Problem number one states, Arnold spelled the word kangaroo with cards showing one letter at a time. Unfortunately, some cards were rotated. By turning the K card back by 90 degrees twice, you can correct the letter K, and by turning the first A card once, you can correct the first A. See the figures. How many times does he need to rotate by 90 degrees for all of the letters to be correct? Alright, so here we have the letters. The problem told us that he has to rotate the K twice and the A once. So now let's take a look at the other letters. For example, he doesn't have to rotate the G at all, since the G is the right way around. He also doesn't have to rotate the R at all, or either of the O's, since they're symmetrical, so it doesn't really matter which way they are around. So now let's go on to the N. For the N, regardless of whether you rotate it clockwise or counterclockwise, if you rotate it once, it'll be in the correct orientation. And for the second A, if you rotate it all the way around, like upside down, so twice 90 degrees, you will get it the right way up. So in total, he had to rotate the letters 2 plus 1 plus 1 plus 0 plus 2 plus 0 plus 0 plus 0 times, or 6 times. So the question asked us, how many times does he need to rotate by 90 degrees for all the letters to be correct? The answer is 6 times. Letter C. Problem number two states, a cake weighs 900 grams. Paul cuts it into four pieces. The biggest piece weighs as much as the three other pieces weigh together. What is the weight of the biggest piece? So we know that the whole cake is 900 grams. And if we know that one piece weighs as much as the remaining three pieces, we know that that one big piece has to weigh as much as a half of the cake. So in this case, half of the cake would be 450 grams. So the left side of the cake would weigh 450 grams and the right side would weigh 450 grams. And that's the only way for one slice to weigh the weight of the remaining three slices. If you'd like to solve this in a different way, you can also solve it algebraically. You can pause the video here and look at how that is done. So the question asked us, what is the weight of the biggest piece? The answer is 450 grams. Letter D. Problem number three states. Two large rings, one gray and one white, are linked together. Peter, in front of the rings, sees the rings as in the picture to the right. Paul is behind the rings. What does he see? So here we have the view from Peter's perspective. Peter is in front of the rings, and this is what he sees. Paul is on the other side of the rings, and we need to figure out what he sees. So Paul's right is going to be to Peter's left, and Paul's left to Peter's right. This means that the white ring is on Paul's right, and the dark ring is on Paul's left. So it would look something like this. Now, up and down doesn't switch depending on whether you look at something from the front or from the back. So Paul should see this just like Peter, only from the other side. So in this case, he would see the dark ring in front of the light ring, which we see in our picture on the right. And on the bottom side, he would see the light ring in front of the dark ring. So let's adjust our picture to account for this. So the picture on the right is what Paul should see from his perspective. The white ring on the right, the dark ring on the left, the white ring crossing in front of the dark ring on the bottom, and the dark ring crossing in front of the white ring on the top. So the question asked us, what does he see? The answer is this, letter D. Problem number four states, in the addition problem to the right, some of the digits have been replaced by stars. What is the sum of the missing digits? So here I've replaced the stars with squares. In order to solve this problem, let's just do simple addition. So first let's start out on the right. 2 plus 3 plus 4 is going to be equal to 9. And that makes sense, since on the bottom we have a 9 as our last digit. We don't carry anything over, and then we move on to the squares. So we know that the sum of the squares has to end in a 0. But is it 0 or 10? Or maybe 20? Well, let's say for a second that it's 10. If it was 10, we would have the 0, and we would bring it down, and that would explain the 0 in the middle of 309. But we would also have to carry a 1. This 1 would then add to the digits on the left, and it would result in a 4. So the answer would be 409, not 309. This means the sum of the digits in the middle has to be equal to 0. So the question asked us, what is the sum of the missing digits? The answer is 0. Letter A. Problem number 5 states. What is the difference between the smallest 5-digit number and the largest 4-digit number? So first, let's start out with the largest 4-digit number, here represented by RRRR. In order to get the largest number possible, let's put in the largest digits that we can. The largest digit that we have is 9. So the largest 4-digit number is 9,999. For the 5-digit number XXXXX, let's put in the smallest digits that we can, in this case, 0. However, there is a problem. If all the 5 digits are a 0, then we would just have a 1-digit number of 0. So we have to put in the next smallest digit, a 1, somewhere in here. If we put it in the least significant part, we would have a value of 1, which is a 1-digit number. So that wouldn't work. So let's move it over. The next one over we would have 10, which would also not work because that would be a 2-digit number. So this shows us that we have to put the 1 all the way to the left to get the number 10,000. Now all we have to do is subtract the two numbers, so 10,000 minus 999 is equal to 1. So the question asked us, what is the difference between the smallest 5-digit number and the largest 4-digit number? The answer is 1. Letter A. Problem number six states, a square with a perimeter of 48 centimeters is cut into two pieces to make a rectangle. See picture. What is the perimeter of the rectangle? Okay, so we know that the perimeter of the square is 48 centimeters. This of course means that each side of the square is 48 centimeters divided by 4 centimeters long. So 48 divided by 4 is 12, which means that each side length of the square is 12 centimeters long. Of course the middle section, the line that you cut across, is also 12 centimeters long because it is the same width as the square. Now on the right, the after figure, after the person arranges the two rectangles together, we can see that the red line and the yellow line are equal, which are the heights of the two rectangles. Since they're equal, that means that the rectangles must be cut in the exact same height, meaning that the 12 centimeters is cut in half into two 6 centimeter sections. So now we know all the lengths of the square and the rectangle. Now let's apply these lengths to the rectangle on the right. So that's what we have. And now let's just add up all the lengths on the outside to get the perimeter of the shape. So we have 12 plus 12 plus 6 plus 12 plus 12 plus 6 centimeters, which in total is a perimeter of 60 centimeters. So the question asked us, what is the perimeter of the rectangle? The answer is 60 centimeters, letter D. Problem number 7 states, Katrina has 38 matches. Using all the matches, she builds a triangle and a square. Each side of the triangle consists of 6 matches. How many matches are used from one side of the square? So we know that she has 38 matches, and we know that she builds a triangle that has a side length of 6 matches. Since a triangle has 3 sides, this means that she uses 3 times 6 or 18 matches in total to make the triangle. 38 minus 18 is 20, which means that she has 20 matches left over to make the square. So she has 20 matches for the square, and there are 4 sides to a square. So 20 matches divided by 4 sides means that she has 5 matches per side of the square. So the question asked us, how many matches are used for one side of the square? The answer is 5, letter B. Problem number 8 states, the pearl necklace in the picture contains dark grey pearls and shiny white pearls. Alex wants to have 5 of the dark grey pearls. He can only take pearls from either end of the necklace, and so he has to take some of the white pearls as well. What is the smallest number of white pearls Alex has to take? So first, Alex can just take the 2 dark grey pearls on either end of the necklace, because there are no white pearls in the way. Next, let's figure out what his ultimate strategy is going to be. First, let's try taking 3 grey pearls from the left side. So we have to remove 4 white pearls to get 3 grey pearls. Now, let's try to do the same thing on the right side. In this case, we also have to remove 4 white pearls to get the remaining 3 grey pearls that he needs. So now, let's try a combination of taking grey pearls from the left side and from the right side. In this case, we can remove 1 white pearl on the left side and 2 white pearls on the right side to get the remaining 3 grey pearls that he needs. We got this by only removing 3 white pearls, which is the minimum that he can take. So the question asked us, what is the smallest number of white pearls Alex has to take? The answer is 3, letter B. Problem number 9 states, Harry participated in a broom flight contest, which consisted of 5 laps. The times when Harry passed the starting point are shown in the picture. Which lap took the shortest time? So let's calculate the amount of time that each lap took him. So first, the first lap took between 9.55 and 10.26. Between 9.55 and 10.26, there was 10 o'clock. Between 9.55 and 10 o'clock, 5 minutes passed. And between 10 o'clock and 10.26, 26 minutes passed. Which means that the first lap took him 5 plus 26 minutes, or 31 minutes. For the second lap, it took him between 10.26 and 10.54. The hours don't change, so we don't have to use an intermediate number. The difference between 54 and 26 is 28. So the second lap took him 28 minutes. The third lap was between 10.54 and 11.28. Between 10.54 and 11.28, there was 11 o'clock. Between 10.54 and 11 o'clock, there were 6 minutes. Between 11 o'clock and 11.28, there were 28 minutes. So in total, it took him 34 minutes to complete the third lap. The fourth lap took him between 11.28 and 12.03. There's 12 o'clock in the middle, and between 11.28 and 12 o'clock, it was 32 minutes. And between 12 o'clock and 12.03, there were 3 minutes. So in total, the fourth lap took him 35 minutes. Lastly, for the fifth lap, the hours don't change, so we can just take the difference. 32 minus 3 is 29 minutes. So here are the times it took him to complete all the laps, and it's easy to see which one took him the least amount of time. So the question asks this. Which lap took the shortest time? The answer is the second, letter B. Problem number 10 states. Ben's digital watch is not working properly. The three horizontal lines in the digit on the far right of the watch do not display. Ben is looking at his watch, and the time has just changed from the one shown on the left to the one shown on the right. What time is it now? So the red horizontal lines will never be displayed. So let's figure out what possible numbers could be displayed on the left and on the right with the three lines not working. So obviously, it has to be a digit. So it can either be 0, 1, 2, 3, 4, 5, 6, 7, 8, or 9 for both of them. On the left, a 0 is not possible, because the two segments on the left would have to be lit. So 0 is not a possible number. 1 is a possible number. 2 is not a possible number, because the segment in the bottom left would have to be lit. 3 is a possible number. 4 is not a possible number, because the segment in the upper left would have to be lit. 5 is not possible either, because the segment in the top left would have to be lit. 6 is not possible either, because the segment in the top right would not have to be lit, and the segments on the left would have to be lit. 7 is possible. 8 is not possible, because all the possible segments would have to be lit. And 9 is not possible, because the top left segment would have to be lit. So the only combinations we have on the left are 1, 3, and 7. Now moving on to the right. 0 is still not possible, because the bottom left has to be lit. 1 is not possible, because the top left would have to be not lit. For the 2, the top left wouldn't have to be lit either, so that's not possible. For the 3, the top left wouldn't be lit either, so that's not possible. The 4 is possible. The 5 is not possible, because the top right wouldn't have to be lit. The 6 is impossible, because the bottom left would have to be lit. The 7 is not possible, because the top left wouldn't be lit. The 8 is not possible, because the bottom left would have to be lit. And the 9 is possible. So in total, we only have a 4 or a 9 on the right. The problem tells us that the times are consecutive, and the only consecutive digits that we have are 3 and 4, which means the time on the left is 1243, and the time on the right is 1244. So the question asked us, what time is it now? The answer is 1244, letter C. Problem number 11 states, which tile must be added to the picture so that the light gray area is as large as the dark gray area? So first, let's analyze each of the tiles individually. Let's start out with this diamond one. We can see that a quarter of it is white, a quarter dark grey, a quarter white, and another quarter dark grey. The quarter that is dark white and the quarter that is dark grey, the both of them, cancel each other out. So we can treat this tile as if it doesn't even exist. We can see the exact same situation right here. Next, let's take a look at this one. In this one, half of the tile is white and half of the tile is grey. Again, the half of the tile that's white cancels out with half of the tile that's grey, so we can treat it as if it's not even there. We can do the same for this one, and this one, and this one. Now we're left with only two white tiles. And the question asked us, what tile should we put there such that the white area is equal to the grey area? Well, we've cancelled out all the white and the grey areas that overlap, and all we're left with is two white areas. So into this square, we should put two grey squares. Obviously it's impossible to put two grey squares into one square, so this question is impossible. So the question asked us, which tile must be added to the picture so that the light grey area is as large as the dark grey area? The answer is, it is impossible. Letter E. Problem number 12 states, Henry and John started walking from the same point. Henry went 1km north, 2km west, 4km south, and finally 1km west. John went 1km east, 4km south, and 4km west. Which of the following must be the final part of John's walk in order to reach the point where Henry ended his walk? Let's let each cell of the grid be 1km by 1km. So let's do a simulation. We'll let the blue dot represent Henry and the red dot represent John. We can't see John in this case because they're both in the same spot because the question said they start off at the same spot. So first the question told us that Henry went 1km north. Then it said that he went 2km west, 4km south, and in the end 1km west. So he ended up here. The problem also told us that John went 1km east, 4 south, and 4 west. In order for John to reach Henry, he should go north 1km. So the question asked us, which of the following must be the final part of John's walk in order to reach the point where Henry ended his walk? The answer is 1km north. Letter B. Problem number 13 states, at the summer camp, 7 children eat ice cream every day. 9 children eat ice cream only every other day, and the rest of the children don't eat ice cream at all. Yesterday, 13 kids had ice cream. How many children will eat ice cream today? So the question told us that there are 7 kids that eat ice cream every day, and 9 that eat it every other day. And we know that yesterday, 13 kids ate ice cream. Of course, this means that the 7 kids that eat ice cream every day ate ice cream that day, as well as 6 kids who eat ice cream on alternate days. So how many kids ate ice cream today? Well, the 7 kids that go every day are still there the next day. But this time, the 3 kids that didn't go yesterday go today for ice cream. So in total, we have 10 kids eating ice cream today. So the question asked us, how many children will eat ice cream today? The answer is 10, letter D. Problem number 14 states, Kangaroos A, B, C, D, and E are sitting in that order clockwise around a round table. Exactly when the bell rings, each kangaroo, except for one, changes its place with a neighbor. The resulting order, clockwise, and starting with the kangaroo A, is A, E, B, D, C. Which kangaroo did not move? Alright, so we know we have the kangaroos A, B, C, D, E, and we know they're sitting at a round table. And we know that after the bell rings, one of them sits still, and the neighbors swap places. And then in the end, we get the order A, E, B, D, C. It's important to note that this order isn't the same reference point as we have here. It is relative to A. So for example, A could move, and it could still be listed as first on the list. I'll show you an example of this later. But right now, let's just try different combinations. So first, let's say that A stays still. If A stays still, the only way that the neighbors can switch is if B and C switch, and D and E switch. If A stays still, we get the order A, C, B, E, D, starting from A, going clockwise. This clearly isn't our answer, because we're looking for A, E, B, D, C, and we got A, C, B, E, D. So let's move on to B. When we keep B still, A and E switch, and D and C switch. And as a result, we get A, E, B, D, C, which is actually the order that we're looking for. So B stays still is an answer, but let's keep on looking just in case. So now we'll say that C doesn't move, and in this case, D and E will switch, and A and B will switch, and we'll get an order starting from A, going clockwise, A, C, E, D, B. Next we'll keep D still, A and E will switch, and B and C will switch, and we'll get an order of A, E, C, B, D. Next let's revert it to our original, and hold E still, D and C would switch, and A and B would switch, and we'd get an order of A, D, C, B. None of these match our pattern except for B, so B must be the answer. So the question asked us, which kangaroo did not move? The answer is kangaroo B, letter B. Problem number 15 states, a square can be formed using four of these five pieces. Which one will not be used? So first let's take a look at pieces D and E. Both pieces D and E have three straight lines. This means that the only piece that can touch them on one of the sides is another straight piece. Of course, a square has straight lines, so the only way that D and E could be included is if both of them are included. For example, we could not make the squares with ABCD or ABCE, because there would be nothing to pair up the straight edge of D or the straight edge with E. So we need to include both of them if we include one of them. And since we can only exclude one shape, we have to include both of them. So we'll start arranging our square with D and E. We'll arrange it like this. So now we've reduced the possible answers so much that we can just guess and check. So let's try B. Let's try putting B right here. Sadly, there is no piece that fits in the upper left corner that works, so B cannot go down there. C cannot go down there as well, because it's got the obtuse section coming out, and D's got that too, so they just don't fit together. So A must fit there. And now the only piece that fits in the upper left corner is C, which means that we can make a square without using piece B. So the question asked us, which one will not be used? The answer is shape B, letter B. Problem number 16 states, a certain natural number has three digits. When we multiply the digits, we get 135. What result do we get if we add the digits? So first, let's find the prime factorization of 135. The prime factorization is what prime numbers we need to multiply together to make the original number. So in order to do this, let's just figure out a number that divides into 135. Obviously I see the 5 at the end, so the first thing I think of is a 5. So if we divide a 5 out, we get multiples of 5 and 27. We can't break up 5 anymore because it is prime, but we can break up 27. 27 is composed of 3 and 9. We can't break up the 3 anymore because it's prime, but we can break up the 9 into 3 and 3. This means that 135 prime factorization is 5, 3, 3, 3. So 135 is going to be equal to 5 times 3 times 3 times 3. So the problem told us that there is a three-digit natural number that had digits who, when multiplied together, are equal to 135. So we could say that one of the digits is 5 and that another digit is 3. And next, we can say that the next digit is 3 times 3, because the problem never told us that the digits have to be primes. And we have two 3s remaining, so if we multiply them together, we get 9. So perhaps the number was 539, or rather just had the digits 5, 3, and 9 in any order. So then the problem asked us for the sum, so 5 plus 3 plus 9 is 17. So the question asked us, What result do we get if we add the digits? The answer is 17, letter D. Problem number 17 states, in a restaurant there are 16 tables, each of which 3, 4, or 6 chairs. Together, the tables which have 3 or 4 chairs can accommodate 36 people. If the restaurant can accommodate 72 people, how many tables are there with 3 chairs? So let's use some algebra to solve this problem. We'll let x equal the number of tables with 3 chairs, y the number of tables with 4 chairs, and z the number of tables with 6 chairs. This means that if we have 3 times x, this will be the number of people that the tables with 3 chairs can accommodate, because each table has 3 chairs so it can accommodate 3 people. We can do the same thing for y and z. We'll have 4y in the case of the tables with 4 chairs, and 6z in the case of the tables with 6 chairs. Now the problem told us that the tables that have 3 or 4 chairs can accommodate 36 people. So algebraically we could represent this as 3x plus 4y equals 36, because the number of people accommodated with the 3 chair tables and the number of people accommodated by the 4 chair tables has to be equal to 36. The problem then told us that the restaurant can accommodate 72 people, so in total 3x plus 4y plus 6z has to equal to 72, for the same reasons stated earlier. And actually there's another bit of information that's very useful here. In the very beginning of the problem it says that the restaurant has 16 tables, so we can say that x plus y plus z equals 16, because x, y, and z are both the number of tables with their respective number of chairs. So now we have 3 equations and 3 unknowns so we can solve. First let's take the bottom equation and multiply both sides by negative 1, to get negative 3x minus 4y equals negative 36. Now let's combine this equation with the equation directly above it by adding them to both sides. So the 3x and the negative 3x cancel, the 4y and the negative 4y cancel, there's nothing to cancel out the 6z, so we get 6z on the left side, and 72 minus 36 is 36. So we get 6z equals 36, dividing both sides by 6, we get z equals 6. So now we can replace z in all of our equations with 6, and this is what we get. x plus y plus 6 equals 16, 3x plus 4y plus 36 equals 72, and then still the original 3x plus 4y equals 36. We can simplify the top equation by subtracting 6 from both sides to get only x plus y equals 10, and we can simplify the middle equation by subtracting the 6 times 6, or 36 from both sides, and getting 3x plus 4y equals 36. This is exactly the same as the lower equation, and there's no point in having two of the same equations, so let's just remove one of them. Now we just have to solve for x and y using these two equations. First let's multiply the top equation by negative 3. So we get negative 3x minus 3y equals negative 30. Now let's combine these two equations by adding them together. So negative 3x plus 3x cancels out, and negative 3y plus 4y cancels out to just y. Negative 30 plus 36 is equal to 6, which means that the number of tables with four chairs each is equal to 6. We can plug in 6 for y into our original equation, and get 3x plus 4 times 6 equals 36. 4 times 6 is 24, and if we subtract 24 from both sides, we get 3x equals 12, or that x equals 4 when we divide both sides by 3. So the question asked us, if the restaurant can accommodate 72 people, how many tables are there with three chairs? The answer is 4, letter A. Problem number 18 states, the points A, B, C, D, E, and F are on a straight line in that order. We know that AF is equal to 35, AC is equal to 12, BD is equal to 11, CE is equal to 12, and DF is equal to 16. What is the distance BE? So here we have all of our points laid out on a number line. And then here's all the information that the problem gives us. And the problem tells us that we're looking for the distance between points B and points E. So in order to do this, first let's take a look at these two. The yellow length and the red length. The difference between the yellow length and the red length will give us the distance of EF. So first let's combine the red length. 12 plus 12 is going to be equal to 24. And now the difference between the yellow line and the red line is going to give us the cyan line. Of course the difference between 35 and 24 is 11, which means that the length of EF is 11. Now let's actually figure out the length of BE. The difference between the red segments and the yellow segments is going to be the green segment, the segment that we want to find BE. So first let's find the length of the red segment. 11 plus 16 is going to be equal to 27. And now if we take the difference of 27 and 11, we'll get the length of BE. The difference between 27 and 11 is 16. So the length of BE is 16. So the question asked us, what is the distance of BE? The answer is 16. Letter D. Problem number 19 states, Marissa set her stones in groups on the desk. After she arranged the stones in groups of three, she found that there were two stones left. When she arranged the stones in groups of five, and again there were two stones left. At least how many more stones does she need so that there won't be any left when she arranges them in groups of three and when she arranges them in groups of five? Okay, so here we have all of Marissa's rocks, and right now they're in horizontal groups of five. And we can see that in the end we have two remaining. We'll say that she has n stones. So if she removes two of these stones, or n-2, we know that this value is going to be divisible by five. Next let's move on to groups of three. When she arranges her rocks into groups of three, here represented in horizontal stripes, she also has two rocks remaining. Of course if we take away those two rocks, then we know that the amount of rocks she has, n-2, is divisible by three this time. So we know that n-2 is divisible by both five and by three. This means that n-2 is divisible by 15. Now we know that the total number of stones is going to be equal to a multiple of 15 plus two. Of course we don't know what multiple of 15, it could be 30 or 1,500. What this also tells us is if we add 13 stones to the multiple of 15 plus two stones, we'll also have a multiple of 15. And this will allow her to arrange her stones in groups of three and five perfectly. So if she adds 13 stones, this can happen. So the question asked us, at least how many more stones does she need so that there won't be any left when she arranges them in groups of three and when she arranges them in groups of five? The answer is 13. Letter E. Problem number 20 states, the faces of a cube are numbered 1, 2, 3, 4, 5, and 6. The faces numbered 1 and 6 have a common edge. The same is true for faces numbered 1 and 5, faces numbered 1 and 2, faces numbered 6 and 5, faces numbered 6 and 4, and faces numbered 6 and 2. Each number is on the face opposite the face with number 4. So in order to solve this problem, let's draw the net of a cube. Since there's really no effective way of showing the faces and all the edges of a cube with the actual cube drawn, we'll have to do it on a net. So now let's locate all the edges that touch each other. So obviously all the ones that touch each other on the picture, but also when the cube net is folded, these edges right here fold together. Same with these two, these two, and then finally when the whole cube wraps around, these two on the sides as well. So any of the two edges with a similar color can have numbers next to each other that share the same edge. So now let's bring up all the numbers that share the same edge. Of course, the number 1 is very common in here, so let's put the number 1 in the very middle. So the first interesting thing is that 1 shares an edge with 6 and an edge with 5, and 5 shares an edge with 6. This means that all of them have to be touching edges together. So the only way we can make it work is if we put it into an order like this. Of course, the exact placement of the 6 and the 5 doesn't really matter, they could be to the left, they could swap places, but this is one possible way. Next let's take a look at the 1 and the 2 and the 6 and the 2. Since the 1 and the 6 are already touching edges, there's only one place to place the 2, right here. Next let's take a look at the 6 and the 4. The 6 already has three neighbors, 2, 1, and 5, and it only has room for one more, the square all the way on the right. So that square has to be a 4. Of course, this means by elimination that the last square has to be a 3. So now the question asked us, what was the face on the opposite side of the face with the 4? So here we have the 4, and on the other side we have a 1. So the question asked us, what number is on the face opposite the face with number 4? The answer is 1, letter A. Problem number 21 states, the 3 by 3 by 3 cube in the picture is made of 27 small cubes. How many small cubes do you have to take away to see the picture on the right as a result from looking from the right, from above, and from the front? Alright so first, let's take a look at the view from the front. In order to get this U shape, if we hollow out these three sections, we get the desired shape. Next let's do the same from the right side. This time we only have to take out two, because one of them has already been taken out by the front section. So now our shape is symmetrical on both the x and the y axes, so we can say whichever face we want is the top. In this case, we'll say the face that's facing us is the top face. Now that we know this, if we hollow out these two, we get the desired shape when looking from the top. So in total, we only had to remove 7 cubes to get that view from the front, the right, and from the top. So the question asked us, how many small cubes do you have to take away to see the picture on the right as a result from looking from the right, from above, and from the front? The answer is 7. Letter D. Problem number 22 states, there are 5 songs. Song A lasts 3 minutes, song B 2 minutes and 30 seconds, song C 2 minutes, song D 1 minute 30 seconds, and song E 4 minutes. These 5 songs are playing in the order A B C D E in a loop without any breaks. Song C was playing when Andy left home. He returned home exactly 1 hour later. Which song was playing when Andy got home? Okay, so here we have the music track A B C D and E. For B it said that it was 2 minutes and 30 seconds, which is 2 and a half minutes, and for D it said 1 minute 30 seconds, so it's 1 and a half minutes. So I just simplified the problem a little bit. Next, the problem tells us that when we start off, song C is playing. That means that it could be anywhere between the white arrow and the yellow arrow. So we'll have to keep track of both of them as we continue. Now, let's take a look at how long all 5 songs take to play. 3 plus 2 and a half plus 2 plus 1 and a half plus 4 is 13 minutes. The problem wanted to know what happens 1 hour from now. So let's take a look at how many times the track loops around in that time. So to do this, we'll divide 60 minutes by 13 minutes. And we get 4 times and a remainder of 8 minutes. So that means that the song loops around 4 times and then goes an additional 8 minutes in that 60 minute time frame. So after it's looped 4 times, we're just back where we started. So 4 times 13 is 52. So at 52 minutes, we're right here. So next, let's move the white and the yellow arrow over 2 minutes to get 54 minutes. Next, let's move them over 1 and a half minutes to get 55 and a half minutes. Next, let's move them over 4 minutes. And then finally, let's move them the 30 seconds that we need. And now it's easy to see that after 60 minutes, no matter where he was during the song, song A is playing. So the question asked us, which song was playing when Andy got home? The answer is song A. Letter A. Problem number 23 states, Dan entered the numbers 1 to 9 in the cells of a 3x3 table. He began by placing the numbers 1, 2, 3, and 4 as shown in the picture. After he was finished, the sum of the numbers in the cells adjacent to, having a common side with, the cell with the number 5 is equal to 9. What is the sum of the numbers in the cells adjacent to the number 6? Okay, so let's just try to place our number 5 somewhere. First let's try to place it here. Of course, right now the sum is 3 and 4, but we need to place another number. So let's place the lowest number that we possibly can to make sure we have the best chance of not overshooting 9. The lowest number that we have still available, we took up 1, 2, 3, 4, 5, so that would be 6. So if a 6 is the neighbor, we would have a sum of 13, which is too high, so 5 cannot be placed there. Let's try placing 5 here. 2 plus 4 is 6, plus 6 is 12, so that's still too high. This doesn't work. Next let's try up here. 1 plus 3 is 4, plus 6 is 10. Closer, but still too high. Let's try the only place we have left. Back here. 1 plus 2 is 3, plus 6 is 9. So this works. The problem told us that he placed the numbers 1 through 9 in the 3 by 3 grid, which means he has to still place 7, 8, and 9. It doesn't matter where these go, since the question asked us for the sum of the numbers adjacent to the number 6. All of them will be adjacent, so it doesn't matter. So the sum is going to be 5 plus 7 plus 8 plus 9, which is 29. So the question asked us, what is the sum of the numbers in the cells adjacent to the number 6? The answer is 29. Letter E. Problem number 24 states. Trees grow on only one side of Park Avenue. There are 60 trees in total. Every other tree is a maple, and every third tree is either a linden or a maple. The rest of the trees are birches. How many birches are there? So this is how we'll represent all of the trees. The maples got red leaves, the linden has green leaves, and the birch has a white bark and green leaves. And now, here we have 30 trees that we need filled up. They're all in a row, but I put them into two rows so it's easier to see them. The reason why we don't have 60 trees and we only have 30 trees is because the same pattern will repeat twice. So let's just do it for 30 and then multiply our answer by 2 at the end. So first the problem tells us that every other tree is a maple. So let's make every other tree a maple. Then the problem tells us that every third tree is either a linden or a maple. So let's go every third one. And if the tree is a maple, then we'll say it's a maple. But if it's not yet identified, we'll say it's a linden. This gives us 5 linden trees. Then the problem tells us that the rest of the trees are birches, so all the remaining unidentified trees will be birches. In total, we have 10 birch trees. But this is only half of the problem, because this is only 30 trees. So we need to multiply our answer by 2, and we get 20 birch trees. So the question asked us, how many birches are there? The answer is 20. Letter C. Problem number 25 states. A thin ribbon is glued on a transparent plastic cube. See the picture. Which of the following pictures doesn't show the cube from any perspective? So first, let's take a look at A. If we arrange our field of view like this, with the orange bars being on the side, the green being the top of our view, and the red being the bottom of our view, we will see this shape. The left will be the little zigzag on the left, the right side will be the zig on the right, and the top will be the pattern on the top. So A works. Now for B, if we arrange our view like this, we will be able to see the hourglass figure. So B works. For C, if we arrange our view like this, we can see the N pattern. So C works. If we rotate our view 90 degrees, we will be able to see the Z pattern in D, so D also works. Now E, I can't show you how to look at it because it's not possible to see it from a perspective, so E is sadly not possible. So A, B, C, D are possible, but E is not possible. So the question asked us, which of the following pictures doesn't show the cube from any perspective? The answer is E. Letter E. Problem number 26 states... What is the time interval between any two consecutive messengers arriving at the castle? So here we have a layout of the map. We'll say that the king right now is at 10 kilometers. Since the problem told us that the interval is going to be the same between any two consecutive messengers, it doesn't really matter. So now, let's just simulate. In one hour, the king is going to travel 5 kilometers, which is one tick mark on our number line. After this one hour, he's going to get a messenger, and he's going to send him back at a speed of 10 kilometers an hour, back to the castle at 0 kilometers. After one hour passes, the messenger moves back two tick marks, and the king moves forward one tick mark. At this point, the king sends back another messenger. After 30 minutes, the first messenger has already made it back to the castle. The elapsed time is two and a half hours. The king has moved forward one half of a tick mark, and the second messenger has moved back one tick mark. In the next 30 minutes, the messenger will move back another tick mark, and the king will move forward another half tick mark. But again, another hour has passed, so the king will send back another messenger. After one more hour, the second messenger will make it back to the castle. It took him only four hours to make it there. Which means we have a difference of four hours and two and a half hours, or one and a half hours, between messengers coming back to the castle. So the question asked us, what is the time interval between any two consecutive messengers arriving at the castle? The answer is one and a half hours, or 90 minutes. Letter D. Problem number 27 states, there were three one-digit numbers on the blackboard. Ali added them up and got 15. Then he erased one of the numbers and wrote the number 3 in its place. Then Reza multiplied the three numbers on the blackboard and got 36. What are the possibilities for the number that Ali erased? So here we'll have our three digits, A, B, and C. We know that if we add them all up, we'll get a value of 15. And we know if we replace the first one with a 3, and then multiply them together, we'll get 36. In this last equation, if we divide both sides by 3, we get that B times C has to be equal to 12. There are only two ways for the product of B and C to be both one digit and have a product of 12. Those two ways are if B and C are 6 and 2, or if B and C are 3 and 4. It doesn't matter whether B is 3 or C is 4, or B is 4 and C is 3, because multiplication is commutative. So now, let's plug these into our first equation to figure out what A could be. First, let's try 6 and 2. Doesn't matter what we plug in for what, because addition is also commutative. So A plus 6 plus 2 equals 15. That gives us A plus 8 equals 15. Subtract 8 from both sides, we get that A equals 7. So that's one possibility, that the first digit could have been 7. But now, let's plug in 3 and 4. So A plus 3 plus 4 equals 15. So A plus 7 equals 15. Subtract 7 from both sides, we get A equals 8. This means that A could have equaled 7 or 8. So the question asked us, what are the possibilities for the number that Ali erased? The answer is either 7 or 8. Letter B. Problem number 28 states, Peter Rabbit loves cabbages and carrots. In a day, he eats 9 carrots only, or 2 cabbages only, or 1 cabbage and 4 carrots only. But some days, he only eats grass. Over the last 10 days, Peter ate a total of 30 carrots and 9 cabbages. On how many of these 10 days did he eat only grass? So first, let's let x represent the number of times Peter eats 9 carrots, and y represent the number of times Peter eats 2 cabbages, z the number of times Peter eats 1 cabbage and 4 carrots, and w the number of times Peter eats grass. The problem told us that this problem takes place over 10 days. So we know that x plus y plus z plus w has to be equal to 10, because he has to eat something every day, and he can only eat x, y, z, or w a day. Next, the problem told us that Peter ate a total of 30 carrots. This means that 9 times x plus 4 times z has to be equal to 30, because every time that Peter ate x, he ate 9 carrots, and every time Peter ate combination z, he ate 4 carrots. So 9 times x plus 4 times z has to equal to 30. Then the problem told us that he ate 9 cabbages. We can do a similar thing here. Since every time he eats y, he eats 2 cabbages, and every time he eats z, he eats 1 cabbage, we can say that y times 2 plus z times 1 has to be equal to 9. Now if we take a look at this bottom equation, the only numbers for x and z that satisfy the expression are if x equals 2 and if z equals 3. You can figure this out quite easily by knowing that x has to be an even number, and x has to be less than 4, yet greater than 0. We can plug in x equals 2 and z equals 3 into all of our equations now, and simplify. The bottom equation is just going to tell us that 30 equals 30, so we can just exclude that one. Now this is what we have left. Now let's solve for y in our bottom equation. Let's subtract 3 from both sides to get y times 2 equals 6, and then divide both sides by 2 to get y equals 3. Now let's plug in y into the top equation, and we get 5 plus 3 plus w equals 10, or 8 plus w equals 10, which means that w must be equal to 2. So the question asked us, on how many of these 10 days did he eat only grass? The answer is 2. Letter C. Problem number 29 states, in Fabuland, every sunny day is immediately preceded by two consecutive rainy days. Also, five days after any rainy day, it is another rainy day. It is sunny today. For how many days, at most, can we predict the weather with certainty? So first, let's go over the rules. They said that if today is a sunny day, then the next two days are going to be rainy days. The problem also said that if on one day, it's rainy, five days in the future, it will also be a rainy day. So now, let's apply these rules. We know that today is a sunny day. If yesterday was a sunny day, this means that today must have been a rainy day, and if two days ago was a sunny day, that means today must have also been a rainy day. This means that neither yesterday nor two days ago could it have been sunny. It must have been raining. Now applying the first rule that following a sunny day are two consecutive rainy days, two rainy days will follow today. Next, five days after this rainy day, there will be another rainy day. And five days after this rainy day, there will also be another rainy day. This is all what we can predict. We can only predict four days into the future. It's a pretty gloomy forecast, but at least we know what it is. So the question asked us, for how many days, at most, can we predict the weather with certainty? The answer is four days. Problem number 30 states, Granny has 10 grandchildren. Alice is the oldest. One day, Granny notices that her grandchildren all have different ages. If the sum of her grandchildren's ages is 180, what is the youngest Alice can be? So in order to solve this problem, let's plug in our answers back into our equation. We know that the ages of all of Granny's grandchildren have to be different and have to add up to 180. So first, let's start out with A. We know that 19 would be the oldest age, so we would do 19 plus 18 plus 17 all the way up until 10. And we get a sum of 145 when we do this. This is not greater than or equal to 180, so A is not possible. Let's move on to B. Now we would be doing 20 plus 19 plus 18 plus 17, on and on and on, all the way until plus 11. Essentially what we're doing is we're just adding 1 for every single number in our previous equation. So rather than redoing the addition, we can just simply say that it's that value plus 10. So in this case, it would be 155, which is still not greater than or equal to 180. Next for C, we'll try 21. We'll use the same trick and we'll add 10, and we get 165, which is not greater than or equal to 180. So let's continue on to D22. Let's add 10 again, we get 175, which is not greater than or equal to 180 either, so that one doesn't work. Of course, this means that E is probably the answer, but let's just check it to make sure. 175 plus 10 will be 185, which is greater than or equal to 180. In order to get that 180 that we need, let's just subtract 5 from the 14 to say that the last grandchild is 9 years old instead of 14. And there we go, it works. So the question asked us, what is the youngest Alice can be? The answer is 23 years old, letter E.

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