This is the Math Kangaroo Solutions Video Library, presenting solution suggestions for Levels 7 and 8 for the year 2020. These solutions are presented by Patryk Lipski. The purpose of the Math Kangaroo Solutions Video Library is to help you learn how to solve math problems, such as those featured in the Math Kangaroo competition. It is important that you make sure to read each problem as well as listen while I read the problem. After reading and listening to a problem, pause the video and attempt to solve the problem on your own before viewing the solution. Problem 1. How many of the following four numbers 2, 20, 202, and 2020 are prime? A prime is a number that is only divisible by 1 and itself. 2 is a prime, since it is divisible by 1 and 2 only. The 20, 202, and 2020 are all even numbers, which means they are divisible by 2. Therefore, 2 is the only prime here, so the answer is just one number, b. Problem 2. In which of the regular polygons below is the marked angle the largest? There is a good formula to know for the size of a regular polygon's interior angle. We can get it by drawing a polygon, here a pentagon, and dividing it into triangles. The polygon with n sides, here 5, will split into n-2 triangles, or 3 here. Each triangle has 180 degrees, so the sum of all the polygons' angles is 180 times n-2. To get the size of just one angle, we have to divide by n. Each angle in the problem is some fraction of 180. The hexagon's angle in A is 4-6 of 180. The two pentagons in B and C each have 3-5 of 180, 2-4 for the square, and 1-3 for the triangle. The biggest portion of 180 here is in the hexagon, or answer A. Problem 3. Miguel solves 6 Olympiad problems every day, and Lazaro solves 4 Olympiad problems every day. How many days does it take Lazaro to solve the same number of problems as Miguel solves in 4 days? To begin, we need to figure out how many problems Miguel solves in 4 days. Since Miguel solves 6 problems every day, he'll solve 6 times 4, or 24 problems in 4 days. Since Lazaro solves 4 problems every day, we need to divide 24 by 4 to get the number of days. 24 divided by 4 is 6, and for Lazaro, this is 6 days worth of problems. That is answer C. Problem 4. Which of these fractions has the largest value? A is 8 plus 5 over 3. 8 plus 5 is 13, so that is 13 thirds. B is 8 over 3 plus 5. 3 plus 5 is 8, so that is 8 eighths. That is 1, which is less than 13 thirds, so B cannot be the right answer. C is also 8 over 8. That is also 1, so that also cannot be the right answer. D is 8 plus 3 over 5, or 11 over 5. Now, 13 is greater than 11, which is a bigger numerator, and 3 is less than 5, which is a smaller denominator. Therefore, D is smaller than A, so D cannot be right. Finally, E is 3 over 8 plus 5, or 3 thirteenths. This is even less than 1, which is definitely less than A. That leaves A as the largest answer. Problem 5. A large square is divided into smaller squares. In one of the squares, a diagonal is also drawn. What fraction of the large square is shaded? Taking a look at our square, we can see that it is cut into fourths. The gray bottom right square is one fourth. Next, the diagonal cuts the top right square in half, so we get half a fourth there, or an eighth. The top left square is cut into fourths again, so we have two pieces that are fourths of fourths, which makes sixteenths. Now, let's simply add our area fractions to get the total area. Starting with the smallest pieces, a sixteenth plus a sixteenth makes two sixteenths, or an eighth. Then using that eighth and the eighth right next to it, we can add them to get a fourth. Finally, that fourth and the bottom right fourth make a half, so our answer must be E. Problem 6. There are four teams in a soccer tournament. Each team plays each of the other teams exactly once. In each match, the winner receives three points and the loser receives zero points. In the case of a tie, both teams receive one point. After all the matches have been played, which of the following total number of points is impossible for any team to have received? There are four teams, which means that each team will play three games, or one, against every other team. Therefore, the point totals will be the points possible to achieve in three games. To start, let's think of the biggest amount of points possible. This happens with three wins, or three sets of three. That is a total of nine. The second highest score would happen with two wins and a tie. That is two threes plus one, or seven. Now we already see that since nine is the highest score and seven is the second highest score, eight is impossible to achieve. Just for good measure, we can show how the other three scores are possible. C is possible with two wins and a loss, B is possible with a win and two ties, and A is possible with a win, a tie, and a loss. Problem seven. The diagram shows a shape made of 36 identical small triangles. What is the smallest number of such triangles that can be added to the shape to turn it into a hexagon? To get a hexagon, we will definitely have to fill in the points of this star-like figure. That's six additional triangles. Now we definitely have to fill in these dents. That makes two triangles per dent, or twelve more triangles. We see that now we have a complete hexagon, and it took 18 additional triangles to draw. Another way to think about it is we added three triangles to each of the hexagon's six sides. The answer is therefore D. Problem eight. Congo wants to multiply together three different numbers from the following list. Negative 5, negative 3, negative 1, 2, 4, and 6. What is the smallest result she can obtain? Since Congo wants a small number, she should aim to achieve a negative. There's two ways to multiply three numbers to get a negative. One is negative times negative times negative, and the other is negative times a positive times a positive. In the first way, there are only three negative numbers. Negative 5, negative 3, and negative 1. Multiplying them together gets negative 15. In the other way, negative times positive times positive, we can multiply negative 5 by the two largest positive numbers, 6 and 4. This gets negative 120. Since negative 5, 4, and 6 are the largest numbers on the list, we cannot get any more negative than negative 120, so the answer must be B. Problem nine. When John goes to school by bus and walks back, he travels for three hours. If he goes by bus both ways, he travels for one hour. How long does it take him if he walks both ways? We have two facts. Taking the bus and walking takes three hours, and taking the bus both ways takes one hour. Dividing this second equation by two, we find out that taking the bus one way should take half an hour. So now we can subtract half an hour from the bus plus the walk to find out that a walk takes two and a half hours. Therefore, walking both ways should take five hours. The answer is D. Problem ten. A number is written in each cell of a three by three square. Unfortunately, the numbers are not visible because they are covered in ink. However, the sum of the numbers in each row and the sum of the numbers in two of the columns are all known, as shown by the arrows in the diagram. What is the sum of the numbers in the third column? The key here is that the sum of the rows equals the sum of the columns, because both sums use all nine numbers on the board. The first row is 24, the second row is 26, and the third row is 40. This makes a total of 90. Now let's look at the columns. The first column is 27, the second column is 20. Now we are missing three numbers from our total. To find the numbers, we have to subtract 90 minus 27 minus 20, which gets 43. Therefore, the sum of the last column must be 43, which is our answer, B. Problem 11. The shortest path from A-town to C-town runs through B-town. The two signposts shown are set up along this path. What distance was written on the broken sign? We have our path from A-town to C-town through B-town. The first signpost has A-town on the left and B-town on the right, which puts it in between the two. The second signpost has both on the left. Distance from the second signpost to A-town is 7, and from the first signpost to A-town is 2. Therefore, the distance between the two signposts is the difference of 7 minus 2, or 5. This distance is split by B-town. Distance from the first signpost to B-town is 4, so the distance from the second signpost to B-town must be 5 minus 4, which is 1. That gives our answer as A. Problem 12. Anna wants to walk 5 kilometers on average each day in March. At bedtime on 16th March, she realized that she had walked 95 kilometers so far. What distance does she need to walk on average for the remaining days of the month to achieve her target? It is important to note that March has 31 days. Therefore, if Anna wants to walk 5 kilometers on average each day in March, she needs to walk 31 times 5, or 155 kilometers. Since she has walked 95 kilometers so far, we can subtract that from her goal to find out that she still has 60 kilometers to go. She has to complete these 60 kilometers from March 17th to March 31st. If we take away the first 16 days from 31 days of March, we see that she has 15 days to complete her goal. Then we can divide the remaining 60 kilometers by 15 days to see that she needs an average of 4 kilometers a day. That is answer C. Problem 13. Which of the following shows what you would see when the object in the diagram is viewed from above? We need to see which of these is arranged the same as the figure. To start, let's look at the two adjacent black diagonals. They are present in every answer. Going clockwise from the black pair, we should see a white and a gray diagonal. In B, C, and E, this is what we see. However, in A and D, those diagonals are flipped. Then we can eliminate A and D. Next, let's look at the gray side in between the two black diagonals. This side is present in B, but not in C or D. So we can eliminate C and E. The only answer remaining now is B. Problem 14. Every student in a class swims, dances, or does both. Three-fifths of the class swim and three-fifths dance. Five students both swim and dance. How many students are in the class? We can represent our class like so, and we know that three-fifths of the class swim and three-fifths dance. There is one-fifth overlap of the student body. This also represents five students. So one-fifth of the students is five students. If we multiply both sides by five, we get that all the students together make 25 students. Therefore, the answer is C. Problem 15. Sasha's garden has this shape shown. All the sides are either parallel or perpendicular to each other. Some of the dimensions are shown in a diagram. What is the perimeter of Sasha's garden? We can start with the edge length 3. If we look at the rightmost imaginary edge length 3, we can split it across the right edges of the figure. Therefore, the green edges add up to 3 times 2, or 6. Now let's look at the edge of length 5. We can create doubles of the edge of length 5 with these pieces. These parts of the perimeter have length 5 times 2, or 10. Finally, let's look at the piece of length 4. We can copy it down below, and this way we complete the figure with two blue pieces of total length 8. Now the total perimeter is the sum of these three sets of pieces. 6 plus 10 plus 8 is 24, which is our answer. Problem 16. Andrew buys 27 identical small cubes, each with two adjacent faces painted red. He then uses all of these cubes to build a large cube. What is the largest number of completely red faces of the large cube that he can make? Since he has 27 small cubes, Andrew's cube must be 3 by 3 by 3. Each small cube has two adjacent faces painted red. Let's look at the corner cubes. Since each corner has three faces exposed, at least one face must be left white, not red. So we're going to get an answer less than 6. And in fact, here we already see that at least two faces are incomplete, so the answer is at most 4. Let's fill in the faces that we've already started by laying the other edge pieces the same way. We can definitely fill in the non-edge pieces, since we just need one red face for each. Now we have the front and top red, but you can imagine we did the same thing in the back and the bottom, giving us four complete red faces, and in this case, two white faces. Again, we can see that there is no way to have the white face turn red without messing up a complete face. So the most faces we can have completely red is 4, or C. Problem 17. A large square consists of four identical rectangles and a small square. The area of the large square is 49 cm2, and the length of the diagonal AB of one of the rectangles is 5 cm. What is the area of the small square? The sides of the crosshatch figure are each 5. It appears to be a square of side 5. To be sure, we can see that the corner angles consist of a pair of opposite angles from same size right triangles, which together make a 90 degree angle. This medium square has an area of 25 since the side length is 5, and it consists of half the gray rectangles and the white square in the middle. In other words, half the rectangles plus the white square has an area of 25. The total area of the figure is 49, which consists of both halves of the rectangles and the white square in the middle. If we subtract the top from the bottom, we see that half the rectangles must have an area of 49-25, or 24. Then the white square must have an area of 25-24, or 1, which is answer A. Problem 18. Werner's salary is 20% of his boss's salary. But what percentage should Werner's salary increase to become equal to his boss's salary? So 20% is 1 5th. Therefore, Werner needs a 5-fold increase in his salary to match his boss's salary. Then if we take Werner's salary, we will need 4 more equivalent salaries to get a 5-fold increase. Each of these is 100% added. Then we are adding 4 times 100%, or 400%. So Werner needs a 400% increase to his salary, that is D. Problem 19. Irene made a city using identical wooden cubes. One of the diagrams shows the view from above the city and the other the view from one of the sides. However, it is not known from which side the side view was taken. What is the largest number of cubes that Irene could have used? Irene could look at the cubes from the north, south, east, or west. If she looks from the south, she would see, as we can see in the figure, two blocks in the first row. This means that the tallest tower of blocks in that row is two blocks tall. In the fourth row, the tallest block is three. One possible arrangement of tower sizes is 3, 2, 1, while another is 1, 2, 3. Both are valid so long as the tallest tower has three blocks in it. To get the most blocks possible, each tower should be as tall as possible. This is similar for the other rows. From each direction, there is a different number of 1, 2, 3, and 4 block towers. We have to count, although the highest number of blocks will likely occur with the most 4s and 3s, as when looking from the north. That arrangement has 24 blocks. And, indeed, when we count the other blocks, we see that it is the largest number possible. Therefore, the answer is 24, or B. Problem 20. Aisha has a strip of paper with the numbers 1, 2, 3, 4, and 5 written in five cells as shown. She folds the strip so that the cells overlap, forming five layers. Which of the following configurations, from top layer to bottom layer, is it not possible to obtain? With three on top, possible arrangements will have 1 and 2, and 4 and 5 folded under. These two arrangements work for C and D. Something to notice is that 1 and 2 stay together, even though their order is reversed, and 4 and 5 stay together. By reversing D, we can get a fold pattern for A. Again, 1 and 2 are together, and so is 4 and 5. In B, we have the same grouping, and so B is also possible. In E, the flaps are not paired up correctly, and so the two pairs would have to intersect to make the arrangement work. Therefore, E is impossible to obtain. Problem 21. Twelve colored cubes are arranged in a row. There are three blue cubes, two yellow cubes, three red cubes, and four green cubes, but not in that order. There is a yellow cube at one end and a red cube at the other end. The red cubes are all touching. The green cubes are all touching. The tenth cube from the left is blue. What color is the sixth cube from the left? We will have to try and arrange the cubes. To start, we know that one end has a yellow cube and the other has a red cube. We also know that all three red cubes are touching. That gives us two possible arrangements to examine. Next, we know that the tenth cube from the left is blue. This is the same as the third cube from the right, and we can try and draw it in. We see that the red cubes have to be on the left, or else they will overlap with this blue cube. Now, we can use the last clue to examine the sixth cube. The last clue is that all four green cubes are touching. Therefore, we must use four adjacent spots for the green cubes. We can put them here or in these two spots as well. No matter what, the sixth cube is always green, which is the answer A. Problem 22. Zeta took a square piece of paper and folded two of its sides to the diagonal as shown to obtain a quadrilateral. What is the size of the largest angle of the quadrilateral? If we unfold Zeta's quadrilateral, we will get back a square. Because she folded these sides to the diagonal, each of the four angles in the folded corner must be the same. We can find the largest angle of the quadrilateral by finding the angle adjacent to it, since they will add up to 180. We can get the adjacent angle by looking at the other angle in the left-right triangle. That small angle is 90 split into 4, which makes it 22.5. Then the adjacent angle is 90 minus 22.5, or 67.5. That means that the angle we are looking for is 180 minus 67.5, or 112.5. That is the answer A. The first fact is that A divided by 2 is twice some whole number. The second is that A divided by 3 is 3 times some whole number. And the third is that A divided by 5 is 5 times some whole number. If we multiply both sides of the first equation by 2, we see that A must be divisible by 4. Similarly, A must also be divisible by 9 and by 25. This means that A must be divisible by 2 squared times 3 squared times 5 squared. In other words, A is divisible by 900. The multiples of 900 with 4 digits are 900 times 2, 900 times 3, and continuing this pattern all the way up until 900 times 11, or 9,900. Since we have 10 numbers in our list, the answer is D. Problem 24. In the finals of the dancing competition, each of the three members of the jury gives the five competitors 0 points, 1 points, 2 points, 3 points, or 4 points. No two competitors get the same mark from any individual judge. Adam knows all the sums of the marks and a few single marks, as shown. How many points did Adam get from Judge 3? We can start filling out the points by looking at Berta. Berta is only missing one score, and to get her sum of 5, she needs a 3 from Judge 3. Next, we can examine Emil. Emil has a score of 11. The only way he could get this score is with two 4s and a 3. Since Judge 3 has already given away his 3, Emil needs a 4 from Judge 3. Then he needs either a 4 or a 3 from Judge 2 and a 4 or a 3 from Judge 1. Let's assume that he gets a 4 from Judge 2. In that case, he will get a 3 from Judge 1. And Judge 1 will have to give Clara or David his 4. If Judge 1 gave David his 4, then to keep David's sum of 4, Judges 2 and 3 would have to each give David a 0. But Judge 2 has already given away his 0. If Clara were to receive Judge 1's 4, that's impossible because her sum is only 3. Therefore, Judge 1 cannot give away his 4 this way. So Judge 2's 4 is incorrect, and it must be a 3 instead. Now, we have enough information to examine Adam's score. Adam is missing two scores, but we can tell that they are definitely not going to be 3, since Judges 2 and 3 already gave away their 3's. Since Adam needs to get a sum of 7, but he already has a 2, he is missing 5 points. Since he cannot have a 3, he cannot make that 5 with 2 plus 3. Therefore, he needs 1 or 4 from Judge 2 and 3. Since Judge 3 has already given away his 4, he must give Adam a 1, and Judge 2 must give Adam a 4. Therefore, Adam receives 1 from Judge 3, B. Problem 25. Sania writes a positive integer on each edge of a square. She also writes at each vertex the product of the numbers on the two edges that meet at that vertex. The sum of the numbers at the vertices is 15. What is the sum of the numbers on the edges of the square? Sania draws four numbers on the square, but we don't know what they are, so let's call them A, B, C, and D. Then, the numbers at the corners are A times B, B times C, C times D, and D times A. When she adds up all these products, she gets 15. We can make this sum nicer by rearranging to group A times B with A times D, and C times B with C times D. Then we can factor the A and C to get A times the sum of B and D added to C times the sum of B and D. Let's group these together one more time to get the sum of A plus C times the sum of B plus D. Once again, this should equal 15. Remember that from the problem, A, B, C, and D are all positive integers, so they're all 1 or bigger. That means that A plus C is greater than 1, and B plus D is greater than 1. The only way we can multiply two numbers to get 15, however, is with 1 times 15, or 3 times 5. Since A plus C and B plus D are each greater than 1, the two sums must be 3 and 5. Therefore, the sum of A plus C plus B plus D, or the four sides of the square as we're looking for, is 3 plus 5, which is 8. That makes the answer C. Problem 26. Sophia has 52 identical isosceles right triangles. She wants to make a square using some of them. How many different sized squares can she make? Sophia is using isosceles right triangles. She can put two of them together to make a square. To make other squares, she can put these squares together like so. To make squares, she needs a square number of square tiles. Each square is two triangles, so we will need twice the square number of triangular tiles. The limit here is Sophia's 52 triangles. We can divide both sides by 2 to get the equation T squared is less than or equal to 26. Since we want a whole number of triangles, our possibilities for T are the squares less than 26, or 1, 4, 9, 16, and 25. This means that we have 5 possibilities for T in our equation, or 5 possible squares. We can also combine right triangles into squares a second way. We can put 4 right triangles together at their right angles. As before, we combine these new squares of 4 triangles into bigger squares. This time, since each square is 4 triangles, we need that 4 times the square number of triangles is less than 52. We can divide by 4 to simplify, and this time our solutions are the squares less than or equal to 13, namely 1, 4, and 9. This gives 3 more possible squares, and brings the total to 8. To see that no other squares are possible, let's give lengths to the triangular pieces. A piece might have legs 1, and then by the Pythagorean theorem, a hypotenuse square root of 2. The triangle's area is then 1 half. Then, the blue squares consisting of 2 triangles have sides of length 1, while the green squares of 4 triangles have sides of square root of 2. Now, suppose that we have a square whose side consists of a one-sided squares and b square root of 2 sided squares. Then the total side length of this new square is a plus b root 2. Squaring the side length to get the area, which we could do by FOIL, we get a squared plus 2b squared plus 2ab square root of 2. a and b are rational, so 2ab square root of 2 makes this total irrational. But since each triangle has a rational area of 1 half, a square made out of triangles will also have a rational area. So since this total is irrational, we cannot have a square that's made out of blue and green squares. Then our answer remains 8, or c. Problem 27. Cleo builds a pyramid with metal spheres. The square base consists of 4x4 spheres as shown in the figure. The floors consist of 3x3 spheres, 2x2 spheres, and a final sphere at the top. At each point of contact between two spheres, a blob of glue is placed. How many blobs of glue will Cleo place? Here we have the base of Cleo's pyramid and the next three layers. We can see that each sphere above the bottom layer rests on top of 4 others. The points of contact are marked in red. That means that each sphere in the top three layers will have 4 blobs of glue attached. This makes 4 times the sum of 9, 4, and 1 blobs of glue, or 56 total blobs. Each sphere is also attached to the spheres next to it in a layer. Here we have a row of 4 attachment points between the top two rows of the bottom layer. There are two more rows of attachment points like this, where we will also need blobs of glue. Likewise, there are 3 columns of 4 attachment points each. This makes 2 sets of 4x3 glue blobs in the bottom layer. We will also have 2 sets of 3x2 blobs in the second layer and 2 sets of 2x1 blobs in the first. This makes 2 sets of 4x3 plus 3x2 and 2x1, or 12 plus 6 plus 2, which is 40 extra blobs. Adding these two totals makes 96, or E. Problem 28. 4 children are in the 4 corners of a 10 meter by 25 meter pool. Their coach is standing somewhere on one side of the pool. When he calls them, 3 children get out and walk as short a distance as possible around the pool to meet him. They walk 50 meters in total. What is the shortest distance the trainer needs to walk to get to the 4th child? Let's draw a picture of the pool and the kids. Here, the kids are at the corners and the coach is on the north side of the pool. The kids are going to climb out and follow these paths to the coach. Let's follow 2 kids at opposite corners. Their paths seem to cover half the perimeter of the pool. If we look at the kids at the other 2 corners, they also cover half the perimeter of the pool. Together, it looks as though the 4 kids walk the total perimeter of the pool. The total perimeter is 2 sides of 10 added to 2 sides of 25, or 70 meters. Since 3 kids walked 50 meters, we have to subtract 50 from 70 to get the distance from the coach to the last child, which comes out to 20 meters. That's answer D. Problem 29. Anne, Boris, and Carl ran a race. They started at the same time and their speeds were constant. When Anne finished, Boris had 15 meters to run and Carl had 35 meters to run. When Boris finished, Carl had 22 meters to run. What is the distance they ran? We need to look at the facts this problem gives us. Carl had 35 meters to run when Boris had 15, but Carl only had 22 meters to run when Boris finished. So, in the time that Boris finished, Carl ran 35 minus 22, or 13 meters. We can say that while Boris ran 15 meters, Carl ran 13. The difference in their running speed is 2 meters per unit time. So, let's see how long it took Boris to get 22 meters ahead of Carl. Since every time unit Boris gets 2 meters ahead of Carl, and he's 22 meters ahead at the end, he must have taken 22 divided by 2, or 11 time units, to get ahead of Carl. Since Boris is running 15 meters every time unit, we can find his total distance run. 15 meters, for 11 time units, multiplies out to 165 meters, or answer D. Problem 30. The statements below give clues to the identity of a 4-digit number. What is the last digit of a 4-digit number? First, we can look at the last clue and immediately eliminate many occurrences of 7, 6, 4, and 2. From the first clue, we know that our number has to contain a 1 and a 3, since those two digits are correct. Let's examine the second clue. Since one digit is correct, it's either the 8 or the 9. Let's see if it might be 8. Then we eliminate the 9s, and we put 8 in our confirmed digit list, as well as its correct place. Now, if we look at the third clue, we see that the 5 and 0 must be the two correct digits. So we know that our number has to contain 5 and 0 as well. Now we have a problem, since we have 5 digits that have to be in our 4-digit number. Since that's not possible, we must take back our assumption about 8 being correct, and say that 9 is the correct digit in the second clue. Since 9 is correct in the leftmost position, the 5 in the third clue cannot be in its correct spot anymore. This leaves the 0 as the other correct digit. Now we know that either 1 or 3 is the last digit. If we look at the fourth clue, we see that the 1 is wrong when it is last, which means that the last digit must be 3 instead. Then the answer is 3, or C.

Math Kangaroo

problem-solving

Patryk Lipski

geometric shapes

prime numbers

logic puzzles

mathematical concepts