Hello and welcome to the Math Kangaroo 2006 level 9 and 10 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 myself every single year and every single year it was a challenge. You're definitely going beyond what your school probably expects of you in terms of learning math and practicing math so you should really be proud of yourself for taking on this challenge. It really is something that's very difficult and very challenging and it's going to help you a lot with your critical thinking and your logic in the future so you should be really really patting yourself on the back just for choosing to take this on. Now the purpose of this video series is to show you a logical path to get to the correct answer of every single question on the exam. Keep in mind the methods that you see in the videos are not necessarily the only methods to solve the problems so sometimes you might get the correct answer by a completely different method and in that case your method might be equally good, it might be better, it might be worse, it might depend on the situation. If you have any questions about this sort of thing please feel free to email me at thomas at math kangaroo org the email listed below and if you have any questions about this exam in general please feel free to email me. Now before moving on to the video series I'd just like to offer two quick pieces of advice. The first piece of advice I have to offer is don't watch any of the videos until you've actually tried solving a problem. It doesn't make sense to watch the video first. You need to struggle with the problem first because if you just watch the video you're going to have the information spoon-fed to you. You're gonna listen, you're gonna watch, and you're gonna go okay, okay, that makes sense. Let's move on to the next one. But you won't have actually learned anything. Instead if you actually struggle with the problem first and you know you actually try thinking through it and there's maybe something that you're missing or maybe you're on the completely wrong track, you have to actually try first because then when you watch the video it'll be like something will click in your head. You'll see oh that's the trick that I missed or oh this is where I made the mistake or oh this is why the answer is different from what I thought it was. You're going to learn from your mistakes and you're going to understand why the solution in the video makes sense, but only if you actually struggle with the problem first. You need to try understanding the problem and only then will the video be a useful tool for you. Now a second piece of advice I have to offer is to please please just write down as much of your work as possible when you're working through these exams. I know a lot of times when somebody's good at math they try way too hard to just sort of keep everything in their heads because they think it'll be faster, they think oh I can keep it in my head it's no problem, but very often this just leads to silly mistakes being done and you might forget some of your work that you did sometime earlier in the problem and then because you didn't write it down you can't revisit it and so you just make everything harder for yourself. And also when you're reviewing the answers in the videos you won't know what you were thinking at that time because you didn't write down your work. If you want to know where you went wrong in a problem you should probably write down your work just so that you could recall your thought process at that given point. So writing down your work will help you reduce overall your silly little mistakes and it will also help you learn from whatever mistakes you do make by allowing you to revisit them as you consult the videos. So with those two pieces of advice in mind I'd just like to say I hope you get a lot out of these videos, I hope you learn a lot, and I hope you have fun. So thanks for watching. Problem number one. What is halfway between 2006 and 6002? To find this number all we have to do is take the average of the two numbers and we do so by adding the two numbers and then dividing by two. So adding them we get 8008 and 8008 divided by 2 is 4004. So the middle value in between these two numbers which is the average is 4004 and our final answer is D. Problem number two. How many four-digit numbers with all different digits are divisible by 2006? Rather than trying to come up with four-digit numbers that are divisible by 2006 it's much easier to divide to multiply 2006 by multiple numbers and see what products we get that have four digits and have all different numbers. So we want to keep the numbers that we're multiplying by 2006 relatively small since 2006 already has four digits. In fact if we multiply 2006 by 5 we're already going to get 10,000 plus so it's going to be 10,030 which is already five digits. So we know we won't check anything beyond five and we also know that it's not going to be 1 times 2006 since that will give us 2006 and that has two zeros in it so it doesn't have all different digits. Now if we check 2 3 & 4 times 2006 we do get three different numbers which all have all different digits and have four digits and based on the fact that we can't multiply 2006 by anything bigger because we'll get more than four digits we know that these are the only three numbers that are four-digit numbers with all different digits divisible by 2006. So our final answer is C. Problem number three. What is the smallest 10-digit number that can be obtained by putting together the following six numbers one after another 309, 41, 5, 7, 68, and 2. To solve this problem we're going to start by looking at the answer choices and looking at which number is the smallest. Now the smallest number will be as small as possible going from left to right. So in other words the first answer choices we should look at should be the ones with very very small leftmost digits. In particular the smallest answer choice we have available is answer choice B which is 1 0 2 3 and some other numbers. It's even smaller than answer choice A which is 1 2 3 4. We know that it's smaller because the 0 is less than the 2 and the ones are equal. So this number is the smallest answer choice we have and normally we could immediately just say okay this is our answer but we need to check to make sure that it's actually possible to make this number and in fact it isn't because 1 can't be the first number in the 10-digit number because 1 is in the number 41 which means wherever we include the 1 4 needs to be directly to the left of it which actually means we can eliminate both answer choices A and B because we're not allowed to have a 1 on the leftmost digit. So now what's our next smallest answer choice? We have two answer choices that start with a 2 and a 3 0 9 3 0 9 and in fact if we check 2 3 0 9 okay so this is possible so far and now we have 4 1 5 and 4 1 6. So the one with the 4 1 5 is the smaller answer choice. Now is it possible to do this? Well we have the 2 we have the 3 0 9 we have a 4 1 and we have a 5. So in fact we can make this and this is the smallest answer choice left. So since this is possible and we could actually go all the way to the end just to make sure. So we have 4 1 5 4 1 5 and then 6 8 7 and we have 6 8 7. So it is possible to arrange this 10-digit number given the numbers that we have and it's the smallest answer choice left. So our final answer is D. Problem number 4. How many times between 0 o'clock and 2359 does an electronic watch show all of the following four digits? 2 0 0 and 6 in any order. For this problem the easiest way to get through it is to brute-force it. There's really no need for any sort of shortcuts you can just imagine going through through all of the times until you reach combinations of a 2 a 6 and two zeros and once you go through it you should come across these five answer choices. There's no real shortcut to it this is just sort of one that you want to take your time go through it and make sure that you don't accidentally skip any of them and these are our only five choices or are only five times so our final answer is C. Problem number 5. A flag consists of three stripes of equal width which are divided into two three and four equal parts as shown. What fraction of the flags area is shaded? So looking at this we see that each of the stripes as given in the problem are of equal width so each one of these stripes represents a third of the area of the flag so we have one third one third and one third. Now here we have two quarters of one third being shaded in that's equal to a half of a third and in the bottom stripe we also have a half of a third being shaded in so we have a half of a third here and a half of a third here. Two halves of a third gives us one third of the flags area being shaded from the top stripe and the bottom stripe. So what about the middle stripe? Well the middle stripe is two thirds of one third being shaded in so that's going to be two thirds times one third so if we add together one third plus two thirds times one third that gives us one third plus two ninths and then we could change one third into three ninths just rewriting the fraction and three ninths plus two ninths gives us five ninths so our final answer is E. Problem number six. My grandma's watch is running fast one minute per hour and my grandpa's watch is running slow one minute per hour. When I left their house I synchronized their watches and told them I would return when the difference between the times on their watches is exactly one hour. After how many hours will I go back to my grandparents house? Now for this one you almost don't need to really write anything down for the math. This is one of those few problems that I think can be done entirely in your head if you wanted to. So we know that one of the watches run slow one minute per hour and one of the watches runs fast by one minute per hour so let's suppose one hour passes. Well in that case one of the watches will be one hour but one minute behind and one of the watches will be one minute ahead and so there's going to be a total difference of two minutes between the watches after an hour passes. So what this tells us is that every single hour we gain a difference of two minutes. Now we want the difference between the watches to be an hour, so all we have to do is divide an hour by two minutes, and that's going to tell us, it's going to be 60 divided by two, how many minutes are in an hour, 60 minutes, divided by two minutes. And that tells us that we want 30 hours to pass, since for every hour we gain a two minute difference. So after one hour there's a two minute difference, after two hours there's a four minute difference, and so on. So after 30 hours, we're going to have 30 times two minutes, or 60 minutes difference between them. So we will come back in 30 hours, and thus our final answer is C. Problem number seven, Jack says that 25% of his books are novels, and one ninth of them are poetry. How many books does Jack have if we know he has between 50 and 100 books? So for this one, there also isn't really that much math to do. We want to find a number that is between 50 and 100, that is perfectly divisible by both four and by nine, because 25% is equal to one fourth. So Jack can talk about one fourth of his books, or one ninth of his books, and it doesn't make sense to be talking about pieces of books. So we assume that these are whole numbers of books. So what number is divisible both by four and by nine that is between 50 and 100? If we look through these answer choices, we find that the only one that is divisible by both four and by nine is 72. So for anyone who doesn't know, the way to check quickly if a number is divisible by nine is all you have to do is add up the digits, and if the digits add up to something that's divisible by nine, then the original number is divisible by nine. So if seven plus two, which equals nine, well that's divisible by nine. So we know that 72 itself is divisible by nine. So that's a good thing to know, a good little reminder. But yeah, since this is the only number between 50 and 100 that is divisible by four and by nine, we know that this has to be it. So our final answer is D. Problem number eight. A circle is divided into four arc lengths of two, five, six, and X. The arc length of two has a measure of 30 degrees. Find the value of arc X. Now arc length corresponds directly to the angle that it forms with the center. So what we're going to do first is we're going to find out how many degrees make up one unit of length. And then we're going to use that to figure out how much of the circle is covered by all of the arcs except for X. And that's going to tell us how many degrees X is, and that's going to be all that we need to know if we want to figure out the arc length. So starting off, we know that an arc length of two corresponds to 30 degrees. That means that an arc length of one would correspond to 30 divided by two, which is equal to 15. Now that means that one unit of length is 15 degrees. Now the three arc lengths that we know the numbers for are two, five, and six. And when we add those together, we get 13. So we have 13 units of length times 15 degrees, and that gives us 195 degrees. And that's the measure of the degrees of these three arcs. Now since a circle has 360 degrees all the way around, we know that arc X has to have a central angle that is equal to 360 minus 195. And that is equal to 165, which happens to be equal to 11 times 15. Notice that a quick way to realize this is that 150 is equal to 10 times 15, or 15 times 10, whichever way you want to write it down. And 165 is just 150 plus 15. So seeing that this is just 150 plus some multiple of 15 is a good way of realizing that you can rewrite it as 11 times 15. Now since it's 11 times 15, and we have 15 degrees per one unit of length, we see that X must be equal to 11. So our final answer is E. Problem number nine. In the diagram shown, each letter represents a different digit, and each digit corresponds to a different letter. What digit does G represent? So one way that we can rewrite what we have on the right-hand side is we can rewrite it as 300 times K plus 20 times A plus 11 times N, because we have an N in the tens place and an N in the ones place, plus 2G is equal to 2006. And now what we're checking here, the reason why this is crossed out right now is because we're checking something. Let's assume that K, which is the thing that we're multiplying by the largest number, let's assume that this will get us as close to 2,000 as possible without going over it. So in that case, K would be 6. Now if K would be 5, then we would have 1,500 being contributed from this. And maybe the rest of this stuff won't quite add up to a high enough number. So there's definitely some trial and error in this method, but basically the idea is start with as big of a contribution from K as possible without going over 2006, and then seeing how the rest of the numbers look after that. So it is a little bit of guess and check, but it turns out that this approach works fairly quickly. So assuming that K is equal to 6 and just seeing how it works out, we know that this will be 300 times 6, which is 1,800. And so we can subtract 1,800 from both sides, and that will give us 20a plus 11n plus 2g is equal to 206. And so that's why we cross this out, because when we cross this out, we basically have what we have left is exactly this. We have 20a, since we have two a's in the tens digits, 11n, since there's an n in the ones and an n in the tens digit, and 2g again. And this adds up to 206, assuming that K is equal to 6. Now if it turns out that the numbers don't make any sense after this, we should probably go back and check maybe K is supposed to be equal to 5. It definitely won't be equal to 7, because then we would have 2,100, which is too big. So we would have to check if it's 5 or maybe less. But now that we see 20a plus 11n plus 2g is equal to 206, it's time for a little more guess and check. In particular, we don't know if up here n plus g plus g is equal to 6, or maybe 16, or maybe 26. Whatever it would be equal to, it would turn out that we would have a 6 in the ones digit, and then we would carry over a 1 or a 2 maybe if it's a 16 or 26. So again, it's going to be a bit of guess and check. But let's assume that this is just going to be equal to 6, that it's not 16 or 26. In what ways can we pick n and g such that they add up to 6? Remember, we need n and g to be different numbers. So it turns out there's only one way to do this. We can't have g equal to 2, because then n has to equal 2, and we get 2 plus 2 plus 2. But n and g are supposed to be two different numbers. No, instead, in this case, assuming that it adds up to 6 and not 16 or 26, we're assuming that g equals 1 here. So that way, n equals 4. So if we were to assume that g is anything bigger than 1, we already know it doesn't work for 2. And we know that g can't be equal to 3, since then we already have 3 plus 3. Well, then maybe we would have something bigger for n, and maybe that would work out. But let's just start with g is equal to 1. We'll pick something and just sort of go with it and see if all of the numbers make sense. Well, if we let g equal 1, then n has to equal 4. So we have 4 plus 1 plus 1 is equal to 6. And then that would also fill in 4 over here. Now let's go ahead and go back to the equation we have on paper. If we let g equal 1 and n equal 4, then we have 20a plus 44 plus 2. And 44 plus 2 is equal to 46, which is equal to 206. Now this 46 is very, very nice, because it subtracts from 206 very cleanly. So if we subtract 46 from both sides, we get 20a is equal to 160. And then we can divide both sides by 20, and we get that a is equal to 8. And now going back and checking everything, we assumed that k equals 6. And from this, we got a equals 8, n equals 4, and g equals 1. The conditions that the problem gave us are not violated. Every single letter corresponds to a different number, and every number corresponds to a different letter. So we found a situation that satisfies the problem. And the case where the problem is satisfied is where g is equal to 1. So we conclude that our final answer is a. Problem number 10. With how many zeros does the product of the first 10 consecutive prime numbers end? So if we were to list out the first 10 consecutive prime numbers, well, here's the list. All you have to do is retrieve these from memory or just go through the list. If you don't know your primes, you should still be able to get through the first 10 fairly quickly, since all of them are less than 30. It doesn't take very long to count this through. And remember that there is sort of a shortcut to know how many zeros a product will end with. You take all of the zeros at the ends of all of the numbers you're multiplying, and you sort of take them away. You multiply out the remaining numbers you have, and then you toss on all the zeros at the end. So that's kind of what we're going to use here, except in a slightly different way. When we look through here, we see that the only numbers that will multiply out to give us anything with a zero at the end are the 2 and the 5, since 2 times 5 equals 10. No other products here are going to give us anything that ends in a zero. You can check as long as you like, but there's not going to be any such number. This is kind of a consequence of there being prime numbers here. This is just sort of a natural behavior for them, especially since you have a lot of 7s, a lot of 3s. You don't really have numbers that multiply nicely into multiples of 10. So we have 2 times 5 giving us 10, and everything else giving us numbers that don't end in a zero at all. So we only have one zero being contributed from the 10 that we're multiplying by, and that means that our final answer is B.

Math Kangaroo

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problem-solving

geometric puzzles

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