Hello my friends, in this video I will show you the solutions for 2017 Mathkangaroo competition for levels 9 and 10. If you have any questions about them you can contact me by an email that you can see on the bottom of this video. You can also check my YouTube channel Let's Solve It Together. I hope these solutions will help you better understand the problems and math in general. So let's jump into them. So let's start with number 1. In the diagram each number is the sum of the two numbers below. Each number must be in the cell marked with an exclamation mark or how do you call it. Well let's look closer, aha, 2039, 20 and 70. So we have something like this. Here we have 2020 and here we have 2017. So that's how it looks. And we want to know what is here in this left bottom corner. So every number is the sum of the numbers below it. So here we have 2039. This is a sum of this number here and this one. This one we know, it's 2020. So what had to be here? Well 19, because 19 plus 2020 will give us 2039. Now we cannot just yet tell what is here, but we can tell what is in the middle. Because we know that here is 2020 and this is a sum of this number and 2017. Which means that here must be 3. And now we can easily solve this one, because 3 plus something equals 19. So this something is just 16. And this is the answer. So question 1st, answer B. Let's check and yeah, here it is. This is this second column from right, it's level 9 and 10. So here we are looking for answers. Okay, so question number 2. Peter wrote the word kangaroo on a piece of transparent glass. What will he see if he turns this piece over around its right side and then rotates it one half turn? So the first time I've seen this, I understood it not right. Well, because I thought that if he turns this piece over around its right side, I thought it's like this. And what it would do? Well, it would do that those O's that were on the end, they now will be on the beginning. And now if he turns half turn over, then it's upside down. So I thought that the answer is D, but actually it cannot be D. If that what I think was the case, then D wouldn't be right, because then R here, you can see R looks like this. But if you do what I did, then R would look like this. So there's no answer for this possibility. And when I thought about this once again, he turns this piece over around its right side, and then it said that he rotates it one half turn. So the first thing is actually that he turns over around, so he makes a full circle. And well, it doesn't change anything. It is still kangaroo, exactly what you can see. And now he rotates its half turn, which makes it upside down. And this is simply answer E. Yes, you can see here, a word kangaroo written upside down. So for the question two, the answer is E. Okay, let's check to E. Yeah, so we can go to the question three. Angela made the decoration with gray and white stars. The areas of the stars are one centimeter square, four centimeter square, nine centimeter square and 16 centimeter square. That is the total area of the visible gray regions. Okay, so as far as I understand, it looks like this, that on the bottom, you have full gray star, but on this lays white star, and on this lays gray star, and on the top, you see white star. So actually, this 16 centimeters is not the part that you see, it's everything. Everything here in the in the center would be 16 centimeters square. And on this lays a nine centimeter squared star that is white. So how much is this shaded area? Well, it's simply 16 minus nine centimeters squared, which will give us seven centimeters squared. And what is this shaded area? Well, this is a star also that has a four centimeters squared area, and on the top of it lays the star that has one centimeter squared of area. So the shaded part is four minus one centimeters squared, and this is three centimeters squared. So now, what is the sum of those shaded areas? Well, this is seven plus three centimeters squared, which give us 10 centimeters squared. This is the answer, 3b. And let's check in the answer key, 3b, yes, so. And the last question in this video for Maria has 25 euros, each of her three siblings has 12 euros. How many euros does she need to give to each of her siblings, so that each of the four siblings has the same amount? Well, so let's start with the question, how much do they have in the beginning? And well, it will not change, because if she will give them their money, in total, they will still have exactly the same amount. So Maria has 24, and her three siblings have 12 each, which means that together they have 24 plus 3 times 12 euros. This is 24 plus 36, which gives us 60 euro. So they have 60 euro. So that if we want to make the situation where they all four have the same amount, well, how much will they have? Well, they will have 60 divided by 4, which gives us 15 euro. So they should have 15 euros, everyone. So she has to give all their three siblings 3 euros to each. This is the answer for C. Let's look at answers for C, yes. So she will give them a total 9 euros, so that she will have 15, but dividing it by 3, it will be 3 euros for each. Now we are starting with question number 5. Which of the following pictures shows the path of the center of the wheel when the wheel rolls along the zigzag line shown? OK, so let's look closer at these pictures. So here we have the terrain is exactly the same, like triangles. And let's think what happens. Well, this wheel turns, the center will surely be still all the time in the same distance from this line. So it has to go parallel. OK, and everywhere goes parallel in the beginning. But then we see the first difference close to this vertex. So what happens here? Well, let's, I will write this, I will draw this. Well, you have this point, yes. The wheel goes, yes, like this all the time in the same distance. But what goes here? So this center has to be in the same distance from this point all the time. And this point doesn't change. Here, the center was not changing and these points were changing. But here, this point is not changing and the center has to go by a circle. So because the set of points that are always in the same distance from center is a circle. So here, in this, close to this point, we have to see circle. And let's look at those possibilities. So it cannot be this one, yes. It's here, the distance between the center and this point would be bigger than it was here. So the wheel would actually levitate over this point. So not this, also not this, because, well, if there would be something like this, then OK, it can go this straight line. But not in this situation. Also not this, because the same problem like with A, but D, D is OK, but also E is OK. So there has to be a difference. And yes, we can see this difference here, where we are close to this point in bottom. So here we see that it goes like this. And here we go, we see it goes by a circle. So which one will happen? Well, this circle will be going like this all the time in the same distance from the points of this straight line. But it will get in some moment to such a situation that it's stuck. And if it wants to go further, it has to turn in a moment. So it will not be a smooth changing direction. It will be with this triangle like change. So this will have to be answer E. Yes, because this, because here you need to change direction in one moment. And it's not the way that of changing direction that it will be smooth like here. No, it will go like in triangle. So this is answer E. 5E, 5E. OK, so now let's go to the question number six. Some girls were dancing in a circle. Antonia was the fifth to the left from Bianca and the eighth to the right from Bianca. How many girls were there in the group? So let's look at this situation. Here we have Antonia, here we have Bianca. And Antonia is fifth to the left from Bianca. So if she has to be fifth, then it means that there are four girls between them by going this way. And eighth to the right, it means that there are seven girls between them in this direction. So if they tell you just that there are, she's like 25th to the left from one person, that this means that between them there are 24 people. So we have four girls here, seven girls here, Antonia and Bianca, which gives us in total 4 plus 7 plus 2, which is 13 girls. And this is the answer. So 6C. Let's check in answer key, 6C. Okay, now 7. A circle with a radius of one rolls along a straight line from point K to point L, where KL equals 11P. See figure. What does the circle look like in the end position at L? Okay, so here we start at point K and our circle looks like this. So there is shaded area, there is black area and there is white area. And it turns this way to the point L. And we know that this distance is 11P. We also know that the radius of this circle is one. So the question, first question that we can answer, how many full rolls will this circle make on this distance? So one full roll is connected with a distance of the perimeter of this circle. So what is the perimeter of this circle? So formula is 2PR, where R equals 1, which means 2P. So one full rotation of this circle will be at the distance of 2P. And we have 11P, which means that we will make five full circles. They will give us 10P and we will still have 1P left here. And what is 1P? Well, this is half turn. So this circle will make a half turn, which first of all means that this point here between shaded and black areas that in the beginning is on the ground, now will be on the top position. So here we have the border between shaded and black areas. And in the bottom, we will have the white area. So we can look and see that the only possible answers are C and E. And what we also know, what we also see, we need to determine if black area is on the left or on the right. So if we turn this circle, black area, the shaded area will always be like before black. I don't know if you understand what I mean, but the shaded area will be here first to touch the ground. And then there will be black area touching the ground. And in every row, every rotation will be the same. So here, well, actually, the black area will be the first to touch the ground here. So it has to be, well, if here is white, here is black, then if we rotate this so that white is here, then black has to be here and here has to be a shaded area. OK, because first will be black touching ground, then will be white and then will be shaded, then will be gray. So here we have white and after this has to be a gray area. So this is answer E. Eight, seven, E. Let's check. Seven, E. Yes. OK, so the question eight. Martin plays chess. He has played 15 games this season, out of which he has won nine. He has five more games to play. What will his success rate be this season if he wins all five remaining games? OK, so he has already played 15 games, out of which he has won nine. So he has nine wins and six loses. He has five more games. What will his success rate be this season if he wins all five remaining? If he wins all five remaining, that it will give him 14 wins and six loses. In total, 20 games. Success rate, well, the number of wins divided by the number of games. And if we want this in a percent, then we have to multiply this by 100%. And we can now simplify this and we get five. 14 times five equals 70, divided by one still is 70. So 70%. And this is the answer. Eight, C. Let's check in the answers. Eight, C. Yes. Question nine. One-eighth of the guests at a wedding were children. Three-seventh of the adult guests were men. What fraction of the wedding guests were women? So one-eighth of all guests were children. So seven-eighth were adults. And three-seventh of them were men. So we have to multiply this by three-sevenths to get what part of all the guests were men. So we can simplify sevens and we get three-eighths. So one-eighth were children, three-eighths were men. So in total we have to add them up, we get four-eighths and we can simplify this by four and we get half. So half people were men or children. So what fraction of the wedding guests were women? Well, the rest. So one minus half and this is half. So here the right answer is A, 9A. So let's check 9A. Yeah, 9A, this is the answer here. Okay, so question 10. My math teacher has a box with colored buttons. There are 203 red buttons, 117 white buttons and 28 blue buttons. Each student is asked to take out and keep a button from the box without looking. How many students have to take out a button to be sure that at least three buttons of the same color are taken out of the box? Okay, so we want to be sure that at least three buttons were taken. So what is the most extreme situation in which it doesn't work? We don't have three buttons. Well, we can have two buttons red, two buttons white and two buttons blue. So we have six buttons in total. But if we get seven, then it has to be one of them. And it will mean that we will have three buttons of the same color. So if we take seven, we are sure that at least in one color, we have at least three buttons, which give us answer C. 10C. Let's check. Let's check 10C. Yes. So, okay, now we are starting four points. So a little bit harder questions. So 11. A, B, C, D is a trapezoid with site A, B parallel to C, D, where A, B equals 50 and C, D equals 20. E is a point on the site A, B with the property that the segment D, E divides the given trapezoid into two parts of equal area. See figure. Calculate the length AE. Okay, let's draw it. So here we have A, B, C, D. We know that this one is 20. This one is 50. And here we also have E. And we know that the area of this part, this triangle is the same as the area of this trapezoid here in the right. So they are both equal half the area of this trapezoid, this original trapezoid. So let's calculate the area of this trapezoid. But we actually cannot give just a number, because we don't know height. But what is nice in this height? The height of this trapezoid is exactly the same as the height of this triangle A, E, D with the basis AE. So the area of this trapezoid would be 50 plus 20 times h divided by 2. And this is 70 h divided by 2, which gives us 35 h. And this is the area of this trapezoid. So the area of this triangle has to be half of this. So 17.5 h. Now let's try to calculate this area. And let's call this this basis x. We want to know x. So the area of this triangle is x times h divided by 2. That's how we calculate the area of a triangle. And we know that it has to be equal to 17.5 h. So first of all, h cancels out. And we can also multiply this by 2. And we get that x equals 17.5 times 2 equals 35. So AE has a length of 35, which is the answer C. So 11C. Let's check in the answers. 11C. Yeah. And the last question in this video, question number 12. How many natural numbers A possess the property, possesses the property that exactly one of the numbers A and A plus 20 is a four-digit number? Okay. So a four-digit number, for example, 1000. So if A would be 980, then A plus 20 will be 1000. And this number is okay. Yes, this is one of the possibilities. We have an A that is three-digit number, but A plus 20 is a four-digit number, which means that exactly one of them is four-digit number. The same will go for 981, 982, up to 999. Still here, still A is three-digit number, but A plus 20 will be 1019. And this will be four-digit number. So here we have exactly 20 numbers, but that's not all. Yeah. Because there's also the other side. The smallest non-four-digit number, the smallest five-digit number is 10,000. And if that would be A plus 20, then A would be 9980. Yeah. And once again, exactly one of those numbers is a four-digit number. So this number is also okay. And up to 9999, all those numbers are okay, because if we add 20, we get five-digit number. So here we also have 20 numbers, and at total, we get 40 numbers that have this property that exactly one of the numbers A and A plus 20 is a four-digit number. So their answer is E, 12E. Let's check. 12E, 12E. Yeah. So that's the right answer here. Now we start with question number 13. Six perpendiculars to the sides are drawn from the midpoints of the sides of an equilateral triangle. What fraction of the area of the initial triangle does the resulting hexagon cover? Okay. So let's draw it. So we have equilateral triangle, which means that all its sides are the same. And now six perpendiculars to the sides are drawn from the midpoints of the sides of an equilateral triangle. Okay. So this is a midpoint. Here we have a midpoint of this side, of this side, and of this side. And here we write line that is perpendicular to this side. The same goes with this one here. But also we can write here and here. And also from this one, we can go here and here. Okay. And we want to know this area here. So maybe I will make it bigger. Ah. So it should be here. Okay. And now let's look at this. So this triangle is equilateral, which means that every angle of this triangle is 60 degrees. And let's look. Here we have a triangle. Here, here, and here. It has an angle of 60, and then an angle of 90, which means that this angle is 30 degrees. And here is one fine thing. If we have a triangle that has angles of 90, 60, and 30, then we can know it from trigonometry. But we can also know it as a simple rule. If we will say that the shortest side of this triangle is a, then the middle is a times square root of 3, and the longest is 2a. I mean lengths of these sides. So when I'm looking at this, so here we have this triangle 90, 60, 30, but actually we have more such angles. Because here we also have 90 degrees. And if here is 30, this means that here we have 60 degrees. Okay. So here we have also triangle 30, 90, 60. So let's call the shortest side of this triangle a. Now, this part is a square root of 3, and this part is 2a. This here. But now, well, the same, exactly the same triangle we have here. So this is also a, this is a times square root of 3, and this is 2a. Which tells us that this line, all this line here, here we have 2a, and here we have a, which means that all this side of this bigger 90, 60 triangle is 3a. And this is the middle, the middle side of this triangle, which means that this, that if this, the shortest one is b, then b times square root of 3 equals 3a. So we can divide it by square root of 3, and we get that b equals a square root of 3. So here we have actually a square root of 3. And this, the longest in this triangle, is 2a square root of 3. But we know that this is the midpoint of this side of the triangle, of this big triangle, equilateral, which means that all the side, all, every side here is 4a square root of 3. And now, we can, we can calculate, for example, the area of this equilateral triangle, because we know its side, and we know that if side would be b, then the area is b square times square root of 3 divided by 4. Our b is 4a square root of 3. So we have 4a square root of 3 squared times square root of 3 divided by 4, which will give us 16a square times 3 square root of 3 divided by 4. And we can simplify 4 and 16, so we can get 4a square times 3 square root of 3, which will give us 12a square square root of 3. This is the area of this triangle. We want to know what fraction of the, of this area is covered by this hexagon in the middle. So first of all, this is a regular hexagon, because here we have 30, here we also have 30, and which means those three angles must have 180, which gives us that this is 120. And the same for all the other angles, so they are all the same, and also all sides are the same, because they are, they are all 2a. So we have a regular hexagon with a side of a length of 2a. How do we calculate this, the area of a regular hexagon? Maybe you know the formula, maybe you know where it comes from. If not, then I will tell you. So let's pretend that this is a regular hexagon. It doesn't really look like a regular hexagon, but let's pretend for a moment. And we can draw its diagonals, the longest diagonals. They intersect each other in one point here. And when we draw them, we cut this hexagon in six triangles. They are all the same, and they are all equilateral triangles. And they have the same side as hexagon. So if hexagon has a side of b, then the area of a triangle is, like here we said, b squared times square root of 3 divided by 4. And we have six such triangles in this hexagon, so we just need to multiply it by 6. And that's the formula for a hexagon with a side of b. Our side is 2a, so we can calculate 6 times 2a squared times square root of 3 divided by 4. And we get 6 times 4a squared square root of 3 divided by 4. Fourth, simplify, and we get 6a squared square root of 3. And now we want to know what part this hexagon is of all the triangles. So we just have to divide the area of this hexagon by the area of this triangle, which we counted here. And this looks very, very pretty. Almost everything cancels out, and we just get that this is half. This is half of this triangle, which means the answer is d, 13d. Let's check in answer key. 13d. Yeah, this is 13d. So now we can go to question number 40. The sum of the squares of three consecutive positive integers is 770. Which is the largest of these integers? Well, okay, so let's say that the smallest is a. Then the next consecutive is a plus 1, and the next is a plus 2. And the sum of their squares, so it will be a squared, here will be a plus 1 squared, here will be a plus 2 squared. And we just need to add them up, and we know that this equals to 770. We can find a, and we can find then the largest one, which will be this. So that's the way just solving a quadratic equation. But actually, here, well, we have answers. It will be faster to just check these answers. Let's say 15, the smallest one. Well, 15 squared, then plus 16 squared, then plus 17 squared. Well, this is 225 plus 256 plus 289. And we need to add this up, so let's do it. 256, 289. So 6 plus 5, 11, plus 9, 20, 0, 2 goes here. And 2 and 8 equals 10, 15, 17, 7, and plus 1 here, 6, 7, 770. Wow, so that was very fast. So the answer is not 15, because 15 was the smallest, and the question is of the largest. The largest is 17, so the answer is C. So actually, I shouldn't, if I was checking answer E, then I should take 13, 14, and 15, and I would get too small number, and I would know that I have to make something bigger. Okay, so let's check. 14C, yeah, 14C. So, okay, now 15. A belt drive system consists of the wheels A, B, and C, which rotate without slipping. B makes 4 full rotations, while A makes 5 full rotations. And B makes 6 full rotations, while C makes 7 full rotations. Find the perimeter of A, if the perimeter of C is 30 centimeters. Okay, so if we have such a system of wheels, let's just take two of them. So, when one of them makes full rotation, this line here is turned by, is moved by the perimeter of this circle. So, we know, let's say that this is C. We know that the perimeter of C is 30. So, in one rotation, this line turns by 30 centimeters. And we know that C makes 7 full rotations, while B makes 6 full rotations. So, if C makes 7 full rotations, that means that this line turns by 210 centimeters, right? 7 times 30. And we know that it corresponds to 6 full rotations of B, which means that one full rotation is 210 divided by 6. So, we can simplify this by 3, which will get 70 divided by 2, and this is 35. So, the perimeter of B is 35. So, now we can take the first statement. B makes 4 full rotations, while A makes 5 full rotations. So, if B makes 4 full rotations, then this line will move by 35 times 4, which will be 140 centimeters and it corresponds to five full rotations of a which means that perimeter of a is 140 divided by five and well multiply this by two 280 and now divide this by 10 28 a simple way to divide by five so the perimeter of a is 28 centimeters so the answer is b 15 b 15 b yeah 15 b so that's how you can solve this and the last question in this video 16 ah i don't really know how to write this name how to read this name taiko taiko wants to prepare a schedule for his jogging over the next few months every week he wants to jog on the same days of the week he never wants to jog on two consecutive days he wants to jog three times per week how many different schedules can he prepare okay so well so three times per week in a week we have monday tuesday wednesday thursday friday saturday and sunday but for our purpose it will be easier to think of them as first second third fourth fifth sixth and seventh day of a week and he wants to run on of on every three of them in every week always the same he also doesn't want to have two consecutive days of running so how many possibilities he have well he may run on monday then he cannot on tuesday but he can on wednesday and what for the third jogging well he cannot run for thursday because it will be consecutive day for a wednesday but he can on friday and on saturday but not on sunday because if he would be right because he if he would be jogging on sunday then the next day on monday he will also be jogging and that would be a consecutive two days so no so if he decides that the first two days are monday and wednesday then he has two possibilities okay so let's uh now check uh monday not he cannot use day not wednesday but thursday well then he cannot run on friday but he can on saturday and still not on sunday because he is running on monday so there's only one possibility and there's also such a possibility monday not tuesday not wednesday not thursday friday and what we get sunday saturday cannot be and sunday cannot be so there's no possibility so these are all possibilities if he is running on monday so that's now let's say that he's not running on monday but he starts on tuesday so no wednesday because it cannot be the consecutive days so the first day he can run is thursday then no friday but he can run on sunday saturday and on sunday now he can run on sunday because monday is his free day so there is no problem so if he is running on tuesday and on thursday then he has two possibilities let's check this next so tuesday no wednesday no thursday but friday well then no saturday but sunday is okay uh so one possibility and if we would turn move this to saturday then there will be no more possibilities uh so now let's check the first day is not monday not tuesday but wednesday so no no thursday can be running on friday no saturday can be running on sunday well he could also run on monday but we will just get to the the one of the previous situations so uh actually the second one i think no the first one so uh so this is there is only one possibility here uh okay so how many possibilities we have here because if we would try to start with thursday then saturday and then nothing so that's all so here we have ah two one two one one uh in total seven so he has seven possible schedules uh 16b 16b yeah 16b so 16b okay uh now question number 17 so four brothers have different heights tobias is shorter than victor by the same measurement by which he is taller than peter oscar is shorter than peter by the same measurement as well uh tobias is 184 centimeters tall and the average height of all the four brothers is 178 centimeters how tall is oscar um okay so let's go to this so first of all tobias is shorter than victor so victor higher than tobias and tobias is also higher than peter and peter is higher than oscar so these are these brothers written from the highest to the lowest to the smallest one and we know that tobias is 184 centimeters and the average and the difference between the brothers is everywhere the same so if we say that this difference is x then this means that victor has a height of 184 plus x peter has 184 minus x and oscar has 184 minus 2x and the mean of the of their average height is 178 centimeters okay so how can we do this so the most straightforward way is just to calculate the average so we need to add their heights divide them by their number so four and this will equal 178 so okay let's do this 184 plus x plus 184 plus 184 minus x plus 184 minus 2x and we have to divide this all by four and we know that this equals to 178 and we have a simple equation with one unknown number x and we can we can calculate this so let's simplify this a little bit because this minus x cancels out with this x and here we get 178 equals i will not uh i will not add this 184 this is four times times 184 minus 2x divided by 4 and now what i can do i can a little bit rewrite this because we can if we have uh we have a subtraction or addition in a numerator and we have one simple one simple number in the numerator then we can create two fractions of this one because subtracting fractions with the same denominator is the same as subtracting their numerators so we can write this as like this way and now we can we can simplify this force so we don't have to calculate what is 4 times 184 which lowers the probability of simple simple mistake in calculations we can also simplify this 2 and 4 so this is 2 this is 1 and we get 178 equals 184 minus x divided by 2 so half of x now we can put x here 178 here so the designs will change so here we have x divided by 2 equals 184 minus 178 which mean x divided by 2 equals 6 now we can just simply multiply this by 2 and we get x equals 12 so the difference between the the brothers is 12 12 centimeters so here we have 196 so a tall guy basketball or volleyball player peter is 172 and oscar is 12 centimeters smaller than peter which means 160 the question was how tall is oscar well 160 centimeters so 17a let's check in the answers 17a yeah so uh okay so this is the most straightforward way but you can do it pretty easily just by knowing some things for example here these numbers are a mathematical sequence their difference is the same between the next numbers and if you have an arithmetical sequence then the mean so the average of these numbers is the same as the median of these numbers and if you want to calculate the medians of four numbers you need to take those that are in the middle and calculate their mean which would look like this 184 plus 184 minus x divided by 2 equals 178 but you can once again simplify this because this is 200 2 times 184 minus x divided by 2 equals 178 which means this is 2 times 184 divided by 2 minus x divided by 2 equals 178 this simplifies you can put x here and 178 here and you get 184 minus 178 equals x divided by 2 and you can say okay but that's exactly what we got here so what's the difference well now you can look at this and remember that if you have for example if you have an even number of numbers that form arithmetical sequence then the difference between one of the middle ones and the mean of the average of all those numbers is the half of the difference between the consecutive numbers in this sequence so if we know that the median is 178 and this brother is second of four then this means that the difference between 184 and 178 which is 6 which is easy to calculate is half of the difference between the heights of the brothers so we can very easily calculate that the difference of the heights of the brothers is 12 centimeters without all this writing and then we can get easily the height of Oscar and one more thing because here well kangaroo comes originally from Australia that's why it's called kangaroo but it became very popular in Europe starting with France and well as you know as you see also in this competition we of course use a system international with meters centimeters kilograms and many others you in America use different measures feet yards inches and so on I know those for the lengths so what I can tell you well 184 it's very close I mean 183 it's almost the same as six feet first of all you may think that it's not fair that you have to use the units that you are not used to but from the mathematical point of view it doesn't matter the calculations will be the same with feet like with like with centimeters so it doesn't matter second well get used to the system international it's really great I will not talk much here why is it great I can make a video someday about this so 18 it rained seven times during our vacation it never rained in the evening and it never rained at night if it rained in the morning it rained only once did that morning and it was sunny in the afternoon if it rained in the afternoon it rained only once that afternoon and it was sunny in the morning there were five sunny days and six sunny afternoons at least how many days did our vacation last so okay so we want to know the smallest number of those days in those vacations so the smallest number will be when the the weather will be rainy or sunny no no other possibilities if that would be the case then well we have seven uh raining days seven rain it was raining seven times in all those days and they it was raining only on mornings or afternoons and only in one of them and we have five mornings six sunny days well there is five mornings and five sunny morning mornings and six sunny afternoons but the number of mornings and afternoons has to be the same so seven times so uh four mornings rainy mornings and three rainy afternoons well and then we would get nine mornings at all and also nine afternoons at all and that's the least number of days that these vacations could take if you would have some other like not raining but also not sunny then it could be more but the least amount of days that this vacation could last is when we have only sunny or only rainy rainy days so this is nine days and the answer here is c okay let's check 18c yeah 18c okay so 90. Jenny decided to enter numbers into the cells of a three times three table in such a way that the sums of the numbers in all four two times two squares are the same the three numbers in the corner cells have already been written in as shown in the figure uh which number should she write in the fourth corner cell marked with this this question mark oh this question mark okay so okay so what we know here is three here is one here is two in these places can be many different numbers but we know that in every four every two on two squares so in this square in this square in this square and in this square there has to be the same sums so if we have like three here and some numbers here uh oh well let's let's let's put here for example zero and let's put here one then what we and let's say that here is five i don't know so the sum in this square two on two is three plus five plus one plus zero which equals nine and there has also to be nine but now we have one plus zero plus one which is two so there has to be seven and also in this one has to be seven nine so now it is one plus two plus five which is eight so here has to be one and also in this part has to be nine in total and now we have seven plus one plus one which already is nine which means that there has to be zero so the answer is zero or impossible to determine huh so surely not five four and one because we got the different different number just just taking some random numbers and calculating so that the rules will be satisfied but thus there has to be zero maybe there's a possibility to get for example one or 58 here and then it will be impossible to determine well then let's think about this the other way so let's say here is three here is x so here is one here is two here is x y z so the sum in this two on two square is three plus x plus y plus z in this four on four we have one plus x plus y plus some a but we know that this is exactly the same as here they have to be the same so x cancels out y cancels out we can also put a 1 on the other side and we get that a equals z plus 2. So here was z and here was z plus 2. So this number always has to be bigger than this number by 2. And we see this here was 5 and here was 7, so bigger than 2. Okay, so now let's think what must be here. Well, we know that z plus y plus 2 has to be exactly the same as 3 plus x plus y plus z. So the same as in this left top corner. And there has to be also some other number, some b, plus b. So this here is b. So y cancels out, z cancels out, and we can put 2 here and we get that b equals x plus 1. So here is x plus 1. And what does it mean? That this number has always to be bigger than this number by 1. And here we got this. Here was 0 and here was 1. And that's kind of obvious. They are common for those two squares. This one is bigger than 1 than this, but the sum of all four has to be the same in both. So this has to compensate for this 1 lost in here. So it has to be 1 bigger than this one. And the same goes for this square. So now we can look at this square. Let's say that here is some c. So we get that c plus y plus x plus 1 plus z plus 2 has to be exactly the same as x plus y plus z plus 3. And now x cancels out, y cancels out, z cancels out, and we have c plus 3 equals 3, which means c equals 0. So no way there has to be 0. This number has to be 0 always, so the answer is really d. So let's check in the answers. 19d, 19d, yeah, 19d. And the last question, the last four-point question, question number 20. Seven natural numbers a, b, c, d, e, f, and g are written in a row. The sum of all of them equals 2017. Any two neighboring numbers differ by plus minus 1. Which of the numbers can be equal to 286? Okay, so first think about this. If we start with odd number and we change, and we change to the, by plus or minus 1, then the next will be even. The next will be once again odd, even, odd, even, odd. And so we get four odd numbers and three even numbers. If we sum them up, we will get even number, but we won't get 2017, which is odd number. So this means that a cannot be odd number, which means that we have to start with even number, then we have odd, even, odd, even, odd, even. And now we have four even numbers and three odd numbers. And then the sum will be odd number, which may be 2017. What does it tell us? 286 that we, that they are asking for, is an even number. So it cannot be on second, fourth, or sixth place. Which means that it cannot be any of them, it cannot be only D, and it cannot be only B or F. So it can be A or G, or C or E. So now let's check, because these numbers at all will be somehow close to each other. And we want to know how much it should be. So let's divide 2007 by 7. This will give us 2, 14, we are subtracting, so 6, 1. So here we have eight sevens, and we subtract 56, which will give us 5, 57, which will be 8, which will be subtracting 56, and we will get 1. What does it mean? That 288 times 7 plus 1 equals 2017. So if we would put seven numbers, 288, we would get 2016. We cannot do this, because they don't differ by plus or minus one, but it tells us where are we, where these numbers will be. So for to have, so I will say that 288, let's say it's like zero for us. Plus one will mean 289, minus one, 287. So our 286 will be minus two. It will be shorter to write and to understand, because what we want to get? We want to get some numbers. If we write in this convention, then we want to have a sum of these numbers plus one. So for example, zero, one, two, three, two, one, two. This notation means 288, 289, 290, 91, 290, 289, 288. And we can see that the sum of those seven numbers is six plus five, which will equal 11, which in real will be 2016 plus 11, which will be 2027. So this is something that doesn't satisfy us, and there's also no 286 here. But it will be easier now, because we want to have a sum of plus one. So I would put as many zeros as I can. For example, let's start with zero. Then let's give here one. Now let's give here zero, but we want to go to minus two. So let's put here minus one, minus two. So we have already five. Now we want to go up, because we want to have plus one. So minus one, and the last one, zero. And well, it's not really okay, we have minus three. This is not satisfying us. Okay, so maybe we should go higher here. So one, two. Now let's give here one, zero, minus one, minus two, minus one, once again. And how many do we have? Oh, we have too many of them already. So three. Okay, so these are seven. So now what is the sum? Well, the sum now is one, which means that 286 can be the last one. So this means that it can be G. Or we can reverse this, start with minus two, minus one, zero, one, two, one, zero. And we also get plus one, which means that it can be A or G. So the answer here is A, 28. Let's check this answer, 28. Yes. It cannot be C or E. Well, because if we would have minus two here, then we would have minus one, and we would have minus one here, zero, zero, one, two, and we will not get to the sum of one. It will be not enough. So it has to be on the edges, so that we can, we will have time to cancel these minuses and get the sum of one. So question number 21. There are four children of different integer ages under 18. The product of their ages is 882. What is the sum of their ages? Well, okay, let's think about this. So the product has to be 882. So let's let's check what I want to say. Prime factor distribution. Okay, let's create the, let's make the prime factor distribution of this number. First of all, it, we can divide it by two. So this is 441. We divide those numbers only by prime numbers. Okay, so 441, the sum is nine, which means that this number is dividable by three. And if we divided this by three, what we get? Well, 147. Now, we can clearly see that we can divide this also by seven, and we will get 21, and now 3, 7, 7, 1. So, so this is, this prime factor distribution is very, very nice for this, for this exercise, because we need to have an integers that are divisors of 882, 882. And all those integers are, or this is one, okay, or maybe those numbers like 2, 3, 7, or they are, they are the result of multiplying those numbers. So, for example, 3 times 7, 21, which means that 882 is dividable by 21. But 21 doesn't interest us, because all those ages has to be under 18. So, they are like puzzles. We have 1, 2, 2 3s, and 2 7s. And from those puzzles, we can get those four numbers that their product is 882. We have to use all of them. Maybe we will, for one of these ages, we will use none of them, and this will result in just one. What we also know, we have to use them in the way that, that we don't exceed this 18. So, for example, we cannot take 7 with 7, because it would be 49. It's way too much. So, also not 7 with 3, because it will be 21. So, we can take 7 with 2, which will give us 14. And we used those two. Now, this 7 has to be alone, because we cannot get, take this with 3. And now, we cannot put these 3s alone, because they will be the same. And it is written here, of different integer ages. So, they have to be different. So, we have to make a pair of them, and 3 times 3 give us 9. And the last one will be 1. So, we're not using any of them. So, this is the possibility. And the sum of those numbers is 10 plus 14, 24 plus 7, 31, which will result in answer D. The question is, is this the only possibility? Well, the sum has to be 31, because we got somehow 31, and we see that it cannot be different numbers. Like, there's no answer. Like, it's impossible to determine. So, but let's think about this. What we start with, is that we took 7 with 2. We couldn't take 7 with 7, and we couldn't take 7 with 3. Did we have to take 7 with 2? Yes, because if we wouldn't take 7 with 2, so the 7 would have to be alone. But also, the second 7 also would have to be alone. And we will get the same integer numbers, but they have to be different. So, no way. 7, one of the 7s has to be with 2. The other has to be alone. And their logical conclusion is that 3 has to be with 3, which results in 9. And the last one is 1. So, that's the only possibility with this information. So, 21D. Let's check this. 21D, yeah. So, now we can go to 22. On the faces of a given die, the following numbers appear. Minus 3, minus 2, minus 1, 0, 1, 2. If you roll the die twice and multiply the results, what is the probability that the product is negative? Okay, so I will show you a nice way to solve such questions when you have some experiment made twice. For example, twice rolling a die. So, the results for the first are minus 3, minus 2, minus 1, 0, 1, 2. And exactly the same for the second. And we build such a square. And all the possible pairs that we can get are somewhere in this square. For example, this point. Well, this means that in the first, there was minus 2. And in the second, there was 1. So, that's one of the possibilities. And first that we can see that there are 36 small squares, which correspond to 36 possibilities at all. Now, we are asking what is the probability that the product is negative. So, when do we have the product negative? Well, simply, we have to multiply negative number with positive number. So, here, here, here, here, here, here, and here, here, here, here, here, here, here. How many are they? 6, 6, 12. 12 at all. And if we... These are our all possibilities. Now, we can just divide this. We can simplify this by 12. And it will give us one-third. So, the probability is one-third. And this is answer E. Let's check. 22E, 22E. Okay. Now, let's go to 23. An arbitrary two-digit number consists of the digits A and B. By repeating this pair of digits three times, one obtains a six-digit number. This new number is always divisible by what? Okay. So, some two-digit number, let's say 60, consists of these digits. By repeating this pair of digits three times, one obtains a six-digit number. So, I think, I understand this, that that's what the result would be if we good, if we start with 16. And this new number is always divisible by what? This number is divisible by 2. But we have chosen 16. Do we have to, do we always have to get the number that is divisible by 2? No. For example, 37, 37, 37. What is the way to find out if the number is divisible by 2? Well, we just simply look at the last number. If it is even, the number is divisible by 2. If it is odd, then it is not. So, this is not. So, this is not. So, we can easily find many numbers that will not be divisible by 2. So, no way for 2. For 5, also no way. Because, once again, the rule of divisibility by 5 is that the last answer is 0 or 5. We can easily take 37 and we don't get 5 or 0 in the end. So, it's not divisible by 5. So, surely not 5. Does it have to be 9? Well, once again, this example shows that not. Because, what is the divisibility by, the rule of divisibility by 9? You have to take all the digits, add them up, and if the sum is divisible by 9, then all the number is also divisible by 9. So, here we have 10, 10, 10. So, at all we get 30. 30 is not divisible by 9. So, all this number is also not divisible by 9. So, we get to the hardest part. 11 and 7. Yeah, well, once again, this example can solve this. But, I would like you to, I would like to show you. Okay, now, let's start with this example. It's not divisible by 11. Why? Because there's a rule of divisibility by 11. So, you take, if you have all the number, you take the digits that are that are on the even places, even numbered places in this number. So, let's say 2nd, 4th, and 6th. You add them up. Here we get 21. You also take the numbers that are on odd numbered places and add them up. 3 plus 3 plus 3 will get to 9. Now, you subtract from one the other, and we get 12. If this number is divisible by 11, then all the number is divisible by 11. This number is not divisible by 11, so it is clearly not divisible by 11. So the only possibility is C, it has to be divisible by seven. Okay, so that's the answer, but why? Why by seven? So let's think about this. So let's say this number is AB, well, I mean 10A plus B, okay? This original number. So if we write this like they said, what we get here? Well, this A is actually A times 10,000. No, 100,000. Well, no, 10,000, sorry. Because it's A times 100,000. Okay, so plus B times 10,000. Okay, then once again, A times 1,000 now, plus B times 100, plus A times 10, plus B. What can we do here? Well, we can take before the bracket from those first two, 10,000. If we do this, then in the bracket, we will get 10A plus B. Then in the second pair, we can take 100 before the bracket and we once again get 10A plus B. And in the last one, we just put this in the bracket, so it will be also 10A plus B. Now, we have everywhere here, we have the same bracket. So we can take bracket before the bracket, yeah? So kinda inception. So 10A plus B. And in the bracket will be 10,000 plus 100 plus 1, which will result in 10,101 times 10A plus B. And this 10A plus B is our original number. It can be everything. So it doesn't really tell us much, but this number, 10,101, this is a concrete number. We know everything about it. And what's more? We know that this thing is the number that we get in the end, is this, this is the six digit number. So everything that this number is divisible by is also, this number is also divisible by that number. So we see, this is not divisible by two, by five, by nine. It's also not divisible by 11, because once again, this rule goes here, one plus one plus one equals three, three, zero plus zero equals zero, three minus zero, well, it equals three, and it's not divisible by three. So there's no reason that it would be divisible by 11. So it has to be divisible by seven. And I would show you a very nice way, very nice rule of divisibility by seven. So how can we check, for example, this number, 10,101? So we cut off the last digit, and now we take this number, 1,010, and we subtract from it two times the last digit. And we get 1,010 minus two, it equals 1,008. And the question is, is it divisible by seven? Well, we don't see it, so what we do? Once again, we cut the last number, and now we take this number, 100. Once again, subtract two times the last digit, which is eight now. This is 100 minus 16, and this is 84. And the question, if this number is divisible by seven? So if you don't see it yet, then once again, you can do the same. And now our number in the front is eight. We subtract two times the last digit, so four, and we get eight minus eight, and this is zero. Is zero divisible by seven? Well, zero is divisible by everything instead of zero. So yes, so this number, 10,101, is divisible by seven. All those numbers, the 1,008 and 84 are also divisible by seven. So that's the first way to check if the number is divisible by seven. And that's the reason why this number will always be divisible by seven. And one last thing, this rule of this divisibility is very nice, because if you really understand it, if you find the proof why it works, you can change this and find the divisibility rule for all the numbers that have in the end, one, three, seven, or nine. And maybe I will in some video show this way, but now I just say this, and you can try your luck to find this. Okay, and the last one. My friend wants to use a special seven-digit password. The digits of the password occur exactly as many times as the digit value. The sum digit, the same digits are always written consecutively. Example passwords are four, four, four, four, three, three, three, or one, six, six, six, six, six, six. How many possible passwords can be choose from? Okay, so what does it mean? So we are using some digits here, and they have to be in this password as many times as is their value. So for example, if we will use five, then there has to be five fives, and they have to be consecutive. And we need to use, well, for example, we can use seven sevens, because we need seven-digit passwords. So if we take seven, then it has to occur seven times, and that's actually seven-digit password, and it's okay, it's pretty completely legit here. So, okay, so that's one of the possibilities. We cannot use eight, because eight would have to show eight times, and this would be more than seven-digit password. So no eight and no nine. And no zero, because zero would have to occur zero times, which will mean that there's no zero. So no way. So, okay, are there other possibilities? Clearly we see, we see four fours and three threes. Also one one and six six. So we can see something here. Four plus three equals seven. And that's what has to be, because if we use four, we took four places in this seven-digit password, and the other three places have to be used by, for example, three threes. So these different numbers have to add up to seven. So we can have four three, five two, six one, and no zero, but seven, okay, we have here also. Is there other possibilities? Well, why just two different digits, maybe three? And yes, there's possibility one, two, four. One one, two twos and four fours. And this will be seven-digit password. And that's the last one. There's no other way to get three different numbers and no way at all to get four different numbers, because if we even put the smallest numbers, it will end up with 10 digits. So way too much. So, nope. So these are the only possibilities, but look at this. We can start with fours or we can start with threes, and there are different passwords. We cannot mix them up. For example, we cannot do something like this because they have to be consecutive. So for example, for four and three, we have exactly two possibilities. The same as for five and two, and for six and one. For just sevens, we got one possibility. And here the last, the most interesting one, two, four. Well, they will be six. Why? Because we can start with one, then two and four, or one, four, two, or two, one, four, And these are six possibilities. You can also find this easy way. You can just see that this is, this is what? This is permutation with three digits. So the possibilities for permutation with a set with three different digits is three factorial, which equals six. So here we have six possibilities. Now let's just add them up. Two, two, two, this is six. Six, one, six, 13. 13, and this is E. 24E, 24E, 24E, yeah. So that's the answer. Question number 25. Paul wants to write a natural number in each box in the diagram in such a way that each number is the sum of the two numbers in the boxes immediately underneath. At most, how many odd numbers can Paul write? Okay, so they are asking about odd numbers and this is not odd, because if you get two even numbers, then they will give you even number. So if they would ask for even number, this question would be obvious because you can get all of them even. You can put even, even, even, even, even, even in the lowest line and all of them will be even. So nothing interesting. But with odd numbers, it's more complicated because two odd numbers will give you even number. But odd number with even number will give you odd number. So you can create new odd numbers, but you can also create even number. So they cannot, for example, there cannot be all of them odd numbers because if we would put here odd numbers, then in the next line, we will get only even numbers and all others will be also even. So there is some maximum number of odd numbers than we can put. And how to find this? Well, all of those numbers depend only of those six that are on the bottom. So we could check all the possibilities for those six on the bottom, and we will find which is the biggest when we will get the most odd numbers. Is this a good idea? No. You know why? Because here we have six places. And for all of them, we have two possibilities, or odd, even, or even. So how many possibilities it will give us? Well, we have to multiply them. We get two to the power of six, which is 64. Well, it's not astronomical number, okay? So it is possible to find this in a lifetime, I think even in an hour, but on a kangaroo, we have 75 minutes for 30 questions. So checking all 64 is clearly not a good idea. It will take way, way too much time to check them. Of course, we can simplify them because three odd and then three even is actually exactly the same as three even and then three odd. It's just, we need to just rewrite this from left to right and we will get exactly the same number. So, okay, but it's still not something that looks very, very nice. So what I propose, let's check smaller blocks of three because there are only eight possibilities and actually even more, even less. So for example, if we would put three even numbers, then we have all even numbers. Well, that's surely not the best. If we put three odd numbers, then we get three even, even three, sounds not bad, but probably there can be something better. So for example, let's check if when the odd number is only one. So two even. So we have actually two possibilities here. You would say three odd, even, even, or even, odd, even, or even, even, odd. So three possibilities. Yeah, but this one will result in exactly the same numbers that this one, because it's just once again, it's just a reversion. So we just need to check those two. So here we get odd. Here even, and here odd, three. So nothing better than here. And what here? Well, we get here odd, odd, and even, three. Okay. So the last thing that we get, two odd, which is the same as one even. So it once again, it will be two possibilities. You may say that three, but once again, this third one will be exactly the same as this first one, just reverse from left to right. So we can have odd, odd, even, odd, even, odd. And the last one, even, odd, odd, we don't check it. So what we got here? Even, odd, odd. So four. So this is more. And how here? Odd, odd, and even. Also four. Okay. So we have two possibilities here that actually it will be important later, but we have such a possibilities where we get the maximum amount for three blocks, for blocks consisting of three puzzles in the bottom. Where we get the maximum number of four. And now, if we take a look at this thing. So it looks like this. Here we have six. Yes. Then in the next line will be five, four, three, no, sorry. Three, two, and one. And you can see three such blocks here. One here, one here, and one here. And also we see this part here that doesn't get into any of these blocks. These two are completely independent. So we can put anything here what we want. This is dependent from those. So we are not sure if we can get anything here, but what we know that there won't be more than four. There can be maximum four. Maybe even less. Maybe it's even impossible to take those two in a way that here we will have also the block that will have four odd numbers. Here we can easily have four. Here we can easily have four. Here we don't know for now, but let's think about maximum, maximum possibility. Let's say that here we have four, four. Here we somehow have also four. So we have here 12 at all. And here are three numbers that we didn't work with them yet. If it would be possible that they will be all odd, then we will have three odd here. And in total, we will get 15 odd, which means that we cannot get 16. We cannot get 17. We're not sure if we can get 15 because they would all have to be odd. And we will have also to get max one of those possibilities here in the top. We're not sure if it is possible, but what we know, no more than 15. So now maybe let's ask ourselves if we can get three odd numbers here. If we cannot, then we cannot even get 15 here. So what are the possibilities? Well, we have to start, those two have to be from those three, one of those three blocks. Let's call it one, this two, and this three. And if we want to have here odd number, then one of them has to be odd and the other has to be even. So what are the possibilities? Well, we can take one and one. So here will be this first block and here also will be this first block. Then we get odd here and even here. This is one possibility. We cannot take one and two because we will have odd and odd, and this will be even. We also cannot take one and three because once again, the same stretch. Let's try with starting with two. If we start with two, we cannot put here one because we would have even here from this place and even here from this. So two, one is not possible. Two, two is possible because then the first one will give us even here and the second one will give us this odd here. So two, two is something that we have to check. Two, three, yes, two, three is possible because here we have even and here we have this odd. So we also have to check two, three. And if we start with three, well, then we can get three, one because once again we have odd even no 3 2 and no 3 3 so there are four possibilities for those uh bottom bottom blocks let's check them so the first one 1 1 so even odd odd even odd odd once again and then also even odd odd and we know what happens here uh here is odd here is even here is odd here is odd now here is odd this means that here will be odd and here will be even so we are we don't get 3 odd here so in this situation doesn't matter what is here we will not get 15 we will get less okay so then let's check possibility 2 2 odd odd and what happens here odd and here we get even here we get hot once again we cannot get 3 so let's check the next one 2 3 starting with 2 and finishing with 3 so odd odd even odd even odd odd odd even odd even odd so once again even odd odd here odd odd even and what we get here well odd oh even even this clearly clearly doesn't look very well we don't even have a possibility to get 14 here because in those three we get only one odd so i don't think so and the last last possibility 3 1 3 1 is exactly the same as 2 3 it's just reverse from left to right here you start here in 2 3 we started odd odd even odd even odd and in 3 1 we will start with odd even odd even odd odd this is the same as the first one just reversed so no reason to check it okay so what does it tell us we cannot go get there three at most we can get two and this means that 15 is also impossible the highest amount is 14 if it's possible okay so now let's check we have two two possibilities here two candidates let's check them what we get with them here we have odd odd even now even odd odd and 1 2 3 4 5 6 7 8 9 10 11 12 13 14 and we get 14 so 14 is possible anything more than 14 is not possible so this means answer is b 14 25 b let's check 25 b yeah so i think that's the fastest way to solve this maybe someone has the better so 26 now lisa tried to calculate the sum of the measures of all the angles of a convex polygon she missed one of the angles and so her result was 2017 degrees the angle she missed measured well if you have a convex polygon convex so you know it doesn't it's not something like this yes so all the lines if you take any two points in this polygon and you connect them with a straight line with a segment you cannot go outside the that polygon so it cannot be this this is not convex oh well let's let's make sure but i think yeah so we need to have something like this this one this here on the left how it will be called let's check um concave yeah so this one is concave we want to have a convex and what we can say about such a polygon that we can cut it to a triangles going from one one vertex vertices so and what is the sum of the angles of this polygon well it is the sum of the sums of the angles in all those triangles and those sums are 180 and we have some integer number of these triangles here which means that the sum of the integers of the angles in each polygon in each convex polygon has to be dividable by 180 actually it's also right for concave polygons but there it's kind of nastier so this sum that she got had to be the had to be dividable by 180 but she got 2017 because she missed one angle so let's check how many 180s we can put here well here we have one 180 and if we subtract it here we have 21 7 1 180 37 which means that 2017 is 11 times 180 plus 37 to the next to the next angle we lose 143 degrees and that's exactly the angle that she missed there would be problem with concave polygon because the fact that it is a convex means that all the angles are smaller than 180 if it could be a concave then we could for example go to not the next but we could have a 323 degrees angle and it will be even more so we would have two possibilities but for convex polygon the one she missed has to be 143 which means answer e and now let's check this 26 yeah e yeah so that's the way so question number 27 there are 30 dancers standing in a circle and facing the center after the command left some dancers turn to the left and all the others turn all the others to the right the dancers who were facing each other said hello it turned out that there were 10 such dancers then after the command around all the dancers turned around made a half turn again the dancers who were facing each other said hello how many dancers said hello this time okay so how to start well let's start with an experiment so let's say that we have six dancers and they are standing in they are facing on like this one this person is facing the actually left because they are there is a center and let's say we have such situation and this one so this person here is turning left this person is turning his or her right and how many people here said hello well this one this one this one and this one so four person or we can also think two pairs now after command turn around after command around what happens let's start with this person now this person is facing this way this person this way this one this one this person this way and this person this way and and the last person is here uh so this person is uh-huh so i've started actually with this one and how many now says hello one two three four and that's all so once again it's exactly the same number so we can have an idea that maybe it's the rule maybe that that's what it has to be or maybe this is a coincidence so no it's not coincidence let's think about this uh if we have a group of dancers then if and we have at least two pairs that are facing each other then we can look at them like standing in a line at least this part between those two pairs so we have one pair facing each other and the other pair facing each other and between them are some people and there are no people facing each other so maybe something like this and an important thing is that there is there has to be a pair that is uh turned back to each other because what situations we can have we can have people that are facing each other we can have people that one is looking on the back of the other person and we can have a pair that is turned back to each other so after command around those this pair will change to those the last pair the third pair so that they will be turned back to each other this uh this pair will not change there will still be a situation where one person looks at the back of the other it will just change the person and this will change into facing each other so actually the important question is if those the number of those pairs has always has to be the same as the number of those pairs because if yes then turning around will not change the number of pairs that are facing each other so if we are in the line it if they are standing in a line it don't have to be like this because for example here we can have a line and we have two pairs that are facing each other and only one pair that is turned back to each other we can also have a different situations like for example here are some more people and here we have this person and now we have exactly the same number of people that are pairs that are facing each other and those that are turned back to each other so here in if in these situations there will be command turn back around then there will be exactly the same number of pairs facing each other it will change to this situation and here we have two pairs facing each other but in this first situation when we had like this oh like this a few people and here here so in this situation the turn back the command around will change this to such a situation and we see we have only one pair facing each other there were two and here is one in this situation we had two and there are also two and there is also a third possibility that okay we have this pair then a few people then this here and we can have like something like this and also here something like this and here we see that we have two pairs facing each other but after command around we will have three pairs because i will write this oh here one second and third so if they are standing in the line there's no such a rule that there can be one less pairs turned back or the same amount of pairs turned back or even one back or even one more pair turned back if then those facing each other but they are standing in a circle and it changes much because if you for example change this to standing to standing in a circle then look what happens this person stands close to this person they are neighbors and this person will look at the back of the other person so here the number of pairs facing each other will be the same as the number of pairs that will be turned back to each other so exactly like in this line and turning back command around in a circle will not do anything but here for example we have two pairs facing each other and just one pair turned back to each other here but once again in a circle this person and this person are neighbors and they are turned back to each other like let's start with this person so here are three person three people looking this way now it's this person so this person is looking this way and on the other hand this person or this person is here then there are three four more person looking people looking at the same direction and there is this one person and look here in a line we had just one pair turned back to each other but in a circle we have two pairs turned back to each other which means that they are exactly the same amount that facing each other and in this situation so here we had two pairs that are two pairs that are facing each other and three pairs that are turned back to each other here here and here but if they will stand in a circle let's once again look at this let's start with this person so this person is looking here and this is looking here and now one more person here now facing each other three more looking at the same direction as that one and then turn back to each other so this person two more looking at the same direction and then facing each other so here we are here and one more person looking the same direction turned back to each other so this person and and it's the last yeah this is the last person looking the same direction and what look what happens those people are neighbors here in a circle and they are facing each other which means that in a circle we have three pairs facing each other one two and where is three and we have also three pairs turned back to each other so here here and here so there is always the same amount of pairs face to each other facing each other that there are pairs that are turned back to each other and this if you think about this this this is obvious because if you have two pairs facing each other then there has to be one pair between those two pairs that is turned back to each other and in a line being between those two pairs means exactly one group this is this group that is between those two pairs but in a circle let's draw a little bit more but in a circle it changes into something like this so here is the first person and this one is uh this one is uh here so here we have two more and facing each other and this one and this one and here being so there are those two pairs that are facing each other and here being between those two pairs can mean this part or this part and they are both equally equally between those two pairs and in each of them there has to be a pair that is turned back to each other because we can look at them like in the line and we showed that there has to be such a pair so for every two pairs facing each other we get two pairs turned back to each other and as we here see in those as we saw in those examples here well where we had like three pairs facing each other it doesn't matter how many pairs so if here we have five pairs facing each other then there has to be also five pairs that are turned back to each other and after command around, all those five pairs turn back to each other, they change into the pairs facing each other, which means that there will be once again, 10 people saying hello. So the answer in 27 is A. So let's check in the answers. So 27A, yes, 27A. So that's the, I think that's the easiest way to solve this. I think it's easy to find the intuition here that it has to be 10, but well, if you want to like prove it to yourself, then it takes some thinking. So let's get to 28. On a balanced scale, three different weights are put at random on each pan and the result is shown in the picture. The weights are 101, 102, 3, 4, 5, 6 grams. What is the probability that the 106 gram weight is on the heavier pan? Okay, so if we have those pans here, then we, first of all, how many possibilities we have? So we just have six pans and we need to put three here on the left. So we have six over three possibilities. Yes, this is a Newton symbol. So we can rewrite this as six factorial divided by three factorial times six minus three factorial, which gives us six factorial divided by three factorial times three factorial. And six factorial, we can write this as three factorial times four times five times six. And here, three factorial times, and three factorial is just one times two times three. So once again, if you have five factorial, it's just one times two times three times four times five. So you just multiply numbers from one to this number and that's what factorial is. Now we can simplify those three factorials. We can also simplify those numbers with six and we get 20. So we have 20 possibilities to put those weights on these pans. But now the question is because if we put one, two, three here, then on this side goes four, five, six. Or we could put here four, five, six and here we'll go one, two, three. It's different situation because here we have one, two, three on the left side and here we have one, two, three on the right side. But actually it is different because here this one in the first situation, this right side pan will be heavier. So it will go down. And in the second situation, this will be heavier and this will go down, which means that actually those two situations are exactly the same from our point of view because once again, one, two, three on B on the lighter and four, five, six on B on the heavier side. And if we want to know what is the probability that 106 gram weight is on the heavier pan, then we are interested how many possibilities are there that some group is heavier. So these two situations are exactly the same. In both of them, four, five, six is heavier. So actually, and for every situation, we have such parts, which means that actually our all situations are just 10. There are 10 different situations that we have some pans here, some weights here that will be on the heavier side. And now is the question, what is the probability that 106 gram weight is on the heavier pan? So in most cases, it will be on the heavier pan. So let's think about those that it won't be on heavier pan. So first of all, how much has to be on the lighter pan that it will be lighter, right? So all of these pans, their sum of weight is 621. You can add them up. Actually, this 100s are not important. Just you can think that they are one, two, three, four, five, six, and the sum of them is 21. So to have a heavier pan, you need to have at least 11. So if we want to have six on the lighter, then we can put with the six, one and two, or one and three, and that's all. If you put six, two, then you have to put three to have a different situation than here, and it's already heavier, so no. So there are only two possibilities in which six lays on a lighter pan, which means that all the other eight are a situation where six is on the heavier pan. There is no situation where they are the same because, well, 21 is not dividable by two, so you cannot make it equal. So eight out of 10, which give us 80% of all the possible situation. Also, the probability is 80%. So let's check. 28B, 28, 28B, yeah. So that's the answer. So we start with question number 29. A and B are on the circle with center M. PB is tangent to the circle at B. The distances PA and MB are integers. PB equals PA plus six. How many possible values are there for MB? Okay, so let's draw this. I will draw it like this. So here is the center of this, let's say, circle. Here we have this line. Here is A. Here is P. Now let's try to write a tangent line. Oh, well, let's see. Oh, oh, not so bad. So here we have B. And here we have also MB. And they told us PA, so this distance, let's call it X, and MB. MB is actually the radius of this circle. And it is the same as MA. So they told us that those numbers are integers. So X and R are integers. And they also told us that PB equals PA plus six. So PA is X, so here we have X plus six. And they ask us how many possible values are there for MB. So first of all, if X is integer, then X plus six is also integer. Second, this is radius and this is a tangent line. And they connect in point B. In such a situation, they have to be perpendicular, right? The tangent line of a circle is always perpendicular to the radius going to the point that is common for this line and this circle. So we have Pythagorean theorem. So X plus six squared plus R squared has to be equal X plus R squared. Okay, so let's rewrite this. X plus six squared, it will be X squared plus 12X plus 36 plus R squared. And this will be X squared plus two XR plus R squared. And now we can simplify this because R squared cancels out, X squared cancels out. And we get 12X plus 36 equals two XR. We can divide this by two and we get six X plus 18 equals XR. Now let's put this here and we get 18 equals XR minus six X. XR minus six X. So, and we can take X before the bracket and we get 18 equals X times four minus six, R minus six. First of all, we'll see that R has to be bigger than six. Second, if R is integer, then also R minus six is integer. And X is integer, which means that X and R minus X has to be divisors of 18. So what are the divisors of 18? How can we get 18? Well, for example, one and 18, two and nine, three and six, six and three. This is different situation, nine and two and 18 and one. So we have six possibilities. Now we can all check them. If X would be one, then our triangle, then R would be, if X equals one, then R equals 18 plus six, which means 24. And here we have one. So X equals to one, then R equals 24. One, really one. If X equals one, yes, then this is 24. And this is 25. Something is not right here. X, ah, no, sorry. X is one, so this is one, but this line here is X plus six. So we would have the sides of our triangle would be seven. R is 24 and X plus R, so one plus 24 is 25. And yes, you can check the Pythagorean formula. This is a right triangle. Also, if X equals two, then this line here is eight. This line here, well, R minus six equals nine, so R equals 15. And X plus R, so 15 plus two equals 17. And that's also right triangle, you can check this. The other possibility, X equals three, which means this side is nine. Then R, R minus six equals six, so R equals 12. And this one, three plus 12, 15. That's also right triangle. For X equals six, we have this side is 12. R, R minus six is three, which means that R is nine. And this side, once again, nine plus six equals 15. This is the same triangle as here, but just a different way around it. This side here would be longer than here. And the next possibility, X equals nine, which means this side is 15. R, then R minus six equals two, so R equals eight. And eight plus nine equals 17. This is the same triangle as the second one. And the last one, if X equals 18, then this side is 24. R minus six equals one, which means that R equals seven. And seven plus 18 equals 25. That's once again, the same triangle as here. So how many possible values are for MB? MB, so for the radius. Well, we have six different possibilities. So 29D, 29, yeah, D. So, and the last one, 30. Oh, I have to admit, it's not easy. I didn't know how to solve it, so I checked the answer. So the answer is A. And well, it helped me find the solution for this, because let's read this. Point D is chosen on side AC of triangle ABC, so that DC equals AB. Points M and N are the midpoints of segments AD and BC, respectively. If angle NMC equals alpha, then angle BAC always equals. Okay, so let's draw this. So this triangle maybe looks like right triangle, but I don't think it has to be right triangle. I wasn't checking this. And here we have A, B, C. And here we have D, such that DC is the same as AB. And here we have this line BD. And here we have point M that is middle point of those two of this AD segment. And we also have N, which is a middle point of this segment BC. So BN is the same as NC, and AM equals MD in the length. And they also written this segment MN, and they told us that NMC, so this angle, is alpha. And they are asking about BAC. So I didn't know how to find this, but I checked the answer. So I know the answer is alpha. So what two alpha? So what does it tell me? Well, how to get to this two alpha? Well, if I have alpha here and two alpha here, well, this sounds like an angles in a circle. Because you have, I don't know for now how these angles are called in English, but if you have an angle that has this vertex in the center of this circle, and you have also some other angle that has a vertex on this circle, and they are connected with the same line here on a circle. So this angle here lies on this line, and this angle also lies on this line. Then this angle is twice as big as this. So if I want to see that this is twice as this, that my thinking was, okay, so we need some circle in which this angle will have a vertices in the middle of this circle. And this angle will have a vertices somewhere on this circle, but there's actually no such a circle. I mean, there could be such circle, but then this angle is completely out of this circle. So that's not the way. What I was also checking about this midpoint. Well, it is a midpoint, but it sounds kind of like Thales theorem. So let's try. So this line is half of this line. So if we would start from this point, a line that is parallel to this line, where will we get? First of all, and we'll also longer length than this line, then this angle will be the same if this is parallel to this line. And what is this distance now? Well, this line is two times bigger than this, which means that this line from here to C, let's call it E. So EC has to be twice as big as MC. And what does it mean? Well, from MD, MD is the same as AM. And DC, this means that DC has to be the same as AE. And that's very good information because now here we have the same sides in this triangle. And now here we can draw such a circle that has a center in A and goes to points B and E. And now this angle BAC has a vertex that lies in the center of the circle. And angle BEC also has a vertex that lies on a circle, and they are connected with the same part of this circle, which means that this one is twice as big as this one. And this one is the same as this one, which means that this one is alpha, which means that this one is two alpha. So I've started here with an answer because without answer, I didn't have a point how to solve this, really. But when I see the answer, then it helped me find the solution. So how you can solve such exercise on an exam with very limited time? Well, training and more training, and more training and a little bit also of luck to have a good intuition. Also, if you solve many different exercises, your intuition is better and you may get to the point where you expect the answer that is really right. So, well, what I can tell you, train, train, and you will get better.

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