Hello, and welcome to the Math Kangaroo 2017 Levels 5 and 6 Solutions video series. First of all, I'd like to congratulate you for participating in the Math Kangaroo. It's a super difficult mathematical challenge, which I participated in myself throughout all of elementary school and high school, so I'm very familiar with the difficulty of the task that you brought upon yourself, and you should be proud of yourself for challenging yourself intellectually this way. Now the purpose of this series is to show you a path to get to the correct answer of every single question on the exam. Now keep in mind, sometimes you might use a different method to get the answer, but you might still get the correct answer. In this case, maybe your method is equally good, or maybe you just got lucky and your method only happens to work for that one problem. If you have any questions on whether your method is equally good or not, or if there's any particular issues with it, please feel free to email me at thomas.mathkangaroo.org, the email listed below, and if you have any other questions about the exam, please do not hesitate to get in touch with me. Now before we actually move on to the videos, I'd like to offer a couple of pieces of advice. The first is, before watching a video for a given problem, try solving it on your own, because if you just watch a video without trying to solve the problem, you'll have the information spoon-fed to you, and you'll listen to all of it, and you'll go, oh, that makes sense, I understand this, I understand the approach that he took, and I know how to do it now, but you won't have actually learned anything. You don't really learn from these kinds of videos unless you struggle with the problem first. You understand why it's difficult, you try a few things out on your own, and you find out from the video where you went wrong, and it helps you improve as a mathematician. But if you just go ahead and watch the video first without trying, you're not going to learn nearly as much. So that's the first piece of advice. Actually try a problem before watching the video for it. The second piece of advice I have to offer, and I can't stress this enough, please, please, please write out as much of your work as possible when working on these problems, because all too often, people who are good at math, the kinds of people who tend to take the math kangaroo exam, they think that they could just keep everything in their heads because it's a little bit faster, because they don't have to write everything down on paper. That's just not true. If you don't write down what you're thinking and what you're working through, it's very easy to lose track of your thoughts. It's very easy to misremember something, to drop a negative sign somewhere, or to make some other silly little mistake. It is much better to write out all of your work, so that way you don't have to constantly be remembering everything and juggling everything in your head. You'll find that you'll have a lot more headspace to work with, and it'll be a lot easier. Also, if you happen to get a problem wrong, and you don't write down your work, you won't know why you got it wrong necessarily. So it's a lot easier to keep track of your previous thoughts if you actually write down all of your thoughts on paper. So those are my pieces of advice to get the best use out of these videos. Again, if you have any other questions, please feel free to get in touch with me. Good luck, and have fun with the series. Problem number one. Four cards lie in a row. Which row of cards can you not obtain if you can only swap two cards? So the cards say 2017. All you have to do is look at each of the cards and read 2017, and mark whichever cards are not in the correct place. If you swap two cards, then exactly two of the cards should be in the wrong location. So if we highlight that, we see that most of them have only two cards swapped with the exception of answer choice B, which has three cards in the wrong location. So that's going to be our final answer, B. Problem number two. A fly has six legs and a spider has eight legs. Together, three flies and two spiders have as many legs as nine chickens, and the question is asking for how many cats. So chickens have two legs apiece, cats have four legs apiece. Let's start by figuring out how many legs three flies and two spiders have. This is a simple matter of multiplication. We have three flies times six legs per fly, plus two spiders times eight legs per spider, which is equal to 18 plus 16, which equals 34. So 34 is supposed to be equal to the number of legs that nine chickens has, nine times two, plus some number of cats times four legs per cat. So 34 is going to be equal to 18, that's nine times two, plus something times four. Well, in fact, we see here that it's 18 plus something times four, and 18 plus 16. These are both equal to each other. So this something times four is supposed to be equal to 16. And again, that's just because 34 minus 18 is equal to 16. So this term right here is equal to 16. What times four is equal to 16? Well, that's going to be four. So we need four cats, which means our final answer is C. Problem number three, Alice has four pieces of this shape. Which picture can she not make from these four pieces? So if we go ahead and go through all of the answers, part A, answer choice A, we can clearly see that there would be three pieces here, three pieces here, three pieces here, three pieces here, or rather like one piece here made of three squares, one piece, one piece, one piece and one piece. So clearly answer A can be made. Answer B is just these four pieces lined up next to each other. So that obviously works. We have one piece here, one piece here, one vertical piece here, and one piece here. So this one obviously works. Answer choice D, we have one piece here, one here, one here, and then one here. So D also works. So that leaves us with answer choice E. And answer choice E doesn't work because, well, we have one piece highlighted here, one piece highlighted here, and one piece highlighted here. And it leaves us with these three extra squares here, which will never line up no matter how we arrange these three other pieces. You can try moving this one to the right by one, or this one to the right by one, or this one to the left by one, but you'll never be able to get these three squares to line up. So E cannot be made with just four pieces of this nature. So our final answer is E. Number four, Cayley knows that 1,111 times 1,111 is equal to 1,234,321. How much is 1,111 times 2,222? Now there are two reasons why I read that number as 1,234,321. And one of them is that I just didn't know whether it was thousands or millions or whatever because I just started reading the problem. And the other is that you don't really need to think about the number as a whole. And here's why. They already did the hard part for you. They did the 1,111 squared. All you have to do now is make a simple observation, namely that 1,111 times 2,222, you can just go ahead and pull out a two from this. So it's equal to 1,111 times 1,111 times two. It's exactly the same as this expression right here. So all you're doing is multiplying this number by two. And if you want to multiply this number by two, all you have to do is double each of the individual digits because it's the same as adding this number to itself. So each of the digits is going to be doubled if you multiply this number by two, since they're all less than five. If there were numbers in this number that were five or greater, it would be a little trickier to multiply it. But because all of these numbers are so low, you can double each of the digits and you end up getting 2,468,642, which appears to line up with answer choice D. So that is our final answer. Problem number five. On a planet, there are 10 islands and 12 bridges. All bridges are open for traffic right now. What is the smallest number of bridges that must be closed in order to stop all traffic between A and B? So notice here that when we're trying to go from A down this route over to B, there's the single path that you take and then it breaks into two other paths. So it doesn't really make much sense to cut off one of these paths because it's just half of the route. That is, if we cut off this bridge or this bridge, it's like we're eliminating both of these connections between A and B. It's kind of like a two for one deal. So if you actually look at A and B, you see that there are three bridges coming out of both of them. In the case of A, it's this one breaking into two, and then there are these two paths which then connect into one. So similarly, as we get rid of one of the bridges from this route for A, we get rid of one of these bridges for B, and that cuts off both of these routes to A. So what you're left with is all you have to do is cut off one of the two bridges here, one of the two bridges here, and that removes all of the connections between A and B. So that means that our final answer is going to be two, which is B. Problem number six. Jane, Kate, and Lynn go for a walk. Jane walks up front, Kate walks in the middle, and Lynn walks behind. Jane weighs 500 kilograms more than Kate. Kate weighs 1,000 kilograms less than Lynn. Which of the following pictures shows Jane, Kate, and Lynn in the right order? So first of all, we know that the order is going to be Jane, Kate, Lynn, or J-K-L, just because that's quicker to write out. And they tell us that J is heavier than K by 500 kilograms. And then this next phrasing is a little bit weird. K is 1,000 less than Lynn. Well, what that means is that Lynn is 1,000 more than Kate, right? We can just flip that around in a more useful way. And if Lynn is 1,000 more than Kate, well, we know Jane is 500 more than Kate, so Lynn is also 500 heavier than Jane from this information. So we have Jane is heavier than Kate, and then Lynn is even heavier than Jane, who is heavier by Kate. So that leaves us with L is greater than J, which is greater than K. So L should be the biggest one, J should be the middle one, and K should be the smallest one. So rearranging that, we should have middle, smallest, largest. And that is consistent with answer choice A, middle, smallest, largest. So that's going to be our final answer. Problem number seven. A special die has a number on each face. The sums of the numbers on opposite faces are all equal. Five of the numbers are 5, 6, 9, 11, and 14. What number is on the sixth face? So in general, if you want to try finding out, well, our strategy here is to sort of pair numbers together until we find two pairs of numbers that add up to the same thing. And then we know that the third pair should also add up to that same thing. But how do we know which numbers to pair together? Well, logically, it doesn't make much sense to take two of the largest numbers to pair them up and two of the smallest numbers to pair them up, because you're clearly going to get different answers. What we want is, if you have a very small number, to pair it up with a very large number. And if you have a number that's closer to the middle, pair it with something that's close to the middle. That's sort of the way to think about it if you want to find the pairs as quickly as possible. And in this particular case, we see 9 and 11 add up to 20, and 6 and 14 add up to 20. So those pair up nicely. That leaves us with 5 adding to something, which equals 20. And that something is 15. So our final answer is E. Problem number 8. Martin wants to color the squares of the rectangle so that one third of all the squares are blue, and half of all the squares are yellow. The rest of the squares are to be red. How many squares will he color red? So for this problem, we could just color in individual squares, or we could try speeding through it a little bit more quickly by doing math. So we're going to try the quick approach. We have 3 rows by 1, 2, 3, 4, 5, 6 columns. 3 times 6 is equal to 18, so there's a total of 18 squares to color in here. Now we know that a third of them are going to be colored a certain way, so one third times 18 is 6. And half of them are going to be colored another way, one half times 18 equals 9. So we have 6, and we have 9 both being colored in. And the rest are going to be colored red. So it's just a matter of taking 18 and subtracting 6 and 9. If you add these together, you get 15, so it's 18 minus 15 is equal to 3. So our final answer is C. Problem number 9. In the time it takes Peter to solve 2 problems for the math kangaroo competition, Nick manages to solve 3 problems. In total, the boys solved 30 problems. How many more problems did Nick solve than Peter? So an observation to make here is that Nick solves 3 times as many problems as Peter, meaning the proportion between the problems they solve is 3 to 2. And that's just restating the information given in the first sentence of the problem. Now we're going to let X equal the number of problems that Peter solved. And we're going to find out how many Peter solved, how many Nick solved, and we're just going to take the difference. So think about it this way. We have X, which is the number of problems Peter solved, and then we have 3 halves times X, which is the number of problems Nick solved. So these are all the ones Peter solved, these are all the ones Nick solved, and if you add them together, you get 5 halves times the number of problems that Peter solved. And that's what this represents. And we know that this number is supposed to be equal to 30, since that's the total number of problems that the boys solved together. So all we have to do now is, if we want this equation to hold true, we can divide both sides by 5 halves, which is the same as multiplying by 2 fifths. That cancels out the 5 halves here and leaves us with X, and it leaves 2 fifths times 30, which is equal to 12 for the other side of the equation. So we get X equals 12. Now that's the number of problems that Peter solved. Remember, we want how many Nick solved. So Nick solved the rest of the 30 problems, and 30 minus 12 is equal to 18. You could also just say 3 halves times 12 is equal to 18. They both amount to the same thing. And 18 minus 12 is equal to 6. So Nick solved 6 more problems than Peter. And that leaves us with a final answer of B. Problem number 10. Bob folded a piece of paper, used a hole punch, and punched exactly one hole in the paper. The unfolded paper can be seen in the picture. Which of the following pictures shows the lines along which Bob folded the piece of paper? So for this kind of problem, you're basically looking for lines of symmetry. For most of these shapes, it's pretty obvious that there's no way you can unfold it along those lines and have the dots spread out this way. We're looking for something kind of diagonal. Of course, not in this sense, because if we would have a triangle-shaped piece of paper from the folds, it would fold out this way, and then this way, and then this way. Then we would have one dot here, one dot here, one dot here, one dot here. And they would be in four separate sides of the square, not in a single line. And this logic holds for most of them. For answer choice D, you can see if we would have a folded piece of paper that would be folded in half along this line, and then folded in half along one of these lines, whichever way you think about it, and then punch a hole in it, then a hole that would start here would transfer over here, over here, and over here, creating this diagonal line. So really, all you have to look for is lines of symmetry. And that's only present in answer choice D, since that's the only one with diagonals. And you can sort of imagine how the hole spreads. If you have difficulty visualizing this, the advice I always give is actually try to do it in real life. So you can very easily take a square sheet of paper, fold it up along these lines, and you'll see that it gets you the same result as they say in the problem. So if you ever have trouble visualizing, always try doing it in real life. But in this case, we know that our final answer must be D. Problem number 11. The modern furniture store is selling sofas, loveseats, and chairs made from identical modular pieces as shown in the picture, including the armrests. So the width of the sofa is 220 centimeters, and the width of the loveseat is 160 centimeters. What is the width of the chair? So all this is really saying is that the cushions are all of identical sizes, and the armrests are all identical to each other. So the sofa armrest, the loveseat armrest, and the chair armrest are all the same. So let's collect the information that we know. If we go ahead and subtract the length of the loveseat from the length of the sofa like so, 220 minus 160, that leaves us with 60. And what we're subtracting is the length of the armrests and two cushions, which if you subtract that stuff from the sofa, you're just left with one square cushion, and we see that that's equal to 60 centimeters. Now if we want to find out how big the armrests are, we can go ahead and either use the loveseat or the sofa. I happen to use the sofa this time around. You can take the full length and subtract the length of three such cushions, 3 times 60, and 220 minus 3 times 60 is like 220 minus 180, which is equal to 40. You could just as easily have used the love seat, 160 minus 120, which is two times 60. Either way works equally well. You find out that the armrests add a length of 40 centimeters. So you have a cushion, which is 60 centimeters, and you have the armrests, which contribute 40 centimeters of length. And if you add those two together, well, that's what gives you the chair. And that's going to add up to a total of 100 centimeters. So our final answer is D. Problem number 12. The five keys fit the five padlocks. The numbers on the keys refer to the letters on the padlocks. What is written on the key with the question mark? So this is going to be a little bit difficult to follow the first time around, possibly. So I suggest watching this video multiple times. That's the first thing I'm gonna suggest. Now, we want to try finding out what each number represents, as far as letters are concerned. That is, which letter is one, which letter is two, which letter is four, and which letter is eight. And we want to start by trying to find a lock with a unique structure to its digits. So what I mean by that is, notice that the DAD, this lock, is the only lock where the left letter and the right letter match, D and D. Here we have B and D, A and D, A and D, H and B. So this is the only one where the left letter and the right letter match, which means that the only key with matching left and right numbers should be the correct key. And that's 414 in this case. Notice, by the way, if none of these four keys would match this DAD, we would know that that's supposed to be the question mark one. But that's not the case here, since this one has the two matching left and right numbers. So we know that DAD is equal to 414, sorry, ignore that. I gave the punchline a little early there. But no, DAD is 414, meaning A corresponds to one and D corresponds to four. And another quick observation we can make, HAB is the only lock other than DAD with an A in the middle, and 812 is the only key with a one in the middle. So these have to match up to each other. So we have DAD is 414 and HAB is 812. And we now know that we have H as eight, one as A, B as two, and D as four. Now I listed this question mark equals AHD or BHD, partially because we didn't know what B was until this point. In fact, you can kind of ignore this line. What's important is DAD is 414 and HAB is 812. So we know what corresponds to what here. Like we already know what each of the four letters represents. So we can conclude from this that ABD is equal to 124, which is this key right here. AHD is 184, which is this key here. And that leaves us with BHD, which if we decipher it using what we know from DAD and HAB, that's going to be 284. So that means that our final answer is C. Problem number 13, Tom writes all the numbers from one to 20 in a row and obtains the 31 digit number, one, two, three, four, five, six, seven, et cetera, et cetera, all the way up to 20. Then he deletes 24 of the 31 digits in such a way that the remaining number is as large as possible. Which number does he get? So it's going to be a seven digit number and we want to make it as large as possible. Now, our first number in that case should definitely be a nine. That is our first goal. So here's the number we're working with and we can immediately cross out all of the numbers that come before the nine. And now that we have this nine, well, we know we're going to get a whole bunch of small numbers here, a whole bunch of ones, twos, threes, a zero. So we know we're going to be moving sort of close to the end where we have an eight, we have a nine. Ideally, our next digit would be a nine as well. And we see that the next nine is almost at the end and we would only have two digits left. So this isn't going to be possible because we need seven digits and this way we'll only have two digits after nine. So it's definitely not going to be a nine. If we were to try an eight, we would have one, two, three, four, five digits and we need six more because we already have this first digit from the nine. So we need to go smaller than eight. So the next largest number we can work with is the seven. And then we can use the eight and then the one, nine, two, zero. So basically what we've determined here is that the largest number we can have for the second digit is going to be a seven because we need to have enough digits left over to complete the number. Of course, we can eliminate the nine and the eight as stated before because there just isn't enough room digit-wise. And so that's going to leave us with the seven, the eight, the one, the nine, the two, and the zero. So that's 9781920, which is answer choice C, final answer. Problem number 14. Morton wants to put the figure shown on the right into a regular box. Which of the following boxes is the smallest he can use? So basically what we need to do is we need to figure out the height, the width, and the length, which are all sort of captured by these lines. So if we want to find out how high it is, we just have to count the number of squares of height. So there's one unit of height here. We see that there's another one here. And then there's a third unit here and it doesn't go any higher, right? So it's a total of three squares of height. So one of our digits is going to be a three. Now, what about the width? We have one, we have two, and then from these boxes, we have three, we have four, and we have five. So it's going to be three by five by something, by the depth. We have one for the depth, then two and three. And this is in the same line as this third one. So this is also three. And then this is going to be four. So it's four deep. So it's three by five by four. Of course, multiplication is commutative. So all we have to do is rotate the box. That is three by five by four is exactly the same as three by four by five. So that means our final answer is going to be C. Problem number 15. When we add the numbers in each row and along the columns, we get the results shown. Which statement is true? Now, if we actually read through all of these, notice that there are three statements that cover all three of the possibilities between the relationship between A and D. That is A can be equal to D, A can be less than D, or A can be greater than D. One of these three statements must be true. There's no way that none of these can be true since there's no other possible relationship between A and D. So from this, we can kind of use that as a mental shortcut to eliminate the other two answer choices. Now, this isn't really necessary to solve the problem, but it helps us focus on what matters slightly quicker in an actual exam setting where every second counts. So we're looking for the relationship between A and D. Now, we can either use A plus C compared to D plus C, or A plus B compared to D plus B, and use those to compare A and D. In this case, I happen to use B. So we know A plus B is equal to two, and D plus B is equal to four. Well, you're adding the same number to A as you are to D, and you're getting a larger number when you add D, which means D must be larger than A by two. So D is greater than A, or if we flip it around, A is less than D, which means our final answer is D. Problem number 16. Peter went hiking in the mountains for five days. He started on Monday, and his last hike was on Friday. Each day, he walked two kilometers more than the day before. When the whole trip was over, his total distance was 70 kilometers. What distance did Peter walk on Thursday? So let's go ahead and let M equal the number of days, or not the number of days, the number of, the amount of distance, the number of kilometers that Peter walked on Monday. So here's how this is going to look. Basically, we know on Monday, Peter walked M kilometers, and then M plus two kilometers on Tuesday, and then M plus two plus two, it's two kilometers more, so M plus four kilometers on Wednesday, M plus six on Thursday, and M plus eight on Friday. And all of this is supposed to add up to 70, according to the problem. So if we actually put this all together, we have 70 equals 5M plus 20. You can subtract 20 from both sides of the equation, and that gives you 50 equals 5M. And so this is only true for M equals 10, because five times 10 is equal to 50. So we know that M is equal to 10, meaning he walked 10 kilometers on Monday. And again, we care about how many he walked on Thursday. So that's Tuesday, Wednesday, Thursday is M plus six, which is 10 plus six, which is 16. So that's going to be E, final answer. Problem number 17. There's a picture of a kangaroo in the first triangle. A side shared by any two triangles acts as a mirror. The first two reflections are shown. What does the reflection look like in the shaded triangle? So the easiest way to think about how reflections would go would be to think of one side as the head side of a triangle, one side to be the leg side, and one side to be the tail side. So this side has the tail next to it, this side has the leg next to it, and this side has the head next to it. And it flips like a mirror image. That is here, since this is the head side for this triangle, it'll be the head side for this triangle. Since this side is the tail side, it mirrors over, and it's going to be this side for the tail side. And similarly, this one will be the leg side, so this one will be the leg side. It's a mirror reflection over this line of symmetry. And if we repeat these reflections over and over again, I know these reflections look kind of gross on the video, apologies that I couldn't make it look prettier, but the idea is to keep track of where the tail, where the head and where the foot goes, and which side is which. And eventually what you should get is that you'll have a foot side here, a tail side here, and a head side up here. And we wanna check which of these is actually consistent with that result of the reflections. And here we have head, foot, and tail, same as here. So our final answer is gonna be E. Problem number 18. Boris has a certain amount of money and three magic wands that he can use only once. This wand adds $1, this wand subtracts $1, and this wand doubles the amount. In which order must he use these wands to obtain the largest amount of money? So it doesn't matter how much money you start off with. All that really matters is that you maximize the pluses and minimize the minuses, if that makes sense. So think about it intuitively. How do we make sure that the addition somehow gets amplified while the subtraction does not? The idea is you want to add one first and then double it because that way you're amplifying the addition. And only then do you subtract. Because if you subtract first and then double, it's like you're subtracting two. So basically all of the stuff that you want to grow should be on the left side of the doubling. And all of the stuff you don't want to grow that you don't want to have too much of should be on the right side of the doubling. Meaning add everything first, then double, so it's like you're adding twice as much, and then subtract. So that way you don't do twice the amount of subtracting. So that would mean that we want to do plus one times two minus one. And you can try this with any value and you'll see that it works. And hopefully that's a sufficient intuitive explanation. But basically you want to amplify the addition and you want to avoid amplifying the subtraction. So that leaves us with the final answer of D. Problem number 19. Raphael has three squares. The first one has a side length of two centimeters. The second one has a side length of four centimeters and a vertex placed in the center of the first square. The last one has a side length of six centimeters and a vertex placed in the center of the second square as shown in the picture. What is the area of the entire shaded figure? So basically the area of the shaded figure is going to be the sum of the three areas of the squares minus these two squares where they overlap. Because we don't want to double count this area from both of this square and this square, and we don't want to double count this area from this square and this square. So we subtract this area and this area from the area of the three individual squares. So since this is a vertex going to the center of the first square, and this is a vertex going to the center of the second square, what that tells us is that this has a side length of one centimeter, and this has a side length of two centimeters since it's half of the side length of the bigger square. So if we go ahead and take a look at all of the areas, it's going to be two squared plus four squared plus six squared, the three squares that we're adding. We're subtracting the double counting of this square. That's minus one times itself. That's going to be this area. And then minus two squared, which is this area that we don't want to double count. And this two squared cancels out with this two squared. And that leaves us with four squared, which is 16 plus six squared, which is 36 minus one squared, which is one. And that's going to be 52 minus one, which is equal to 51. So our final answer is B. Problem number 20. Four players scored goals in a handball match. All of them scored a different number of goals. Among the four, Mike was the one who scored the least number of goals. The other three scored 20 goals in total. What is the largest number of goals Mike could have scored? So for this, we want to find out what three numbers add up to 20 that are as close to each other as possible. And why do we want it this way? Well, if they were super far away from each other, for instance, one, two, and 17, well, those do add up to 20. But remember, we need someone who scored less goals than the smallest of those three numbers. And the smallest of those three numbers is gonna be one. So that's kind of a problem because that means he scored no goals. So the goal is to get as close to the center as possible. By center, I mean like all of the numbers as close to each other as possible. So that way, that minimum number is as large as possible. So how do we find what a good number is to get close to? Well, the first thing to notice is what numbers can you multiply by three to get close to 20? Because we want three numbers that are close to each other. So it's kind of like multiplying by three. So we know six times three is 18, and seven times three is 21. So maybe seven and six would be good numbers to add together and then figure out some third number that we can add. Well, seven plus six is 13. So we would need to add seven again to get 20. And the problem with that is that we can't have repeats. We can't have two people scoring the same number of goals. So it takes a little bit of trial and error. But what we basically determine is that the largest minimum number you can get is five. And there's sort of two different ways to do that. You have eight plus seven plus five, or six plus nine plus five. And both of those are gonna give you 20. There's no way to get anything greater than five as the minimum number. And Mike needs to score at least one goal less than five because he scored the least number of goals. So that's gonna be four for Mike. So our final answer is C. Problem number 21. A bar consists of two gray cubes and one white cube glued together as shown in the figure. Which figure can be built from nine such bars? So basically the approach here is to try to see if it is possible to make each individual cube using just nine bars. And the moment that you find that there's some sort of logical conflict, like it just isn't possible to have bars set up in such a way to make that cube possible, we can eliminate that answer. So the answer, I'm gonna tell you right away, it's going to be answer choice A. But I'd like to show you why a couple of the other ones might be wrong in case it isn't clear. So for example, what we can have, answer choice B, this is a pretty obviously bad one because so here this row won't work. If this happens to be one such bar, then clearly that doesn't work because you need two gray bars and one white bar. Well, you might be saying, wait, what if they're going this way? What if they're sort of going out the page? But the problem here is that we have three white bars in a row. So that also isn't possible. So we would either need two gray and one white here or two gray and one white here. Since it isn't the case for either of these lines, answer choice B isn't possible. Answer choice C definitely isn't possible. If you look at all this white plus this one right here for this line, if you think about it, you will very clearly see why this doesn't work. For answer choice D, you have two whites and a gray, two whites and a gray, two whites and a gray. This just isn't gonna fly. You can't have two whites and a gray going in every single direction. This is sort of the point of conflict. And that's why some of them don't work. Now let's focus on why answer choice A does work. Basically, what I've highlighted here are the first three bars that work, And of course, this first top one is the same as this first top one, right? This first bar we've already accounted for, so we just care about these two back bars. And they're both grey, grey, and white, so they clearly work. And we don't really care about these four here, because we don't see most of the bars. We just see one square from each bar. So we can assume for this white one, this white one, and this white one, that there's going to be two greys in the back. And for this grey one, there's going to be another grey, and then a white in the back. Since we can't see it, we can assume that it's right, because the rest of it works. And if you check any of the other answers, you'll see that the visible bars do not work. So that means, since we've shown that it works for answer choice A, that's going to be our final answer. Problem number 22. The numbers 1, 2, 3, 4, and 5 have to be written in the five cells in the figure in the following way. If a number is just below another number, it has to be greater. If a number is just to the right of another number, it has to be greater. In how many ways can this be done? So another way of phrasing this is that the largest numbers are out here, and then you sort of shrink towards this corner. Now let's make a quick observation, or two. The first one is that this corner has to have 1, because 1 can't be to the right or below any number, because 1 is the smallest number available to us. So 1 is definitely going to go in this corner. Now let's make a second observation. If we have 1 here, let's pick any number here that isn't 5, basically, because 5 is big and you can't have anything bigger than 5 in this slot. So let's say it's just going to be 2. And then that this number is going to be 3. Well, in this case, we only have one option here, it has to be 4 and then 5, because there's only one way to organize the remaining two numbers, it has to be smaller than larger. So basically, whenever you pick three numbers here that work, there's only one version of that. So what I mean is, for 1, 2, 3, there's only 4, 5 here, for 1, 2, 4, there's only 3, 5. It's not like we can have 1, 2, 3, and then two different possibilities here. So we can really just think about this in terms of what three numbers we put in this column. That is the way I would like you to think about this. So here are our possibilities. All of these are just how to list from top to bottom. So we could have 1, 2, 3, in which case the numbers on the right would be 4 and then 5. 1, 2, 4, in which case we'd have 3 and 5. 1, 2, 5, in which case the numbers on the right would be 3 and 4. 1, 3, 4, which would give us 2 and 5 on the side. 1, 3, 5, which would give us 2 and 4 on the side. And then 1, 4, 5, which would give us 2 and 3 on the side. So all we're really thinking about are these three numbers going down. And these are all of the possible combinations we can have, where we have a smaller number and then a larger number. So there's three possibilities with the 2, because there's three numbers bigger than 2. There's two possibilities with the 3, since there's two numbers bigger than 3. And with the 4, there's only one number bigger than 4, so there's only one possibility there. So that gives us a total of six possibilities. And that means our final answer is D. Problem number 23, eight kangaroos stood in a line as shown in the diagram. At some point, two kangaroos standing side by side and facing each other exchanged places by jumping past each other. This was repeated until no further jumps were possible. How many exchanges were made? So basically our goal is to take all of the kangaroos that are facing left and jump them all the way to the right. So we have all the ones facing left on this side, and all the ones facing right on the other side. You could also go the other way. You could make all of the right facing kangaroos jump to the right. But basically all the kangaroos want to jump to the direction that they're facing. So let's start with the left most kangaroo. We have this kangaroo, which is the left most one that is facing left. And we see he would need to make one, two, three jumps, or three exchanges. Then the next left most facing kangaroo would be this one. And we see he would need to make one, two, three, four, five jumps. He doesn't need to jump here since there's already a left facing kangaroo here. So there's five from this. And similarly, there's five from the last one, one, two, three, four, five. So if we add all of those exchanges together, that's gonna be three plus five plus five, which is 13. So our final answer is D. Problem number 24. Monica has to choose five different numbers. She has to multiply some of them by two and the others by three, in order to get the smallest number of different results. What is the least number of results she can obtain? So basically, since we know we're multiplying by two and three, two and three should be a couple of the numbers that fall into this list. Because you can multiply two by three, and you can multiply three by two. And that gets you just one result. And then you want another couple of numbers that pair up that way, where if you double one and triple the other, they give you the same product. And then you're gonna have that fifth number, which doesn't pair with any of the other ones. And that's gonna create its own third result. So an example of this would be 2, 3, 12, 18, and some other number. 2 by 3 and 3 by 2 are both 6, 12 by 3 is 36, 18 by 2 is 36. And we have some other number which can multiply by 2 or 3. It's not gonna lead to 6 or 36, since we already multiplied by 2 or 3 to get those numbers. So we're gonna get some third result from this. Let's say it's like 10 or something, so you'll get 20 or 30. But the point is, by pairing these two numbers and pairing these two numbers so they get the same result, we're guaranteed to get a minimum of three results, three different results from multiplying by choosing the twos and the threes. Correctly, as well. So that gives us three different answers, three different products. So our final answer is C. Problem number 25, the square floor in the picture is covered by triangular and square tiles in gray or white. What is the smallest number of gray tiles that need to be exchanged with white tiles so that the pattern looks the same from each of the four directions shown? So looking around, there's a possibility here to notice, and really, there is no shortcut to seeing this. It's just trying to hope for pattern recognition. In particular, you want it to look the same from all four directions. So you want it to be a shape that you can sort of rotate, and it looks the same however you rotate it. Well, you see this triangle in this square here, it's sort of like a fish going in this direction, it's going upwards. On the left side, if you were to look at this triangle in this square, it looks like a fish on the left side of this face going upwards. So maybe this fish thing is a decent thing to aim for. And again, this is just something that you can sort of notice and aim for. Well, we can swap this triangle tile with this triangle tile to get a fish going this way, and we can swap this square tile with this square tile to get a fish going this way. So if we highlight those, we swap these two square tiles and these two triangle tiles. And in other words, we make a switch of one triangle and one square. And that's going to get us a shape that we can rotate that's gonna look the same from all directions. So our final answer is C. Problem number 26, a bag contains only red and green marbles. For any five marbles we pick, at least one is red. For any six marbles we pick, at least one is green. What is the largest number of marbles that the bag can contain? So let's start by unpacking the first part of this statement. For any five marbles we pick, at least one will be red. So let's assume that, what this means is, in the five marbles we pick, even if we pick all of the green ones in the bag, we will still end up with one red one. So that means there has to be less than five green marbles. In particular, what this tells us is that we have four green marbles. It's always going to be one less than this number, because you're thinking of it like a worst case scenario, like you pick green after green after green after green, and then, oh, there's a red. So in this case, you would pick a green marble, another green, another green, another green. You have four green marbles. And then you would have to pick a red one based on how this statement is phrased. The only way that can be true is if you have only four green marbles. By similar logic, for any six marbles we pick, at least one is green. That means, at most, five of them are red. So that means there are four green ones and five red ones. And if we add that together, that gives us nine. So that's the final answer of C. Problem number 27. Ala likes even numbers. Bata likes numbers divisible by three. And Selena likes numbers divisible by five. One by one, the girls walked up to a basket containing eight balls with numbers written on them. And each one took out all the balls of the numbers she likes. It turned out that Ala collected balls of the numbers 32 and 52. Bata collected balls with numbers 24, 33, and 45. And Selena collected balls with numbers 20, 25, and 35. In what order did the girls approach the basket? Let's write out the numbers each girl took a little bit more clearly. What we are looking for is some kind of overlap where a girl took a number that she liked and also a number that another girl liked because that would show us which girl went first. Ala likes even numbers, so numbers divisible by two. I'm going to circle all the even numbers in blue. So Ala's numbers are even. Bata also has one even number. And Selena has an even number. Which means that Ala had to go after both Bata and Selena because those girls were able to get the even numbers before Ala did. This means that Ala went third. Bata likes numbers divisible by three. She has numbers divisible by three, 24, 33, and 45. Selena does not have any numbers divisible by three. Now let's see if we know who went first based on numbers divisible by five. Bata has a number divisible by five. And Selena obviously has all the other numbers divisible by five. This means that Bata went before Selena. If Selena went before Bata, she would have taken the ball with 45 because that's divisible by five. We already know that Ala went third. We're trying to figure out who went first and who went second. Since we now see that Bata went before Selena, we know that Bata went first and Selena went second. So the order is Bata, Selena, Ala. This is answer D. Problem number 28. John wants to write a natural number in each box of the diagram in such a way that each number above the bottom row is the sum of the two numbers in the boxes immediately underneath. What is the largest number of odd numbers that John can write? So a big part of this strategy is to maximize the number of odds we have down here. Now the problem is if we do all four of these as odd numbers, then all of these are going to be even. I guess a thing to observe here is that if you add two odds or two evens, you're always going to get an even number. You can try it. Name any two odd numbers or any two even numbers, and they will add up to an even number. That is a well-known property in math. But if you add an odd and an even number, you're going to get an odd number for the sum. So that's something to keep in mind when you're trying to figure out how to structure this pyramid. Now if we have four odds down here, then we're going to get an even, and even, and even. And all of these are going to be even numbers above this row. So yeah, we get all odds down here. But that doesn't really seem ideal because you're going to get no odds in any of the other rows. So what if we try using three odds? Well, if we use three odds, it's a question of where should the even number be. Should the even number be one of the outer ones or one of the inner ones? Well, it should be one of the inner two ones because if it's an outer one, then you're going to have even and odd, and then odd and then odd. And what that will do, like pretend this even is here and this odd is here. Pretend these two switch places. Then these two still add up to an odd number, but you're going to get an even number here and an even number here because it's these three odds adding up. So that's going to be even and even. And even on the outside will only give us one odd number in the third row. And even on the inside will give us two odd numbers in the second row. And then that will give us one more odd number in the third row and then one more at the top. Now, you can go ahead and try it the other way and you'll see that you get fewer odd numbers. And if you try only using two odds, then you're going to get fewer odd numbers. You can go ahead and check that for yourself. The issue there is that you're just starting with too few odd numbers on the bottom. And if you only have one odd number on the bottom, then clearly you're just setting yourself up for lots of additions of even numbers. And that's going to result in very, very few odd numbers. So this is sort of the intuitive explanation for why you want three odds and an even. And the even should be one of the inner two blocks. Also, you could just do a lot of guess and check if you have the time, but this is the way to think about it. This is sort of a reason why you can be confident that this is the correct bottom setup. It's that you're maximizing the number of odds in the bottom row, and you're also setting it up so that you get two odds, not just one, in the third row. And if we add up all of the odds that we get from here, it's 1, 2, 3, 4, 5, 6, 7. So our final answer is D. Problem number 29. Julia has four different colored pencils and wants to use some or all of them to color the map of an island divided into four countries, as shown in the picture. If two countries with a common border cannot be the same color, in how many ways can she color the map of the island? So let's check how many colors we can choose for this one, then how many for this one, then how many for this one, and how many for this one, and multiply all of those possibilities together. That gives you all of the different possible combinations of colors. So we have four choices for this country, since we have four colors and we haven't pre-committed to anything yet. So there's four choices for this one. Now, you can't match this color with this color, so there's only going to be three choices here. Now, this country is bordering with both of these countries, so you can't use this color, you can't use this color. You have only two colors left to choose from. And now for this country, it overlaps with these two countries, so these two colors are restricted, so you get only two choices here. This color can match this color, since the countries don't border each other. But basically, you have four times three times two times two. And by the way, the order that I picked to look at these countries in didn't really matter. You could have picked a different order. You should still end up getting the same multiplication if you end up going around in a row. So you could have started from here and said, oh, this is going to be four, so this is going to be three. And this neighbors with these two, so there's only going to be two. And this neighbors with these two, so there's only going to be two. It doesn't matter which direction you start in or where you finish. You will find that it's always going to be four times three times two times two, because that's how combinations work in a nutshell. And I won't go super deep into that. All you need to know is you just need to multiply the number of choices that you have for each decision. And in this case, the decision is what color do you fill in for a country? And the number of choices you have is what you're multiplying by. So it's going to be four times three times two times two, which is equal to 48. So our final answer is E. Problem number 30. In each cell of a six by six board, there is a lamp. We say that two lamps on this board are neighbors if they lie in cells with a common side. At the beginning, some lamps are lit. And each minute, every lamp having at least two lit neighboring lamps is lit if the three cells have a common vertex. What is the minimum number of lamps that needs to be lit at the beginning in order to make sure that at some time all of the lamps will be lit? So if you're reading this, hopefully you're imagining that you probably need to fill in an entire diagonal in the grid. Let's start by drawing out a grid. That's an ugly grid, but it'll have to do. So basically, the idea is you're going to need to either fill in this diagonal or this diagonal. Because if you want, for instance, this candle to be lit, you would either need these two or these two. If you want this candle to be lit, it would need to be this two, this two, this two, or this two. You basically need two neighboring cells which share a corner to also be lit. And the most natural way to do this isn't to do something like, oh, we'll have two lit here, we'll have two lit here, we'll have two lit here. No, it makes sense to just line all of them up. You'll find that it takes up less candles that way. And it doesn't matter which diagonal you pick. So basically, thanks to these two, you're going to light this one and this one, and this one and this one. And since this one and this one is lit, this one will be lit. And that will sort of spread throughout the entire square. You can go ahead and keep track of that. But the key here is to realize that if two neighbors that share a vertex with each other are lit, that's what makes one of these squares light up. So it makes sense to just fill out an entire diagonal if you want to be optimal. And it could be either this diagonal or this diagonal. But in either case, you have a total of six candles lit in the beginning. So our final answer is C.

Math Kangaroo 2017

Levels 5 and 6

problem-solving strategies

number manipulation

geometry

logical reasoning

exam solutions

learning strategies

mathematical skills