of the network shown are labeled with the numbers from 1 to 10. The sum s of the four labels of each square is the same. What is the least possible value of s? So let's label the vertices that we've reused a and b, and we know that the sum of all the squares is going to be s, and we know that we have to use all the numbers from 1 through 10. The sum of all the numbers from 1 through 10 are 55. We know that the value of 3s is going to be equal to 55 plus a plus b. The reason for the plus a plus b is because a and b are used twice. So now we know that the whole right side of the equation must be divisible by 3. The closest number divisible by 3 would be 57. However, in order for this to be true, a and b would have to both be 1, which is not possible. We have to use different numbers. So the next number is 60. For 60, we can say that a and b are 4 and 1, for example, and it works. Now let's make sure it works by filling out the remainder of the network with the remaining numbers. And sure enough, it does. We get a sum of 20 on every single square. So the question asked us, what is the least possible value of s? The answer is 20, letter C. Problem number 24 states, how many planes pass through at least three vertices of a given cube? So the vertices of a cube are all the corners of a cube where the edges meet. So those are these. Let's name them all by letters. So we'll name them A through H. And now let's first start off by finding all the vertices that pass through exactly three points. So we'll start off with A, F, H. In order to get another plane, we can just rotate it by 90 degrees to get plane D, E, G, again to get plane C, F, H, and again to get plane B, E, G. Now in order to get more planes, we can just flip this over the vertical axis to get plane A, C, F, and then keep on rotating it around 90 degrees to get planes B, E, D, A, C, H, and B, D, G. So in total we have eight planes that pass through exactly three vertices, but the question asked us to find at least three vertices, so we have to look for more planes. Now let's look for planes that pass through four points. So here we have one of them, B, C, E, H. Let's again use the previous technique by rotating it 90 degrees to get A, B, H, G, A, D, F, G, and finally C, D, E, F. This time we can't flip it over the vertical axis because we'll just get a plane that we've already had before. For example, if we flip the one that we have now, we will get plane A, B, H, G, which we already had before. Another possibility, however, is if we go to plane A, C, E, G, and again we can also rotate this one 90 degrees to get a sixth combination, B, D, F, H. So in total here we have six. So now actually the last one is pretty simple, it's just the faces of the cube. So I'm not even going to write down the names of the planes, but I mean it's pretty easy to see that there are six faces, so there are six planes. So ultimately we have 8 plus 6 plus 6 possible planes, that is 20 planes. So the question asked us, how many planes pass through at least three vertices of a given cube? The answer is 20, letter E. Problem number 25 states, four distinct straight lines pass through the origin of the coordinate system. They intersect the parabola y equals x squared minus 2 at 8 points. What can be the product of x coordinates of these 8 points? So first let's graph the function they're talking about, x squared minus 2, and then we know there are four lines. Each of them have an equation of y equals mx, but m is different for all of them. We know the line is now vertical, because if the line was vertical, then it would only cross the parabola at one point, and we know that all of them cross at two points. So now let's try to figure out where a line generally passes through the parabola. So let's set the two equations equal to each other, and then bring them all to one side. Now let's use the quadratic formula to find the zeros of the function. Now the problem told us that all the x values would be multiplied together. So let's multiply these two x values together. We get the following. Luckily a lot of it cancels out, and we get negative 8 fourths, or negative 2. Let's keep in mind that we have four lines, and we have to multiply them all together. So we have to take negative 2 to the fourth power. Negative 2 to the fourth power is 16. So the question asked us, what can be the product of the x coordinates of these 8 points? The answer is only 16, letter a. Problem number 26 states, for how many integers n, is the absolute value of n squared minus 2n minus 3 a prime number? So first we see we have this quadratic, so let's factor it. After factoring we get n minus 3 times n plus 1. Since we know the product has to be positive, we can move the absolute value signs like this to make all the members of the product positive, which doesn't change the outcome at all. In order for a number to be prime, a number that is not 1 or 0 has to be multiplied by 1. So either n minus 3 or n plus 1 has to be equal to 1, and the other number has to be equal to not 1 or 0. So on the right side, we would get n equals 0 and n equals negative 2. In both of these cases, the right hand expression would be equal to 1. On the left, for n minus 3, we would have n equals 4 and n equals 2. Luckily there is no overlaps in the sets, so we will not get a product of 1. So now let's plug in the numbers for n. For n equals 4, we get 5, which is prime. For n equals 2, we get 3, which is prime. For n equals 0, we get 3, which is prime. And for n equals 2, we get 5, which is prime. Although the 5 and the 3 repeat, it doesn't matter, because the problem just asks for the number of integers n that result in a prime number, not a unique prime number. So the answer is 4. So the question asked us, for how many integers n is the absolute value of n squared minus 2n minus 3 a prime number? And the answer is 4, letter D. Problem number 27 states, the path DEFB with DE perpendicular to EF and EF perpendicular to FB lies inside the square ABCD as shown. Given that DE equals 5, EF equals 1, and FB equals 2, what is the length of the side of the square? So first let's draw a line that will be the diagonal of the square. And let's mark the point of intersection on the line FEI. At this point it is clear to see that the length of FI plus the length of EI are going to be equal to 1. We can find what these lengths actually are by using similar triangles. By saying that the ratio of 2 to FI and the ratio of 5 to EI are equal, we can solve for what FI and EI are. First let's solve for FI, 1 minus EI, and then plug that into our ratio. We get that EI equals 5 7ths. This means that FI equals 2 7ths. Now we can just find the hypotenuses of our two triangles using Pythagoras' theorem to figure out the length of the diagonal. So let's do the top one first. The length of BI is going to be the square root of 2 squared plus 2 7ths squared. So we'll have 4 plus 4 49ths. When we add them together with a common denominator, we get 240 9ths, which is 10 root 2 over 7. For the red hypotenuse, we have the square root of 5 squared plus 5 7ths squared, which simplifies down to 25 plus 25 49ths, which then with a common denominator is 1250 49ths. And finally, we have 25 root 2 over 7. In order to find the total length of the diagonal, let's add these two up. We get 35 root 2 over 7, which simplifies down to 5 root 2. So we know the length of the diagonal is 5 root 2, and we know that angle BDC is 45 degrees, since BD is the diagonal of the square. So therefore, the side length of this square is 5. So the question asks us, what is the length of the side of the square? And I don't see our answer here, so the answer is none of the above. Letter E. Result number 28 states, the sequence a1, a2, a3 starts with a1 equals 49. For n is greater than or equal to 1, the number an plus 1 is obtained by adding 1 to the sum of the digits of an, and then squaring the result. Thus, a2 is equal to 4 plus 9 plus 1 squared, which is 196. Question a2019. So they told us that a1 was 49 and that a2 was 196. So let's keep on finding more and more terms of the sequence until we can find some kind of a pattern. So a3 is going to be 1 plus 9 plus 6 plus 1 squared, which is 17 squared, so 289. a4 is going to be 2 plus 8 plus 9 plus 1 squared, which is going to be 20 squared, so 400. a5 is going to be 4 plus 0 plus 0 plus 1 squared, which is going to be 5 squared, so 25. a6 is going to be 2 plus 5 plus 1 squared, which is going to be 8 squared, so 64. a7 is going to be 6 plus 4 plus 1 squared, which is going to be 11 squared, so 121. a8 is going to be 1 plus 2 plus 1 plus 1 squared, so it's going to be 5 squared, so 25. So we can see that our pattern repeats every 3. So now let's generalize this. We can say that 64 happens whenever the term is a multiple of 3. In this case, n will be a whole integer. 25 will occur when there's 3n minus 1, and 121 will occur when there's 3n plus 1. n is restricted in our case, n is greater than or equal to 2, but that really doesn't matter since we're going all the way up to 2019. So in order to find out which one 2019 is, let's divide 2019 by 3. And it actually works, we get 673, a whole number, which means if we plug in 673 for n, for n, and then solve all the expressions, we get that 2018 is equal to 25, 2019 is equal to 64, and 2020 is equal to 121. So the question asked us, determine A, 2019. The answer is 64, letter C. Problem number 29 states, three different numbers are chosen at random from the set 1 through 10. What is the probability that one of them is the average of the other two? So first, let's figure out how many possibilities there are for drawing numbers. We cannot draw the same number twice, and it doesn't matter what order we draw them in, because the question just asked, what is the probability that one of them is the average of the other two? So for this, let's use combinations. We have 10 numbers to choose from, and we're going to choose three of them, so let's compute 10 choose 3. 10 choose 3 is the probability that one of them is the average of the other two. 10 choose 3 is defined like this. So we get a total of 720 over 6 possible combinations, or 120 combinations. The probability that two of the numbers are the average of the third is going to be the successful outcomes over 120. So now, let's just figure out how many successful outcomes there are. If one number is equidistant from the other two numbers on the number line, it is the average of those two numbers. Here all the numbers are separated by 1. 1 plus 1 is 2, and 2 plus 1 is 3. We have 8 of these possible combinations. Now let's separate them by two numbers on the number line. For these, there are only 6 possibilities. If we separate them by 3 numbers, there are only 4 possibilities. And if we separate them by 4 numbers, there are only 2 possibilities. We cannot separate them by 5 numbers, that is too much. So we have 8 plus 6 plus 4 plus 2, which is equal to 20. So we have a 20 out of 120 chance that one of the numbers is the average of the other two, or in other words, a 1 out of 6 chance. So the question asked us, what is the probability that one of them is the average of the other two? The answer is 1 sixth, letter B. Problem number 30 states, the square shown is filled with numbers in such a way that each row in such a way that each row and each column contains the numbers 1, 2, 3, 4, and 5 exactly once. Moreover, the sum of the numbers in each of the three regions within bold borders is equal. What number is in the upper right corner on the square with the question mark? So we know that each row is going to have 1, 2, 3, 4, 5, which means that the sum of the row is going to be 15. Now we know we have 5 rows, which means that the total sum of all the numbers in the square is going to be 5 times 15, or 75. We have three sections and the sum of all of them has to be the same, so we have to distribute the 75 equally amongst them to get 25 points in each section. Let's start out by trying to fill out the middle section first. The middle section is a lot larger than the other two sections, so we need to ideally fill it out with the smallest numbers we can. So let's start out with the bottom row. We already have a 2, so let's fill in the remaining squares with a 1 and a 3. Let's continue filling out this section by filling out a 1 and a 3, staircasing all the way up. Now the last one in the top is forced to be a 3 because of what's in the column. Now we have 5 3's and 4 1's. Again, with so many spaces, we really need to make sure that we minimize the numbers in the squares. So rather than doing it this way, let's flip all the 1's and the 3's. So now we have 4 3's and 5 1's, for a total of 17. We need 8 more, and we have 4 empty squares, so let's fill them up with 2's. Now we have a total of 25. Perfect. Now let's fill in the bottom left. So far we only have a total of 2 in that region. In this situation, we have quite the opposite problem of the previous one. In this situation, we have quite the opposite problem of the previous one. Since we have so little squares, we want to maximize the efficiency of every square and use as high numbers as possible. So let's use the 3 numbers as 5's. So now we have 17, and we need 8 more to make a sum of 25. So let's use 2 4's. Now we have 1 2, 3 5's, and 4 2's, for a sum of 25. For the top section, it's going to be quite easy to fill out. It's just going to be like a game of Sudoku, figuring out what goes where. And in the very top right corner, we get that there's a 3. Now let's make sure that the section actually adds up to 25. We have 1 3, 2 5's, and 3 4's, which adds up to 25, so it works. So the question asked us, what numbers in the upper right corner add up to 25?

geometry

algebra

probability

logic

mathematical principles

ratio

equations

triangles

Sudoku