This is the Math Kangaroo Solutions video library, presenting solution suggestions for levels 9 and 10 from the year 2015. These solutions are presented by Lukasz Nalewskowski. The purpose of the Math Kangaroo Solutions video library is to help you learn how to solve math problems, such as those presented in the Math Kangaroo Competition. It is important that you read the problem, as well as listen as I read the problem. After reading and listening to the question, pause the video and try to solve the question. Question 1. Which of the following numbers is closest to 20.15 times 51.02? To do this problem, we first can simplify the problem by multiplying 20, instead of 20.15, by 50, instead of 51.02. Now when we do this, we first multiply 2 times 5, and we get 10. Next, we have to notice that there are two zeros that we have to add. So, we get 10, add an additional zero, and another one. And with that, we get our solution, which is B, 1000. Question 2. Mother did the laundry and hung t-shirts in line on a clothesline. She left a gap between each t-shirt and the one next to it. Then, she asked her children to put a single sock in every gap between two t-shirts. Now there are 29 pieces of clothing on the line. How many t-shirts are there? So first, we know that mother hangs up t-shirts, which we will represent with Ts. And in between each t-shirt, she has her children hang up a single sock. So, if there were two t-shirts, there would be one sock, with a ratio of 2 to 1. Now, if there was an additional t-shirt, there would be an additional sock. Now we have a ratio of 3 to 2. So, keeping this in mind, we know that there is always one more t-shirt than there are socks. And this can be represented by S plus 1 equals T. Now, we know that the total amount of clothing is 29. So, the total amount of t-shirts plus the total amount of socks is equal to 29. If we substitute with the formula that we already know, T equals S plus 1, we can write down S plus 1 plus S equals 29. And we simplify the formula, and we get 2S equals 28. So, we know that there will be 14 socks. The question asked us how many t-shirts there are. So, we go back to our first formula, S plus 1 equals T, and we substitute 14 for S. And we get 14 plus 1 equals T. So, our answer is going to be E, 15. First, let's look at our problem. Now, since we are multiplying the ones digit, we can just erase and ignore the rest of the number. And we get 5 squared plus 5 to the power of 0 plus 5 to the power of 1 plus 5 to the power of 5. Now, we can solve these. 5 squared is 25, 5 to the power of 0 is 1, 5 to the power of 1 is 5, and 5 to the power of 5 will be a number ending in 5. Now, we just have to add these numbers together. So, we can just add in 5 plus 1 plus 5 plus 5. Since we only care about the ones digit, we can ignore everything else. With this, we get a total of 16. So, again, we are only looking at the ones digit, and this gives us our solution, which will be C, 6. Question 4. We can take a closer look at our figure. Now, we know that there is a side length of A, so we can label it as such. Now, since we know that there is a semicircle, and it has a diameter of A, since it is equal to the side length of the square, the radius will be equal to A half. Now, this is a square, so the other side will also be equal to A halves. Now, we know that since we have a semicircle and two quarter arcs, their area together will be equal to half of the area of the square. Now, we know that the area of the square is A times A, and all we need to do now is just take half of that, and we get A times A divided by 2. We see that as a solution. The correct answer will be B, A squared divided by 2. Question 5. Ann paid 80 cents, Beth 50 cents, and Cindy 20 cents. If they had divided the cookies proportionally to the price each paid, how many more cookies should Ann have received? So first, we should find out the total amount that the cookies cost. We can do this by adding 80 cents plus 50 cents plus 20 cents, and this gives us the total of 150 cents. We know that Ann paid 80 cents, so she paid 80 out of 150, which can be simplified to 8 out of 15. Now, we know there were 30 total cookies, so we can just multiply this fraction, 8 divided by 15, times 30. And when we do this, we get our solution, which is 16. Now, first, they all received 10 cookies, and the question wants to know how many more Ann receives. So we do 16 minus 10 and get our final answer, which will be E, 6. Question 6. Mr. Hyde wants to dig up a treasure that he buried in his garden years ago. He only remembers that he buried the treasure at least 5 meters away from the hedge and at most 5 meters away from the trunk of the old pear tree. Which of the following pictures shows the region where Mr. Hyde should look for the treasure? So first, we draw out the hedge and the tree. Now, we know that the treasure will be at least 5 meters away from the hedge, so we can draw out this red area, which signifies he should not look for the treasure here, and this leaves this whole yellow area for him to search. But we also know that the treasure is at most 5 meters away from the trunk of the tree, so we can draw a circle that stays within the yellow area, but does not overlap with the red area, like so. So Mr. Hyde should search for this green circle minus the little section cut off top. And we notice that this is something that looks like the solution of B, so that is the correct answer. Question 7. There are 33 children in a class. Each of them likes computer studies or physical education. Three children like both subjects. There are twice as many children who like only computer studies as those who like only P.E. How many children like computer studies? So first, we know that there are 33 children total, but three of them like both subjects, so we can subtract them for now and get a total of 30 students who only like one subject each. Now, we know there are twice as many children who like only computer studies as those who like P.E., so there are two-thirds and one-thirds, respectively. If we multiply those by 30, we get 20 and 10 students. This means there are 20 children who like only computer studies and 10 children who only like P.E. And this satisfies the need that there are twice as many children who like only computer studies versus the children who only like P.E. Now, the question wants us to find out how many children like computer studies, so we have to add in the three children that like both subjects. And with this, we get our answer, which will be E, 23. Question 8. Three lines intersect at one point. The picture to the right shows the measures of angles between two pairs of these lines. What is the measure of the smallest angle between any of these two lines? So first, let's take a closer look at our diagram. We see that we have three lines, and we know the angles of two parts of this. We have 120 degrees and 105 degrees. And we know that a circle always measures 360 degrees, so the missing angle will be equal to 135 degrees. Next, we also know that any single line will have a measure of 180 degrees, so what we can do is do 180 degrees minus 135 degrees to find out an actual angle. And we find that both of these, since they are close to the 135, will have angles of 45 degrees. Now we just need to repeat this process for all of the angles. For 120, we know that 60 degrees is what's left, so we can write down 60 degrees on both sides. And finally, 180 degrees minus 105 degrees gives us 75 degrees, so these two will be marked here. Now the question asks us for the smallest angle between two lines. We see that the smallest angle is 45 degrees, so the answer is B. Question 9. Which of the following is neither a square nor a cube of a natural number? Let's start working backwards from E. 2 to the power of 9. Now we can break this number down to 2 to the power of 3 times 3, which means that this number is a cube, so the solution cannot be E. Next, we have 3 to the power of 10, and this can be broken down to 3 to the power of 2 times 5, which means it is a square, so we can eliminate this possibility as well. Next, we have 4 to the power of 11. Now 4 can be broken down to 2, and this gives us 2 to the power of 11 times 5. 2 to the power of 11 times 2, so we can cross this off as well. Next, we have 5 to the power of 12, and similarly, like in D, we can just simplify this to 5 to the power of 3 times 2 times 2, so this number could be a square or a cube, so it cannot be B. We are only left with A, 6 to the power of 13. And now we know that this is the solution. So if we take a closer look at our diagram, we see that each of the little squares has a side length of 1, and there are 8 total squares. We want to get from start to finish in the least amount of moves. Now, we know that the fastest way to travel is through a diagonal, so we want to use this to our advantage, so we can travel two diagonals, and once we do this, we travel two straight lengths. Now, to calculate this distance, we need to remember how to calculate the hypotenuse of a triangle, since each of these squares can be divided into two right triangles. Now, we know that A squared plus B squared is equal to C squared, and in this instance, each of the squares has a side length of 1, so we will have 1 squared plus 1 squared equals C squared, and this simplifies to 1 plus 1 equals C squared, or 2. Now, we just have to take the square root, and we know that C equals the square root of 2. Now, we know the hypotenuse of one of these diagonals that we traveled. To get the total distance that we traveled, we just need to add together all the numbers. So, we traveled one diagonal, which is root 2, plus another diagonal, root 2, plus one length, plus another length. When we add all of these together, we get our answer, which will be C, 2 plus 2 times the square root of 2. Question 11. The number of right angles in a certain convex pentagon is n, which is the complete list of all the possible values of n. So first, let's start off by drawing a regular pentagon. Now, we notice that there are no right angles, so right away, we can cross off answers A and E, since those require at least one right angle, and we see that we can make a pentagon with no right angles. Next, we can make another pentagon starting off with one right angle. Now, we have two right angles, and now we finish this pentagon off, and we notice that we made a pentagon with three right angles, so we can cross off answer D. Now, we have B and C left, and the only difference between the two of them is that B includes a pentagon with four right angles. Now, if we try to draw this, we start off with one right angle, add another right angle, add an additional right angle, and the only way to make a fourth right angle is like so, making a rectangle or a square. So with that, we know that it is impossible to make a pentagon with four right angles. So our answer is going to be C, 0, 1, 2, and 3. Question 12. One of the letters A, B, or C appears on each of the faces of a certain die. The picture shows this die in two different positions. On how many faces of this die does the letter A appear? Now, we notice the diagram gives us two pictures of the same die from different perspectives. On the first picture, we have two A's and a C. And on the second picture, we have one A and two B's. Now, we can recognize that these are two opposite sides of the die. And we know this because if we look at the A, where there are two A's and a C, there is no way to combine the die that will make sense as a complete die because it will overlap on top of one of the B's. So we know that each of the A's in both pictures is unique. So the answer will have to be C, 3. Question 13. Mr. Candle bought a hundred candles. He burns one candle every day and always makes one new candle from the wax of seven burnt candles. After how many days will he have to go and buy new candles again? So first off, we know that we have a hundred days and a hundred candles. Now, after these a hundred days, we know that Mr. Candle will use the remains of the other candles to make new ones. So we can divide by seven and get 14 candles with a remainder of two. So this means that Mr. Candle can burn candles for another 14 days. Now, out of these 14 days worth of candles, again, we can divide by seven and get two candles with a remainder of two. So those were two days of candles once more. Now we have a remainder of four. So Mr. Candle cannot make any more candles. So now we just have to add up the amount of days that he was able to burn the candles for, which will be a hundred plus 14 plus two. And with that, we will get our answer, which is D, 116. Question 14. Every inhabitant of the winger planet has at least two ears. Three inhabitants named Eemee, Deemee, and Treemee met in a crater. Eemee said, I can see eight ears. Deemee said, I can see seven ears. Treemee said, that's strange. I can see only five ears. None of them could see his own ears. How many ears does Treemee have? So first we can set up an equation. Deemee and Treemee together will have eight ears. Since we know that Eemee sees both of their ears in total, which is eight. Eemee and Treemee have a total of seven ears. And Deemee and Eemee have a total of five ears. Now, if we combine all of these and do Deemee plus Treemee minus the sum of Eemee and Deemee plus Eemee and Treemee we will get this number, which will be 8 minus 5 plus 7, which is the total amount of ears, 10. Now, all we have to do is rewrite this formula and simplify without plugging in the numbers. And we get T plus T or 2T. Now both of these formulas are equal to each other. So we just have to set 2T equal to 10. And this means when we divide by 2 we know how many ears Treemee has. The answer is C, 5. Question 15 A container in the form of a rectangular prism whose base is a square with a side length of 10 cm is filled up with water up to a height of 8 cm. A solid cube with an edge of 2 cm was put in the container and the water level rose to a height of 2 cm. The smallest value of h for which this is possible is? So first, let's figure out the volume of the rectangular prism. Now, we know that the base is a square with a side length of 10 cm and the height is h, which is unknown. The volume will be equal to 10 times 10 times h or 100 h. Next, we look at the volume of the cube which has an edge of 2 cm. Simply, we just do 2 times 2 times 2. That gives us our volume of the cube, which is 8. Next, we are told once this cube is put into the prism the water level rises to a height of 2. So we can replace h with 2. And our new volume will be 10 times 10 times 2 which gives us 200. Now, we know that 100 h plus 8 equals 200. So we simply just simplify this and get 100 h equal to 192. And when we divide both sides by 100 we will get our answer, which will be a, 1.92 cm. Question 16. Today, the product of the ages of a father and his son is 2015. What is the difference of their ages? Now, there are multiple ways to divide up two numbers. We know the product is 2015. Of course, the simplest way is to multiply 2015 by 1. But it is ridiculous to assume that the father is 2015 years old. The next set of numbers that gives us this product is 403 and 5. But again, it is ridiculous to assume the father is 403 years old. Next, 155 and 13. This is still a tad ridiculous to assume the father is 155 years of age. So our next number will be 65 and 31. For a father and son, to be 65 and 31 is a reasonable assumption. So this is likely the ages of the two. All we have to do is to subtract the son's age from the father's age to get the difference. When we do this, we get our answer, which will be D, 34. Question 17. The square ABCD has an area of 80. Points E, F, G, and H lie on the edges of the square. And segments AE, BF, CG, and DH are all equal to each other. If segment AE is equal to 3 times segment EB, what is the area of the shaded region? So first, let's take a closer look at our diagram. We know that the area of the square is equal to 80. So the side length will be equal to root 80. Next up, let us divide the square in half. Since we know that both of these shaded end regions are equal to each other, dividing this square in half will make solving this problem easier, if we can just focus on one rectangle. Now, we know that segment AE is equal to 3 times EB. So that means segment EB will be equal to 1 fourth times root 80, and segment AH will be equal to 3 fourths times root 80. And we know that segment AH is equal to EB. So that is also 1 fourth times root 80. And since we divided the square in half, the side length of our new rectangle is 1 half root 80. Now, we have a rectangle that we can focus on. Now, we know that the area is equal to root 80 times 1 half root 80. Next, let's start looking at the different segments. We can see that this triangle, AEH, will have a total area of 1 half times 3 fourths root 80 times 1 fourth root 80. Since this is the formula for a right triangle. We have this triangle, which will have the area of 1 half times 1 fourth root 80 times 1 half root 80. Next, we can make an imaginary line dividing this total square, or the rectangle, in half again. And we can divide it on the 1 fourth line. We know that this new rectangle of segment EB will have the area of 1 fourth root 80 times 1 half root 80. In the right triangle, to the left of it, we'll have the total area of 1 half times 1 fourth of root 80 times 1 half of root 80. Now, we can stop looking at the diagram since we have all of the information that we need. We have the top formula, root 80 times 1 half root 80, and then we have the total areas of all of the white spots. So we can simplify these, and we get 1 half times 80, 1 half times 3 fourths times 1 fourth times 80, 1 half times 1 fourth times 1 half of 80, 1 fourth times 1 half times 80, and 1 half times 1 fourth times 1 fourth times 80. Which again, we can simplify to 5, 5, 10, and 5. Now, we just need to remember that this was the rectangle that we made. 40 is equal to the total area of the rectangle, and all of the other values are equal to the white spaces in the rectangle. So we take these together and subtract them, and we get 12.5. This is the value of the shaded triangle in the rectangle that we drew. We just need to remember that the rectangle that we calculated was half of the area of the entire picture we are looking at. So to find the total shaded area, we will multiply by 2. And with this, we get our answer, which is B, 25. Question 18. The side lengths of a given rectangle are prime numbers, and its perimeter is 70. What is the sum of the digits of the product of the two prime numbers? First we can draw a rectangle and label the side lengths x and y. Next, since we know the perimeter is 170, we can make the formula 2x plus 2y equals 170. If we simplify this formula by dividing both sides by 2, we get x plus y equals 85. So we know that the two numbers together, which have to be prime, will equal 85 when added together. So I've listed out all the prime numbers from 1 to 85. Now the first step that we can do to eliminate some of these possibilities is to erase any number that ends with a 1 or a 9. Because if the number ended with a 1, then our sum of 85 would end up with a number 4 at the end, so it could not be a prime number. The same would be true if we subtracted a number ending with a 9. We would end up with our sum being 85 minus a number ending with a 9, so a 6 would be at the end of that number, and that would not be a prime number. Next we can eliminate the 5, since 85 minus 5 will be 80, so the other number could not be a prime number. Now we are left with all of these numbers, and we can eliminate, again, any number that ends with a 7, since if we subtract from 85 a number ending with 7, we will end up with a number ending in 8, and that cannot be a prime number. Now all of these numbers are numbers that we can consider, but since we know that x plus y has to equal to 85, we notice we have the numbers 2 and 83. If we add these two numbers together, we get 85. Now the question wants us to multiply these numbers and get the product, which is 166. Finally, we are to add the digits of this product, so we will do 1 plus 6 plus 6, and that will give us our answer, which will be B, 13. Question 19. How many three-digit positive integers are there in which any two adjacent digits differ by 3? So let's name one of the digits in our three-digit integers A. If we assume that A is either 1 or 2, there are a number of ways that this could play out. We could have a number A, then a digit A plus 3, and then the third digit be A again. Or we could have A, A plus 3, and then A plus 6. So there will be four different options, because the numbers could be 141, or 147, or 258, or they could be 252. Next, if we have A either be 4 or 5, then our numbers could be the first digit A, then A minus 3, and the third digit A. So for example, if the number A was 4, we would have 414. Or if it was 5, we would have 523. We could also have A, A plus 3, and A. So again, we have four different options. Finally, if we have A be a number between 7 and 9, we could have A, A minus 3, and A. A, A minus 3, and A minus 6. This gives us six different options. However, we need to consider that A could be 3 or 6. In this case, our number options would be A, A plus 3, and A. A, A minus 3, and A. And this would give us four options total. There's one more scenario in which we have 369 and 630. This gives us two final options. So in total, there will be 14 plus 4 plus 2 options. And getting the sum of that gives us our answer, which will be D, 20. Question 20. Which of the following is a counterexample to the statement, if n is prime, then at least one of the numbers n minus 2 and n plus 2 is prime? Since we want a counterexample to the statement, we need to prove the statement false. So we can start off with answer A, n equals 11. However, if we add 2 to 11, we get 13. And that means this number is prime. So this would not be a counterexample to the statement. We can cross off A. Next, we can look at B, n equals 17. However, again, if we add 2, we get a prime number. So we can cross off B. Now, if we have C, where n is equal to 19, if we subtract 2, again, we have a prime number, 17. So that cannot be the answer. In D, where n is equal to 29, if we add 2, we get 31, which is a prime number. So it cannot be answer D. And finally, when n is equal to 37, if we add 2, we get 39, which is not a prime number. And if we subtract 2, we get 35, which again is not a prime number. So our answer is going to be E, n equals 37. Question 21. The figure shows seven regions formed by three intersecting circles. A number is written in each region. It is known that the number in any region is equal to the sum of the numbers in all neighboring regions. We call two regions neighboring if their boundaries have more than one common point. Two of the numbers are shown. Which number is written in the central region? We can take a closer look at our diagram, and we know 1, 2, and we're trying to find out the question mark. To start, we can label the other regions with A, B, C, and D. And start off by making the formula A equals 1 plus 2, since A is equal to the two neighboring regions. So we all know that A equals 3. Next, we can write the formula B equals 1 plus D. And then we can have C equals 2 plus D. With these two formulas, we can also write down C equals B plus 1. Next, we can write down the formula D equals C plus B plus question mark. Now, since we know that C equals B plus 1, we can put that into that formula and get D equals B plus 1 plus B plus question mark, which simplifies to D equals 2B plus 1 plus question mark. Now, the question mark will equal 3 plus D, since it is adjacent to 1, 2, and D. This can be rewritten as question mark minus 3 equals D. Since this is equal to D, and the formula above it is also equal to D, we can set them equal to each other and get question mark minus 3 equals 2B plus 1 plus question mark. When we simplify this, we get negative 4 equal to 2B. Dividing both sides by 2, we get B equals negative 2. Next, we look at B equals 1 plus D. Since we now know that B is negative 2, we can write down negative 2 equals 1 plus D, or negative 3 equal to D. So, D is negative 3. Now, the question mark is equal to all the adjacent numbers 1, 2, and negative 3. So, with that, we know that the question mark is equal to 3 minus 3, and this gives us our answer, which is A, 0. Question 22. Petra has three different dictionaries and two different novels on a shelf. How many ways are there to arrange the books if she wants to keep the dictionaries together and the novels together? So, when we have a question like this, where there are two types of things, and then one type of thing has three possibilities, and the other has two, we can write down a formula 2 times 2 factored times 3 factored, and this breaks down to 2 times 2 times 1 times 3 times 2 times 1, which we can then simplify to 4 times 3 times 2. With this, we know that our answer will be B, 24. Question 23. How many two-digit numbers can be written as the sum of exactly six different powers of two, including 2 to the power of 0? So, we can start off with 2 to the power of 0. That gives us 1. Now, we do 2 to the power of 1, which gives us 2. 2 to the power of 2 gives us 4, and we go so on, getting 8, 16, 32, 64, and 128. We can stop here. Since we know that 2 to the power of 7 gives us 128, there is no way that adding this number with five other numbers will give us a two-digit solution. So, we can eliminate that number, and we are left with seven different numbers. Now, when we add these all together, we get 127. Now, we know that we want the sum of six different powers of 2, so we can get rid of one of these seven numbers, but we need to get a two-digit number. Now, if we got rid of 2 to the power of 0, which equals 1, we would only be down to 126, which is still a three-digit number. So, instead, we need to get rid of the biggest number, 64. If we get rid of this number, we will be down to a two-digit solution. The same is true if we got rid of 32, which would also be a two-digit solution. Now, if we got rid of 16, we would still have a three-digit solution. So, there will be only two possible answers. So, the answer will be C. Question 24. In the triangle ABC, we can draw a line parallel to its base AC through point X or Y. The areas of the shaded regions are equal. The ratio BX to AX is 4 to 1. What is the ratio of BY to YA? Take a closer look at triangle ABC with line X. We know that the ratio is 4 to 1, so we can label BX as 4 and XA as 1. Now, we can use this formula to find out the area. We do 4 divided by 5, all to the power of 2, and we get this number by taking line segment BX, dividing it by BA, and this will give us 16 divided by 25. Now, since we want the shaded region of the triangle, we have to do 1 minus this number, and this gives us 9 25ths. Now, we know that when we have line segment A, and we have this new shaded region, the areas are the same. So, again, 9 divided by 25. So, simply, we just have to reverse this formula. What we do is root 9 divided by 25 and get 3 divided by 5. Now, with that, we will know what the ratio is. 5 minus 3 is 2. So, our answer of the ratio of BY to YA will be 3 and 2. So, the answer is D, 3 to 2. Question 25. A bicyclist rode the distance of 84 kilometers at a constant speed. Each hour, he rode 2 kilometers farther than he planned, which shortened his whole ride by 1 hour. What was the bicyclist's speed? So, to start this problem, we can write down the formula speed times time equal to 84. You also need to write down s plus 2 multiplied by t minus 1 equal to 84, since he, in actuality, rode 2 kilometers farther and 1 hour faster. Now, we can set these two equal to each other. So, the sum of s and 2 multiplied by the difference of t and 1 is equal to s times 2. Now, if we further this out, we can get s times t minus s plus 2t minus 2 equal to s times t. And if this is simplified, we get negative s plus 2t minus 2 equal to 0, or 2t equals s plus 2. Now, if we take a look at our original formula of s times t equal to 84, we can multiply both sides by 2, and this gives us the value of 2t. And with this, we can substitute 2t for s plus 2 and get s times the sum of s and 2 equal to 2 times 84. Simplifying this, we get s squared plus 2s equal to 168. If we break this down and add 1 to each side, we will get the sum of s plus 1 squared equal to 169. If we take the square root of both sides, we get s plus 1 equal to 13. Subtracting 1, we get s equal to 2. However, it is important to recognize that the bicyclist rode 2 kilometers farther than he planned, so we actually have to add 2 more to s to get our solution. And with this, we do 12 plus 2, which will give us our answer of d, 14 kilometers an hour. Question 26. In a certain right triangle, the angle bisector of an acute angle divides the opposite side into segments with the lengths of 1 and 2. What is the length of the bisector? To start off this question, there is a formula that we should keep in mind. We can draw out a right triangle and label the sides as A, the hypotenuse is B, and the remaining sides that are divided by the bisector as X and Y. Now we can keep in mind that A divided by B is equal to X divided by Y. Now moving on to the actual problem, we draw out our triangle with the bisector, and we have side of H, and then we have the two segments, which will be 1 and 2, together 3, and the hypotenuse will just be Pythagorean theorem, which is root of H squared plus 3 squared, and then our bisector makes these segments be 1 and 2. Using the formula that we just discussed, we can set these equal to each other, and we get H divided by the square root of H squared plus 9 equals 1 half. Simplifying this, we get 2H equals the square root of H squared plus 9. We can further simplify this until we get 3H squared equals 9, or H squared equals 3. Now we want to get the hypotenuse of the smaller triangle that we made with the angle bisector, so we just simply have to do H squared plus 1 squared to get the length of the bisector. Doing this, we get the square root of 3 plus 1. We now have the solution, which is C, the square root of 4. Question 27. After one of the numbers 1, 2, 3, and so on, n minus 1, n was eliminated. The mean of the remaining numbers was 4.75. What number was eliminated? So first off, we need to start with a formula that determines the mean. We can start off with n times the sum of n plus 1 divided by 2 minus r, and now we need to divide this whole thing by n minus 1, and the question tells us that the mean of the numbers will end up being 4.75. Now, we know that 1 will be less than or equal to r, which will be less than or equal to n, since the other values wouldn't allow this to make sense. Now, if we write this down and replace the values as 1, r, and n, we can get an equation that looks like this, which we can simplify to n halves is less than or equal to 19, 4 is less than or equal to n plus 2 divided by 2. Or, we can simplify this even further and get 2n is less than or equal to 19, which is less than or equal to 2n plus 4. Now with this, we know that 2n is less than or equal to 19, so n is less than or equal to 9, and since we know that 19 is less than or equal to 2n plus 4, we can simplify this and get 8 is less than or equal to n. With these two values of 8 and 9, we can go back to our original formula and input 9. So, we get a formula that looks like so. When we simplify it, we get 45 minus r over 8 is equal to 4.75, and we just have to simplify this formula even further. When we multiply both sides by 8, we get 45 minus r equals 38. So with this, we can figure out that our solution will be B, 7. Question 28. The number 12 has six factors that are natural numbers, 1, 2, 3, 4, 6, and 12. What is the largest number of factors that are natural numbers that a two-digit number can have? Let's start this problem off backwards with the largest number. So, we can start off with answer E, 16, and test if it is correct. With this, we can set two sides, both having eight values. So, to maximize our numbers, we can start off with the smallest numbers possible, 1 through 8. Now, we know to get a big number, we will multiply this 8 by the smallest number on the other side. So, if we, for example, multiplied 8 times 9, the highest number that we could multiply 8 by, while still having a two-digit number, would be 12. And the numbers in between would be 10 and 11. So, 8 times 9 would be 72. Now, this would not be a possible answer, because it would not include the 5. We cannot multiply 5 by any value to get us 72. If we multiply 8 by 10, we get 80. This, again, has the same issue, but instead of 5, we look at the number 3. There is no value that we can multiply 3 by to get 80. If we multiply 8 by 11, we come back to the issue with the 5. And if we multiply 8 by 12, again, we have no value to multiply 5 by. So, with this, we can cross out answer E. And now, we can get rid of two more slots on each side, and test out D, 12. So, now we have the values 1 through 6. Now, we can look at the values that we can multiply 6 by, while still having a product that is two digits. We can start off with multiplying by 7, and the biggest number we can multiply by is 16, and all the values in between. Now, when we multiply 6 by all of these values, these are all of the answers that we can get. Right off the bat, we can cross out most of these, since they would not be able to multiply by 5. We are only left with 60 and 90. When we multiply 6 by 10, we get 60. And when we multiply 6 by 15, we get 90. So, if we look at 60 as our two-digit number, we can see 6 times 10 gives us 60. 5 times 12 gives us 60. 4 times 15, 3 times 20, 2 times 30, and 1 times 60. With this, we find our solution. The largest number of factors that are natural numbers that a two-digit number can have will be D, 12. Question 29. Ten different numbers are written down. Any number that is equal to the product of the other nine numbers is then underlined. How many numbers can be underlined at most? Let's start off with A and B. Now, the product times A must equal B, and the product times B must equal A. We can rewrite this formula and get P times A divided by P times B equals B over A, or A squared equals B squared. Now, with this, we can write down A equal to negative B, since when a value is squared, it will lose its negative sign, even if it is a different number. So this means that P can equal to negative 1. With this, we know that our solution will be 2. We will have numbers that are negative and positive, and the most that we can underline will be two values. Question 30. Several points are marked on a line, and all possible line segments are constructed between pairs of these points. One of the points lies on, and is not the endpoint of, exactly 80 of these segments. Another point lies on, and is not the endpoint of, exactly 90 segments. How many points were marked on the line? Whenever we label a point, we can label it as X, and it will be between points A and B, for example. Now, we can call the number of points marked on the line N, and we have the formula of A plus B equals N minus 1. Now, since the question tells us that the first point lies on 80 segments, and the other point lies on 90 segments, we can put down these formulas. A times B equals 80, and C times D equals 90. And we can set these equal to each other, A plus B equals C plus D, because this line has the same length. Now, 80 can be broken down like so, and 90 like so. To keep our formula of A plus B equal to C plus D, we can get the given values for A, B, C, and D. 5, 16, 6, and 15. Now, with these, we can put them back into this formula, and we get 5 plus 16 equals N minus 1. And when we simplify this, we get our answer, which is B, 22.

Math Kangaroo

2015 Competition

Lukasz Nalewskowski

problem-solving

mathematical techniques

geometric reasoning

algebraic problems

multiplication estimation

international competition