Problem number one states, Basil is writing the word kangaroo one letter each day. He wrote the first letter on Wednesday. On what day will he finish writing the word? So here we have a sort of calendar, Sunday, Monday, Tuesday, Wednesday, Thursday, Friday and Saturday. We know he writes the first letter on Wednesday. So on that day, he writes a K. Then on the next day, on Thursday, he would write the next letter, A. On Friday, the next one, N. On Saturday, G. On Sunday, A. On Monday, R. And on Tuesday and Wednesday, O. This means that he finishes the word as he started it, on Wednesday. So the question asked us, on what day will he finish writing the word? The answer is on Wednesday, letter C. Problem number two states, a motorcyclist drove a distance of 28 kilometers in 30 minutes. How many kilometers would he drive in one hour if he drives at the same speed? So first, let's take a look at what speed is. Speed is a measure of how much you move and how much time it takes you to move that distance. So in our example, we know that the motorcyclist moved 28 kilometers in 30 minutes. 30 minutes is the same thing as half of an hour, so we can replace that with 0.5 hours. So the motorcyclist was moving at a speed of 28 kilometers per half hour. Of course, we want to know how far the motorcyclist moved in one hour, so we have to change the denominator. Of course, if we multiply our value 28 over 1 half by 1, it won't change anything. And the way we can multiply it by 1 is by multiplying it by 2 over 2, which is 1. So now let's multiply our fractions across. 28 times 2 is 56, and 1 half times 2 is 1. This gives us 56 kilometers over one hour, which means that in one hour, the motorcyclist can drive 56 kilometers. So the question asked us, how many kilometers would he drive in one hour if he drives at the same speed? The answer is 56 kilometers, letter B. Problem number three states, how many faces do six cubes have altogether? So first, let's count the number of faces on one cube. From this angle, we can see that the cube has three faces, and if we flip it over to the other side, we can see that it has another three faces. So in total, each cube has six faces. The problem wanted to know how many faces six cubes have. So if we have six cubes with six faces, that will give us six times six faces, or 36 faces. So the question asked us, how many faces do six cubes have altogether? The answer is 36. Letter D. Problem number four states, a square piece of paper was cut into two pieces along a straight line. Which of the following shapes cannot be the result of the cut? Okay, so here we have a square and let's just try cutting it in different ways. So for example, first let's try to cut it vertically like this. When we cut it vertically, we have two rectangles that come off of this one square. This means that a rectangle is possible to achieve with one straight cut. Next, if we cut the square from one corner to the other corner diagonally, we will get a triangle. One of the angles of this triangle will be the angle of a square, so 90 degrees, so we will have a right triangle. Two of the side lengths will also have the same length. For example, the green side length and the blue side length will be the same. Therefore, it will also be an isosceles triangle. An isosceles triangle is a triangle that has two edges that are the same length and one that is different. Next, if we cut the square like this, we will get this as the resulting shape. This shape has five sides which makes it a pentagon. It isn't a regular pentagon, but it is still a pentagon, so we can also make a pentagon. The only object that we cannot make is a square. So the question asks us, which of the following shapes cannot be the result of the cut? The answer is a square, letter A. Problem number 5 states, In Crazy Town, the houses on the right side of Number Street have odd numbers. However, Crazy Towners do not use numbers containing the digit 3. The first house on the right side of the street is numbered 1. What is the number of the 15th house on the right side of the street? So on the right side of the street, we can only use odd numbers. So these are the odd numbers from 1 through 29. And the problem told us that these Crazy Towners cannot have a number that has a 3. So for example, the number 3 could not appear, 13, or 23. So we have to discard these. So these are the numbers that remain. In total, here we have 12 numbers. And the question asked us for the number of the house that's the 15th house. So we have to continue checking these odd numbers. So let's add some more. Of course, all of the 30s will get cut out because they all have 3s. So 31, 33, 35, 37, and 39 are also out, as well as 43. So now we have this many. Let's check how many numbers we have. And it looks like we have 15 numbers, so that's perfect. So the 15th house would have to have number 47. So the question asked us, What is the number of the 15th house on the right side of the street? The answer is 47, letter E. Problem number 6 states. A system of pipes connects the upper container Z with two lower containers X and Y. See the picture. At each fork in the pipes, the water flowing down divides into two equal amounts. 1000 gallons of water were poured into the empty container Z. How many gallons of water will reach container Y? So first, let's label all the junctions. A, B, C, and D. And now, let's just simulate the water flowing from container Z. So in the very beginning, we know there are 1000 gallons of water. When these 1000 gallons reach junction A, half of them will go to junction B and half will go to junction C. So this means that 500 gallons will go down the left side and 500 gallons will go down the right side. Now let's analyze junction B. In junction B, there is 500 gallons of water flowing into it. This means that it will be split up evenly, half of it going into container X and half going into junction D. This means that 250 gallons will go into container X and 250 gallons will go into junction D. Now we know how many gallons go into container X. We know it's 250 gallons. And we know that in container Z, we started out with 1000 gallons. We know that none of this water disappeared, which means that the amount of water that we poured in must have ended up inside of containers X and containers Y. So in other words, the amount of water in container X plus the amount of water in container Y must be equal to 1000 gallons. Now that we know the amount of water in container X, we can say that X is 250. So 250 plus Y equals 1000. In order to solve this, let's subtract 250 from both sides and get that the amount of water in container Y is 750 gallons. So the question asked us, how many gallons of water will reach container Y? The answer is 750. Letter B. Problem number 7 states, square ABCD has a side equal to 5 inches. Point P is 5 inches away from point A, and 1 inch away from side BC. What is the area of triangle APD? So first, let's label all of our nouns. We know that the side length of the square is 5 inches. So we'll say that AB is 5 inches, but really AD, DC, and CB are also 5 inches. Next, the problem told us that point P is 5 inches away from point A, so we know that the length of AP is 5 inches. Next, the problem told us that point P is 1 inch away from edge BC, so we can draw a line there and say that it is 1 inch. Because we know that the distance between point B and line BC and the distance between point P and line AD must be equal to 5 inches, we know that the distance between point P and point AD must be 4 inches. Again, as stated before, length AD will be 5 inches. And now we can easily figure out what the area of triangle APD is. The area formula is 1 half base times height. The base of this triangle is 5 inches, and the height is 4 inches. 1 half times 5 times 4 is going to be 10 inches squared. So the question asked us, what is the area of triangle APD? The answer is 10 inches squared, letter B. Problem number 8 states, in triangle ABC, AC is equal to BC, and angle ACB is equal to 30 degrees. AD is the height of the triangle as marked in the picture. What is the measure of angle BAD? So the first thing that the problem tells us is that AC is equal to BC. This means that triangle ACB is an isosceles triangle. Next, the problem tells us that angle ACB is equal to 30 degrees. Then the problem tells us that AD is the height of the triangle. This means that AD meets line CB at a 90 degree angle, meaning that ADB and ADC are both 90 degrees. Since we know that triangle ACB is isosceles because lines AC and BC are equal to each other, this means that we know that some angles are also equal to each other. We know that angle CAB and angle CBA are equal to each other. Of course, we know that in triangle ABC, the sum of all the angles has to be equal to 180 degrees. So angle CAB plus angle CBA plus angle ACB have to be equal to 180 degrees. Now of course, we know that CAB and CBA are equal to each other, so we can substitute that into our equation to get angle CBA plus angle CBA plus angle ACB is equal to 180 degrees. Now let's combine like terms. Now it looks like we're stuck, but we forgot to put in that we know what angle ACB is. The problem told us that ACB is 30 degrees, so we can replace that with 30 degrees. Now let's subtract 30 from both sides to get 150 degrees equals 2 times angle CBA, and then divide both sides by 2 to get 75 degrees is equal to angle CBA. Now that we know this, it's quite easy to figure out what angle BAD is going to be. Since we know that the sum of all the angles of triangle ABD has to be equal to 180 degrees, we can do 180 degrees is equal to angle DBA plus ADB plus angle BAD. Of course, we know two of these. We know DBA is 75 degrees and ADB is 90 degrees. Let's combine like terms, and then subtract 165 from both sides to get that angle BAD is 15 degrees, which is what the problem was asking for. So the question asked us, what is the measure of angle BAD? The answer is 15 degrees. Letter D. Problem number nine states, which of the pieces below is needed to make the solid shown to the right into a prism? So in order to make the solid into a prism, we need a shape that looks like this. So let's take a look at each of these shapes and see which one is the exact same as this shape. Right now nothing is jumping out, but that's probably because the pieces rotated somehow. The first two pieces in the top left cannot be the answer, because the green piece sticks out in all three directions, X, Y, and Z. Both of these pieces are flat along one of their axes, so they cannot be the answer. Next let's try to consider the next one to the right. If we rotate this piece 90 degrees like this, we can see that it's close, but the piece at the very top is the wrong way around. So the top right piece cannot be the answer. For the next one, let's flip it the other way 90 degrees, and we can see that we have a perfect match. That means that this must be the answer. So the question asked us, which of the pieces below is needed to make the solid shown to the right into a prism? The answer is this shape. letter E Problem number 10 states. Maria's cat drinks 60 milliliters of milk each day. However, if the cat catches a mouse, he drinks one third more milk. The last two weeks, the cat caught a mouse each day. How much milk did the cat drink in the last two weeks? So here we have a schedule of the two weeks. Let's start out on Sunday as the first day, although it could have been any day. But it really doesn't matter. So we know that the cat would drink 60 milliliters if it did not catch a mouse on that day. But we know that every day it did catch a mouse, so it had to drink one third more milk. One third of 60 is 20, and 20 plus 60 is 80. Which means the cat drank 80 milliliters of milk that Sunday. Now the cat caught a mouse every single day, which means that the cat drank 80 milliliters of milk every single day for those two weeks. There are seven days in a week, and we have two weeks, so in total we have 14 days. 80 milliliters times 14 days is equal to 1,120 milliliters drank by the cat. So the question asked us, how much milk did the cat drink in the last two weeks? The answer is 1,120 milliliters, letter D. Problem number 11 states, Andrew wrote the letters of the word kangaroo in a table with two rows and four columns, with one letter in each cell. He can write the first letter in any cell. He then writes each of the following letters into a cell that has no letters in it. He can write the first letter in any cell. He then writes each of the following letters into a cell that has at least one point in common with the cell in which the letter before it was written. Alright, so let's analyze these one at a time. Let's start out with the first one. So obviously we have to start out with the first letter of kangaroo, so let's start out with K. From K we can only move down to A, and then the next letter would be N, so we have to move up and to the right. Then the next one is G, so we gotta go down to the right. Then we gotta go right for A, up for R, left for O, and then finally, down to the left for O. All of these moves, moving up, down, left, right, and diagonally are okay, because all of the cells share one point. So the first one works. Now for the second one, let's again start with the K. Now we have two choices of where to go. We'll go to the left to A. We could have gone down and to the left. Let's go to the left N, then down for G, then to the right for A, then to the right for R, to the right for O, and then back up for O again. If we chose the other way at the beginning, we just would have zigzagged diagonally around, and it would have still worked. And even if it didn't work, it doesn't matter, because we only need one way for it to work, not two ways. So now let's start out with the K, and again we're faced with two A's, so we can choose whether to go right or down and to the right, and for now we'll choose to the right, and if it doesn't work out, we can always backtrace. So now let's go down to the right for N, up for G, down to the left for A, and then to the left for R, to the left for O, and then up for O. And it works. Again, if it didn't work, we could have tried the other A, but since one way works, we don't have to try the other one. Now let's try out the fourth one. Again, let's start out with the K. Now we can only go one way to the A, so we have to go down, then up and to the right for N, then down for G, and then the next letter would be A, but G and A do not share a common point, only the one to the left, but we had to use that one, so this one is actually impossible. So we've had three possible ones and one impossible one. This means that the fourth one must be the one that doesn't work. The last one, I mean, we can check it quickly and we can see that it does work. So the question asked us, Which of the tables shown below cannot have been filled in by Andrew? The answer is table D, letter D. Problem number 12 states, 2011 all four digit numbers that can be made by rearranging the digits 2, 0, 1, 1 were placed in increasing order. What is the difference between the two numbers on either side of 2011? So we have the digits 2, 0, 1, 1 and let's try to make all the possible four digit numbers. So first let's start out from the left most side. Either we can start with a 2 or a 1. We cannot start with a 0 because then it would be a three digit number. So from the 1 the next number could either be another 1, a 0 or a 2 and from the 2 we could either have a 0 or a 1. Now in the top left we have 1, 0 so we could either continue with a 1 or a 2. With 1, 1 we could either continue with a 2 or a 0. With 1, 2 we could continue with another 1 or a 0. In 2, 0 we have to continue with a 1 since our only numbers left are 1's. In 2, 1 we can either continue with a 0 or with a 1. And now there is only one digit that we can place at the end of all of these numbers in order for it to have all of the digits. So I won't go through that, but here they are. Now we have to arrange them in numerical order. And then if we take a look at what numbers surround 2011 we have 2,101 and 1,210. If we take the difference of those numbers we get 891. So the question asked us, what is the difference between the two numbers on either side of 2011? The answer is 891, letter B. Problem number 13 states, four of the numbers in the box on the left were moved to the spaces in the box on the right, in such a way as to make the addition correct. See the picture. Which number remains in the box on the left? Alright, so first, let's try to figure out what number should go into the bottom box. What is the sum of these three numbers? So in order to do this, let's add up the three smallest numbers. The three smallest numbers we have is 17, 30, and 49. And this will give us a minimum of what should be in the red box. 17 plus 30 plus 49 is equal to 96. 96 is one of our numbers, so we've basically solved the problem. We know that it's 17 plus 30 plus 49 is 96. But let's say that it wasn't 96. Then what would we do? Well, if it was less than 96, we knew that either 96 or 167 would be inside of the red box. And then we would just have to figure out three numbers that would add up to those numbers. If we had a number that was greater than 96, then we would know that the number in the red box would have to be 167, and we would have to search for three numbers that add up to 167. So it wouldn't be that bad either way. But we found the answer kind of accidentally, so we'll take it. So the question asked us, which number remains in the box on the left? The answer is 167, the letter E. Problem number 14 states, Agatha used 36 identical cubes to build a fence around a square region. The cubes are all connected. How many cubes identical to the cubes she used to build the fence does Agatha need to fill in the enclosed region? So we know that she has 36 cubes. And we know that a square has 4 sides. So let's divide the number of cubes she has by the number of sides in a square. So 4. 36 divided by 4 gives us 9 cubes per side. So let's construct a square where we have 9 cubes on each side. And as you can see, we have this here. But the number of cubes we used isn't 36. It's actually 32, because all the cubes in the corners are reused twice. So what we can do is we can increase the side length by 1 to make the side lengths 10, and use those additional 4 cubes. The interior region of the square that we created is 8 cubes by 8 cubes. This means that in order to fill it up, she would need 64 cubes. So the question asked us, How many cubes identical to the cubes she used to build the fence does Agatha need to fill in the enclosed region? The answer is 64. Letter C. Problem number 15 states, Paul wanted to multiply a certain number by 301, but instead he multiplied it by 31. He got 372 as the result. What would the result have been if he had multiplied the number he wanted to? So first, let's figure out what number he wanted to multiply by 301. So in order to do this, let's see how many times 31 goes into 372 by using long division. 31 goes into 37 once, so we put a 1 on the top. 31 times 1 is 31. Then we subtract, and we have 6. Then we carry down the 2 to get 62. 31 goes into 62 twice, so we put a 2 on the top. 31 times 2 is 62. 62 minus 62 is 0. So 31 goes into 372 12 times. This means what Paul did is he multiplied 31 by 12 to get 372. And the problem told us that what he really wanted to do was multiply 12 by 301. And 12 times 301 is 3612. So the question asked us, what would the result have been if he had multiplied by the number he wanted to? The answer is 3612, letter B. Problem number 16 states, there are four identical right triangles inside a rectangle, as shown in the picture. The lengths of the two sides of the rectangle are 28 centimeters and 30 centimeters. What is the sum of the areas of all four triangles? Okay, so we know that the width of this shape is 30 centimeters, and the height of it is 28. That was the point of putting a rectangle around it, but we can just remove it and we'll have the same problem. So the most interesting thing to me is that the height of this shape is 28 centimeters. This is interesting because 28 centimeters is the length of the height of two of these triangles. Since we know they're exactly identical, we know that each one of them has a height of 14 centimeters. Now if we apply this knowledge back to the original, we can take a look at the width, which is 30 centimeters. It consists of two heights of triangles and one width of the triangle. So in other words, it consists of two segments of 14 centimeters, and then one segment of the width of the triangle. So 14 centimeters plus 14 centimeters is 28 centimeters, and in total we know that length was 30 centimeters, which means that the width of the triangle must be 2 centimeters. So now if we take a look at just one triangle, we can figure out the area of that triangle. So the area equation for a triangle is 1 half base times height. The base of the triangle in this case is 2 centimeters, and the height is 14. 2 times 14 is 28, divided by 2 is 14. So the area of one triangle is 14 centimeters squared. But of course that's not the answer yet, because we have four triangles. Four triangles each having an area of 14 centimeters squared means that we have an area of 56 centimeters squared in total. So the question asked us, what is the sum of the areas of all four triangles? The answer is 56 centimeters squared. Letter D. 17. In a tournament, the soccer team FC Barcelona scored a total of three goals and had one goal scored against it. It is known that this team won one game, tied one game, and lost one game. What was the score of the game which FC Barcelona won? Here we have a table. The rows of this table are going to be the matches played by FC Barcelona, and then the columns are going to be what the score was. So at the bottom we know that the total goals scored by FC Barcelona was three, and on the right we know that the total number of goals scored by the opponents was one. So first let's start at the very bottom, when they lost a match. The only way FC Barcelona could have possibly lost a match is if they scored zero goals and their opponent scored one goal. In the rightmost column, the total number of goals scored by the opponent has to add up to one. This means in that column the remaining entries must be zero. Since the match tied in the middle row, we know that the score must have been the same, so it must have been zero to zero. In the left column, we know that the sum of all of those values has to be equal to three, and right now we have a sum of zero. So when they won a match, they must have scored three goals. It must have been three to zero. So those are the scores of all the matches, and now we know how the score looked when they won a match. So the question asked us, what was the score of the game which FC Barcelona won? The answer is three to zero, letter B. Problem number 18 states, there are three points that form a triangle. In how many ways can a fourth point be chosen in order to create a parallelogram? So first, let's talk about what a parallelogram is if you forgot. It is a four-sided figure which has straight lines and the opposite sides are parallel. So here we have an example. Okay, so the problem told us that we have three points. And if we connect them all together, we have a triangle. Okay? And then the problem asks us, how many ways can we put the fourth point in order to create a parallelogram? So for example, we could create a fourth point right here. And this would make it a parallelogram. Another place where we could place the point is right here. This would also make a parallelogram. And the only other way we could do it is if we put a point down here. So all three of the gray points are possible places we could add another vertex to make our shape a parallelogram. So the question asked us, in how many ways can a fourth point be chosen in order to create a parallelogram? The answer is three ways. Letter C. Problem number 19 states 8 points are connected with several segments, as shown in the picture. Each of these points needs to be labeled with number 1, 2, 3, or 4. The number on each end of any given segment needs to be different. Three of the points are already labeled, see the picture. How many points will be labeled with 4? So first, let's try to figure out what this point is. This point is directly connected with these three. Well, it's actually connected with one other one, but we're not sure what it is, so it doesn't matter. This one is connected with a 1, a 2, and a 3, which means this one must be a 4. Next let's try to figure out what this one is. The vertices that we know are connected via these ones, and it has neighbors 1, 2, 3, so this one is also going to be 4. Next let's try to figure out this one. It has again 3 neighbors, all of its neighbors are 1, 2, and 3, so it must be 4. Next this one, we know it has neighbors 1, 2, 3 as well, so it will have a value of 4. Then let's take a look at this last one. This last one has neighbors of 4, 4, 4, and 4, so its neighbors could either be 1, 2, or 3. But this doesn't really matter what it actually is, because the problem asks for how many 4s there are, and 1, 2, or 3 cannot be 4, so in total we have 4 4s. So the question asked us, how many points will be labeled with 4? The answer is 4 points, letter D. Problem number 20 states, there are 10 students in a dance class. One of the boys brought 80 pieces of candy. If he gives each of the girls in his class the same number of pieces of candy, there will be 3 pieces of candy left. How many boys are there in the class if we know that there are at least 2 girls? Alright so we know we have 10 students in the class, and we know that one boy brought 80 pieces of candy. After distributing them, he had 3 pieces of candy left. This means that he distributed 77 pieces of candy. The number 77 is divisible by 1, 7, 11, and 77. These are the number of possible students he could have distributed it to. 77 and 11 are not possible because we only have 10 students, so he couldn't have distributed it to 11 or 77 girls. And he could have not distributed it to 1 girl as well, because the problem told us there are at least 2 girls, so he must have distributed it to 7 girls. He gave each of them 11 pieces of candy. This means there are 7 girls in the class and 3 boys. So the question asked us, how many boys are there in the class if we know that there are at least 2 girls? The answer is 3. C. Problem number 21 states, Eve wants to make a square using only pieces like the one in the picture. What is the smallest number of pieces she needs to make a square? So first, if we take a look at this piece, we can see that if we add another piece, rotated at 180 degrees, we can make a rectangle. This rectangle is 2 by 5 squares. The smallest square we can make with these rectangles is a 10 by 10 by stacking these shapes into a 5 by 2 pattern. Let's see how this is done. That's one row, and then we flip it over exactly the same on the other side. And this is how our shapes look. We've used 20 shapes to do this, and this is the smallest square we can make with this L shaped piece. So the question asked us, what is the smallest number of pieces she needs to make a square? The answer is 20 pieces, letter E. Problem number 22 states, by arranging four puzzle pieces, different shapes can be made. Which of the five shapes below cannot be made out of these four pieces? So first, let's start out with the first shape, and let's try to make it. First, let's take a look at the top half of the shape. It seems to be a long rectangle. The only way we can get these flat ends of the rectangle is by using the two pieces on the far right. If we combine them like this, we can make a rectangle. There's no other way to make a rectangle like this. Then in order to make the bottom section, with the piece coming out on the left and the piece cutting in on the right, we have to combine the remaining two pieces like this and put them below the rectangle. So the first piece is possible to make with these shapes. Now for the second one, a very long rectangle. As we established before, in order to get the flat ends of the rectangle on the sides, we need to use the two pieces on the far right. So let's use those as the very end caps of our shape. In order to fill it in, we need a shape that has a rounded point on the left side and a cut-in rounded point on the right side. We can make this with the remaining two pieces. So a long rectangle is possible. Next, for this bent piece. First, let's focus on the left section. On the left, we have a piece coming out, and then it's flat at the bottom and flat on the right side. The only way we can do this is by using the piece all the way on the left and the piece all the way on the right and stacking them on top of each other. Then for the piece on the right, we need to use the remaining two pieces like this. This is the only way to make this shape. For the fourth shape, it's quite easy to see how it's made. We just take all the pieces and stick them together horizontally, and that's the shape. Now, we've been able to make four pieces, which means we probably won't be able to make the fifth piece, but let's just make sure. So this is how the piece looks. And as we can see, there are three rounded sections that come out on it. And from our piece set, we only have three of these sections, which seems to be OK, because we would have to put three of them into there, but it really doesn't make sense, because two of these rounded sections are on one piece. And that piece is quite short, which means it would be impossible to use both of the sections that go out and around it. So this piece is impossible to make. So the question asked us, which of the five shapes below cannot be made out of these four pieces? The answer is this shape, letter E. Problem number 23 states, in how many ways can we choose four of the numbers 2, 3, 5, 6, 10, 15, and 30 in such a way that any two of the four chosen numbers will have a common factor greater than one? So first let's take a look at our numbers. And let's list all of the factors in each of the numbers. So the problem told us that they have to have a common factor greater than one. So we can just cancel out all of these ones because we're not going to use them anyway. So that kind of simplifies down our problem because now we have a lot less numbers. So first, in order for two out of the four chosen numbers to have a common factor greater than one, let's make sure that all of the numbers in the group have the same common factor. So for example, let's start out with 30. No other numbers have a common factor of 30, so it's impossible for 30 to be the common factor of all four numbers. Same thing with 15 because there are only two numbers, we need four numbers. Same thing with 10, we need four numbers, we only have two. Same thing with 6, but with 5 we have a different story. We have four numbers with a common factor of 5, which means if we make a group of 5, 10, 15, and 30, no matter which two numbers you choose, there will be a common factor of 5. Now let's cross out all the 5's and now move on to the 3's. Same story with 3, 6, 15, and 30. If you choose any two of them, they will have a common factor of 3. Now onto the 2's, we have 2, 6, 10, and 30, and if you choose any of them, they will also have a common factor of 2. So now we're done, right? Wrong, we're not, because we assumed that all of the common factors have to be the same for all four numbers, but that is not necessarily true. If we take a look at, for example, 6, 10, 15, and 30, let's say they were all in a group, and you chose any two numbers. So let's say you chose 6 and 10, they would have a common factor of 2. If you chose 10 and 15, they'd have a common factor of 5. If you chose 15 and 30, they would have a common factor of 3. And you would have a common factor for whichever two you chose. So this one would also be a group. Luckily, 6, 10, 15, and 30 are the only numbers in our set that have more than two factors. So this is the only way it can work. So 6, 10, 15, and 30 is our last set, and we have four of these possible groups of four. So the question asked us, In how many ways can we choose four of the numbers 2, 3, 5, 6, 10, 15, and 30 in such a way that any two out of the four chosen numbers will have a common factor greater than one? The answer is 4. Letter C. Problem number 24 states, we have several square grids with odd numbers of rows and columns. All the small squares in the grid which are either in a row or in a column with an even number are painted white. The rest of the small squares are painted black. Grids that are 1x1, 3x3, and 5x5 are shown in the picture. How many small white squares will there be in a grid that has 25 small black squares? So first, looking at the examples they gave us, let's count how many black squares there are. In the first one, there's only one black square. In the second one, there are four black squares. And in the third one, there are nine black squares. There seems to be a pattern where whatever index we have, if you square it, that is how many black squares we have. So for the first one, one square. For the second one, two squared. For the next one, three squared. The problem is asking about a square that has 25 small black squares. So in order to get 25 small black squares, we would need to do five squared. So the fifth index, which would be a 9x9 square. Here's that square, and as you can see, it has 25 black squares. Then it asked us to count all the white squares in the square. Now I don't want to count all of them individually, so I'll just count the number of white squares in each row, multiply, and then add it up. So first, in these long rows, we have nine white squares. And we have four of them. So, nine times four. And then, every other row, we have less white squares. We only have four white squares, but we have more of these rows. We have five of these rows. So plus five times four. Four times nine is 36, five times four is 20. 36 plus 20 is 56 white squares in the 9x9 square. So the question asked us, how many small white squares will there be in a grid that has 25 small black squares? The answer is 56. Letter D. Problem number 25 states. Rectangle ABCD is divided into 9 squares. The areas of 2 of the squares are 64 inches squared and 81 inches squared as shown in the picture. What is the length of side AB? So first let's take a look at the square which has an area of 81 inches squared. Since it's a square and it has an area of 81 inches squared, we know that it must be 9 by 9. For the 64 inches squared, we know it must be 8 by 8. We'll be putting all of our numbers inside of this notation because it's easier then to figure out what the side lengths are. So if we take a look at the square to the left of the 8 by 8 square and above the 9 by 9 square, the little tiny square marked in yellow, we can see that the square is the difference between the side length of the 9 by 9 square and the 8 by 8 square. So 9 minus 8, so 1. So the yellow square is a 1 by 1 square. Next, let's take a look at this square. This square's side length is the yellow square's side length shorter than the 8 by 8 square's side length. So 8 minus 1 is 7, so the red square is 7 by 7. Now the red square is moved over here, and we can see that one of its side lengths consists of the 9 by 9 square's side length and of the 1 by 1 square's side length. So its side length is 10 by 10. Next, let's take a look at this one. This square's side length consists of the 7 by 7 and of the 8 by 8 square. So it must be 7 plus 8, so this one must be 15 by 15. Next, let's take a look at this one. This one's kind of a little bit tricky because the side length of this one consists of the side length of the 10 by 10 plus the 1 by 1 minus the 7 by 7. So we would have to do 10 plus 1, which is 11, minus 7, which is 4. So this square is 4 by 4. Next, let's take a look at this one. This one consists of the 10 by 10 and the 4 by 4. So this one's area is 14 by 14. The question asked us for the side length AB. So AB consists of 14, 10, and 9. So if we add them all up, we get a value of 33. So the question asked us, what is the length of side AB? The answer is 33 inches. Letter B. Problem number 26 states, in a certain month there were 5 Saturdays and 5 Sundays, but only 4 Fridays and 4 Mondays. In the following month there will be... So let's start out with the calendar. Here we have Sunday, Monday, Tuesday, Wednesday, Thursday, Friday, and Saturday. And the first thing that the problem told us is that in this month there were 5 Saturdays. So let's make 5 boxes below Saturday. Of course we can't fill them in with numbers because we don't know when the month started. The next thing the problem told us is that there were 5 Sundays in this month as well. So we'll put another 5 boxes below it. The next thing was that there were 4 Fridays. So 4 boxes below Friday. And then also that there were 4 Mondays. So 4 boxes below Monday. If there were only 4 Mondays this month and 4 Fridays this month, this means there must have been 4 Tuesdays, 4 Wednesdays, and 4 Thursdays as well. Now in order to fill in the numbers, let's bring Saturday over to the left. This will make it easier to fill out the numbers because the gap between the right side of the calendar and the left side of the calendar will result in a row shift down. So we know that the first day happened on Saturday and that the last one happened on Sunday. We know that this month had 30 days. But enough with this month. The question asked us about the following month. So let's move on to the next one. Now we know that the next month started on Monday, so let's just fill out all the numbers. And here we have them. Now let's take a look at all the statements that we had before to see which one of them is true. And it looks like none of them are true. This is because we assumed that this month is going to have 30 days. But every month that follows a month that has 30 days has 31 days. So therefore this month must have 31 days, which means that 31st lands on Wednesday. Which means that Wednesday happened five times this month. So the question asked us, In the following month, there will be five Wednesdays. Problem number 27 states, On a six-sided die, number of dots on any two opposing faces adds up to seven. The picture shows three six-sided dice stacked one on top of the other, in such a way that the sum of the number of dots on any two faces that touch is five. In the picture, the number of dots on one of the faces is shown. How many dots are there on the top face of the solid? So first let's name all of our dice. Let's name the dice on the top A, the one in the middle B, and the one on the bottom C. First let's take a look at the top dice. We'll say the value on the top of the dice is X, and the value on the bottom of the dice is B. We know that X plus B must be equal to seven. Because the problem told us that the number of dots on any two opposite faces adds up to seven. Now let's solve for B. We can do this very simply by subtracting X from both sides, and getting that B is going to be equal to seven minus X. Next let's take a look at dice B. We'll say the top value of dice B is going to be C. The problem told us that the sum of the number of dots on any two faces that touch is five. So we know that seven minus X plus C must be equal to five. Now let's solve for C by subtracting seven minus X from both sides. So we subtract seven from five to get negative two, and negative X to get X minus two equals C. So C is equal to X minus two. Next for the bottom portion of dice B, we'll call it D. We know that X minus two plus D must be equal to seven. Let's subtract X minus two from both sides to get that the value of D is equal to nine minus X. Now let's take a look at dice C. We'll call the top of dice C, E. And again we know that two faces that touch have to add up to five, so nine minus X plus E has to equal to five. Subtract nine minus X from both sides, and we get E equals X minus four. So now we know a value on C. We know that X can either be equal to one, two, three, four, five, six. Those are the numbers of dots that appear on a dice. We can't, for example, have zero or seven. And we know that we have X minus four on the top of C. So if we plug in X, all the values of X, into X minus four, we get these values. Negative three, negative two, negative one, zero, one, and two. Of course, if the value on the top of dice C cannot be negative three, it cannot be negative two, negative one, or zero, that doesn't make any sense. So it can only be one or two. This means that the value of X has to be equal to either five or six. But how do we figure out which is which? Well, the number on a dice can only appear once. For example, the one dot that we have here can only appear once on the dice. It would make no sense for there to be another one on the top of the dice. So that is how X minus four cannot be equal to one. So X minus four must be equal to two, and X must be equal to six. So these are the values of all of the dice faces. So the question asked us, how many dots are there on the top face of the solid? The answer is six, letter E. Problem number 28 states. Adam is drawing sets of four circles in such a way that in each set any two circles have exactly one point in common. In each set he counts the number of points which lie on at least two circles. The greatest number of such points in a set is... So first let's take a look at how he's drawing these circles. He's drawing these circles such that they have one point in common, just like these two circles have. And the one way that he can draw them is like this. And in this combination we have six points touching each other. And in this way we have six circles touching each other. And in this combination we have six points where two circles touch each other. But how do we prove that this is the greatest amount we can have? First let's name all the circles. A, B, C, and D. Then let's take a look at all the possible ways they can touch each other. Now I'll color code it here for you to see that we have every single one of these combinations in this set. So therefore this is the maximum number of collisions we can have between two circles. Six. So the question asked us, the greatest number of such points in a set is... Six. Letter D. Problem number 29 states, Alex said that Thomas was lying. Thomas said that Mark was lying. Mark said that Thomas was lying. Tony said Alex was lying. How many of the boys were lying? So here we have all the boys, and let's see how they're throwing blame at each other. First, Alex said that Thomas was lying. We'll represent the arrow as saying that the person at the base of the arrow is blaming the person at the end of the arrow of lying. Next, Thomas said that Mark was lying. And then Mark said that Thomas was lying. And then finally, Tony said that Alex was lying. So in order to solve this problem, let's either believe or not believe Tony. First, let's start out by saying that we believe Tony. We believe that what he is saying is true. And what is he saying? He's saying that Alex is a liar. So we'll represent green people as people who tell the truth, and red people as liars. So in this case, Alex would be lying. And what's Alex lying about? He's lying that Thomas is a liar, meaning that Thomas is telling the truth. And what's Thomas telling the truth about? That Mark is a liar. And Mark is then again lying that Thomas is a liar, when in reality we already know he's not, and that just proves it further. So in this scenario, we have two liars, Alex and Mark. We made the assumption that Tony was telling the truth. Now let's make the assumption that Tony is a liar. Tony is lying about Alex being a liar, so Alex must be telling the truth. He's telling the truth that Thomas is a liar. And Thomas is lying that Mark is lying, so Mark is telling the truth. And Mark is telling the truth about Thomas being a liar. So it all checks out. So in this case, we also only have two liars. So whether or not we believe Tony is telling the truth, or he is lying, there are two liars in either case. So the question asked us, how many of the boys were lying? The answer is two of them. Letter C. Problem number 30 states, How many 5-digit numbers, which use the digits 1, 2, 3, 4, and 5, with all different digits, have the following properties, when looking at the number from the left to the right? The first two digits form a number divisible by 2, the first three digits form a number divisible by 3, the first four digits form a number divisible by 4, and the 5-digit number is divisible by 5. So here we have all the rules that were stated in the problem. So first let's take a look at the last rule, that the 5-digit number is divisible by 5. The only way that this number can be divisible by 5 is if it ends in a 5 or if it ends in a 0. However, 0 is not one of the digits we can use, so we can only use 5. So we know the number is going to be x, x, x, x, 5. We don't know what the values of x are. Next let's take a look at these two rules, that the first two digits form a number divisible by 2, and that the first four digits form a number divisible by 4. In order for a number to be either divisible by 2 or divisible by 4, the number must be even, meaning that it has to end in an even number, and we only have two even numbers, 2 and 4. So in the second digit, or in the fourth digit, we either have to have a 2 or a 4, and there are two ways of doing this. Now we have two places where we could place two different numbers in two combinations, which means we only have four numbers that we have to check. So rather than going through these rules, let's just get all the combinations of numbers that we can, and see if they follow the rules. So these are all the combinations of numbers that we have, and let's begin checking them. Since the second digits are all even, the first two digits make an even number for all of them, so they all pass that test. Next for the three-digit numbers, 143 and 341 are not divisible by 3, but 123 and 321 are. Next for the four-digit test, 1234 is not divisible by 4, and 3214 is also not divisible by 4. So we have no numbers left. There is no number that contains the digits 1, 2, 3, 4, and 5 that has these properties. So the question asked us, The answer is there are no such numbers. Letter A.

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