Problem number 21. Rhea wants to write a number in each of the seven regions of the diagram to the right. Two regions are neighbors if they share part of their boundary. The number in each region needs to be the sum of the numbers in all its neighboring regions. Rhea has already written in two of the numbers as shown. What number does she need to write in the center region? First, notice that the space with the 2 and the space with the negative 4 are both adjacent to almost exactly the same set of regions. The only difference is that the space with 2 is also adjacent to the question mark. But here, the negative 4 is adjacent to the outer two spaces and these two slightly more in the middle spaces. Whereas this space with 2 is also adjacent to the same outer two spaces and the same inner two spaces, the difference is that the 2 is also adjacent to the question mark. Since the negative 4 is the sum of the numbers in these four spaces, we know that the sum of the four spaces is negative 4 and that by adding a question mark, negative 4 plus question mark, that's what makes this space a 2. So we write down the equation negative 4 plus question mark equals 2 and what we do, we add 4 to both sides of the equation and this gives us 6 equals question mark. So the center most space must be a 6. One possible way to fill out the circle completely is by filling in all of the outer spaces with a negative 4 and all of these middle spaces with a 2 and of course the central space being a 6. There's more than one way to do it, but this is one possibility. In any case, question mark equals 6, so our final answer is C. Problem number 22. In a group of kangaroos, the two lightest kangaroos weigh 25% of the total weight of the group. The three heaviest kangaroos weigh 60% of the total weight. How many kangaroos are in the group? First, notice that we can split up the weight distribution of the kangaroos into three units, one that is 25% of the weight, one that is 15%, and one that is 60%. Furthermore, we know that the 25% consists of two kangaroos and the 60% consists of three kangaroos. So what we're really trying to find out is how many kangaroos are in the group with 15% of the weight. And the logic for this is as follows. There can only be one kangaroo in this group because if we would have two kangaroos or more, but assuming we have two, then here we have a group of two that makes up 15% of the weight, but the lightest two kangaroos make up 25% of the weight. This is a contradiction because if we have two kangaroos here that make up a smaller portion of the weight, then they must be lighter, but we have already mentioned that these two are the lightest. Therefore, there can only be one kangaroo in the 15% group. So we know that 2 plus 1 plus 3 equals 6, and thus our final answer is A. Problem number 23. In trapezoid ABCD, the sides AB and DC are parallel. The measure of angle CDA is 120 degrees, and CD equals DA, which equals 1 3rd AB. What is the measure of angle ABC? When we sketch out a trapezoid that actually fits all the conditions given in the problem, it should look something like this. Notice that we have labeled two additional points, E and F. Furthermore, we have marked some equilateral triangles within the area of the trapezoid. Here we have CD equal to DA as given in the problem, and then we have three equal line segments, AE, EF, and FB must also be equal to satisfy the part of the problem that says that DA is equal to 1 3rd AB. That means that we need each of these three segments to be equal to this DA here, so that way this segment as a whole is three times as long. Also note that we satisfy the condition that we have 120 degrees here by the point D because equilateral triangles all have angles of 60 degrees, so 60 degrees plus 60 degrees is 120. When we look at this angle right here, CFE, we know that that is also 60 degrees, which means that since this right here is a straight line, that the angle CFB must be equal to 120 degrees. Next we note that since FB is equal to the length of all of the other equilateral triangle lengths, it must be equal to the length CF. Since CF equals BF in length, that means we have an isosceles triangle here, which means that not only are the side lengths equal, but the angles by B and C are both also equal, and that means we can calculate them fairly easily because we already know the angle CFB is equal to 120 degrees. So if we want to find angle CBF, it would be equal to 180 minus 120 divided by 2. Since the angles all must add up to 180 degrees, and we have 120 accounted for here, we have 60 degrees split equally among these two remaining angles, so each angle is 30 degrees. Now the question originally asks, what is the measure of angle ABC? Well, it is exactly the same as angle FBC or angle CBF. They're both the same. The angle ABC is simply equal to 30 degrees. So our final answer is B. Problem number 24. Five positive integers, not necessarily all different, are written on five cards. Peter calculates the sum of the numbers on every pair of cards. He obtains only three different totals, 57, 70, and 83. What is the largest integer on any of the cards? We've labeled five cards A, B, C, D, and E, and we are assuming that they are in order, meaning that A is the lowest value integer and E is the highest value integer. By this construction, we know that A plus B must equal 57, the lowest sum, and D plus E must equal 83, the highest sum. We also know that since these two sums are odd numbers, that A cannot equal B and D cannot equal E because only even numbers can be the sum of two of the same number because that's the same as multiplying by two. Now we move on to the only other sum that is mentioned in the problem, which is 70. B plus C and C plus D must both equal 70. And the reason for this is that if B plus C were to equal something else, such as 57, then that would mean that C equals A. However, that would also mean that C is lower than B, and that would mean that these cards are out of order. But if we would flip these cards back around each other so that C would be here and B would be here, then A plus C would sum up to some number that's even lower than 57, and clearly this is not possible based on the information that is given in the problem. Therefore, we still have all these cards in the same order, and we know that B plus C must equal 70. By similar logic, except going from the greater side, we know that C plus D must equal 70. And to avoid any other sums popping up, we must make sure that B, C, and D are all equal. So what two equal numbers would add up to 70? Well, that would be 35 and 35. So we know that B equals C equals D, which equals 35. Now we can calculate out what A and E actually are by simple subtraction. A is equal to 57 minus 35, which equals 22, and E is equal to 83 minus 35, which equals 48. E is, of course, the highest integer by construction, and that means that our final answer is C. Question 25. Mary wrote down the remainders obtained by dividing the number 2015 by each of the numbers 1, 2, 3, and so on, up to and including 1,000. What is the largest of these remainders? We see that as long as the quotient is 2, decreasing the divisor by 1 increases the remainder by 2, such as 2015 equals 2 times 1,000 plus 15, or 2015 is equal to 2 times 999 plus 17, or so on, such as 2015 equals 2 times 998 plus 19. Since we know that 2015 divided by 3 is equal to 671 plus 23, we know that the last time the integer quotient is 2 happens for 672, since 2015 equals 2 times 672 plus 671. So 671 is the highest remainder. Therefore, the correct answer is C, 671. Problem number 26. A square with an area of 30 is divided in two by a diagonal and then into triangles, as shown to the right. The areas of some of these triangles are given in the diagram. Which part of the diagonal is the longest? This is the diagonal that splits the square into two equal halves. This diagonal can be used to identify the height of each of the triangles. As a matter of fact, they all share the exact same height, which is H, which is the distance from a corner to the center of the other diagonal. All of these triangles having the same height is useful when we try to use the equation for the area of the triangle, which is 1 half base times height. First, we must find out what the areas are of the remaining two triangles. Note that since the square is split in half by this diagonal, we have one large triangle, which has an area of 15, and another large triangle, which also has an area of 15. So all we have to do is subtract the areas of the other triangles, like so. Here we have 15 minus 5 and minus 4, which gives us 6 for the remaining area of this triangle. And similarly, 15 minus 9 and minus 2 gives us a remaining 4 for this remaining triangle. Now that we have the areas of all the triangles, we can move on to find out what each of the values of the segments are. And we do that as follows. Remember, the area of the triangle is equal to 1 half base times height. And so we use algebra to find out what A equals and what A plus B equals and so on. And that is all listed here. So for instance, we have segment A here. And we know that the area is equal to 2, which is equal to 1 half base times height. And through a little bit of algebra, we're able to prove that A is equal to 4 over H. And similarly, just through simple algebra, we are able to show that a plus b equals 10 over h, b plus c equals 8 over h, c plus d equals 12 over h, d plus e equals 18 over h, and e equals 8 over h. Our next step is going to be to find a way to split up all of the segments as follows. We know that a equals 4 over h, since a is the base of one of these triangles. And what we can do, we can subtract 4 over h from 10 over h, which is equal to a plus b. 10 over h minus 4 over h would be equal to 6 over h, which just equals b. We can apply a similar method to each of the segments that are two portions long, b plus c, c plus d, and d plus e. And when we do all of that math, we should get the following listed values. Notice that the largest value we get is d equals 10h, which means that our final answer is d. Problem number 27. Every positive integer is to be colored according to the following two rules. One, each number is either red or green. Two, the sum of any two different numbers of a given color is that same color. In how many different ways can this be done? First, let us assume that only one number is colored differently from the others. It is easiest to start with the number 1 before looking at other examples. And in this case, we have 1 being green and the rest being red. Of course, the opposite color scheme is also possible. 1 can be red and the rest of the numbers can be green. But we'll get back to that later. This is one example of a series of numbers that actually follows the two rules given. Now, let's go ahead and check if we can do the same thing, except with two numbers. What if we try doing this with 1 and 2? If 1 and 2 are green, then we know that 1 plus 2, which is 3, must also be green. And then 1 plus 3 equals 4 must also be green. And by similar logic, 5, 6, 7, and so on, all of those numbers have to be green if both 1 and 2 are green. So another possibility is that everything is the same color, green. So we've looked at what if 1 is green and what if 1 and 2 are green. What about 1 and 3? Well, in this case, we would have a situation where 2 is the only red number. But 1 plus 3 equals 4, all of that is still green. 1 plus 4 equals 5, still green. All of it works out. So it would appear that we would be able to use this same approach for pretty much every integer if we don't look at it any further. For instance, why not have 1 and 4 be green and then 5, 6, 7, 8, and so on? And if we look at it that way, it looks like there's an infinite number of possibilities for this pattern. But notice that there's a little problem that we run into if we try using a number that is greater than 3. If we try making 1 and 4 green, then OK, 1 plus 4 equals 5. And since 1 and 4 are green, that means 5 is green. However, 2 and 3 are red. And 2 plus 3 equals 5, which means 5 also has to be red. We're not allowed to have a number that is both green and red at the same time, which means that this pattern is not possible. We run into a similar problem when we try having a number 1 and any number higher than 3 as the green numbers. There's no other way of organizing the numbers other than the three ways that you see displayed here. No other way works out mathematically because you'll have a sum somewhere farther down the line that must be both green and red. So we have these three possibilities, as well as their counterparts, where the green and red are flipped. This adds up to a total of six different possibilities. So our final answer is D. Problem number 28. Five points lie on a line. Alex finds the distances between every possible pair of points. He obtains, in increasing order, 2, 5, 6, 8, 9, K, 15, 17, 20, and 22. What is the value of K? First, let's go ahead and label all of the points that we'll be working with, A through E. It follows that the segments AB plus BC plus CD plus DE add up to the length of the entire line, AE, which is equal to the longest length Alex found, which is 22. Our next step is to try to find out what the lengths of these primitive segments are, because then we can look at all of the sums that they add up to and find out which one is missing for K. How do we find out what the four primitive segments are? Well, what we have to do is look at the smaller sums that Alex found and identify which four of them can add up to exactly 22. And as a matter of fact, there's only one set of numbers available here that adds up to exactly 22. And that is 2 plus 5 plus 6 plus 9 equals 22. Next, we have to try to find out what order these distances are placed in, or if the distance even matters. Perhaps it doesn't. Notice that 2 plus 5 equals 7. And 7 is not in the list of distances that Alex found. It would be right here between the 6 and the 8, since the list is in numerical order. This means that when we mark down all of the distances on the line, 2 and 5 cannot be next to each other. Also notice that 6 plus 9 equals 15, which is one of the distances that Alex found. This means that 6 and 9 can and should be together. So we now have some guidelines for what our sketch of the line should look like. 6 and 9 should be together, and 2 and 5 should be separate on the outer edges of the line. So our sketch can look something like this. Notice that the 2 and 5 are separate. The 6 and 9 are together. And when you try adding up any of the distances, you should get every single number that Alex found. It all works out, with the exception of 9 plus 5. 9 plus 5 equals 14. And 14 is a value that is greater than 9, but less than 15, which means that 14 is the perfect value to fill in for K. Since all of the other sums are accounted for, and this can, of course, be checked mathematically, we know that K is equal to 14. So our final answer is E. Problem number 29. The figure shows a sheet with points marked. The distance from one point to the next is the same, both horizontally and vertically. Let four points at a time become vertices of different squares. How many squares with different areas is it possible to make? We have two diagrams listed below. One will be used to point out the easier-to-find squares, and one will be used to point out the slightly trickier-to-find squares. So first, looking at the diagram on the left, we will use this to point out all of the squares that can be possibly made that are at the same tilted angle as the diagram, like so. We have a standard 1 by 1 square, a 2 by 2, and a 3 by 3 square that are possible. And each of these obviously have different areas. Next, we will move on to the diagram on the right, where we reveal two slightly harder-to-find squares that have actually different areas from the non-tilted squares in the diagram to the left. And they look like this. As you can see, this square is larger than a 1 by 1, but not quite as large as a 2 by 2. And similarly, this square right here is larger than a 2 by 2, but certainly not as large as a 3 by 3. These five types of squares are the only five that can possibly be drawn on these diagrams. It is not possible to draw any larger squares, because we don't have enough points. Therefore, our final answer is D. Problem number 30. We need to write each of the numbers from 1 to 9 in the boxes of the diagram to the right in such a way that the result of each operation indicated is in the box to which the arrow points. Which number do we need to write in the box marked with the question mark? There are actually multiple ways to fill in this diagram, and all of them end up filling the question mark box with the same number. So we will focus on one specific solution for the diagram. However, all of them can actually be found using basically the same logic, as we will demonstrate. We'll start by working from the top and go downwards towards the bottom. So starting with the multiplication operation, we first note that we need to use small numbers for the two multiplication boxes, because we want to keep our result below 10. Furthermore, we are not allowed to use the number 1 here, because that would lead to a repetition of numbers, since 1 times any number is equal to that number, and we're not allowed to use a number twice here. So it's logical to use the number 2, since it's the smallest number available to us aside from 1. The other number must be either 3 or 4, since anything higher would give us a product of 10 or more, which is certainly not possible, given the problem. If you try to use the number 3 in the second box, you will run into complications later in the diagram, and I urge all of the viewers watching from home to try using a 3 at this point in the diagram and convince themselves that it doesn't work. However, we know that it's not going to work, so we're going to move on and use 2 times 4 for our boxes, and 2 times 4 equals 8. So now we've used three of the numbers available to us. Next, we have to pick a number to subtract from 8. As a matter of fact, we can use pretty much any number here that's still left for us to choose from, except for the number 6, since if we perform the operation 8 minus 6, we get 2, which has already been used earlier in the diagram. It doesn't really matter which of the odd numbers we use. All of them will work out for some sort of solution. So let's just go ahead and pick the number 5 here to subtract. 8 minus 5 equals 3. Now, the remaining four spaces, there isn't really a systematic way to perform this problem under a practical test condition. It makes more sense to just sort of try logicing this out, in the sense that you try plugging in the numbers where they make sense in the boxes and see if you can make them work and fiddle around with them for a few seconds. If you can make it work, then great. If not, then you should probably consider putting in a different number earlier in the diagram. This is basically what makes these sort of problems difficult for tests, because it's difficult to come up with a systematic approach in such a short period of time. However, here, there is one way to put the remaining four numbers so that the entire diagram is filled, satisfying the conditions of the problem. And it would look like this. 7 minus 1 equals 6, and 3 plus 6 equals 9. Here, we've used one number, one of each number in the diagram, and we've satisfied all the conditions of the problem. And so we filled in the number 6 for the question mark box. Therefore, our final answer is d.

mathematical problems

geometry

logic puzzles

trapezoid angles

triangle areas

permutation calculations