Hello, and welcome to the Math Kangaroo 2016 level 5 and 6 solutions video series. First of all, I'd like to congratulate you for participating in the Math Kangaroo. When I was in elementary school and in high school, I used to participate in the exams myself every year, and it was definitely a lot of hard work. So definitely pat yourself on the back for going through this mathematical challenge. It's going to benefit you a lot later in life just by helping you solve these logical problems. The purpose of this video series is to show you a logical path to get to the solution for every single question on the exam. Now keep in mind, the solutions that are presented in these videos are not necessarily the only methods to get to the correct answer. If you used an entirely different method, or maybe just a partially different method, and still got the correct answer, then your method might be equally good. Although there might also be reasons why one method is better than the other. If you have any questions about something like this, please feel free to email me at thomas at mathkangaroo.org, the email which is listed below. Before moving on to the actual videos, I'd like to offer two pieces of advice. First of all, I highly recommend that you try working through each problem before looking at the video for a given problem. So the reason for this is that if you look at a video right away without even trying the problem, then you'll see, oh, here's the solution, here's how you get it, and you're done. And you're having the information spoon-fed to you, but you don't actually learn anything from it. You'll just hear my explanation and think, oh, okay, that probably makes sense. Whereas if you actually try to work on the problem beforehand, and struggle with it, and think about it, then the video serves as something that will help fill in the gaps, and it's going to help you understand maybe where you went wrong, maybe something that was wrong in your thinking, or it might just help put in that last piece of the puzzle for whatever you were thinking about. The point is, struggling over a problem first is going to help you actually learn. So definitely try the problems before watching the videos. And my second piece of advice, and this is something I really can't stress enough, is to show as much of your work as possible. And the reason for this is that a lot of times people who are generally smart mathematicians, people who are good at math, they tend to like to do a lot of it in their head, because they think it will save them time. However, it really doesn't make sense to go through a problem quickly if you're going to make a silly mistake every once in a while, just because you didn't want to write down 2 plus 2 on a sheet of paper. Really, what you should do is write down as much of what you're thinking as possible, so that way you're forced to think about what you're thinking about, and in that way, you're going to be more careful and watch out for any mistakes you might make. Basically, people who try doing too much math in their head, they make more mistakes. I myself am definitely guilty of trying to do too much mental math at a time, and then getting confused, maybe switching a plus and a minus here or there, and having an answer that's completely wrong. Where if I'd have just written out my work, it would have been actually quicker and more accurate. So with these two pieces of advice in mind, I hope that you find my solutions very useful, and I hope you learn a lot from them. And again, if you have any questions about these solutions or about the exam in general, please feel free to email me. I'm more than happy to answer any questions. Anyway, that's enough chit-chat. I hope you enjoy all the problems. Good luck! Problem number one. Which of the following traffic signs has the largest number of lines of symmetry? Looking at answer choices B and D first, we see that because these shapes are so complicated looking, they they look very different from the left and right side and from the top and bottom side. We can't actually find any lines of symmetry as a result of how strange these signs look. So B and D are definitely going to be eliminated since there's no line of symmetry. Now looking at A, C, and E, your intuition should tell you that C is probably the right answer because it's the simplest of the three signs. After all, sign A has an arrow pointing in one direction, which means that you're not going to have a top-down line of symmetry. Or rather, you're not going to have a left-right line of symmetry because the right side looks different from the left. And similarly, a car looks different on the top than it does on the bottom. Whereas this sign, if you were to reflect it horizontally or vertically, it would look exactly the same. So comparing the lines of symmetry, we have one line of symmetry on A and on E, but shape C has two lines of symmetry, and thus that is our final answer. Problem number two. Mike cuts a pizza into quarters, then he cuts every quarter into thirds. What part of the whole pizza is one piece? We can solve this problem in two ways. The long way is to actually draw out a pizza. I know that's a beautiful pizza there. And we can go ahead and cut it into quarters, which is signified by these black lines. And now if we cut each slice into thirds, then it should look something like this. Where we put two lines in each of the quarters, and that will give us three pieces in each of the quarters. Now if we count how many slices we have here, we're going to get a total of 12. And that's exactly what one slice of pizza is in this case. It's a twelfth of the whole pizza. However, the quicker way is to simply realize that you can perform the operation 1 4th times 1 3rd equals 1 12th. Which is a sort of standard approach for these kinds of word problems. What's 1 3rd of 1 4th? It's 1 4th times 1 3rd. And so this will also give us an answer of 1 12th. And thus our final answer is E. Problem number three. A thread with a length of 10 centimeters is folded into equal parts as shown in the figure. The thread is cut at the two marked places. What are the lengths of the three parts? This problem might be a little tricky to think about at first, because you have to wonder how are we going to divide up this thread? How are we going to compare the lengths of these threads? Like are we just going to look at these vertical portions? How are we going to consider these curvy portions here? And so the way that I went about it was I marked endpoints with the following circles. So we kind of break up the string into multiple S shapes. Again, every circle is an endpoint. So we have 1 2 3 S's on the first portion of this string before this first cut. And then we have 1 2 3 4 5 portions on the second cut. So 3 and 5. And then finally we have 1 and 2 of these portions on the right side. So marking down all of these we see we have a ratio of 2 portions to 3 portions to 5 portions. And this is consistent with answer choice A, which is 2 centimeters, 3 centimeters, and 5 centimeters. So that's our final answer. Problem number 4. On Lisa's refrigerator, 8 strong magnets, the black circles in the picture on the left, hold some postcards. What is the largest number of magnets that she can remove so that no postcard falls to the ground? Looking at the picture up close, we want to look at magnets that we can remove. And in particular, we would like to avoid removing magnets which hold multiple postcards together. So for example, if we just remove this magnet on the right side here, then both of these postcards will stay up. But if we remove this magnet, which is holding two of them together, then this postcard with the flowers is going to fall down. But anyway, moving along, we see that we can remove this this magnet and both of these postcards will be fine. So this is one that we can eliminate. And then looking here, we can eliminate this one as well since the postcard will still be hanging by this magnet, which holds two of them together. And of course, we can't remove this because this postcard will drop. We also can't remove this magnet because the flower postcard will drop. Although we can remove this magnet since this postcard is being held up by this magnet. So we can remove this magnet and either one of these two magnets, but we need to leave at least one for this postcard to stay on the fridge. So marking all the ones that were allowed to remove, we see that four magnets can be taken off the fridge without a single postcard falling off. Thus, our final answer is C. Problem number five. Kathy draws a square with a side length of 10 centimeters. She joins the midpoints of the sides to make a smaller square. What is the area of the smaller square? Looking at this green square, what we can do first is we can take this square and break it up into four triangles of equal dimensions. And the side lengths of these triangles, we don't know this side length, but we do know that each of these side lengths are equal to one half of the larger side length and thus they are each equal to 5 centimeters since 5 is half of 10. Now if we want to find the area of such a triangle, we perform the operation just half of base times height, which is what you always use to find the area of a triangle. The height in this case is 5 centimeters, and the base is equal to 5 centimeters. So we'll have 5 times 5 times 1 half, which is equal to 25 halves when we multiply this out. And then since this square, this green square, is equal to 4 of these triangles with areas of 25 halves, we just take 25 halves and multiply by 4 and this should give us 50 centimeters. And that means our final answer is E, 50 centimeters squared. Problem number 6. Alice's mother wants to see a knife on the right side of each plate and a fork on the left side. How many interchanges of a knife and a fork does Alice need to make in order to please her mother? Looking at the picture, we can see that there are a few pieces that are out of place. The fork on the right side, the knife on the left side, and then both of the utensils on this plate are all out of place. So we have four utensils total out of place. We cannot make all of them go into the correct place with a single move, so we're going to need more than one. Now one way that we could try switching these utensils is by taking this knife and this fork and swapping them. So we'll have a fork on the left and a knife on the right and we're going to have a knife on the right over here. And then the other switch will be this knife on the left with this fork on the right. Once we make both of these switches, all of the plates will have a fork on the left and a knife on the right. So we know that we need more than one switch, but two switches is enough. And therefore our final answer is B, 2. Problem number 7. A centipede has 25 pairs of shoes. It needs one shoe for each of its 100 feet. How many more shoes does the centipede need to buy? This problem is, I think, just checking to make sure you're not reading too fast. In particular, it says that there are 25 pairs of shoes rather than 25 shoes. So if you read this too quickly, you might think, oh 100 feet, 25 shoes, 100 minus 25 is 75, done. But, you know, that's not correct because there are 25 pairs. So be sure to read carefully when you're actually taking the exam because we see now that 100 minus 25 pairs, or 25 times 2, equals 100 minus 50, which equals 50. And that's actually the number of shoes that the centipede needs. Again, don't try rushing through questions on the test, even if they look easy. Because even though this is just a simple subtraction problem, basically, if you try going through it too quickly, you can miss a small word like pairs, which completely changes your calculation. So again, our final answer is D. Problem number 8. Tom and John are building rectangular boxes using the same number of identical cubes. Tom's box looks like this. The first level of John's box looks like this. How many levels will John's box have? First, we want to find out how many cubes Tom and John are each working with. And since we see Tom's entire box, we are able to calculate how many cubes there are total. We see that there's a height of 2 cubes, a depth of 2 cubes, and a length of 6 cubes. So multiplying these dimensions out, we get 2 times 2 times 6 is equal to 24. Now that means that we need John's box to have 24 cubes as well. And looking at his base, we have 3 cubes times 2 cubes equals 6. So the number of levels that are going to be on this box needs to be the number such that 6 times this number is equal to 24. And in that case, based on what we know about our multiplication tables, that's going to be 4 layers. So again, we have 2 times 3 equals 6. And then we have 6 times 4 layers gives us 24. And so our final answer is C. Problem number 9. In the room shown in the figure to the right, there are four beds with pillows placed as shown by the dark ovals. A girl is sleeping in each of the beds, either on her right side or on her left side. On the left side of the room, Bea and Pia are sleeping with their heads on their pillows and facing each other. On the right side of the room, Mary and Karen are sleeping with their heads on their pillows and with their backs to each other. How many girls are sleeping lying on their right side? Going through this problem, let's start by looking at the left side of the room. And keep in mind, we're looking at the room from above. Since the two girls are facing each other, that means that the girl in this top bed is facing this direction, down. And this girl is facing in the upward direction. Now, if we try to imagine what it's like to be in these positions, looking at them from above, we can see that this girl is sleeping on her right side, and this girl is sleeping on her left side. Now, moving over to the right side of the room, looking from above, they're facing away from each other. This girl is facing upward, so she's sleeping on her right side. And this girl is facing downward, so she's sleeping on her left side. So she's sleeping on her left side. So if we want to mark which girls are sleeping on their right side, we would circle the top two beds, and that gives us a final answer of C. Problem number 10. The piece of paper shown on the left is folded along the dotted lines to make an open box. The box is put on a table with the top open. Which face is at the bottom of the box? For this problem, if you have difficulty visualizing how the box folds, it might be helpful to actually write out the letters on a strip of paper that is shaped in the same way, and fold it in the same way as they describe in the problem. However, if you can imagine this all on your own, that's fine, too. But if you're really not convinced by this explanation, go ahead and try folding the box for yourself. Now, we can imagine that the side with A on it folds up 90 degrees. And then along these dotted lines, we have E folding up 90 degrees, and then D folding up towards C. And then C folding up towards B. And so, if we're looking at this from above, B is on the bottom of the box, A is on the left face, C is on this bottom face, D will be on the right face, and E will be on this top face. And then the top of the box will be open, and at the base, we will have the letter B. Again, if this is difficult to imagine, feel free to actually try to fold a box in this way with the top open. And once you've convinced yourself, then you will know that the final answer is B. Problem number 11. Which of the following figures cannot be formed by gluing these two identical squares of paper together? Looking at each of these shapes, it's not immediately obvious which one is not possible. Although a good hint is noticing that A and E both look kind of similar, in that you have a square, and then on top of it, or in this case on the bottom of it, this is rotated 180 degrees. On top of it, you place another square tilted on a 45 degree angle. So if we ask ourselves, which of these makes more sense, E or A? Well, remember that these two squares are identical, which means that the diagonal length of one of these squares will be longer than the side length of one of these squares, since their side lengths are the same. So again, this diagonal should be longer than this side length. And this length right here is both the side length of the square and a diagonal length of a square, since we have this square on a 45 degree angle. But this doesn't make any sense. The square that's on its angle should be sticking out more. And then when we look at answer choice E, we see that this is sort of the more logical, more correct version of what we're seeing with answer choice A, because the diagonal here is clearly longer than the side length of the square. So looking at answer choice A, we see that it cannot be possible because this diagonal is too short. So that is our final answer. Problem number 12. Mary, Anne, and Nata work at a kindergarten. Each day from Monday to Friday, exactly two of them come to work. Mary works three days per week, and Anne works four days per week. How many days per week does Nata work? First, let's write out the five days of the week. We're just gonna number them one through five, and try to fill in some days for Mary and for Anne. So Anne works four days per week, and Mary works three days per week. Notice that I did not put these three Ms over, for example, on days one, two, and three, because then day five wouldn't have Mary or Anne present, and we need two people present on each day. So I could have just as easily put an M at one, at two, and at five, because then we could have two people in each day. But we can't have, we need to have at least one person already present in day five if we want to fill this in correctly. Otherwise, we can pretty much organize the As and Ms however we want. Now, if we try filling in the blank spaces, we see that there's one, two, three spaces left. And so our final answer is C. Problem number 13. Five squirrels, A, B, C, D, and E, are sitting on the line. They are going to pick up the six nuts, each marked with an X. At the same moment, each of the squirrels starts running to the nearest nut at the same speed. As soon as a squirrel picks up a nut, it starts running to the next closest nut. Which squirrel will get two nuts? Here, we're going to check the path that each squirrel is going to take and see how many steps it takes to collect two nuts. So in the case of squirrel A, we see that there's going to be one step and then two, three, four to get this nut. However, squirrel B is going to catch this nut, which is the closest, in one, two, three. And three steps is less than four. So squirrel A is not going to catch this second nut. Squirrel B, we don't have to worry about either because going all the way to catch this nut and then going all the way to catch this nut, which is right next to C, is just not realistic. So we know that answer choice A and answer choice B can be eliminated. Now we look at answer choice C. It runs first one step and collects a nut and then two, three, four, and it would collect a second nut. Now looking at squirrel D, we're going to have squirrel D run one to get this nut, which is the closest, and then two, three, four, five to get this nut. But squirrel C can collect both of these nuts in four steps, whereas D needs five steps. So squirrel C is going to be able to capture both of these nuts. Now if we check answer choice E, well, we're going to pick up the first nut with this squirrel but then this nut will already be taken by squirrel D. So squirrel E and squirrel D are both also out of the running and squirrel C is going to be the one that captures two nuts. Thus our final answer is C. Problem number 14. There are 30 students in a class. They sit in pairs so that each boy is sitting with a girl and exactly half of the girls are sitting with a boy. How many boys are there in the class? This problem tells us that there are twice as many girls as there are boys, since half the girls are sitting with a boy but every boy is sitting with a girl. In other words, the number of girls is equal to two times the number of boys. And now what we're going to do next is we're going to note that the number of boys plus the number of girls equals 30. But since girls is equal to two times boys, we can replace girls here with two times boys. So boys plus two times boys is just going to be equal to three times the number of boys. Since three times the number of boys is equal to 30, we divide both sides of the equation by three and this gives us that the number of boys is equal to 10. So our final answer is D. Problem number 15, the number 2581953764 is written on a strip of paper. John cuts the strip two times and gets three numbers. Then he adds these three numbers. Which is the smallest possible sum he can get? Now, looking at our answer choices and the fact that we need to split this set of numbers into three numbers, we know that we can't leave ourselves with, for instance, a five-digit number because all of our answer choices are only four digits. And if we have a five-digit number, then our sum is going to be bigger than 10,000. So our goal is to split this strip of paper into one four-digit number and a couple of three-digit numbers. Or perhaps two four-digit numbers and a two-digit number, if we want to try it that way. Now, our most natural attempt at trying to cut up the numbers is by trying to minimize the four-digit number, since that's the biggest number that we're working with. And we do this by making the thousands digit as small as possible, which in this case would happen if the thousands digit were a one, since one is the smallest number available here. Now, if we were to cut out this four-digit number, 1953, 1953, and leave ourselves with these two three-digit numbers, then their sum would be equal to 2,975. If you go ahead and add it up, this is what we will get. Now, if you were to try, for instance, cutting out 2,581, you would be left with 953 and 764, which is going to be much more than 1,000. So you're adding 2,500 something with something that's more than 1,000. And that means that your answer is going to be bigger than 3,000. So that's the other natural attempt, because you have a four-digit number that's starting with a low thousands digit, the number two. And if our four-digit number starts with anything bigger, three or more, then that's obviously going to be higher than answer choice B, because B is less than 3,000. So we conclude that our final answer must be B. Problem number 16, Bart is getting his hair cut. When he looks in a mirror at the reflection of the clock behind him, the clock looks like this. What would he have seen if he had looked in the mirror 10 minutes earlier? Now, we can draw out a picture of the clock as it would look reflected in a mirror. And in particular, when you look in a mirror, the left and right side of an image switch. So instead of having 12, one, two, three, and so on, it actually goes counterclockwise, 12, one, two, three, like so. So looking at the minute hand and the hour hand on the clock, if we were to go backwards 10 minutes, the hand would move from the three to the one. And so our shape would look like that of answer choice E. And so that is our final answer. Problem number 17, grandmother bought enough cat food for her four cats to last for 12 days. On her way home, she picked up two more cats from the shelter. If she gives each cat the same amount of food every day, how many days will the cat food last? First, we check how much food there actually is for the cats. And we can find this by multiplying four cats by 12 days. And what we get, four times 12 equals 48. This tells us that there are 48 portions of cat food to be distributed. And if we have six cats to distribute these 48 portions to, well, then we perform the operation 48 divided by six. And this gives us eight, which is the number of days that the cats will be able to last if you split the portions among six cats. So our final answer is A. Problem number 18, each letter in Benjamin represents one of the digits one, two, three, four, five, six, or seven. Different letters represent different digits. The number Benjamin is odd and divisible by three. Which digit corresponds to N? To solve this problem, we first need to remember the fact that for a number to be odd, it's ones digit must be odd, which means that it's going to be either one, three, five, or seven in this case, since nine is not an option. And for a number to be divisible by three, the sum of all of its digits needs to also be divisible by three. So with this in mind, we're going to rewrite the digits of Benjamin as follows. So we're just pairing the two ends together. And if we count how many letters there are, we see that there are seven different letters, which means if we were just to take all of the letters and add them together without one of the Ns, we would actually have one plus two plus three plus four plus five plus six plus seven is equal to 28. Again, this is if we are excluding one of the Ns, then we have seven different numbers that we're adding together. So what we can gather from this is that the second N that we're adding has to make this number divisible by three once we add it. And furthermore, this number that we're adding needs to be somewhere from one to seven. So let's consider, which of these numbers can we add an extra one of so that 28, once we add that number, will be divisible by three? And looking at this, we see that we can add two, and that'll give us 30, or we can add five, which will give us 33. However, we can't use two as the N because N is in the ones digit of Benjamin, and we are told that this is an odd number, which means the first digit needs to be odd, or the ones digit needs to be odd. So from this, we conclude the letter N is represented by the digit five. So our final answer is D. Problem number 19, Tim, Tom, and Jim are triplets while their brother Carl is three years younger. Which of the following numbers could be the sum of the ages of the four brothers? First, let's go ahead and define A as the age of any of the triplets. Now, if we add together all four of the ages that are being considered, this is going to be equal to three times A because we have three triplets, plus A minus three, since Carl's age is three years less than A. And if we go ahead and group these terms together, we have three times A plus A, this is going to be equal to four times A, and then we're gonna have this minus three. So what we have now is the sum of the four ages of the brothers, and what we see is that it's going to be a four-digit number minus three. So one of these answer choices has to be a four-digit number minus three, and that's going to be our correct answer. Looking at the answer choices, we see that 53 is actually equal to 56 minus three, and 56 is in fact divisible by four. We easily know this because 60 is divisible by four, and 60 minus four is equal to 56. So again, since 53 is a number divisible by four minus three, our final answer is A. Problem number 20. The perimeter of the rectangle ABCD is 30 centimeters. Three other rectangles are placed so that their centers are at the points A, B, and D. See the figure. The sum of their perimeters is 20 centimeters. What is the total length of the thick line? When we actually look at the figure, we need to realize that what's actually being added from these three new rectangles on the vertices is half of their total perimeter, and here's why. Since each of the rectangles is centered at one of the vertices, when we compare a point, or rather a segment, like this one right here, going from the center over to the right, this segment is equal in length to the new segment that's being added by this rectangle at vertex A, and similarly, this segment is being replaced by this segment, so these segments cancel each other out. For instance, these two segments that are circled cancel each other out, and when we take this into account, we see that all of this portion of the rectangle and all of this portion of the rectangle are canceled, and so we are left with two sides of the rectangle at vertex A, and this means that we are actually only left with half of the perimeter of A, and we can apply similar logic to the perimeter of the rectangle at D and the rectangle over by B. Each of these perimeters are being partially canceled out so that only two sides are remaining, and thus half of the perimeter is remaining, so we have 30 centimeters, the original perimeter of the rectangle, plus half of the perimeters of A, B, and D, these extra rectangles, which is half of 20, so 30 plus half of 20 is equal to 30 plus 10, which is equal to 40, so our final answer is C. Problem number 21, Anna folds a round sheet of paper along the middle line, then she folds it once more, and then one last time. In the end, Anna cuts the folded paper along the marked line. What is the shape of the middle part of the paper when unfolded? This is another problem that tests your visualization, and again, if you're not convinced by the explanation, I strongly recommend that you actually cut out a circle, follow the instructions that are given in the problem, and cut out a section just like the problem instructs you to, and see that it actually all checks out. If you're having difficulty visualizing this, the best way to do it is to actually try it in real life. Now, what happens when we cut a portion as described in the problem, is that when we unfold the portion after something is cut out, the cutout is reflected along a line of symmetry, and this is demonstrated as follows. At first, we have this cut, and this cut is reflected over by this dotted line when we unfold it, so we have a part missing here, and a part missing here. Now, if we fold it out one more time to get this figure, we have this portion being reflected over to this side, so we have this big triangle here, and then this cutout is mirrored by this cutout, and finally, once we unfold and are left with our original sheet of paper, we see that this big triangle is reflected down here, and the two smaller triangles on the left and right are reflected and made into bigger triangles, so our shape should look like that of figure D. Again, if this is unclear, please go ahead and find a circular sheet of paper and follow the instructions and actually cut it out. You will find that the cutouts are reflected along the folded lines, so again, our final answer is D. Problem number 22. Richard writes down all of the numbers with the following properties. The first digit is one. Each of the following digits is at least as big as the one before it, and the sum of the digits is five. How many numbers does he write? Fortunately for this problem, since all of the digits have to add up to five, we can actually just brute force this problem. We can list out all of the different ways that we can add up numbers to get five, such that the numbers are in numerical order. So if we make a list, it would look like this. Of course we can add five ones. We can add three ones and a two, a one and two twos, two ones and a three, or a one and a four. And again, we need to make sure that all of these numbers, all of the digits are in numerical order, so we can't, for example, have one, two, one, one, because that goes against the rules given in the problem. The trick here is to not go too quickly through the problem and make sure you consider all of the possibilities for all of the numbers that are less than five. For example, some people might think that they can just start with straight ones, five ones, and then get rid of the last digit, and then replace the second to last digit with a two, and then get rid of the two, and then replace it with a three, and then get rid of the three, and then replace the second digit with a four. And by repeating this process, you'll have four numbers, but this process doesn't give you 122. So the trick in this problem is to actually consider all of the possibilities and not speed through it. But because five is such a small number, it is relatively easy to make a full comprehensive list. So these are all the possible numbers that follow the rules of the problem, and thus our final answer is B. Problem number 23. What is the greatest number of shapes of the following form that can be cut out from a five-by-five square? First of all, note that this figure has one, two, three, four little squares, and a five-by-five square, five times five, is equal to 25. So how many times can we fit four into 25? Well, the answer to that is six, since four times six is equal to 24, which is just slightly below 25. So in the best-case scenario, we're going to be able to fit in six such shapes. Now, we actually need to work through the problem and see how we can fill in the boxes to make sure that there isn't some problem with this shape. If there's a problem with the shape that prevents us from filling in six of them into the box, then that's going to be an issue. So we actually have to double-check and make sure the math works out. We can fill in the box as follows. I'm sure that there's more than one possibility for filling in this box but this is the solution that I came up with and In fact, we do see that there's one two, three, four five six Different figures of this form that are fitting inside the box Also notice that I am NOT ever flipping this form. I am only rotating it 90 degrees, 180 degrees, 270, and so on. I'm never flipping it horizontally or vertically Because the problem doesn't actually say that you're allowed to do that. So all of these are rotations, which is something to look out for Anyway, we've confirmed that we're able to fit six of these shapes inside the box Which is clearly the maximum since we are only left with one square afterwards. So our final answer is D Problem number 24 Luigi started a small restaurant. His friend Giacomo gave him some square tables and chairs. If he uses all the tables as single tables With four chairs each, he will need six more chairs. If he uses all the tables as double tables with six chairs each He will have four chairs left over. How many tables did Luigi get from Giacomo? From this question we can write down two equations and I'm going to write them down first and then explain what they mean First of all C is equal to the number of chairs that we have and T is equal to the number of tables that we have and Now I'm going to explain this notation here since I learned what it meant when I was in sixth grade And I do expect that there are some fifth graders who are going to be using these videos And so it would be a little unreasonable to Expect them to know this and this is just a quicker way to write things down. So The number four here is a constant it's a number and we know what it is the letter T here represents some number in Particular the number of tables, but we don't know what this number is So we call this a variable again four is a constant because we know what it is and T is a variable because we don't know what it is and You can write down a constant and a variable next to each other So in this case 4T and this is exactly the same as saying 4 times T so 4T and 4 times T are the same and Why do we write it this way? Well because it's shorter and it's quicker and Again, I didn't know about this until I was in sixth grade. So I thought I would explain that first Now, what do these equations actually mean? C equals 4T minus 6 that means that the number of chairs is equal to 4 times the number of tables minus 6 Because we will need 6 more chairs if we assign 4 chairs to each table So therefore almost 4 times as many chairs as there are tables. There's actually 6 less than that so that's what this equation stands for and Now this equation says that the number of chairs is also equal to 6 times T over 2 plus 4 and this is because We have 6 chairs at every double table where a double table is 2 tables put together So since every double table is 2 tables put together The number of those is half of the number of tables that you have. So if you have 8 tables you have 4 double tables So that's why we have 6 times T over 2 and Since we will have 4 chairs left over in this case. We're having this plus 4 here So the number of chairs equals these two different terms Now our next step is going to be to notice that both of these terms are equal to C Which means that both of these terms 4T minus 6 and 6 times T over 2 plus 4 They have to be equal to each other since they're both equal to C and if we set them equal to each other Then we can work it out as follows. First of all 6 times T over 2 is Exactly the same as just 3 times T because it's 6 times T times 1 half and 6 times 1 half is 3 So this top term here is 4T minus 6 and this bottom term here is 3T plus 4 Now we want to find out what T equals since that's the number of tables so what we want to do is we want to get all of the T's on one side of the equation and all of the constants and other terms everything that isn't a T on the other side of the equation and In this case we add 6 to both sides of the equation that cancels out this part and it turns this into a 10 and Because we're doing the same thing on both sides of the equation that stays equal, you know if this is equal to this then this plus 6 is equal to this plus 6 and We do the same thing for the T's we subtract 3T from both sides that cancels out this 3T And it gives us 4T minus 3T is just equal to T so subtract 3T from both sides and add 6 to both sides and we get T equals 10 So our final answer is B Problem number 25 Clara wants to construct a big triangle using identical small triangular tiles She has already put some tiles together as shown in the picture What is the smallest number of tiles she needs to complete a triangle? Looking at this my first instinct was actually to try to build this bigger triangle over here where we fill in this as the base of the triangle and work our way towards the Top or the tip of the triangle this top vertex, which would be somewhere over here but then There's actually a quicker way to do this with fewer triangles we start by filling in three triangular tiles over here and if you tilt this on an angle we can consider this the top of the triangle and then we can fill in the rest of the base here and That's going to be a total of nine triangular tiles There's no way that you can do this with five tiles you can try but you'll find that you will need a lot more than five and A lot more than nine if you want to try to do this any other way So our final answer is B Problem number 26 a big cube was built from eight identical small cubes some black ones and some white ones Five faces of the big cube are as follows What does the sixth face of a big cube look like? this problem is actually a little bit tricky because You don't need to think about what the cube looks like But instead you need to notice a very critical fact about which faces are black and which faces are white in a 2 by 2 by 2 cube You should for on each of the cubes be able to see three of the faces So if you have one black cube inside this big 2 by 2 by 2 cube Three squares should be black when you add up all the faces If you look at a Rubik's Cube and look at each of the pieces on the corners of the cube or the vertices of the Cube you will see that there are three stickers on them the same logic applies here so what we really need is for the total number of black squares to be divisible by three and Also the number of white squares to be divisible by three so if we go ahead and Look at all these options first of all we can obviously eliminate C and E just because we we can't even begin to imagine how we could have So much black on one face and have so little black on any of these faces none of these faces are completely black which means that it would be impossible for C to be the case and For similar reasons E would not be possible either. So these can just be eliminated sort of by common sense But now we go back to the fact that the total number of black squares that we see Needs to be divisible by three. So we add these one two, three, four five six Now if we add one more face We want to make sure that this number will be divisible by three well six plus two is eight which isn't divisible by three six plus one is seven, which isn't divisible by three and Six plus zero is Equal to six which of course is divisible by three So again because corner pieces have three of their faces visible We should have a multiple of three equal to the number of black squares that we have and also to the number of white squares That we have so if you'd like to double-check this we can actually do this right now One two, three, four five six seven eight nine ten eleven twelve thirteen fourteen Fifteen sixteen seventeen eighteen eighteen is also divisible by three six times three is eighteen so This absolutely confirms that D is going to be our final answer since it satisfies this property Problem number 27 Kristen wrote numbers in five of the ten circles as shown in the figure She wants to write a number in each of the remaining five circles such that the sums of the three numbers along each side of The Pentagon are equal which number will she have to write in the circle marked by X? Notice first of all that This circle here is shared by two sides The side with the three and the seven and the side with the one now we don't know what this circle should contain, but we know that three plus seven is equal to ten and Ten plus this circle equals something and that something needs to equal one plus this circle plus this circle since these two circles plus this circle and These two circles with the three and seven plus this circle should be equal We can conclude that three plus seven needs to be equal to one plus whatever goes in here three plus seven is equal to ten and One plus nine is equal to ten. So we know that this needs to be a nine Because again, this circle is shared. So we need these two to equal these two Now we apply a similar logic here since this circle is shared by 2 and X and 6 and 9 we have 6 plus 9 equals 15 and 15 minus 2 is equal to 13 So even though we don't know what we have in this circle We know that X needs to be equal to 13 since 13 plus 2 equals 6 plus 9 Therefore our final answer is D. Problem number 28 The symbols circle square and triangle represent three different digits If you add the digits of the three digit number circle square circle The result is the two digit number square triangle If you add the digits of the two digit number square triangle you get the one digit number square Which digit does circle represent? First of all notice from the problem that square plus triangle is equal to square that means that triangle must be equal to zero and Furthermore notice that since each of these are one digit numbers Well consider the three biggest one digit numbers 9 9 and 9 just 9 plus 9 plus 9 would be equal to 27 so if you add the three largest possible one digit numbers the tens digit is going to be a 2 and otherwise the tens digit is going to be a 1 so square is either going to be equal to 2 or 1 and Triangle is definitely equal to 0 since square plus triangle equals square as given in the problem Thus we have 10 or 20 from square triangle and now we want to consider What number? Rather how will it work when you try adding two of the same number to square to get either 10 or 20 in? particular 1 plus 2 circles equals 10 and 2 Plus 2 circles is equal to 20. What are the two circles equal to in each case? Well, if we write it out, we want these two numbers to be equal 2 plus 9 plus 9 So the circles are each 9 is equal to 20 Whereas 1 plus 9 halves plus 9 halves is equal to 10 But 9 halves is not a single digit number. So we shouldn't be thinking about this This isn't a single digit. So circle cannot be represented by this that means that the square must be equal to 2 and that the circles must be equal to 9 and Therefore our final answer is e. Problem number 29 a little kangaroo is playing with his calculator He starts with the number 12. He multiplies or divides the number by 2 or 3 if possible 60 times total What which of the following results can he not obtain? Well, let's go ahead and just look at these answer choices one by one starting with answer choice a 12 Now if we want to get 12, all we have to do is Multiply by 2 and then divide by 2 over and over again Since times 2 and then divided by 2 is an even number of moves and we could just repeat these moves over and over again 60 times total will just be left with 12 So we can eliminate answer choice 12 since it's very easy to come up with a way to get this number Of course, we could also do times 3 divided by 3 times 3 divided by 3, but that's that's just another example Now moving on to answer choice B To get to answer choice B. We can find 12 divided by 2 Times 3 12 divided by 2 is 6 6 times 3 is 18 And then we do a similar thing of times 2 divided by 2 times 2 over and over again And we'll be left with 18 so we can eliminate answer choice B Now we're gonna move on to answer choice D just to show why it works For 72 what we can do is we can multiply 12 by 2 and then by 3 12 times 2 times 3 is equal to 12 times 6 which is equal to 72 and Then 72 times 2 divided by 2 and so on repeating this trick of times 2 divided by 2 over and over again will give us 72 at the end and This is keep in mind because we have an even number of inputs left over so This is input number 1 then input number 2 We have 58 more inputs and 58 is an even number So we have times 2 divided by 2 over and over again if we would have an odd number of inputs we would have a problem because times 2 divided by 2 times 2 divided by 2 we won't have a Divided by 2 to cancel out every times 2 we're gonna have an extra times 2 at the end If we were to have an odd number of inputs here Now if we move on to answer choice e 108 That's going to be equal to 12 times 3 times 3 or 12 times 9 Which is 108 and then we repeat the same process as usual now looking at 36 the problem with this is that 12 times 3 gives us 36 but then we have 59 more inputs and 59 is an odd number which means we can't use this repeating trick of times 2 divided by 2 times 2 divided by 2 and This is why we know that our final answer must be C Problem number 30 two three-digit numbers are made using six different digits The first digit of the second number is twice the last digit of the first number What is the smallest possible sum of two such numbers? So what we're asking for here is ABC plus DEF where each of the letters is a different digit and C times 2 is equal to D. So for instance if we have C is equal to 4 D will be equal to 8 Now looking at our answer choices. We quickly see that all of them are less than 600 Which means that we need to have D is less than 6 Because you know if D is 6 Then we're definitely gonna have something more than 600 and that's not going to be in our answer choices. So that's clearly too large So C equals 3 would be too big because D would equal 6 so we should consider two cases one where C equals 2 and D equals 4 and one where C equals 1 and D equals 2 and We will go ahead and set up the rest of the digits so that we get as small of a sum as possible and That will determine what what we have in each case So for instance assume that C is equal to 2 and D is equal to 4 in This case we have a b 2 plus 4 e f and then if we fill in these earlier digits to minimize the sum Well, the smallest number we can have in the hundreds digit is going to be 1 so we're gonna fill in a 1 for 102 here and For B, we could put in 0 or 3 and then for E we could put in 3 or 0 It doesn't matter where we put each one since they add up to the same thing. That's it's not gonna make a difference and Then F is going to be equal to the next smallest number that's left over which is going to be 5 so we get 102 plus 435 or 132 plus 405 either one is the same and that's going to give us 537 Now again, we minimized this sum by making the bigger digits like the hundreds digits and the tens digit as small as possible Now what if we have C equal to 1 and D equal to 2? well in this case we can have the smallest possible hundreds digit as being the number 3 and again, we're gonna fill in a 0 for B a 4 for E and a 5 for F of Course the E and B can be switched just like before so 301 plus 245 gives us 546 and somewhat surprisingly The case with 1 and 2 actually give us a bigger sum than the case with 2 and 4 But I think this is because we have the availability of a 1 for the hundreds digit here Whereas we have a 3 for the hundreds digit here So even though this might appear bigger it does make sense that the sum can be smaller since we're allowed to have a much smaller hundreds digit here 1 instead of 3 So we compare these two cases and we see that 537 is the smaller of the two sums and Therefore our final answer is E

Math Kangaroo 2016

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