Problem number one states, in the picture, the big triangle is equilateral and has an area of nine. The lines are parallel to the sides and divide the side into three equal parts. What is the area of the shaded part? So since we know the lines divide the sides into three equal parts, we know that if we draw lines like this, all the triangles will have the same area and be equilateral. We have six shaded parts and three more unshaded parts, making a total of nine parts. Since the total area of the equilateral triangle is nine, we know that each triangle has an area of one. Therefore, the area of the shaded region of the triangle is six. So the question asked us, what is the area of the shaded part? The answer is six, letter D. Problem number two states, it is true that 1,111 over 101 equals 11. What is the value of 3,333 over 101 plus 6,666 over 303? So first, let's deal with the left side. Let's factor out a 3 from the numerator to get 3 times 1,111 over 101. Now for the right side. Let's factor out a 3 from both the numerator and the denominator. In the numerator we have 3 times 2,222 and in the denominator we have 3 times 101. The 3's in the numerator and the denominator cancel, and then we can factor out a 2 from the numerator. Now finally, the problem told us that 1,111 over 101 equals 11, so we can substitute that in for that fraction to get 3 times 11 plus 2 times 11, which of course is equal to 55. So the question asked us, what is the value of this sum? The answer is 55, letter D. Problem number three states, the masses of salt and freshwater in the seawater in Protaras are in the ratio 7 to 193. How many kilograms of salt are there in 1,000 kilograms of seawater? So we know the mass of the salt in the seawater over the mass of the water in the seawater is going to be 7 over 193. Now let's call the mass of the salt in the seawater x, and we know that the total mass of the seawater is 1,000, so if we do 1,000 minus x, we'll get the mass of the freshwater. Now let's substitute these in our equation, and now let's just solve for x. Let's do this first by cross-multiplying, then distributing the 7, and then combining like terms. Then let's divide both sides by 200 to get that x, the mass of salt, is equal to 35 kilograms. So the question asked us, how many kilograms of salt are there in 1,000 kilograms of seawater? The answer is 35, letter A. Problem number four states, Ann has this square sheet of paper shown to the right. By cutting along the lines of the square, she cuts out copies of the shape shown further to the right. What is the smallest possible number of cells remaining? By saying that she cuts out this shape, it's basically the same thing as saying, how many of these shapes can we use to cover up the grid, but not have any overlapping squares? And that problem is easier to show visually how to solve, so let's solve that one. First, let's try to cover up the bottom left corner. We can start out with this move. As she cuts out these pieces of paper, it really doesn't matter the orientation that she cuts it out in. We can flip it over the x or the y axis as we please, because she can just flip the piece of paper over. So let's have our next move be this one. As you can see, there's a dead space in the top left corner, but we cannot prevent that. The only remaining move we can do is this one. After this, we have four squares that are not covered. It is impossible to create a combination of these shapes such that there are less than four empty squares. So the question asked us, what is the smallest possible number of cells remaining? The answer is 4, letter C. Problem number 5 states, Ru wants to tell Conga a number with the product of its digits equal to 24. What is the sum of the digits of the smallest number that Ru could tell Conga? So we want the number AB to be as small as possible. We know that the digits of AB have to be such that A times B is equal to 24. The only possibilities for the numbers of A and B, since A and B are integers, are A equals 1, B equals 24, A equals 2, B equals 12, A equals 3, B equals 8, A equals 4, B equals 6, and all of them, just A and B switched. The reason why we don't show any negative answers here is because digits can only be positive, and digits can also only have one digit, which means that these are not possible. So now we have reduced our possibilities down to four. We could have either 38, 46, 64, or 83. Now we want the number to be a minimum, so obviously 38 is going to be as small as possible. This of course occurs when A equals 3 and B equals 8. And the question wanted the sum of the two digits, so 3 plus 8 is going to be 11. So the question asked us, what is the sum of the digits of the smallest number that Roo could tell Conga? The answer is 11, letter E. Problem number six states, a bag contains balls of five different colors. Two are red, three are blue, ten are white, four are green, three are black. Balls are taken from the bag without looking and not returned. What is the smallest number of balls that should be taken from the bag to be sure that two balls of the same color have been taken? So here's a possibility of all the balls in the bag. To solve this problem, let's try choosing balls such that there are not two alike colors drawn for as many draws as possible. So first, let's draw a red ball. The order that we draw the color doesn't matter as long as we don't have two that are the same color. So next, let's just say that we draw a white, and then we'll draw a green, and then maybe a blue, and then a black. Now we have drawn every single color, and if we draw another ball, we would have drawn two balls that are the same color. This would take six draws. So the question asked us, what is the smallest number of balls that should be taken from the bag to be sure that two balls of the same color have been taken? The answer is six, letter E. Problem number seven states, Alex lights a candle every ten minutes. Each candle burns for 40 minutes and then goes out. How many candles are alight 55 minutes after Alex lit the first candle? So here's a timeline of all the relevant candles. Right now we are at the 55 minute mark. Everything before the 55 minute mark was alit. At this point, some of them may have burned out. We know that candles stay alight for 40 minutes, so 55 minus 40 is 15 minutes. So every candle between 15 minutes and 55 minutes has not yet had the chance to be burned out. But every candle before 15 minutes has been burned out. This leaves us with four candles that are still lit. So the question asked us, how many candles are alight 55 minutes after Alex lit the first candle? The answer is four, letter C. Problem number eight states, the average number of children in five families cannot be. So we know in finding the average of something, we have to add up all of those values and then divide it by the number of values there are. So in this case we have five families, so we'll add up the number of children in each family, A, B, C, D, and E, and then divide it by five. We know that the sum A, B, C, D, E has to be a whole number because a family can't have a fraction of a child. We'll call this sum X. So our average is equal to X over five. Now let's multiply both sides by five, and get that five times the average has to be equal to a whole number. So let's multiply all of our possibilities by five, and see if we get a whole number for all of them. 0.2 x 5 is 1. 1.2 x 5 is 6. 2.2 x 5 is 11. 2.4 x 5 is 12. And 2.5 x 5 is 12.5. It is impossible to have a sum of 12.5, therefore it is impossible to have an average of 2.5. So the question asked us, The average number of children in five families cannot be 2.5. Letter E. Problem number nine states, Mark and Lisa stand on opposite sides of a circular fountain. They then start to run clockwise around the fountain. Mark's speed is nine-eighths of Lisa's speed. How many circuits can Lisa complete when Mark catches up with her from the first time? So here we'll mark Lisa in green, and Mark in red. When Lisa runs around once, Mark will run around nine-eighths of a turn. This means that now Lisa and Mark are separated by only three-eighths of the circle. After the second turn, Lisa and Mark are only separated by a quarter of the fountain. After three, they are only separated by one-eighth. And after four, they cross each other. During each of these times, Lisa ran one time around the circle. And the question asked us, How many circuits has Lisa completed when Mark catches up with her for the first time? So Lisa ran four times around the circle. So the question asked us, How many circuits has Lisa completed when Mark catches up with her for the first time? The answer is four. Letter A. Problem number 10 states, The positive integers x, y, and z satisfy x times y equals 14, y times z equals 10, and z times x equals 35. What is the value of x plus y plus z? So the problem told us that x, y, and z are positive integers. This means that x and y only have a certain number of values such that x times y is equal to 14. x can be 14, 7, 2, or 1, and y can only be 1, 2, 7, or 14. For y times z equals 10, y can either be 10, 5, 2, or 1, and z can either be 1, 2, 5, or 10. Of course, y has to be the same value both times, and the only values that repeat are 2 and 1. So y can either have a value of 1 or a value of 2. Of course, this also restricts the value of x and the value of z. Now x can only be 14 or 7, and z can only be 5 or 10. In order to figure out which one, let's look at the last equation, z times x equals 35. In this case, x can either be 35, 7, 5, and 1, and z can either be 1, 5, 7, or 35. The only number that repeats in x is 7. Therefore, x must be 7. All the remaining numbers are forced. Since x is 7, and x times y has to equal 14, y has to be equal to 2. Since y times z has to be equal to 10, and y is equal to 2, therefore z has to be equal to 5. So in the end, we have x being equal to 7, y being equal to 2, and z being equal to 5. The problem asks for the sum of all of these, so 7 plus 2 plus 5 is 14. So the question asked us, what is the value of x plus y plus z? The answer is 14, letter C. Problem number 11 states, Karina and a friend are playing a game of battleships on a 5x5 board. Karina has already placed 2 ships, as shown. She still has to place a 3x1 ship so that it covers exactly 3 cells. No two ships can have a point in common. How many positions are there for her 3x1 ship? So first, let's look at all the positions where she cannot place her battleship. These will be all the positions that are adjacent to her battleships that she already placed. So we cannot place the battleship covering up any of the red squares. So let's just try counting combinations. So here's one possible combination. If we shift it to the left, we have 2. Then let's shift it up to get 3. Shift it right to get 4. Shift it back up to get 5. And shift it left to get 6. We have 6 combinations in the vertical orientation. Now let's flip it to the horizontal orientation. Here we have an example of 1, so that's 7 total combinations. And we can shift it up to get 8 combinations. There are no more possible combinations without repeating ourselves. So the question asked us, how many positions are there for her 3x1 ship? The answer is 8 combinations. Letter E. Problem number 12 states, In the diagram, alpha equals 55 degrees, beta equals 40 degrees, and gamma equals 35 degrees. What is the value of delta? So first let's put in the information that the problem gave us. Alpha is 55 degrees, beta is 40, and gamma is 35. In order to solve for delta, let's try to find the remaining angles in the quadrilateral that have the angles delta and 40 degrees. First, let's start out with this angle. We'll call it x. In order to find its value, let's take a look at this triangle. We know that the sum of the angles of a triangle have to be equal to 180 degrees. So we know that 35 plus 40 plus x has to be equal to 180. Using this, we can solve for x. We get that x is 105 degrees. Now let's solve for the other unknown angle in the quadrilateral. We'll call it angle y. We'll use this triangle and the same trick. We know in this case that 55 degrees plus 40 degrees plus y has to be equal to 180 degrees. This means that y must be 85 degrees. Now we know all the angles in the quadrilateral except for delta. We know that all the angles in the quadrilateral add up to 360 degrees. So we know that 85 degrees plus 105 degrees plus 40 degrees plus delta must be equal to 360 degrees. Using this, we can solve for delta and get that delta is equal to 130 degrees. So the question asked us, what is the value of delta? The answer is 130 degrees, letter e. Problem number 13 states, the perimeter of a trapezoid is 5 and the length of its sides are integers. What are the measures of the two smallest angles of the trapezoid? So here we have a trapezoid and we know that the side lengths have to be integers. In this case we'll call the top A, we'll call the sides B, and we'll call the bottom base C. We know that the perimeter is equal to 5, so we know that A plus C plus 2B has to be equal to 5. Due to this restriction, A, B, and C can only be A equals 1, B equals 1, and C equals 2. Now that we know the lengths of all the sides of the trapezoid, let's figure out which angles are the smallest. The question asked us for the two smallest angles, and those are obviously going to be the acute angles at the bottom. They'll have the same value because our trapezoid is symmetrical horizontally. So now let's split up the base of our trapezoid. We know that the middle section has a length of 1, and that both of the sides have a length of 1 half. We know this since the trapezoid is symmetrical. Looking at the triangle on the left, we can see that it is a 30-60-90 triangle. We know this because the hypotenuse is twice that of the base, and that is typical of a 30-60-90 triangle. In this case, the angle between the 1 and the 1 half is going to be 60 degrees. The angle on the other side of the trapezoid is exactly the same as this one because the shape is symmetrical. So both the smallest angles of the trapezoid have a value of 60 degrees. So the question asked us, what are the measures of the two smallest angles of the trapezoid? The answer is 60 degrees and 60 degrees, letter B. Problem number 14 states, one of the following nets cannot be folded to form a cube. Which one? So let's first start out with the first net. When we fold the net over this line, these two edges would meet. The same edges on the bottom would also meet. Therefore, the triangles on the right would look like this with respect to the red square. This of course would make the net of a cube, so this one works. For the second one, we can see the same thing happening at the top. So let's just fold it over, and now let's deal with the triangles at the bottom. When we fold up our shape, the two red lines would connect each other. Instead of making the break there, let's move the break over a little bit. In this case, we move everything to the right. The triangle that was in the bottom right corner overflows and now goes to the left. The red line stays in the same spot both times. Now that this is simplified, it's easy to see if we fold it over the red line, we get this shape. Which is a net of a cube, so when folded, it would be a cube. So the second net also works. Now for the third net. Let's do the same thing that we did last time. These are our red lines, and let's shift everything to the right. This is what we get. For the top it's pretty easy, but then for the bottom, when we fold it over, we get that the triangles are on top of each other with respect to the square above them. This is not the net of a cube, so this one doesn't work. At this point, we've found the answer. So every test-taking strategy is telling you to stop now, but we're going to keep on going. This is the fourth net. In this case, when we fold it over the yellow line, the top triangle will look like this with respect to the square below it, and the bottom triangle will look like this with respect to the square above it. Of course, at this point you may be asking yourself, where is this triangle attached to? And I would like to remind you that this triangle is still attached to the yellow square. We are simply showing how this triangle looks with respect to the red square. So now let's fold the other two sides over. The top side is trivial, we've been doing it for a while. And then the bottom side would look like this folded over. And this again makes the net of a cube, so it works. For this last one, the exact same thing happens as the previous one before it, only the bottom is shifted slightly. And this again forms the net of a cube, so this one also works. So the question asked us, which one of these cannot be folded into the net of a cube? The answer is this one, letter C. Problem number 15 states, Asya wrote down several consecutive integers. Which of the following cannot be the percentage of odd numbers among them? So let's start off at the beginning. Let's start out with 40%. 40% represented as a fraction is going to be 40 over 100, which simplifies down to 2 fifths. In order for 40% of the numbers to be odd, we need there to be two odd numbers and three even numbers. This of course is possible. For example, we could have 4, 5, 6, 7, 8. There are five numbers, and 5 and 7 are both odd. So A works. Let's move on to B. 45 over 100 simplifies down to 9 over 20. In this case, we would need 9 odd numbers and 11 even numbers. This is not possible, because the difference of odd to even numbers in consecutive integers can only vary by one at the very most. In this case, we would need it to vary by two. So B is not possible. Now let's try C. 48 over 100 simplifies down to 12 over 25. In this case, we would need 12 odd numbers and 13 even numbers. Since the difference only varies by one, it is possible. Here is an example. 2 through 26, 13 are even and 12 are odd. So C works. For D, we have 50%, which of course simplifies down to 1 half. We would only need one odd number and one even number, and it's pretty easy to see this is possible. Simply 2 and 3, one odd number, one even number. D works. Now E. 60% is the same as 3 fifths. So we would need three odd numbers and two even numbers. And this is again possible. For example, 3, 4, 5, 6, 7. Five numbers, 3, 5, 7, are both odd. Therefore, E also works. So the only one that doesn't work is B. 45%. So the question asked us, which of the following could not be the percentage of odd numbers among them? The answer is 45%. Letter B. Problem number 16 states. The edges of rectangle ABCD are parallel to the coordinate axis. ABCD lies below the x-axis and to the right of the y-axis, as shown in the figure. The coordinates of the four points, A, B, C, and D, are all integers. For each of these points, we calculate the value y-coordinate over x-coordinate. Which of the four points gives the least value? So the problem told us that for all the vertices, A, B, C, and D, we calculate a value, which was the y-coordinate of that point over the x-coordinate of that point. The problem told us that the rectangle is in the fourth quadrant, which means that y is less than zero and x is greater than zero. This means that the value of y over x will always be negative. This means that we want the absolute value of y over x to be as large as possible, because then when we put the negative sign in, it will be as small as possible. In order to get the largest number possible, we need to make sure that x is as low as possible, and the absolute value of y is as great as possible. So for the greatest value of the absolute value of y, we have a tie between A and B, since they are on the same y-level. We can shorten our list of possible vertices by taking into account which x-value is lower. In this case, the x-value of A is lower. A will always have the lowest value. Here are a couple of examples. So the question asked us, Which of the four points gives the least value? The answer is A. Letter A. Problem number 17 states, All four-digit positive integers with the same four digits as in the number 2013 are written on the blackboard in an increasing order. What is the largest possible difference between two neighboring numbers on the blackboard? So here are all the numbers that are written on the blackboard. We don't include any numbers that begin with a zero, because that would only be a three-digit number, and the problem specified that only four-digit numbers were written down. Now the largest gaps happen when the first digits change. If you noticed this, you wouldn't even have to write down all the numbers. You could have just figured out what the largest number is that starts with a 1, and what the smallest number is that starts with a 2, as well as finding the largest number that starts with a 2, and the smallest number that starts with a 3. So now let's actually compare the differences. 2013 minus 1320 is equal to 693. And 3012 minus 2310 is 702. So the largest difference in this case is 702. So the question asked us, what is the largest possible difference between two neighboring numbers on the blackboard? The answer is 702, letter A. Problem number 18 states, in the 6x8 grid shown, 24 of the cells are not intersected by either diagonal. When the diagonals of a 6x10 grid are drawn, how many of the cells are not intersected by either diagonal? So here we have a grid that is 6 cells tall and 10 cells wide. The tricky part of this problem is going to be drawing the diagonals accurately. So first, let's figure out the slope of each of the diagonals. The slope is defined as the rise over the run. In this case, the rise is 6 and the run is 10, so 6 over 10. 6 over 10 gets simplified down to 3 over 5. First, let's draw the diagonal from the top left corner to the bottom right corner. Each cell that we move over, we will move down 3 fifths of a cell. So for the first one, we would look like this. The end point would be at x equals 1 and y equals 3 fifths. And then we would continue. Now y would be 6 fifths, 9 fifths, 12 fifths, and 15 fifths. 15 fifths makes sense because 15 fifths is 3, and as we can see, it is 3 down. So now we just continue this process for the remainder of the diagonal to draw an accurate diagonal. Once that is done, we will draw the other diagonal. And now, let's mark all the squares that intersect the diagonal. Now all that is left to do is to count up the squares that do not intersect the diagonal, and there are 32 of them. So the question asked us, how many of the cells are not intersected by either diagonal? The answer is 32. Letter E. Problem number 19 states, Andy, Betty, Kathy, Danny, and Eddie were born on February 20th, 2001, March 12th, 2000, March 20th, 2001, April 12th, 2000, and April 23rd, 2000. Andy and Eddie were born in the same month. Also, Betty and Kathy were born in the same month. Andy and Kathy were born on the same day of different months. Also, Danny and Eddie were born on the same day of different months. Which of these children is the youngest? So here are all the dates, and let's try to figure out which date belongs to which child. So the first statement told us that Andy and Eddie were born in the same month. There are two pairs of the same month, the two pairs of March and the two pairs of April. So either it could be either Andy or Eddie. The next statement told us that Betty and Kathy were also born in the same month. So the same is true for them. Betty and Kathy could have either been born in March together or in April together. The only person that hasn't been spoken about yet is Danny. Since we know that Andy, Betty, Kathy, and Eddie were all born in the months of March and April, we know that Danny had to be born in February. So now let's move on to the next statement. Andy and Kathy were born on the same day of different months. The only day that is the same and that there are different months and both Andy and Kathy share a possibility are these two, which means we can cross out Betty and Eddie from each of these. This of course changes the possibilities of the other two dates as well. Now let's move on to the next statement. Danny and Eddie were born on the same day of different months. The only dates that Danny and Eddie share are these two. So therefore, Eddie must have been born on March 20th. Because of this, all the other dates are also figured out. So the child who is the youngest is the one that was born the latest. And the latest date is April 23rd, 2001. So the question asked us, which of these children is the youngest? The answer is Betty, letter B. Problem number 20 states, John made a building of cubes standing on a 4x4 grid. The diagram shows the number of cubes standing on each cell. When John looks from the back, what does he see? So we know that John is going to look at the structure from the back. This means that his right is going to be our left. So the leftmost column is going to appear on his most right side. What he sees, it doesn't depend on how close it is to him, but rather on what's the highest there. So in each column, let's take the maximum value. In this case, it's 4. So on the rightmost side, he would see 4 blocks. Then on the column to our right, to his left, the highest number is 3. So in this case, he would see a stack of 3 blocks. Then moving one more over, the highest number is 3, so he would again see 3 blocks. And then moving over one last time, the highest number is 2 blocks, so he would see 2 blocks. So this is the shape that John would see. So the question asked us, when John looks from the back, what does he see? The answer is he sees this, letter C. Problem number 21 states, the diagram shows a shaded quadrilateral, KLMN, drawn on a grid. Each cell of the grid has sides of length 2 centimeters. What is the area of KLMN? So in order to solve this problem, let's first draw a liberal square encompassing the quadrilateral. Now, in order to find the exact area of the quadrilateral, let's subtract all the triangles that fall outside of the quadrilateral, but still are inside of our rectangle. Before we do this, let's actually find the area of the rectangle. In this case, it's 7 by 4, so we're going to subtract 7 by 4, In this case, it's 7 by 5 units, so it's going to be 35 units squared. So first, let's take a look at this rectangle. The area is obviously going to be 1 half base times height, so we're going to have 1 half 3 times 4, which just simplifies down to 6. Now for this one, we're going to have 1 half 1 times 7, which is going to simplify down to 7 halves. The area of this rectangle is going to be 1 half 1 times 4, or just 2. Now we have a little bit of a rectangle here, and that's quite easy. It's just base times height. Its base is 1 and its height is 1, so the area is 1. Lastly, let's take a look at this orange triangle. Its area is 1 half base times height, so 1 half 1 times 3, or 3 halves. So the area of the quadrilateral is going to be this difference. When evaluated, we get a value of 21. The problem doesn't end there, because we know that the quadrilateral takes up 21 squares. Each square has a length and width of 2 centimeters, so therefore we need to multiply the number of grid squares we have by 2 centimeters and then again by 2 centimeters to get 84 centimeters squared as our final answer. So the question asked us, what is the area of KLMN? The answer is 84 centimeters squared, letter B. Problem number 22 states, Let s be the number of squares among the integers from 1 to 2013 to the 6th power. Let q be the number of cubes among the same integers. Then... So we have 2013 to the 6th power, and we can break this up into 2013 squared and then cubed, or we can also break it up into 2013 cubed and then squared. Let's first try to find out how many cubes there are. So we know that 2013 squared and then cubed is going to be the largest cube we can have. The next smallest one will be 2013 squared minus 1 cubed, and then 2013 squared minus 2 cubed, and it will keep on going on and on and on until we get the base of the exponent to be equal to 1, and then we would have 1 cubed, which is our minimum range. This means we have 2013 to the 2nd power cubed. For squares, it's a very similar process. The highest square we can have is 2013 cubed squared. The next highest one we can have is 2013 cubed minus 1 squared, then 2013 cubed minus 2 squared, and on and on and on until the base of the exponent is 1. And then we would have 1 squared, which is our minimum number. So the number of squares we have is 2013 cubed. So s is equal to 2013 cubed, and q is equal to 2013 squared. Let's bring back up our relations and see which one works. D looks like the right one, s is equal to 2013 cubed. So the question asked us, what was the relationship between s and q? The answer is s equals 2013 cubed, letter D. Problem number 23 states, John chooses a 5-digit positive integer and deletes one of its digits to make a 4-digit number. The sum of this 4-digit number and the original 5-digit number is 52,713. What is the sum of the digits of the original 5-digit number? So here we have our 5-digit number, ABCDE, and then the 4-digit number, WXYZ. Of course, WXYZ shares some similar digits with ABCDE, but we don't know which. And we know that the sum of those two numbers is 52,713. What's interesting about this number is that the number is odd. If you add two numbers that are identical to each other, you'll get an even number. Let's look at this a little bit closer. Here's an example of the number 23,541 added to itself without each of its digits. As you can see, when the last digit is the same, a 1 in this case, the sum is a 2. But when the last digit isn't the same, we get a different number. If the last digit of ABCDE and WXYZ were the same, we would get an even number. In this case, we didn't get an even number, so we know that the digits must be different. This means that John must have removed the last number. This means that WXYZ is actually ABCD. Now using this, we can solve for what ABCD and ABCDE are. First, let's break up ABCDE and ABCD into the following mathematical expression. Now let's factor out a 10 from the first four terms, and now let's combine like terms. Now let's call our number ABCDX. So we have 11X plus E equals 52,713. We know that E is a digit, so we know it's between 0 and 9 inclusive, so let's try to figure out what it is by dividing 52,713 by 11. We get 4,792 and a little bit spare. This means that X is 4,792. 11 times 4,792 is 52,712, which means that E must be 1. Of course, this means that ABCD is equal to 4,792 and E is equal to 1. That of course means that A is equal to 4, B is equal to 7, C is equal to 9, and D is equal to 2. The question asked for the sum of ABCDE, so the sum of 4 plus 7 plus 9 plus 2 plus 1 is 23. So the question asked us, what is the sum of the digits of the original five digit number? The answer is 23, letter C. Problem number 24 states, A gardener wants to plant 20 trees, maples and lindens, along an avenue in the park. The number of trees between any two maples must not be equal to 3. Of these 20 trees, what is the greatest number of maples that the gardener can plant? So we'll represent an orange tree as a maple, and a green tree as a linden tree. So now let's just clarify the rules. The number of trees between any two maples must not be equal to 3. In this case, the number of trees between the two maples on the ends is equal to 3. So this could not work. So now let's try to maximize the number of maples. In order to do this, let's first start out with a maple. Let's keep on adding maples until we cannot add any more maples. If we add one more, the number of trees between the maples will be equal to 3. So instead, now let's add linden trees. If we add a maple now, there will still be three trees between these two maple trees. So we need to keep on adding linden trees in order for our pattern to reset. Let's make some more space. Now if we add a new maple tree, we can see that the number of trees between these two trees is equal to 4. This does not break our rule, so we can continue. Again, we will add four maple trees, and now we can see a pattern forming. We will add four more linden trees, and then four more maple trees. We have five sets of four trees, which is 20 trees in total, which is the number the problem asked for. Of these 20, 12 are maple. The question asked us, of these 20 trees, what is the greatest number of maples that the gardener can plant? The answer is 12 trees, letter C. Problem number 25 states, Andrew and Daniel recently took part in a marathon. After they had finished, they noticed that Andrew finished ahead of twice as many runners as finished ahead of Daniel, and that Daniel finished ahead of one and a half times as many runners as finished ahead of Andrew. Andrew finished in 21st place. How many runners took part in the marathon? So here is how they finished. On the right is first place, on the left is last place, and in between there is Andrew and Daniel. The number of runners between last place and Andrew, including last place, is X. The number of runners between Andrew and Daniel is Y, and the number of runners between Daniel and first place, including first place, is Z. We were told that Andrew was in 21st place. This means that Y plus Z plus 1 is equal to 20. The reason for the plus 1 is because we have to account for Daniel as well. The next thing that the problem told us is that Andrew finished ahead of twice as many runners as finished ahead of Daniel. This means that the number of runners between last place and Andrew, including last place, is equal to twice that of the number of runners between Daniel and first place, including first place. So X equals 2Z mathematically. The next statement gets a little more complicated. Daniel finished ahead of one and a half times as many runners as finished ahead of Andrew. So in this case, the number of people between last place and Daniel, including last place, is going to be equal to one and a half times the number of people between Andrew and first place, including first place. Mathematically represented as X plus Y plus 1 equals one and a half times Y plus Z plus 1. The plus 1 on the left is because we have to account for Andrew, and the plus 1 on the right is because we have to account for Daniel. Now all that's left to do is to solve the equations. First, let's distribute the one and a half on the bottom equation, and then let's put them all to one side, and now, finally, let's solve. First, let's replace X with 2Z in our bottom equation, and now in our top right equation, let's solve for Y, and now let's substitute this value for Y in for Y in our left equation, and now finally, let's solve. We get that Z equals 10. Using this, we can quite easily solve for the remaining variables. We get that X equals 20, and that Y equals 9. This means that the number of people between last place and Andrew was 20, including last place, the number of people between Andrew and Daniel was 9, and the number of people between Daniel and first place, including first place, were 10 people. This means there were 20 plus 9 plus 10 plus Andrew plus Daniel number of people, or 41 runners. So the question asked us, how many runners took part in the marathon? The answer is 41, letter B. Problem number 26 states, four cars enter a roundabout at the same time, each one from a different direction, as shown in the diagram. Each of the cars drives less than once around the roundabout, and no two cars leave the roundabout in the same direction. How many different ways are there for the cars to leave the roundabout? So let's label the roads that the cars enter and exit from the roundabout A, B, C, and D. Now, let's count possibilities. First, let's start out with the blue car. At the beginning, the blue car has three options. It can either exit through B, C, or D. In this case, we'll let it exit through A, B, C, or D. No more cars can leave through road C. Next, let's consider the car that entered through road C, the green car. The green car, again, has three options. It can either go through road B, A, or D. In total, we have had nine options so far. Let's say that it decides to exit through A, B, C, or D. We have had nine options so far. Let's say that it decides to go through D. No other car can exit now through road D. Now, let's consider the red and the yellow cars. If the yellow car exits through road A, the red car will be forced to exit through road B. This cannot happen, because the car cannot exit from the same spot that it entered. So instead, the yellow car will be forced to exit through road B, and the red car forced to exit through road A. In total, we had three times three combinations, or nine in total. So the question asked us, How many different ways are there for the cars to leave the roundabout? The answer is nine ways. Letter A. Problem number 27 states, A sequence starts with 1, negative 1, negative 1, 1, negative 1. After the fifth term, every term is equal to the product of the two preceding terms. For example, the sixth term is equal to the product of the fourth term and the fifth term. What is the sum of the first 2013 terms? So here we have our pattern, 1, negative 1, negative 1, 1, negative 1. So let's try expanding it a little bit. 1 times negative 1 is going to be negative 1. Negative 1 times negative 1 is going to be 1. 1 times negative 1 is going to be negative 1. 1 times negative 1 is going to be negative 1. So we can see a pattern that repeats every three values. It goes 1, negative 1, negative 1. This means that two-thirds of the numbers in the sequence are going to be negative, and one-third are going to be positive. Half of the negative numbers, so one-third of all the numbers, are going to cancel out with the one-third that's positive. And then we'll only be left with one-third of the numbers that's negative. Which means that the total sum of the sequence, the 2013 digits, is going to be 2013 over 3, negative. So 2013 over 3 is 671. So the sum is going to be negative 671. So the question asked us, what is the sum of the first 2013 terms? The answer is negative 671, letter B. Problem number 28 states, Ria bakes six raspberry pies, one after the other, numbering them 1 to 6 in order, with the first being number 1. While she is doing this, her children sometimes run into the kitchen and eat the hottest pie. Which of the following could not be the order in which the pies are eaten? So we'll represent these dots as pies that she baked. So let's just try to go in order and see which combinations are possible. So let's start at the beginning, 1, 2, 3, 4, 5, 6. So first she bakes pie 1, and then the kids eat it. Then she bakes pie 2, and the kids eat it. Bakes pie 3, kids eat it. Bakes pie 4, they eat it. Bakes pie 5, they eat it. Bakes pie 6, and they eat it. So 1, 2, 3, 4, 5, 6 is possible. Next, let's try 1, 2, 5, 4, 3, 6. So she bakes the first two pies, and they're going to eat the hottest pie first, so they're going to eat 2 first, and then they're going to eat 1, and then she's going to bake 3 more pies, all the way up to number 5, and they're going to eat 5, 4, 3, because they're going to eat the hottest first, and then she's going to bake 6, and the kids are going to eat the 6th one. So 1, 2, 5, 4, 3, 6 is possible. Next, 3, 2, 5, 4, 6, 1. First she bakes 3 pies, the kids eat the last one, and then the second one, and then she bakes 2 more, they eat the 5th one, then they eat the 4th one, then she bakes another one, they eat the 6th one, and then finally they eat the first pie that she baked, pie number 1. So 3, 2, 5, 4, 6, 1 is possible. Next, 4, 5, 6, 2, 3, 1. First she bakes 4 pies, the kids eat the 4th pie, she bakes the 5th one, they eat the 5th one, she bakes the 6th one, and then they eat the 6th one. Next, the kids eat the 2nd one. This is obviously not possible, since the hottest pie would be the 3rd pie. So 4, 5, 6, 2, 3, 1 is not possible. Now let's just check the last one. She bakes all 6 pies, and then the kids just eat them in order from hottest to coldest. So they eat 6, 5, 4, 3, 2, and 1. So the only combination that doesn't work is 4, 5, 6, 2, 3, 1. So the question asked us, which of the following could not be the order in which the pies are eaten? The answer is 4, 5, 6, 2, 3, 1. Letter D. Problem number 29 states, Each of the four vertices and six edges of a tetrahedron is marked with one of the ten numbers 1, 2, 3, 4, 5, 6, 7, 8, 9, and 11. Number 10 is omitted. Each number is used exactly once. For any two vertices of the tetrahedron, the sum of two numbers and these vertices is equal to the number on the edge connecting these two vertices. The edge PQ is marked with the number 9. Which number is used to mark edge RS? So in order to solve this problem, let's find edges that are not connected by a common vertice. For example, PQ and SR are a pair. Then we also have PR and SQ, as well as PS and QR. We note the sum of any of the two edges that do not have a common vertice will be P plus Q plus S plus R. This is because the value on the edge is the sum of the vertices that make up the edge. So in this case, the edge PS would have the value of P plus S, and the edge QR would have the value Q plus R. Same thing occurs with the other pairs. We also know if we get the sum of all the vertices' values, we will also get P plus Q plus R plus S. This means that in total, the whole tetrahedron will have a value of 4 times P plus Q plus R plus S. Since we know what values there are, we can calculate the total. The total is going to be equal to 1 plus 2 plus 3 plus 4 plus 5 plus 6 plus 7 plus 8 plus 9 plus 11. This is equal to 56, so we can substitute that in. Now let's divide by 4, and we get that P plus Q plus R plus S is equal to 14. If we substitute this in to the equation where PQ plus SR are, if we plug in the known value of PQ, which is 9, and plug in 14 for P plus Q plus R plus S, we get that 9 plus SR is equal to 14, or that SR is equal to 5. So the question asked us, which number is used to mark edge RS? The answer is 5, letter B. Problem number 30 states, a positive integer n is smaller than the sum of its three greatest divisors, naturally excluding n itself. Which of the following statements is true? So in order to solve this, let's try to find some counter examples. By simply guessing and checking, we can see that 30 is less than its three greatest proper divisors, 15, 10, and 6, which is 31. 30 is not a multiple of 4, so a cannot be correct. For B, it's even easier to find a counter example. We can just look at 12. 12's greatest proper divisors are 6, 4, and 3. 6 plus 4 plus 3 is 13, and 12 is less than 13. 12 is not divisible by 5, so B cannot be true. Using the same example, we can also disprove D. Since 12 is less than 13, and 12 is not a multiple of 7, D is not correct. Since we were able to find numbers such as 12 and 30, E is also not correct. This only leaves us with answer C. Of course, this isn't a concrete proof, but the concrete proof for this is quite complicated, so it's just easier to solve the problem this way. So the question asks us, which of the following statements is true? The answer is all such integers n are divisible by 6. Letter C.

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