Problem number 1 states, 4 chocolate bars cost $6 more than 1 chocolate bar. What is the cost of 1 chocolate bar? So the problem told us that 4 chocolate bars cost $6 more than 1 chocolate bar. This means that mathematically represented, 4x is equal to 6 plus x, where x is the cost of 1 chocolate bar. Now let's subtract x from both sides, and then divide both sides by 3. We can see that 1 chocolate bar costs $2. So the question asks us, what is the cost of 1 chocolate bar? The answer is $2, letter B. Problem number 2 states, 11.11 minus 1.111 equals... So in order to solve this problem, let's just do standard subtraction. There are different numbers of digits before and after the decimal, so let's fill in the gaps with zeros. And now, let's just subtract. We can't do 0 minus 1, so we need to borrow from the left side. So that number turns into a 0, and then we have a 10, so 10 minus 1 is 9. We again reach the same problem. We can't subtract 1 from 0, so we have to borrow from the left number. That number turns into a 0, and we have a 10, 10 minus 1 is 9. We have the same problem, so we have to borrow from the left side. That number becomes a 0, and we have a 10 now. 10 minus 1 is again 9. We have the same problem again, we have to borrow from the left side. That number becomes a 0, and we have a 10 again. 10 minus 1 is 9. So in the end, we have 9.999. So the question asks us, what is the value of 11.11 minus 1.111? The answer is 9.999, letter D. Problem number 3 states, a watch is placed face up on a table so that its minute hand points northeast. How many minutes pass before the minute hand points northwest for the first time? So here we have a watch, and we know that the minute hand points northeast. So it's going to point up and to the right. And we know that eventually we want it to point to the northwest, which is up and to the left. So let's try to simulate it moving by 15 minute intervals. So after 15 minutes, it will be pointing southeast. This is because every 15 minutes, the minute hand moves one quarter around the circle, because a full circle is 60 minutes. After another 15 minutes, it moves to southwest. And then after another 15 minutes, it moves to the northwest. So it would do this after 45 minutes. So the question asks us, how many minutes pass before the minute hand points northwest for the first time? The answer is 45, letter A. Problem number 4 states, Mary has a pair of scissors and five cardboard letters. She cuts each letter exactly once along a straight line, so that it falls apart into as many pieces as possible. Which letter falls apart into the most pieces? So as we do this, let's try to cut through each letter. So as we do this, let's try to cut through the major sections to try to shatter the shape into as many parts as possible. Sadly, this is quite hard for the O. The only way we can really cut it is this way, and this only splits it up into two parts. For the F, we can cut it this way, and this splits up the F into four parts. For the S, the most efficient way would be this way, and this cuts up the S also into four parts. The most efficient way for the H would be this way, and it would also be cut into four parts. And if we cut the M simply horizontally, we will cut it up into five parts. Which makes M fall apart into the most pieces. So the question asks us, which letter falls apart into the most pieces? The answer is the letter M, letter E. Problem number five states, a dragon has five heads. Every time a head is chopped off, five new heads grow. If six heads are chopped off one by one, how many heads will the dragon have at the end? So here we have a dragon. The bottom is the dragon's body, and the five things on the top are its heads. So once somebody cuts off one head, it goes away, and then five more come back in its place. The problem said we have to cut off six heads, so let's keep on going. We cut off the second head, that head disappears, we have five more. Then we cut off the third head, that head disappears, and then we get five more. Cut off the fourth head, it disappears, five more appear. The fifth head is then cut off, it disappears, and five more come back. And then finally, the last head is cut off, it disappears, and five more come back in its place. So in total, now the dragon has 29 heads. So the question asked us, If six heads are chopped off one by one, how many heads will the dragon have at the end? The answer is 29. Letter C. Problem number 6 states, In which of the following expressions can we replace each occurrence of the number 8 by the same positive number other than 8 and obtain the same result? So let's calculate the value of each of these, and see if we can find something interesting. So for the first one, 8 plus 8 is 16, divided by 8 is going to be 2, plus 8 is going to be 10. The next one, 8 times 16 is going to be 8 times 16 divided by 8, so it's going to be 16. Then we're going to have 8 plus 8 is 16, minus 8 is 8, plus 8 is 16. And then for the next one, we have 8 plus 8 is 16, minus 8, which is 8, times 8 is 64. Then for the last one, 8 plus 8 minus 8 is 8, divided by 8 is 1. Other than the last one, all of these depended on the 8. So let's investigate this one a little bit further. Instead of an 8, let's replace it with an x. So we have x plus x minus x, divided by x. Of course, let's combine like terms. And now we have x divided by x, which is simply 1. Which means that no matter what you plug in for x, as long as it's not 0, because then you'd be dividing by 0, you will get 1 as an answer. So the question asked us, In which of the following expressions can we replace each occurrence of the number 8 by the same positive number other than 8 and obtain the same result? The answer is 8 plus 8 minus 8, divided by 8. Letter E. Problem number 7 states, Each of the 9 paths in a park is 100 meters long. Anne wants to go from A to B without going along any path more than once. What is the length of the longest route she can choose? So since Anne is going to be going into B, it is impossible for the two paths highlighted in blue to be both used. Since once she's there, she's already completed her goal, and therefore she can't go the other way. A similar thing happens around point A. Anne can only walk through one of the green points. Because if she walks through both of them, then she would be back at point A. And this means that she could not continue unless if she traced back her steps, which isn't allowed. So we know that only one of the blue lines can be used, and one of the green lines, which puts up the maximum number to 7. So let's try to make Anne go from point A to point B in 7 moves. So of course, she'll start at point A, and then let's have her move over here, 1, over here 2, like this 3, 4, 5, 6, and then finally 7. So it is possible for her to go from point A to point B in 700 meters. So the question asked us, what is the length of the longest route she can choose? The answer is 700 meters, letter C. Problem number 8 states, the diagram shows two triangles. In how many ways can you choose two vertices, one on each triangle, so that the straight line through the vertices does not cross either triangle? So first, we can identify this line. This line passes through two vertices, and the line does not cross through either triangle. Of course, we can flip this over vertically, and get this line as well. There's another line that looks like this, that also does the same thing. It crosses through two points, and does not cross through either triangle. We can again flip this over the vertical axis, to get this one as well. So in total, we have four lines that cross through two vertices, but do not cross through a triangle. So the question asked us, in how many ways can you choose two vertices, one on each triangle, so that the straight line through the vertices does not cross either triangle? The answer is 4, letter D. Problem number 9 states, Werner folds a sheet of paper, as shown in the figure, and then makes two straight cuts with a pair of scissors. He then opens up the paper again. Which of the following shapes cannot be the result? So first, let's take a look at the first one. If Werner makes cuts that look like this, and then he unfolds the piece of paper, he will get the desired shape. So the first one works. The second one does not. The third one does not. The fourth one does not. So the first one works. For the second one, let's say Werner makes these two cuts, unfolds it, and it also makes the desired shape. So it also works. Next one, let's say that he makes these cuts, unfolds it, and it looks good. So the third one also works. For the fourth one, let's say he makes these cuts, unfolds it, and it looks good, so it works. But it actually doesn't. Because, in this case, we made four straight cuts, and the problem said that he can only make two straight cuts, meaning that this one is impossible to make with two straight cuts. Now let's just take a look at the last one to make sure that it's possible. Let's say that he cuts it like this, and when he unfolds it, he gets something that looks like this, so it works. So all of them work, except for the fourth one. So the question asked us, Which of the following shapes cannot be the result? The answer is this one, letter D. Problem number 10 states, A rectangular prism is made up of four pieces as shown. Each piece consists of four cubes and is a single color. What is the shape of the white piece? So first, let's take a look at the very dark gray piece all the way to the right. We know that the last piece, the fourth one that's hidden, has to be at the bottom. This is the only way that it can work. Now we kind of know the position of all of the pieces except for two, which are the remaining white pieces. So if we cut away this and this, we can see that the remaining two white squares are all the way at the bottom left. So the shape created is this Z pattern. So the question asked us, What is the shape of the white piece? It's this shape, letter D. Problem number 11 states, Conga forms two four-digit natural numbers using each of the digits 1, 2, 3, 4, 5, 6, 7, and 8 exactly once. Conga wants the sum of the two numbers to be as small as possible. What is the value of the smallest possible sum? So first, let's take a look at a more simple example. Let's take a look at a problem with only two digits, the digits 1 and 2. If we put the lower digits into more significant places, the number is lower. In this case, 12 is lower than 21. So now let's solve the actual problem. We have 1, 2, 3, 4, 5, 6, 7, and 8. We'll want to put the least digits in the most significant spaces, so we'll want to put 1 and 2 in the thousands space, the 3 and 4 in the hundreds, the 5 and 6 in the tens, and the 7 and 8 in the ones. So let's just alternate. Let's say that the 1, the 3, the 5, and the 7 goes to the first number, and that the 2, the 4, the 6, and the 8 goes to the second number. Now let's add them together. 1,357 plus 2,468 is 3,825. It does not matter what digit we put into which number. For example, we can switch the 1 and the 2, or the 3 and the 4, or the 5 and the 6, or the 7 and the 8, and we would still get the same answer. So regardless of the order, our answer is 3,825. So the question asked us, what is the value of this smallest possible sum? The answer is 3,825. Letter C. Problem number 12 states, Mrs. Gardner grows peas and strawberries. This year, she has changed the rectangular pea bed to a square by lengthening one of its sides by 3 meters. As a result of this change, the area of the strawberry bed was reduced by 15 meters squared. What was the area of the pea bed before the change? So the first thing the problem tells us is that this year, she's changed the rectangular pea bed to a square. This means that we know the pea bed on the right has a width x and a height x. The next thing the problem told us was that she did this by lengthening one of its sides by 3 meters, which means that on the left, the height of the pea bed is x-3. Then they told us that the area of the strawberry bed was reduced by 15 meters squared. That means that this rectangle right here with a height of 3 meters has an area of 15 meters squared. Using this information, we can figure out that the whole width of her garden is 5 meters. Of course, the width of the garden didn't change, so on the right, it is also 5. Now that we have solved for x, we can plug it in back on our left side to get that that length is 2. Finally, the question asked us for the area of the pea bed before the change. So that would be 5 times 2, or 10 meters squared. So the question asked us, what was the area of the pea bed before the change? The answer is 10 meters squared. Letter C. Problem number 13 states, Barbara wants to complete the diagram by inserting three numbers, one in each empty cell. She wants the sum of the first three numbers to be 100, the sum of the three middle numbers to be 200, and the sum of the last three numbers to be 300. What numbers should Barbara insert in the middle cell of the diagram? So the first thing the problem tells us is that the sum of the first three numbers has to be equal to 100. So mathematically, 10 plus x plus y is equal to 100. Next they tell us that the sum of the three middle numbers has to be equal to 200. So x plus y plus z equals 200. And then that the sum of the last three numbers has to be equal to 300. So y plus z plus 130 equals 300. We have three equations and three unknowns, so it is trivial to solve. First on the left, let's subtract 10 from both sides to get that x plus y equals 90. We can substitute this into our middle equation and get 90 plus z equals 200. And then subtract 90 from both sides to get that z equals 110. Now let's plug z in on the right. So we have y plus 110 plus 130 equals 300. And then combine. And then subtract 240 from both sides to get that y equals 60. Now let's plug that into our left equation. So we have x plus 60 equals 90. Subtract 60 from both sides, and we get that x equals 30. This means that the three numbers you should insert into the diagram are 30, 60, and 110. The middle number being 60. So the question asked us, what number should Barbara insert in the middle cell of the diagram? The answer is 60, letter B. Problem number 14 states, in the figure to the right, what is the value of x? So first, let's take a look at this triangle right here. We'll call the unknown angle y. We note that the sum of the angles of a triangle have to be equal to 180 degrees. This means that 180 degrees is equal to 58 degrees plus 93 degrees plus the angle y. Now we can combine like terms and move them all to one side, to get the value of y is 29. Next, let's take a look at this triangle. Again, the sum of the angles of a triangle have to add up to 180 degrees, so we know that 180 is equal to 100 degrees plus 29 degrees plus x. Solving for x, we get that x is equal to 51 degrees. So the question asked us, in the figure to the right, what is the value of x? The answer is 51 degrees, letter C. Problem number 15 states, four cards each have a number written on one side and a phrase written on the other. The four phrases are divisible by 7, prime, odd, and greater than 100. And the four numbers are 2, 5, 7, and 12. On each card, the number does not correspond to the phrase on the other side. What number is written on the same card as the phrase greater than 100? So here we have all the phrases and all the numbers. So let's figure out which numbers belong to which phrases. So 2 is only prime, 5 is both prime and odd, and 7 is prime, odd, and divisible by 7. So in order for on the other side of the card of 7 to not be true, the only statement we could put is greater than 100. So that answers our question. So the question asked us, what number is written on the same card as the phrase greater than 100? The answer is 7, letter C. Problem number 16 states, three small equilateral triangles of the same size are cut from the corners of a larger equilateral triangle with sides of 6 centimeters as shown. The sum of the perimeters of the three small triangles is equal to the perimeter of the remaining gray hexagon. What is the side length of the small triangles? So first, the question told us that the length of this large triangle is 6 centimeters. We'll call the side length of the small triangles x, making this length of the hexagon 6 minus 2x. Then the question told us the sum of the perimeters of the three small triangles is going to be equal to the perimeter of the remaining hexagon. So the sum of the perimeters of the three small triangles is going to be 3 times 3x, because there are three triangles, and each triangle has three sides. And that's going to be equal to 3 times 6 minus 2x plus 3x, because the hexagon has three sides of length 6 minus 2x, and three sides of length x. Now let's just distribute and combine like terms, and then divide both sides by 12, to get that x is equal to 1.5. So the question asked us, what is the side length of the small triangles? The answer is 1.5 centimeters, letter D. Problem number 17 states, a piece of cheese is cut into a large number of pieces. During the course of the day, a number of mice came and stole some pieces, watched by the lazy cat Ginger. Ginger noticed that each mouse stole a different number of pieces, each of which was less than 10, and that no mouse stole exactly twice as many pieces as any other mouse. What is the largest number of mice that Ginger could have seen stealing cheese? The number of cheese that each mouse could have stolen, is between 1 inclusive and 10 exclusive. So let's just try figuring out how much each mouse stole. Let's start out with 1. We'll say that the first mouse stole one piece of cheese. Then we'll say that the second mouse stole two pieces of cheese. Only this is not possible, because 2 is twice that of 1, and no mouse could have stolen twice that of what another mouse stole. So this is not possible. So instead, we'll say that there were 3. Then we'll say that the next mouse stole 4. 4 is not twice 3 or twice 1, so it works. The next one is 5. 5 is odd, so it works. Then 6. 6 doesn't work, because 6 is twice that of 3. So instead, let's put in a 7. 7 works, then let's try 8. But 8 doesn't work, because 8 is twice that of 4. So instead, let's do 9. And that's all we can do. So we have 1, 3, 4, 5, 7, and 9. Which means there were 6 mice. Now of course, we could have done it starting from 9. And we would have gotten 9, 8, 7, 6, 5, 1. But it doesn't matter, because we would have gotten 6 mice anyway. So the question asked us, What is the largest number of mice that Ginger could have seen stealing cheese? The answer is 6 mice. Letter C. Problem number 18 states, At the airport, there is a moving walkway 500 meters long, which moves at a speed of 4 kilometers an hour. Anne and Bill step on the walkway at the same time. Anne walks at a speed of 6 kilometers an hour on the walkway, while Bill stands still. When Anne comes to the end of the walkway, how far ahead of Bill is she? So here we have our walkway, with Anne and Bill. And we know that both of them are traveling on the walkway forward at a speed of 4 kilometers an hour. But we also know that Anne is walking at a pace of 6 kilometers an hour forward, which means that relative to the ground, she is moving at a speed of 10 kilometers per hour. If we take the reciprocal of this, we get that she travels 1 kilometer in 1 tenth of an hour. In order to get the amount of time it takes her to travel the length of the runway, we just have to multiply this by the length of the runway. The length is 500 meters, but we need to keep our units consistent, so 0.5 kilometers. When we multiply it, the kilometers cancel out, and we get 0.05 hours. Now we need to figure out how far Bill traveled in this time. So we know he traveled at 4 kilometers an hour, so we can multiply it by that. The hours cancel, and 0.05 times 4 is 0.2 kilometers, meaning that when Anne was at the very end, Bill was only 0.2 kilometers along the path, which means that the distance between them was 0.3 kilometers, or 300 meters. So the question asked us, when Anne comes to the end of the walkway, how far ahead of Bill is she? The answer is 300 meters, letter E. Problem number 19 states, a magical talking square originally has sides of length 8 centimeters. If he tells the truth, then his sides become 2 centimeters shorter. If he lies, then his perimeter doubles. He makes four statements, two true and two false, in some order. What is the largest possible perimeter of the square after the four statements? So here we have a square, and we know that each of its sides is 8 centimeters long. So we know that in this case, the perimeter is 32 centimeters. So first let's assume that he tells the truth, and that each of his side lengths is 2 centimeters shorter. So now he's 6 centimeters by 6 centimeters, and has a perimeter of 24 centimeters. Now instead, let's say that he doubles. So now he would have a side length of 16 centimeters, so now his perimeter would be 64 centimeters, which is obviously greater than 24 centimeters if he told the truth. So it's better for him to lie first. Now let's check again. Let's assume he tells the truth, so now he'd be 14 centimeters by 14 centimeters, and have a perimeter of only 56. But if he lied, his perimeter would double to 128, so his side lengths would be 32 centimeters by 32 centimeters, which would grow the perimeter significantly, as opposed to it shrinking if he told the truth. So now he lied twice, so now he has to tell the truth twice. So each of the side lengths is going to get smaller by 2, so it's going to be 30 by 30 for a perimeter of 120, after the first thing he tells the truth about. And then again, it would be 28 by 28, so his final perimeter is 112. And this is the maximum size that he can be. So the question asked us, what is the largest possible perimeter of the square after four statements? The answer is 112 centimeters, letter D. Problem number 20 states, A cube is rolled on a plane so that it turns around its edges. It begins at position 1, and is rolled so that one of its faces touches the plane in positions 2, 3, 4, 5, 6, and 7 in that order as shown. Which two of these positions were occupied by the same face of the cube? So first, let's start out by assuming that at the very beginning, the bottom face is touching the ground. We'll use D for down. Then it goes to the right, so then the right face would be touching the bottom. We'll use R to denote right. And then it would turn over to the right again, so now the up face would be touching the ground. And then it would turn over to the right again, so now the up face would be touching the ground. Then the cube would roll forward, so the forward face would now be touching the ground. And then it would roll over to the right again, so now the left side would be touching the ground. So now the cube would roll forward again, and it would be back again on the face that it originally started. Which means that the tiles with the numbers 1 and 6 represent the same face of the cube. So the question asked us, which two of these positions were occupied by the same face of the cube? The answer is 1 and 6, letter B. Problem number 21 states, Rick has five cubes. When he arranges them from smallest to largest, the difference between the heights of any two neighboring cubes is 2 centimeters. The largest cube is as high as a tower built from the two smallest cubes. How high is a tower built from all five cubes? So here we have all the five cubes. And if we say that the smallest cube has a height x, we know then that the next largest one has a height x plus 2, and the next one x plus 4, and then x plus 6, and then x plus 8. Then the problem told us that if we built a tower of the two smallest ones, it would be the same height as the largest one. So mathematically, x plus x plus 2 equals x plus 8. Now let's combine like terms. Subtract x from both sides. Then finally subtract 2 from both sides. We get that x equals 6. Now if we substitute in 6 for x, we get the heights of all of the cubes. 6, 8, 10, 12, and 14. So now if we built a tower out of all of them, it would have a height of 6 plus 8 plus 10 plus 12 plus 14, or just simply 50. So the question asked us, how high is a tower built from all five cubes? The answer is 50 centimeters, letter E. Problem number 22 states, in the diagram, ABCD is a square. M is the midpoint of AD, and MN is perpendicular to AC. What is the ratio of the area of the shaded triangle MNC to the area of the square? So first, let's try to find the area of the shaded triangle. We'll say that the width of square ABCD is x. This means that its height is also x, since it is a square. And this means that the area of ABC is 1 half x squared, because the area of a triangle is 1 half base times height, and the base and the height are both x. Now let's try to find the area of triangle CDM. So we know DC is x, and we know that point M splits DA into two equal parts. So we know that MD is 1 half x. Now using our triangle area formula, we get that the area of CDM is 1 fourth x squared. Now let's try to find the area of MAN. This is a little bit trickier. So we know that the length of MA is 1 half x, and the problem told us that MN is perpendicular to AC, so we know that that is 90 degrees. We also know that angle MAC is 45 degrees, because AC is the diagonal of square ABCD. This of course also means that AMN is also 45 degrees, since all the angles of a triangle have to add up to 180 degrees. So now we know that we have a 45-45-90 triangle. In a 45-45-90 triangle, if the sides are a, then the hypotenuse is a times the square root of 2. Or, if we know the hypotenuse, we know that the sides are a over the square root of 2. In this case we know that the hypotenuse is 1 half x, so we know that the sides will have a length of 1 over 2 root 2 times x. Now that we know the base and the height of the triangle, we can just do 1 half base times height, to get 1 sixteenth x-squared. Now let's add up all the areas of the square that are not in the shaded triangle. So we have 1 sixteenth x-squared plus 1 fourth x-squared plus 1 half x-squared, which gets us 13 sixteenths x-squared. Since the width and the height of the square is x, we know that the area of the square is x-squared. So if we subtract the unshaded region from the area of the square, we will get the shaded region. So x-squared minus 13 sixteenths x-squared is going to be 3 sixteenths x-squared. Now as stated before, we know the area of the square is going to be x-squared, so we'll have a ratio of 3 sixteenths x-squared to x-squared. Converting this into a proper ratio, we get 3 to 16. So the question asked us, what is the ratio of the area of the shaded triangle MNC to the area of the square? The answer is 3 to 16, letter D. Problem number 23 states, the tango is danced in pairs, each consisting of one man and one woman. At a dance evening, no more than 50 people are present. At one moment, three-fourths of the men are dancing with four-fifths of the women. How many people are dancing at that moment? So let's represent the number of men at the evening to be m, and the number of women at the dance evening to be w. We know that three-fourths of the men are dancing, and that four-fifths of the women are dancing. Since every man has to be dancing with a woman, we know that these two values are equal to each other. Now to multiply both sides by four-thirds to get that m is equal to sixteen-fifteenths w. This of course means that for every 15 women, there are 16 men at the dance evening. This of course means that the total number of people at the dance has to be a multiple of 31, since for every 15 women, there are 16 men. The problem also told us that the total number of people at the evening has to be less than 50. So the only possible value for m plus w is 31. Of course, since we know that for every 16 men, there are 15 women, the number of men has to be equal to 16, and the number of women has to be equal to 15. Now if we plug these values into the equation at the top, we get that there are 12 men dancing with 12 women. So in total, there are 24 people dancing. So the question asked us, how many people are dancing at that moment? The answer is 24. Problem number 24 states, David wants to arrange the 12 numbers from 1 to 12 in a circle, so that any two neighboring numbers differ by either 2 or 3. Which of the following pairs of numbers have to be neighbors? The first pair of numbers has to be 2. The second pair of numbers has to be 3. So on our left, we have our possible pairs, and on the right, we have David's circle. So first, let's start off with the number 1. In order to have a difference of either 2 or 3, the number would either have to be negative 1 or negative 2, or it could be positive 3 or positive 4. Of course, negative numbers are not possible, so 3 and 4 are the only possibilities. The number 1 would have two neighbors, and we only have two possibilities, so we know both of its neighbors have to be 4 and 3. The order doesn't matter, we could put the 3 on the left or the 4 on the right. Now let's talk about the number 2. In order to have a difference of 2, the numbers would either have to be 0, negative 1, 4, or 5. 0 and negative 1 are not possible answers, so 4 and 5 are the only possible answers. We already have a 4, so we know where to place the 2 relative to the other numbers. Now, as we can see, the 3 and the 5 cannot be next to each other, and the 4 and the 6 cannot be next to each other because the 4 is already surrounded by a 1 and a 2. So these two combinations cannot happen. Next, let's talk about the 3. The possible neighbors for the 3 are 0, 1, 5, and 6. 0 is not a possible number, a 1 is already one of its neighbors, and 5 is used, so the other neighbor must be a 6. So now we're kind of stuck, so now let's do the same thing from the other side. Rather than starting from the smallest number, we'll start with the largest number, which is 12. In order to have a difference of 2, 12 can either have 14 or 15, or 10 or 9. Of course, 14 and 15 are not possible. It has two neighbors, so its two neighbors are 10 and 9. We don't know where these fit with respect to the remaining numbers, so let's just put them to the side. Now let's consider the number 11. In order to have a difference of 2 or 3, it would either have to be 13, 14, or 9 and 8. 13 and 14 are not possible, so its two neighbors have to be 9 and 8. 9 is already used, so we know where it is relative to the bottom numbers. Next, let's take a look at the 10. The 10 has possibilities of 12, 13, 8, and 7. The 8 is already used, so 12 is one of its neighbors, and the 13 is impossible to get, so we know that the remaining one has to be 7. Okay, so now we've used up all of the numbers, so now let's just try to figure out how they fit into one another. So if we put the 7 next to the 6, the difference would only be 1, and that doesn't work, so the 7 has to go next to the 5. Now that we know where the 7 goes, we know where all the other numbers go as well, and the differences between all the numbers are either 2 or 3. So now it is clear that the 6 and the 8 have to be next to one another. The 7 and the 9 aren't next to one another, and the 5 and the 8 aren't next to each other either. So the question asked us, which of the following pairs of numbers have to be neighbors? The answer is 6 and 8. Letter D. Problem number 25 states, some three-digit integers have the following property. If you remove the first digit of the number, you get a perfect square. If instead you remove the last digit of the number, you also get a perfect square. What is the sum of all the three-digit integers with this curious property? So the problem told us that if we have a number with three digits, ABC, and we remove the last one so that we only have AB, it is going to be a square. And if we remove A and we're left with BC, it is also going to be a square. Now there are only so many two-digit squares, right? We have 16, 25, 36, 49, 64, and 81. So let's try to find all the combinations of squares such that the last number of one of the squares is the first number of another square. So first let's try to look at AB being equal to 16. We're looking for a number that begins with a 6, and luckily we have one, 64. So the number 164 works. No other number works with 16, so let's move on to 25. No number begins with a 5, so let's move on to 36. The only number that begins with a 6 is 64, so another possible number is 364. No other numbers work for 36, so let's move on to 49. No two-digit squares begin with a 9, so let's continue on to 64. For 64, 49 begins with a 4, so we have 649. And no other numbers work, so let's move on to 81. Only 16 begins with a 1, so we have 816 as another number. And no other number works for that one. So we have the numbers 164, 364, 649, and 816. The question asked for the sum of all of them, and if we add them up, we get the value of 1,993. So the question asked us, what is the sum of all the three-digit integers with this curious property? The answer is 1,993, letter D. Problem number 26 states, a book contains 30 stories, each starting on a new page. The length of the stories are 1, 2, 3, through 30 pages. The first story starts on the first page. What is the largest number of stories that can start on an odd-numbered page? So first we gotta start off on page 1. Now if we add to it an odd number, we'll end up on an even-numbered page. So let's not do that. Instead, let's add an even number. So if the story is two pages long, then we'll start on page 3. If the stories are constantly even-numbered pages long, we will always start the story on an odd number. Of course, we can only do this until our stories are 30 pages long. At this point, we have 16 stories that start on an odd-numbered page. 16, not 15, because we have to count for the first story. Now, we only have odd-numbered stories remaining, so let's just use those. So after a one-page story, we would be on an even-numbered page. After a two-page story, we would again be on an odd-numbered page. And it would alternate like that, on and on and on, until we got to page 29. And we would end up on an even page. So in total here, we have 7 stories that start on an odd-numbered page. Which means that in total, we have 23 stories that start on an odd-numbered page. So the question asked us, what is the largest number of stories that can start on an odd-numbered page? The answer is 23, letter E. Problem number 27 states, An equilateral triangle starts in a given position, and is rotated into new positions in a sequence of steps. At each step, it is rotated about its center, first by 3 degrees, then by a further 9 degrees, then by a further 27 degrees, and so on. At the nth step, it is rotated by a further 3 to the n power degrees. How many different positions, including the initial position, will the triangle occupy? Two positions are considered equal if the triangle covers the same part of the plane. Ok, so first let's start out with 0 degrees. Then we know that the object is rotated an additional 3 degrees. 3 to the first, so 0 plus 3 to the first is 3 degrees. Next, the equilateral triangle is rotated an additional 3 squared degrees, so that's 9 degrees, so that would be 12 degrees. Then the triangle is rotated again, this time it's 3 to the third, so 27 degrees plus 12 is 39 degrees. After this it is rotated again, this time 3 to the fourth, 3 to the fourth is 81 plus 39 is 120. Now if you have an equilateral triangle, and we rotate it 120 degrees, it is exactly the same as if it hasn't moved. And the problem told us that two positions are considered equal if the triangle covers the same part of the plane. In this case, 120 degrees is exactly the same as 0 degrees. So whenever we have a number larger than 120 degrees, we have to take away the largest multiple of 120. So then we add 3 to the fifth, 3 to the fifth is 243. Again, we have to take away the largest multiple of 120, so in this case 240, so we're only left with 3 degrees. Now I see a pattern forming, but let's just continue one more time. Then we have to add 3 to the sixth, which is going to be 729, so we have 732. Again, take away the largest multiple of 120, and we're left with 12. So we can quite easily see that the pattern repeats every four, which means we have four unique positions. So the question asked us, how many different positions, including the initial position, will the triangle occupy? The answer is four positions, letter B. Problem number 28 states, a rope is folded in half, then in half again, and then in half again. Finally, the folded rope is cut through, forming several strands. The lengths of two of the strands are 4 meters and 9 meters. Which of the following could not have been the length of the whole rope? Alright, so here are the possible combinations for the length of the rope. And here's the rope that's folded four times over, so it has eight strands going back and forth. We'll call each of the strands length X, and we know that somebody cuts the rope. We'll say they cut it along this red line. When they cut the rope, one part of the rope will be of length C, Therefore, this length is called X minus C. So this blue section would have a length of 2 times X minus C, since it's folded two times over. And there would also be this length, which would be a length 2C. Now let's go through all the lengths of the rope, and see if we can get the desired form. Now let's go through all the lengths of the rope, and see if we can get the desired 4 and 9 meters. So first, let's start out with 52 meters. There are 8 coils, so 52 divided by 8 is 6.5. So each strand would be 6.5 meters long. So now let's plug in 6.5 for X. Now we get the lengths of the cut rope to be 2C, C, and 13 minus 2C. The 13 minus 2C is just a distributed version of the blue part. And now again, our goal is to make one of them 4, and to make one of them 9. And we can see that if we plug in 2 for C, we can get 4, 2, and 9. 4 and 9 are there, so it works. 52 is possible. Now for 68 divided by 8 is 8.5, so our strand length is 8.5. Again, let's plug in 8.5 for X. And in this case, if we plug in 4, we can see that we get 8, 4, and 9. So this one also works. Now let's go to 72. 72 divided by 8 is 9. So let's plug in 9 for X, so we get 2C, C, and 18 minus 2C. And we're trying to get 4 and 9. And in this case, no matter what you plug in, you cannot get 4 and 9. So 72 is not possible. So now we kind of know the answer, but let's just do 88 just to make sure. So 88 divided by 8 is 11, so our strand length is 11. So we have 2C, C, 22 minus 2C, and we need them to equal 4 and 9. And if we plug in 9 for C, we get 18, 9, and 4. So this one works. So the only one that didn't work was when the rope length was 72 meters. So the question asked us, which of the following could not have been the length of the whole rope? The answer is 72 meters, letter C. Problem number 29 states, a triangle is divided into 4 triangles and 3 quadrilaterals by 3 straight line segments. See the figure. The sum of the perimeters of the 3 quadrilaterals is 25 centimeters. The sum of the perimeters of the 4 triangles is equal to 20 centimeters. The perimeter of the whole triangle is 25 centimeters. The perimeter of the whole triangle is equal to 19 centimeters. What is the sum of the lengths of the 3 straight line segments? So the problem told us that all of the yellow region, the perimeter of the quadrilaterals, is equal to 25 centimeters. Then they told us that all the perimeters of the triangles are 20 centimeters. Here they're represented in thin red lines, so you can see where the lines overlap. And then finally, they told us the perimeter of the big triangle is 19 centimeters. That's the big blue outline. Now, what has become pretty clear is that the whole perimeter is covered in either yellow or red. The yellow and the red do not overlap on the perimeter. And every point inside of the triangle overlaps yellow and red exactly twice. This means if we add up the lengths of the yellow and the red, we will have the perimeter of the triangle in addition to twice the length of all the 3 line segments. In order to get just the lengths of the line segments, we would have to subtract out the perimeter, which we know, and then divide it by 2. So let's add up the yellow and the red region, and then subtract out the blue region to get 26. Now 26 is twice the length of all the 3 line segments, so if we divide it by 2, we get 13 centimeters, which is the length of the line segments inside the triangle. So the question asked us, what is the sum of the lengths of the 3 straight line segments? The answer is 13 centimeters, letter C. Problem number 30 states, a positive number needs to be placed in each cell of a 3 by 3 grid shown, so that in each row and in each column, the product of the 3 numbers is equal to 1, and in each 2 by 2 square, the product of the 4 numbers is equal to 2. What numbers should be placed in the central cell? So first, let's count how many times each of the cells are used in the 2 by 2 square. We can see that each of the corners are used once, each of the edges are used twice, and the center is used 4 times. We know that the product of each of these 2 by 2 squares is 2, so the product of all 4 of the 2 by 2 squares is 16. Next, we know that the product of all of the rows and the columns is 1. This means that if we take the product of all of the squares, we get a product of 1. Now we know that the product of the 4 edge squares with the center square 3 times is also 16, but the product of the middle column and the middle row is also 1. Therefore, the center square must be 16. Here is a combination of how it could look. So the question asked us, what numbers should be placed in the central cell? The answer is 16, letter A.

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