Problem number one states, the diagram shows a shape made from 10 squares with side lengths of 1 centimeter. What is the length of its perimeter? So before we tackle this shape, let's tackle a much simpler shape, a simple square. The problem tells us that each of the side lengths are 1 centimeter long. And since it's a square, we know that all of the side lengths are 1 centimeter. The perimeter is defined as the path that surrounds the 2D shape. So in this case, we would add up all of these individual lengths. So we would get 1 centimeter plus 1 centimeter plus 1 centimeter plus 1 centimeter, and that of course equals 4 centimeters. So the perimeter of this shape is 4 centimeters. But that's not what the question was asking. The question was asking about this shape. So first, let's start at the top. The top edge is 2 centimeters long. And then the edge to the right of that is another 2 centimeters long. To the right of that, 1 centimeter. And then we go on and on counting 2 centimeters, 1 centimeter, 2, 1, 4, 1, 2. Again, the perimeter is the path that surrounds the 2D shape, so we have to add up all of these individual lengths. And that ends up equaling 18 centimeters. So the question asked us, what is the length of its perimeter? The answer is 18 centimeters. Letter B. Problem number two states, when the answers to the following calculations are put in order from smallest to largest, which will be in the middle. So in order to solve this problem, let's just compute each of these values and then put them in order from smallest to largest and then see which one ends up in the middle. Luckily this addition is pretty easy. So 1 plus 2345 is 2346. 12 plus 345 is 357. 123 plus 45 is 168. 234 plus 5 is 1239. And 12,345 is 12,345. So now let's put all of these numbers in order, and we can see the middle value is 1239, or 1234 plus 5. So the question asked us, which will be in the middle? The answer is 1234 plus 5, letter D. Question number three states, who is the mother of the daughter of Anne's mother's mother? So first let's draw a diagram. At the very bottom we'll have Anne, and then going up we'll have her ancestors. So first let's try to figure out who Anne's mother's mother is. So Anne's mother is obviously Anne's mother, and Anne's mother mother will be Anne's grandmother. So now we've figured out who this is, who Anne's mother's mother is, Anne's grandmother. And the question asked us, who is the mother of the daughter of this person? So let's take a look. The daughter of that person would be Anne's mother, and the mother of that person would be Anne's grandmother. Another way we could do this is by realizing the fact that whenever we have a mother and a daughter, those kind of cancel out. So when they ask who is the mother of the daughter, we can kind of ignore that, and then turn this the question to who is Anne's mother's mother, which is obviously her grandmother. So the question asked us, who is the mother of the daughter of Anne's mother's mother? The answer is Anne's grandmother, letter E. Question number four states, when Cosmo wears his new shirt properly as shown on the left, the horizontal stripes form seven closed rings around his waist. This morning, he buttoned his shirt incorrectly as shown on the right. How many closed rings were there around Cosmo's waist this morning? So now when you first look at this question, the answer seems obvious. Obviously it's six, since six halves line up on the left side with six corresponding halves on the right side. This wouldn't mean there are six rings, but this is not true, because if we look at the back of the shirt, we can see that there are no rings at all. It is just one big helix. So therefore, there are no rings that go around Cosmo's waist. So the question asked us, how many closed rings were there around Cosmo's waist this morning? The answer is zero, letter A. Problem number 5 states, in the calculations shown, each letter stands for a digit. These digits make up two digit numbers. The two numbers on the left have a sum of 79. What is the sum of the four numbers on the right? A lot of these answers seem oddly high. So first let's take a look at what the maximum value can be if A, B, C, and D are all 9. We want to see what the maximum number we can get if all of our two digit numbers are maximized, so if they are 99. So let's replace all of them with 9s. And when we add them up, we get 396. This means that C, D, and E, which are all greater than 396, are not possible. So now let's solve for the missing sum. Let's write down what we know. A times 10 plus B plus 10 times C plus D equals 79. A and C are multiplied by 10 because they are in the tens place. B and D are not because they are in the ones place. So that's the left equation in math form. Now let's write the right equation in math form. That would give us 10 times A plus D plus 10 times C plus D plus 10 times A plus B plus 10 times C plus B. And that equals our answer that we're trying to get. If you look carefully, you can recognize part of the top equation in the bottom equation. And since we see this, we can actually do a substitution and just replace that whole chunk there with 79. And now if you rearrange the bottom equation, we can see that we can do yet another substitution for another 79. So 79 plus 79 is 158. In case you were wondering what the values of A, B, C, and D are, there are a lot of possibilities. Actually there are 48 possibilities, but we didn't actually have to solve for any of them. So the question asked us, what is the sum of the four numbers on the right? The answer is 158, letter B. Problem number 6 states, the sum of four consecutive integers is 2. What is the smallest of these integers? So here we have a number line, and let's just start at 0 and add the four consecutive numbers and see what sum we get. So 0 plus 1 plus 2 plus 3 is going to give us 6. 6 is too large, so let's move one number down the number line. So we have negative 1 plus 0 plus 1 plus 2, which gives us our desired value of 2. If we keep on going down to negative 2, negative 1, 0, 1, we would get negative 2 as the sum. So that would be going too far. As we continue to go down, the sum will only grow smaller. So this is our answer. And the question asked, what's the smallest of these integers? So that's negative 1. So the question asked us, what is the smallest of these integers? The answer is negative 1, letter C. Problem number 7 states, the years 2020 and 1717 both consist of a two-digit number repeated twice. How many years after 2020 will it be until the next year which has this property? So in order to solve this problem, let's work backwards from the answers. Let's start out with 2020 and add each of these numbers to see if that year has this property. So 2020 plus 20 is going to be 2040, which does not have this property. 2020 plus 101 is 2121, which does have the property. We kind of know this is the answer because this is the smallest number that has this property. But let's just keep going. So 2020 plus 120 is 2140, which does not follow the property. The next one is 2020 plus 121, which is 2141, which also does not have that property. And then the last one is 202, which would give us 2222, which actually does have the property. But as stated before, 101 is smaller, so that is the answer. So the question asked us, how many years after 2020 will it be until the next year which has this property? The answer is 101, letter B. Problem number eight states, Mary has ten pieces of paper. Some of these are squares and the rest are triangles. She cuts three squares diagonally from corner to corner. She counts the total number of vertices, the 13 pieces of paper she now has, and gets an answer of 42. How many triangles did she have before making the cuts? So here we have the ten pieces of paper. They are circles because we don't know whether or not they're triangles or squares. So let's say that the number of squares that she initially had is X. That means that the number of triangles that she initially had is 10 minus X, since X plus 10 minus X has to be equal to 10, because that's the number of pieces of paper she initially had. Then she cut three of the squares diagonally from corner to corner. This means that she had three less squares since she cut them up and she had six more triangles since each square turned into two more triangles. Now we know that each square has four vertices and each triangle has three vertices. This means that the total number of vertices is four times the number of squares plus three times the number of triangles. We know the number of squares and the number of triangles with respect to the initial number of squares. So let's substitute those in. And this is what we get. We also know the number of vertices at the end, which is 42. So now all we have to do is solve for X. So first let's distribute the 4. Next let's distribute the 3. Then let's distribute the 3 again. And now let's drop all the parentheses. And now let's combine all the like terms. So 4X minus 3X is X. And then minus 12 plus 18 plus 30 is 36. So we have 42 equals X plus 36. Subtract 36 from both sides, we get X equals 6. If we remember initially, X was the number of squares. And the question asks us about the number of triangles. So we do 10 minus X or 10 minus 6 for the number of triangles, which is equal to 4. So the question asked us, how many triangles did you have before making the cut? The answer is 4. Letter E. Problem number 9 states, Martin made a kite by cutting a straight wooden pole into 6 pieces. He used 2 of them of lengths 120cm and 80cm as the diagonals. The remaining 4 pieces connected the midpoints of the sides of the kite as shown. How long was the pole before it was cut? So first let's label all the points on the kite. We'll label them A, B, C, D. Next let's label all of the midpoints on the kite. We'll do W, X, Y, Z. In order to solve this problem, you have to know about the triangle midsegment theorem. You don't actually have to know what it's called, you just have to know what it is. So the triangle midsegment theorem implies that there has to be some triangle, but all I see here is a kite. So first let's define the triangle ABC. The triangle midsegment theorem states that if W is the midpoint of line segment AB, and X is the midpoint of line segment BC, then the length of WX is equal to half the length of AC. And we were given the length of AC, we know that it's 80cm. That means that it's easy to solve for the length of WX, it's half that of AC, which is 40cm. Next let's try and solve for ZY. So let's just flip the triangle over, and since Z is the midpoint of line segment AD, and Y is the midpoint of line segment CD, then the length of ZY is equal to half that of AC, which is again 40cm. Now let's do that for line segment XY and WZ. In this case line segment XY is equal to half the length of line segment BD. We were given that BD is 120cm, so XY is 60cm. And the same thing for WZ, it is again 60cm. So now we have all of the individual lengths of the pole. So now if we add them all up, we will get the total length of the pole. Adding them up, we get 400cm. So the question asked us, how long was the pole before it was cut? The answer is 400cm, letter C. Problem number 10 states, four points are marked on a grid of squares with side lengths of one. Forming a triangle using three of these points, what is the smallest area that we can obtain? So since there are four points and we have to choose three of them, there are four possible combinations. This is how all of them look. Out of all of these, I think we can agree that the two on the left side are bigger than the two on the right side. So let's not even consider those, since they obviously have a bigger area. For the top one, if we have to find the area, it's 1 half base times the height. The base of this triangle is 1, and the height is 3. So 1 half times 1 times 3 is going to be 3 halves. For the bottom one, the area is also 1 half base times height. In this case, the base is 1, and the height is also 1. So we get 1 half times 1 times 1, or just 1 half. So clearly the triangle on the bottom is smaller and has an area of 1 half. So the question asked us, what is the smallest area that we can obtain? The answer is 1 half. Letter A. Problem number 11 states, Helen wants to spend 18 consecutive days visiting her grandma. Her grandma reads storybooks on her on story days, which are Tuesday, Saturday, and Sunday. Helen wants to spend as many story days with her grandma as possible. On which day of the week should she start her visit? So first let's start out by drawing a calendar and then marking all the story days, Tuesday, Saturday, and Sunday. Problem told us that she has to visit for 18 consecutive days. And 18 consecutive days is 2 weeks and then 4 days, since each week is 7 days. So 2 times 7 is 14, plus 4 is 18. Let's first see how many story days she can be at if she starts on Monday. She would be at 3 on Tuesday, 2 on Saturday, and 2 on Sunday. Which is 7 for Monday. Next let's move it over to Tuesday, and it's the same story. So 7 days. When she arrives on Wednesday, she has one fewer story day on Tuesday, but one more on Saturday. So this is also 7. If she arrives on Thursday, she is there an additional day, because there is an additional Sunday story day. So that is in total 8 story days. If she arrives on Friday, nothing changes. And if she arrives on Saturday, she is there for an additional Tuesday story day. So 9 for Saturday. And for Sunday, she would have one less Saturday day. So that would be 8 story days. So the maximum number of story days here we can see is on Saturday, with 9. So the question asked us, on which day of the week should she start her visit? The answer is Saturday. Letter D. Problem number 12 states, the integers a, b, c, and d satisfy a times b equals 2 times c times d. Which of the following numbers cannot be the value of the product a times b times c times d? So let's consider all the answers. So let's start off with 50 and start by breaking it up. So 50 is equal to 25 times 2. We need to break this up even more though because we need 4 numbers to fit a, b, c, and d. So let's break up 25 more into 5 and 5. We only have 3 numbers but we can get another one by realizing that any number times 1 is that number. So we can have 5 times 5 times 2 times 1 equals 50. And with this, we can plug this into our A times B equals 2 times C times D equation, because one of the values is twice that of another, and all the other values repeat. So in this case, 2 is twice that of 1. So for example, C and D would be 5 and 1, and A and B would be 5 and 2. So 50 works. Next, let's try 100. 100 is 25 times 4. We can break up both 25 and 4 into 5 times 5 and 2 times 2. And from here, we can't really do anything. And even if we initially broke it up to 10 times 10, this wouldn't help either, because we would have 5 times 2 times 5 times 2, which would ultimately be 5 times 5 times 2 times 2. So 100 will not work. We kind of know the answer at this point, but let's keep on going to verify our answer. Let's try 200. 200 is 100 times 2. And then we already know what 100 is. 2 times 2 times 5 times 5. In this case, we have 5 numbers, but we can combine 2 of the 2's into a 4. And then the 4 is twice as big as the 2, so we can put it into our equation. In this case, 2 and 5 would be C and D, and 4 and 5 would be A, B. Next, for 450, we can break it up into 10 and 45. And then we can break up 45 and 10 into 9 times 5 times 5 times 2. In this case, we can break up the 9 again into 3 by 3. And now we can multiply either the 3 or the 5 by the 2 to get twice the number. In this case, we'll multiply the 5, and we'll get 3 times 3 times 5 times 10. So in this case, A, B would be 3 times 10, and C, D would be 3 times 5. So 450 works. For 800, we would break it up into 8, 10, and 10. And then we could break up the 8 into 2 by 4. And 4 is twice as great as 2, so it works. C, D could be 2 and 10, and A, B could be 4 and 10. So 800 also works. So the only one that didn't work was 100. So the question asked us, Which of the following numbers cannot be the value of the product A, B, C, D? The answer is 100. Letter B. Problem number 13 states, the shortest path from A-town to C-town runs through B-town. Walking on this path from A-town to C-town, we should first find the signpost shown on the left. Later, we would find the signpost shown on the right. What distance was written on the broken sign? So first, let's make a map of A-town, B-town, and C-town. We'll mark them A, B, C. We know the first sign appears between A-town and B-town, because the sign points one way to A-town and the other way to B-town. This means this sign must be in the middle of A-town and B-town. And we know exactly where it is. We know it's 3 kilometers from A-town and 1 kilometer from B-town. This means that between A-town and B-town, it is 4 kilometers. Using the same logic, we can figure out where the other sign is. Since one way it points to C-town and the other way it points to B-town, we can conclude that it is between B-town and C-town. We know it is 6 kilometers away from C-town. We also know the distance between this signpost and A-town, which is 6 kilometers. Now we can set up an equation. The fact that the distance between A and X, the right signpost, is equal to the distance between A-town and B-town, plus the distance between B-town and the signpost. Plugging in our numbers, we get 6 is equal to 1 plus 3, or 4, plus the distance between B-town and the signpost. Subtracting 4 from both sides, we get that the distance from B-town to the signpost is 2 kilometers. So the sign should say that B-town is 2 kilometers away. So the question asked us, what distance was written on the broken sign? The answer is 2 kilometers, letter B. Problem number 14 states, an isosceles triangle has a side with a length of 20 centimeters. Of the other two side lengths, one is equal to two-fifths of the other. Which of the following values is the perimeter of this triangle? So here we have an isosceles triangle. Two of the side lengths are the same. And first, the problem tells us that one side length is 20 centimeters. Let's assume that's the bottom one. And then it says that one of the other two is two-fifths of that. So we'll say that this side is two-fifths of 20. Two-fifths of 20 is 8. Since this is an isosceles triangle, that means that the other side would also be 8. So it seems that we've solved the problem. We just have to add up all these values. As a matter of fact, we did not. Because this triangle is impossible. It is impossible to make a triangle of side length 8, 20, and 8. This triangle wasn't drawn to scale, and we can see in this drawing how it would look. The two ends could just not meet in a point, since 8 plus 8 is less than 20. So a triangle like this is not possible. So where did we go wrong? We went wrong by assuming which side was 20 centimeters. Let's assume that the other side is 20 centimeters. That means that this other side will also be 20 centimeters, since they are equal. And this bottom one will be two-fifths of the other one. Two-fifths of 20 is again 8. This triangle looks more possible. So the perimeter would be 20 plus 20 plus 8, which is 48. So the problem asked us, which of the following values is the perimeter of this triangle? The answer is 48 centimeters. Letter B. Problem number 15 states, Tom wants to write a number in each of the nine cells of the triangle. He wants the sum of the three numbers on each diameter to be 13, and the sum of the eight numbers on the circumference to be 40. What number does Tom have to write in a center cell? So first let's give all of these cells names. We'll name them A, B, C, D, E, F, G, H, and then the middle one, X. Now let's start representing each of these numbers. We'll start with the Now let's start representing all of our constraints mathematically. So we know that all the diameters, the sums of the diameters, have to be equal to 13. So one side of the diameter plus X plus the other side of the diameter has to equal 13. So A plus X plus E equals 13, B plus X plus F equals 13, and on and on. And then we also know that if we add up A, B, C, D, E, F, G, and H, we should get 40. So now in order to solve this problem, let's add up all of the diameter equations together. So we'll have A plus B plus C plus D plus E plus F plus G plus H plus 4X equals 52. And now we can do a substitution in the circumference equation. We can substitute A plus B plus C plus D plus E plus F plus G plus H with 40. So let's do that. So now we have 40 plus 4X equals 52. Now let's just solve for X. So subtract 40 from both sides to get 4X equals 12, and then divide both sides by 4. So we get X equals 3. So that means that the middle value is 3. In this case, we didn't actually have to solve for A, B, C, D, E, F, G, or H. If you were curious as to what they might be, here's a possible solution. So the question asked us, what number does Tom have to write in the center cell? The answer is 3. Letter A. Problem number 16 states, Masha put a multiplication sign between the second digit and the third digit of the number 2020 and noted that the resulting product, 20 times 20, is a square number. How many numbers between 2010 and 2099, including 2020, have the same property? So first let's take a look at 2020, and let's see what makes it so special. It's not exactly that the number 20 is multiplied by the number 20, that they're perfect squares. This can also appear in other ways. If we take a look at their prime factorization, we can see that the prime factorization of 20 is 5, 2, and 2. For every single prime factorization, there's an even number of every single value of prime factorization. So in this case, there are 4 2's and 2 5's. This means they can be split up evenly, and we have a square number. But we have a lot of numbers to look at for the prime factorization. So maybe we can figure something out to cut out most of these. And what we can realize in every single one of these numbers, there's a 20 in the front. That means that every single one of them will have the prime factorization 5, 2, 2, and then some other numbers. There's only one 5, though, and we need an even number of 5's, which means that this number has to be a multiple of 5. So luckily, we can toss out 80% of the numbers for not being multiples of 5. An example of a number working here would be 2005, but we're only asked to look for the numbers between 2010 and 2099. So that one doesn't work, sadly. So first, let's start out with 2010. 20 times 10. 10 breaks up into 5 and 2. There's an odd number of 2's, so it doesn't work. 2015 is 5 times 3. There's an odd number of 3's, so it doesn't work. 2020, we already know works. There's an even number of 5's and an even number of 2's, so it works. 2025 is 5 times 5, which gives us an odd number of 5's, so that doesn't work. 2030 is 5 times 2 times 3, which gives us an odd number of 2's and 3's. 35 gives us an odd number of 7's. 2040 gives us an odd number of 2's. 2045 gives us even numbers for both of them. 2 5's, 2 2's, and 2 3's. So we can split up into 5 times 2 times 3 and 5 times 2 times 3, which is 30. So 20 times 45 is 30 squared. For 2050, that breaks up into 5 times 5 times 2, which gives us an odd number of 5's and 2's. For 55, that gives us an odd number of 11's. For 60, it gives us an odd number of 3's. For 65, it gives us an odd number of 13's. For 70, it gives us an odd number of 2's and 7's. For 75, it gives us an odd number of 5's and 3's. For 2080, it gives us an even number of both 2's and 5's, so 2080 works. We can break it up into 5 times 2 times 2 times 2, which is 40. So 40 squared. So 20 times 80 is 40 squared. For 85, there's an odd number of 17's. For 90, there's an odd number of 2's. And for 95, there's an odd number of 19's. So the only numbers that work are 2020, 2045, and 2080. So the question asked us, how many numbers between 2010 and 2099 have the same property? The answer is 3. Letter C. Problem number 17 states, two squares of different size are drawn inside an equilateral triangle. One side of one of these squares lies on one of the sides of the triangle, as shown. What is the size of the angle marked by the question mark? So, first thing that the problem tells us is that this is an equilateral triangle. That means that all of these values are 60 degrees. Next, let's draw a construction line parallel to this square's side. Since it is parallel, we know that this angle is 90 degrees. Now we have a quadrilateral with one unknown value. We'll call it x. Let's solve for value x. So we know that all the angles of a quadrilateral have to add up to 360 degrees. And we know that the angles are x, 90 degrees, 60 degrees, and 70 degrees. So x plus 70 plus 60 plus 90 equals 360. That means that x is 140 degrees. Next, let's try and solve for this angle. We'll call it y. Since we know that 140 and y are supplementary, we can say that 140 plus y has to be equal to 180. Subtracting 140 from both sides, we get that y equals 40. Now we've solved for two of the three angles in the triangle, which means we can finally solve for the question mark. We know that the sum of the angles of a triangle is 180 degrees. So we know that 180 is equal to 40 plus 90 plus the value of our mystery angle. If we do 40 plus 90, that's 130. Subtract it from both sides, we get that the mystery angle is 50 degrees. So the question asked us, what is the size of the angle marked by the question mark? The answer is 50 degrees. Letter E. Problem number 18 states, Luca began a 520 kilometer car trip with 14 liters of gas in the car's fuel tank. His car consumes one liter of fuel per 10 kilometers. After driving 55 kilometers, he reads a road sign showing the distances from that point to five gas stations ahead on the road. These distances are 35 kilometers, 45 kilometers, 55 kilometers, 75 kilometers, and 95 kilometers. The capacity of the car's fuel tank is 40 liters, and Luca wants to stop just once to fill the tank. How far is the gas station that he should stop at? So here's his whole trip, 520 kilometers. And we know that he at first travels 55 kilometers before reading that road sign. And the road sign tells him that the first gas station is 35 kilometers away. And then the next one is 45 kilometers away, which is 10 more than 35. And then the next one is 55 kilometers away, which is 10 more than 45. The next one is 75 kilometers away, which is 20 more. And the last one is 95 kilometers away, which is again 20 kilometers more. Let's solve for the final leg of the trip as well. We'll call that x. And we know that the whole trip is 520 kilometers. And we know that it consists of 55 plus 35 plus 10 plus 10 plus 20 plus 20 kilometers plus our mystery x value. So adding them all up and subtracting, we get 520 equals 150 plus x, or that x is equal to 370 kilometers. So these kilometer values are great, but it would be much more useful if we converted these to the amount of liters of fuel it takes to travel that distance. Luckily, the problem gave us a conversion. They said that the distance that he travels divided by 10 is equal to the liters of fuel consumed. So we can simply put a decimal place, one to the left of where it already is, to see how many liters of gas it'll take to travel that distance. So now let's just simulate how Luca's car trip goes. So he starts out at the beginning with 14 liters of fuel. And he travels the first 55 kilometers and uses up 5.5 liters of fuel. So now he has 8.5 liters. Then when he arrives at the first gas station, he only has 5 liters of fuel. At the next gas station, he only has 4. At the next one, he only has 3. And then he only has 2 at the next one. In order to get to the next gas station, he would have to use up another 2 liters of gas. However, he only has 1. So he should refuel here and then continue on his journey. He'll be able to finish his journey with 1 liter of fuel left. If he refueled one gas station earlier, he would have to refuel again at the last gas station. Because in order to do the last leg of the trip, he needs 37 liters of fuel. But he only has 36. So he'd have to refuel again. And the problem told us that he only wants to refuel once. So the gas station that he should refuel at is 75 kilometers away. So the question asked us, how far is the gas station that he should stop at? The answer is the one at 75 kilometers. Letter D. Problem number 19 states, let 17x plus 51y equal 102. What is the value of 9x plus 27y? So in order to solve this problem, let's try and solve it by elimination. So let's try and make the x values equal one another. So let's multiply the numerator by 9, and the denominator by 17. If we multiply the left side, though, we have to multiply the right side. After multiplying it, we get 153x plus 459y equals 918 on the top, and 153x plus 459y equals 17z. z is the unknown value that we're solving for. We can see that the left side of the equation is exactly the same. This means that they can perfectly cancel out. So if we multiply the bottom equation by negative 1, and then add them together, we get 0 equals 918 minus 17z. So now it's easy to solve for z. Add 17z to both sides, and then divide both sides by 17. We get z equals 54. So the question asked us, what is the value of 9x plus 27y? The answer is 54. Letter A. Problem number 20 states, a square-shaped stained glass window is 81 decimeters squared, is made out of 6 triangles of equal area. See figure. A fly is sitting exactly on the spot where the 6 triangles meet. How far from the bottom of the window is the fly sitting? So the problem tells us that the square is 81 decimeters squared. Since it is a square, it has equal side lengths. So we can take the square root of this number to figure out the side lengths, which are 9 decimeters. Since each of the triangles has an equal area, we can divide the total area by the number of triangles, in this case 6. So 81 divided by 6 will give us the area of each individual triangle, in this case 13.5 decimeters squared. So now let's consider this triangle. Its total area is 13.5 plus 13.5, or 27 decimeters squared. We know that the area of a triangle is 1 half base times height. We know the area of this triangle is 27 decimeters squared. And we know the length of its base, 9 decimeters. So we can plug all these in, and then we can solve for the height. So divide both sides by 9, and we get 3 equals 1 half the height. Then multiply both sides by 2, to get that the height is 6 decimeters. So the point at which the fly is sitting at is 6 decimeters above the bottom of the window pane. So the question asked us, how far from the bottom of the window is the fly sitting? The answer is 6 decimeters, letter D. Problem number 21 states, the digits from 1 to 9 are randomly arranged to make a 9 digit number. What is the probability that the resulting number is divisible by 18? So before we start this problem, let's answer the question, how do we know if a number is divisible by 9? Then we'll figure out how to know if a number is divisible by 18. So the rule for knowing if a number is divisible by 9, for example, we'll choose the number 187,569, is to add up all of its digits, 1 plus 8 plus 7 plus 5 plus 6 plus 9, and then see if that value is divisible by 9. If you don't know, you can add up the digits again, so for example, if I don't know if the number 36 is divisible by 9, I can add up the digits again, so 3 plus 6 would be 9, and obviously I know that 9 is divisible by 9. So in this case, our number is divisible by 9. So now we know how to figure out if a number is divisible by 9. So how do we know if a number is divisible by 18? Well that's simple, that's if the number is divisible by 9, and if the number is divisible by 2. So in other words, all the digits summed up must be divisible by 9, and the rule for 2 is that the number must end in either a 0, a 2, a 4, a 6, or an 8. So let's take all of our digits, which are 1 through 9, and let's add them all up. In this case, we get 45. 45 is divisible by 9, so we know that any combination of 1, 2, 3, 4, 5, 6, 7, 8, and 9 in a 9-digit number will be divisible by 9. Next, let's take a look at point 2, whether or not the number ends in a 0, 2, 4, 6, or 8. We have 4 of these numbers in our 9 digits. That means that the chance of the last digit being a 0, 2, 4, 6, or 8 is 4 out of 9. The chance of all the digits summed up being divisible by 9 is 100%. That means in total, our chances are 1 times 4 out of 9, or just 4 out of 9. So the question asked us, what is the probability that the resulting number is divisible by 18? The answer is 4 out of 9. Letter B. Problem number 22 states, a hare and a tortoise competed in a 5 kilometer race along a straight line. The hare is 5 times as fast as the tortoise. The hare mistakenly started perpendicular to the route. After a while he realized his mistake, turned and ran straight to the finish point. He arrived at the same time as the tortoise. What is the distance between the hare's turning point and the finish point? So here we have our race, at the bottom is the start line, at the top is going to be the end line. We know that this race is 5 kilometers long. And we know that at the very beginning, the hare started running in the wrong direction, perpendicular to the race track. We'll say that the amount that he ran is x. And then we know, after he ran those x kilometers, he figured out that he was going in the wrong direction and ran straight for the end. We'll call that distance y. In this case, y is the hypotenuse of our right triangle. That means that if we use Pythagoras' theorem, a squared plus b squared equals c squared, where a and b are the lengths of the sides of the triangle and c is the length of the hypotenuse, and then we plug in our values, so 5 for a and x for b, and then for c we'll plug in y, the hypotenuse, we'll get 5 squared plus x squared equals y squared. So now let's just solve for y, take the square root of both sides, and we get that y is equal to the square root of 5 squared plus x squared, which, by the way, is not equal to 5 plus x. That is not how radicals work. Instead, all we can do is square the 5 and we'll get the square root of 25 plus x squared. Let's replace that for y, and now we have all of our variables in terms of x. So now let's say that the turtle walks at a rate of t, or maybe I guess runs at a rate of t, and we know that the hare runs 5 times faster than the turtle. So the rate at which it runs will be 5t. Exactly the units of t don't matter since it'll end up canceling out at the end, and I'll show you that. So we know that time is equal to the distance over the rate, and if you don't know how I got this equation, you can sort of think about it like this. If we take a rate unit like kilometers per hour, and for time plug in hours, and for distance plug in kilometers, we will see that everything ends up canceling out, and we get hours equals hours. So time equals distance per rate. So let's turn all of our distances into times. So first let's start out with the 5 kilometers. We know that the distance is 5 kilometers, and the rate is the rate at which the turtle traveled, so t. So the time it takes the turtle to travel that 5 kilometers is 5 over t. Now let's look at x. For x, the distance is going to be x, and the rate is going to be 5t, since the hare is traveling this distance. So the time it takes the hare to travel this distance is going to be x over 5t. Same thing for the hypotenuse, except this time for the distance, we'll use square root of 25 plus x squared. So it will be square root of 25 plus x squared over 5t. Now the problem told us that the turtle and the hare reached the finish line at the same time. That means the total time it took the turtle to finish is equal to the total time it took the hare to finish. So the sum of the times of the green segments is equal to the sum of the times of the yellow segments. So mathematically, 5 over t is equal to x over 5t plus the square root of 25 plus x squared over 5t. And if we multiply both sides by t, we can see that t cancels. So exactly what unit it was, it didn't really matter. Next let's multiply both sides by 5, and we get this. Now let's subtract x from both sides to get 25 minus x equals the square root of 25 plus x squared. Let's square both sides. So in this case it would just be 25 minus x times 25 minus x. Let's FOIL that out to 625 minus 50x plus x squared equals 25 plus x squared. Let's subtract 25 from both sides, and then let's subtract x squared from both sides. And now let's add 50x to both sides, and finally let's divide by 50. We get that x equals 12 kilometers. So now we can plug that in. But the question didn't ask for that distance. The question asked for the distance between the hare's turning point and the finish point. So that would be this distance. But we know x, so we can just plug it in. Square root of 25 plus 12 squared. 12 squared is 144. Square root of 25 plus 144 is going to be 13. This is a 5-12-13 triangle. So the question asked us, what is the distance between the hare's turning point and the finish point? The answer is 13 kilometers. Letter C. Problem number 23 states, There are some squares and triangles on the table. Some of them are blue and the rest are red. Some of these figures are large and the rest are small. We know the following two facts are true. If the figure is large, then it is a square. And, if the figure is blue, then it is a triangle. Which of the statements A through E must be true? So first, let's start out by drawing all of these shapes. One of each. So big, small, blue, red, squares, and triangles. In every single combination. And here we have them all. So now let's go through the rules. If the figure is large, then it is a square. So let's toss out all the large triangles. So now we only have six shapes left. And the second rule states, If the figure is blue, then it is a triangle. So let's toss out all the blue squares. So now we just have these four shapes remaining. And let's take a look at each of the answer choices and see if they're true. The first one says that all red figures are squares. That is false, since we have a red triangle. The second option says, all squares are large. That is not true, since we have a small square. The third one says, all small figures are blue. That is not true, since we have a small red square and a small red triangle. The fourth option says, all triangles are blue. That is not true, since we have a red triangle. And the fifth one says, all blue figures are small. Which is true. So the question asked us, Which of the statements A through E must be true? The answer is, all blue figures are small. Letter E Problem number 24 states two identical rectangles with side lengths 3 cm and 9 cm overlap as shown in the diagram. What is the area of the overlap of the two rectangles? So we can see that we have a parallelogram here where the two shapes intersect. In order to find the area of this we need to figure out what the base of it and what the height of it is. Luckily the problem tells us that one of the side lengths of the rectangle is 3 cm. So we already know what the height is, we just need to figure out what the base is. So first let's label a bunch of relevant points A, B, C, D, and E. Next let's recognize that these are rectangles, so they have right angles. Next let's recognize that angle ACB and ECD are vertical angles, which means they have the same value. Now through side angle-angle we can prove that triangle ABC and triangle EDC are congruent. So therefore we know that all of these sides are equal to one another. So now let's start solving for this hypotenuse. Let's call length of line segment CDX and we'll call the hypotenuse Y. Let's use Pythagoras' theorem which states that A squared plus B squared equals C squared. A and B are the side lengths of the triangle and C is the hypotenuse. So for A we'll sub in 3 cm, for B we'll sub in X, and for C the hypotenuse we'll sub in Y. So we have 3 squared plus X squared equals Y squared. Solving for Y we take the square root of both sides and we get the square root of 3 squared plus X squared. Now if we remember, line segment AD has a length of 9 cm. Since we know that CD has a length of X, we know that line segment AC has a length X less than 9 cm, so 9 minus X. And we know that line segment AC and CE are equal to one another. So we can set them equal and solve for X. So first let's square both sides. Let's FOIL it out. Subtract X squared from both sides. Compute what 3 squared is, 9. Subtract 81 from both sides. Multiply both sides by negative 1. And then finally divide by 18. And we get that X equals 4. If we substitute in X everywhere here, these are the values we get. So we get that the base of our parallelogram is 5 cm. The area of the parallelogram is the base times the height. And we know both of those values. So we substitute them in. So area is equal to the base, 5, times the height, 3. Which would give us 15 cm squared. So the question asked us, what is the area of the overlap of the two rectangles? The answer is 15 cm squared. Letter D. Problem number 25 states. Kanga labeled the vertices of the square-based pyramid using 1, 2, 3, 4, and 5 once each. For each face, Kanga calculated the sum of the numbers on its vertices. Four of these sums are 7, 8, 9, and 10. What is the sum of the numbers at the vertices of the fifth face? So first, let's label all the vertices. We'll label them A, B, C, D, and E. So first, let's figure out what the total sum of the vertices is. We know that each of them has to have a value of 1, 2, 3, 4, or 5, and none of them repeat. So we know that if we add them all up, we should get the total sum. So 1 plus 2 plus 3 plus 4 plus 5 is 15. Next, if we consider the face B, C, D, E, we know that its value is going to be B plus C plus D plus E. If we substitute that into our equation, we get that B, C, D, E is equal to 15 minus A. So if we know what A is, we know the value of face B, C, D, E. So now we're kind of stuck, so let's just make some assumptions. First, let's just assume that A equals 5 and see where that takes us. If we assume that A equals 5, we know the value of B, C, D, E. That would be 10. But more importantly, let's try and get the minimum value that we can out of a triangular face. So in this case, we'll substitute B and C with the smallest values that we still have, which are 1 and 2. So the value of this face would be 5 plus 1 plus 2, which would be 8. This is a problem, since the problem told us that the face values were 7, 8, 9, 10, and then one of the answers, either 11, 12, 13, 14, or 15. If we can't make any value lower than 8, then how are we supposed to get 7? We can't get it from B, C, D, E, since we already know it's 10. So our assumption that A is 5 cannot be correct. So we know that A must be greater than or equal to 1 and less than or equal to 4. If it's 4 here, we can see that we can make a 7. So now let's try and make the minimum value possible for B, C, D, E. In this case, we're going to try and maximize A. So we'll use 4. So B, C, D, E is going to be equal to 15 minus 4, or 11. Well, what does this tell us? Since the only face values we know are 7, 8, 9, or 10, and we know that B, C, D, E has to be greater than or equal to 11, that means that face B, C, D, E has to be the unknown face that we are solving for. So now let's go back to assuming. Let's try and figure out more from A. If we assume that A equals 4, that means B, C, D, E will be equal to 11. And the only way that we can make 7 the minimum value is if 1 and 2 are touching each other. So that would give us a face value of 7, 4 plus 1 plus 2. Now if you look on the other side, to the 4, 5, and 3, A, D, E, we would get a value of 12. This creates some possibilities because we cannot have two answers greater than 10. So that means our initial assumption that A equals 4 is wrong. Let's instead assume that A equals 3. In this case, B, C, D, E would be 12. Let's say that we want to make 7. So A plus some number, either 1, 2, 4, or 5, plus another number, 1, 2, 4, or 5, must be equal to 7. However, this is not possible. No matter what you do, you cannot get a sum of 7. So our initial assumption that A equals 3 is wrong. So let's assume that A equals 2. In this case, B, C, D, E would be equal to 13. First, let's try and make 7. The only way we can make 7 is 2, 1, and 4. So let's move those next to each other. And in this case, this triangle right here would make a sum of 7. The next one over here would make one of 9. The next one would have 10. And the last one would have 8. This means we did it. We were able to get every single face value in the top row and one from the bottom row, which means that we've answered the question. A equals 2, and B, C, D, E equals 13. So let's now see what would happen if we assume that A is 1. That would mean B, C, D, E was 14. So now let's try and make the value 10. The only way we can do this is if the 5 and the 4 are next to each other. Now we can do 1 plus 4 plus 5 and make a triangular face have a sum of 10. But that means the triangle on the other side would have a value of 6. And 6 is not one of our valid values. So this cannot work. So A truly is 2, and B, C, D, E is 13. And we know that B, C, D, E is the missing face. So the question asked us, what is the sum of the numbers at the vertices of the fifth face? The answer is 13, letter C. Problem number 26 states, a large cube is built of 64 smaller identical cubes. Three of the faces of the large cube are painted. What is the maximum possible number of small cubes that have exactly one face painted? So the problem tells us that we have 64 cubes. Let's figure out how long each of the edges are of this big cube. In order to do this, we can take the cube root of the total number of cubes. Taking a cube root of a number is pretty hard, so it's probably easier to just guess and check. We know it's probably going to be a round number, and if we try 4, 4 cubed is 64. So each of the edges are 4 cubes long. So here we can see the net of the cube. Each of the squares is one face of a smaller cube. This cube net is just a representation for the cube. So let's actually mark all the edges that meet up with one another. So for example, the edge all the way on the left and the edge all the way on the right. When the cube is built, they wrap around and connect up to one another. Same thing with all of these corner ones. And same thing with this top one and with this bottom one. So now let's start painting our cube. We need to start somewhere, so let's first paint this face. At this point, we have 16 cubes with one side painted. So now, which face will we paint next? Well, we have 5 choices, but 4 of them are identical, so we only have 2 choices. If we pick one of the one question mark faces, then that will share an edge with the face that is already painted. So some of those cubes will be painted with 2 sides. So rather than doing that, let's take the face with 2 question marks. So let's paint that one. And here we have 32 cubes with one side painted. Now to paint the next cube face, it doesn't really matter which one we pick, because all of them are basically identical. They will share 2 edges with the faces already painted. So in order to prevent us from going outside the cube net and then back in, let's just pick this one. And as you can see, the number of cubes with one side painted hasn't changed. That is because all of these cubes right here are actually one cube, and they have 2 sides painted, so they don't count for the value. So on the left side we have 12, on the right side we have 12, and in the middle we have 8. That is a total of 32 cubes painted with one side. So the question asked us, what is the maximum possible number of small cubes that have exactly one face painted? The answer is 32. Letter C. Problem number 27 states, Anna wants to write a number in each of the squares of the grid so that the sum of the four numbers in each row and the sum of the four numbers in each column are the same. She's already written some numbers as shown. What number does she write in the shaded square? So first let's start out by representing all the rows and columns mathematically. So first we'll start by giving all of the unknown squares variable names. So a, b, c, d, e, f, g. So now let's figure out the equation for the first column. So now let's figure out the equation for the first column. We'll say that x will be the sum of the rows and the columns. So in this case x equals 1 plus b plus c plus e. And we'll move on to the next column x equals a plus 2 plus 7 plus f and so on for all the columns. Now we'll move on to the rows. So in this case it would be x equals e plus f plus 7 plus g. For the next one it would be x equals 7 plus c plus d plus 4 and on and on for all the rows. So now looking at these equations, let's take a look at both of the equations that take a into account. That would be these two. And now let's set them equal to each other since both of them equal x. And now let's solve for f. If we subtract a from both sides the a's drop out and now we just have a bunch of numbers which all drop out to give us that f equals 1. So now let's solve for f. So we figured out what f is. Now we can do a similar thing with all of the equations that involve d. We can do a similar thing with all the equations that involve d. So let's again set them equal to one another since both of them equal x. And now if we subtract d from both sides they drop out and if we subtract all the other numbers we get that c equals 4. So we can substitute that in. Now let's take a look at these two. If we set them equal to one another, so if we solve for a with d we get a equals 5 plus d. We can replace that now in our equations. And now let's go ahead and actually simplify some of our equations. I don't know why we kept them in that form for so long. And we can see that a bunch of them are just saying the same thing that x is equal to 15 plus d. There's no point in having four equations tell us the same thing so we can just remove three of them. Then let's simplify all the other equations. And this is what we get. The thing that glares out to me here is that x is equal to 15 plus d and x is equal to 15 plus g. If we set them equal to one another and subtract 15 from both sides we will see that d equals g. So we can substitute all the g's with d's. Now we have two equations telling us that x equals 15 plus d so let's get rid of one of them. And now let's consider these two equations. If we set them equal to one another the d's drop out and then if we subtract 8 from both sides we get that e equals 7. And e happens to be the variable that represents the value in the shaded square. So we can replace it. So the question asked us what number does she write in the shaded square? The answer is 7. Letter C. Problem number 28 states, Alice, Belle, and Kathy had an arm wrestling contest. In each game, two girls wrestled, while the third rested. After each game, the winner played the next game against a girl who had rested. In total, Alice played 10 games, Belle played 15 times, and Kathy played 17 times. Who lost the second game? So first, let's take a look at how many times each of the girls played. The problem told us that Alice played 10 times, Belle played 15 times, and Kathy played 17 times. That means that in total, there were 42 games, if we add them all up. But that's not really true, because that's how many times each girl played, and since two of them played per game, we should divide that number by two. So actually, in total, there were only 21 games. Next, let's figure out how many times somebody didn't play in a game. So for example, Alice didn't play in 21 minus 10 games, because 21 is the total number of games, and she played in 10 games, so that means that she didn't play in 21 minus 10 games. Let's do that for all the girls. And we get that Alice didn't play in 11 games, Belle didn't play in 6 games, and Kathy didn't play in 4 games. But rather than saying that each of the girls didn't play in that many games, we can turn this around and say that the other two girls played each other in those games. So for example, Alice didn't play in 11 games, we can instead say that Belle and Kathy played together in 11 games, and we can do that for all of them. So the thing that glares out at me is that Belle and Kathy played together in 11 games. 11 games is over half of the total games. Let's represent Belle and Kathy playing together as X, Alice and Kathy playing together as Y, and Alice and Belle playing together as Z. Here we have the total number of games, 21. And in order for Belle and Kathy to play together in 11 games, they had to play the first one, and then every other one after that, because they couldn't have played in two consecutive games, because the losers sat out in the next game. So it was impossible for neither Belle or Kathy to lose in the game. So now we know who played in the third game. That means that in the second game, Alice must have lost in order for Belle and Kathy to play together in the third game. So the question asked us, who lost the second game? The answer is Alice. Letter A. Problem 29 states, a zigzag line starts at point A at one end of the diameter AB of a circle. Each of the angles between the zigzag lines and diameter AB is equal to alpha as shown. After four peaks, the zigzag line ends at point B. What is the size of angle alpha? So first let's finish up the drawing and then mark all the relevant points. In this case we only mark half the points because we can see that if we take a line from O to the top, we can see that it perfectly symmetrically cuts our shape. So that means that whatever we do to the left is also applicable to the right, but in this case we'll only show it on the left for simplicity. So we know that O is going to be the center of the circle, and we know that any line going from the center of the circle to the edge of the circle will have a length of R, the radius of the circle. So in this case OE will have a length of R, and if we draw a line OD that will also have a length of R. Next let's consider triangle ECO. It is isosceles because two of its angles are exactly the same. Therefore edge EO and edge EC are exactly the same, and we know the value of EO is R, so therefore edge EC must also be R. Next let's consider these two triangles. Angle AOD and triangle OEC. We can realize that line segment AO and OE have the same value, R, and that angle DAO and angle COE have the same value, alpha, and line segment DO has the same length as line segment CE. So therefore we can prove that these two triangles are congruent. So that tells us that their bases are equal, CO and DA are equal to one another. It also tells us that angle ADO is alpha, and since we know the triangle ADC is isosceles because two of its angles have the same value, we also know that DC is equal to DA. But back to the triangles, we can see that ADO has the same value as ECO of alpha. And next let's consider this triangle, triangle ODC. We can see that two of its side lengths are the same, which means that it's isosceles. The angles on the other sides of the side lengths must be the same as well. We'll call them beta. And now let's solve for beta. We know that the angles of a triangle have to add up to 180 degrees. We know that two of the angles are beta, and we know that the other one is 180 minus alpha. We know the sum of all of these have to be equal to 180. So if we subtract 180 from both sides, we get beta plus beta minus alpha equals zero. Combine the betas, and then add alpha to both sides, we get two beta equals alpha. Dividing by two, we get beta equals alpha over two. So beta has half the value of alpha. So next let's consider this angle. We'll call it gamma. We know that gamma plus alpha over two has to be equal to alpha. This means that we can solve for gamma in terms of alpha. And simply subtracting alpha over two from both sides, we get that gamma equals alpha over two. So now let's take a look at triangle ADC. We know that the sum of its angles have to be equal to 180 degrees. So angle DAC plus angle CDA plus angle ACD have to be equal to 180 degrees. And we know all of those values relative to alpha, which means that we can solve for alpha. We know that DAC is alpha, CDA is alpha over two, and ACD is alpha. Combining all of those, we get five alpha over two equals 180 degrees. And then multiplying both sides by two-fifths, we get that alpha equals 72 degrees. So the question asked us, what is the size of angle alpha? The answer is 72 degrees, letter B. Problem number 30 states, a consecutive three-digit positive integers have the following property. Each of them is divisible by its last digit. What is the sum of the digits of the smallest of the eight integers? So here we have the three-digit number, we'll represent all the digits as variables, so a, b, c, so a, b, c, and the value of a, b, c would be 100 times a plus 10 times b plus c. And the problem tells us this number should be divisible by its last digit. So 100 a plus 10 b plus c over c should be equal to a natural number. We can break up the fraction to 100 a plus 10 b over c plus c over c, and c over c is obviously 1. So we get 100 a plus 10 b over c plus 1, and that must equal a natural number. What we see here is the number doesn't have to be divisible by c, it just has to be divisible by 100 times a plus 10 times b, and we can use that to our advantage. But first let's try and restrict c a little bit. If we say that c equals 0, we would be dividing by 0, which obviously cannot happen, so c cannot be equal to 0. And there are only two possible strings of eight numbers between 1 and 9. We could have 1, 2, 3, 4, 5, 6, 7, 8, or 2, 3, 4, 5, 6, 7, 8, 9. So one of those has to be the value of c. So we know that 100 a plus 10 b must be divisible by one of these sets of numbers. So in order to do this, we could easily take them all and multiply them together, but sadly that number would be too large, and we can only have a three-digit number. So we need to cut out some of these numbers. Luckily, if we know the number is divisible by 8, it does not have to be divisible by 1, 2, or 4, because a number divisible by 8 is automatically divisible by these numbers. And if we know a number is divisible by 2 and is divisible by 3, it doesn't have to be divisible by 6. So if we just find a number that's divisible by 8, 3, 5, and 7, and we'll talk about the 9 later, we'll be good. So in order to do this, let's multiply all the numbers together. So 3 times 5 times 7 times 8 gives us luckily a three-digit number, 840. So now we know what a and b are, but let's figure out what c is. So these could be our possible numbers. Let's see if 849 is divisible by 9. In order to do this, let's add up all of its digits. 8 plus 4 plus 9, and that gives us 21. 21 is not divisible by 9, so 849 is not divisible by 9. So c has to be equal to 1, 2, 3, 4, 5, 6, 7, 8. And here we can prove that. 841 over 1 is a natural number, over 2 is a natural number, over 2 is a natural number, and it all works out. But the question asked us for the sum of the digits of the smallest of the 8 numbers. Luckily, that's easy. So this is the smallest of the 8 numbers, 841. 8 plus 4 plus 1 is 13. So the question asked us, what is the sum of the digits of the smallest of the 8 integers? The answer is 13. Letter D.

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