This is the Math Kangaroo Solutions Video Library, Levels 3 and 4 from the year 2006, presented by Agata Gazzal. The purpose of the Math Kangaroo Solutions Video Library is to help you learn how to solve math problems such as those presented in the Math Kangaroo Competition. Let me start with a few tips of how to best utilize these solutions. First, make sure that you read the problem, and in this case, listen to it carefully. This will help you understand what kind of information you need to solve the problem. After reading the problem, I will wait for a few seconds. Please use this opportunity to pause the video and try solving the problem on your own. If you get stuck, don't worry. I will start the presentation of the solution with a hint of how to arrive at the correct answer. You can pause the video again at this point and try to finish the problem. My presentation of the solution will follow. Keep in mind that there might be ways of solving the problem other than what I present. I hope that you will learn something new from each problem you attempt, and most importantly, that you will have fun. Problem number one. During a summer math camp in the city of Zakopane in Poland, the students take part in a trip to the top of Mount Giewont. It takes three hours to get to the top of the mountain. They stay at the top of the mountain for half an hour. Afterwards, it takes two and a half hours to come down the mountain. What time in the morning at the latest does the trip need to start so that everybody is back at the camp for a meal at 3 p.m.? To solve this problem, we will first find how long the whole trip would take, and then we will find what time the group needs to leave in order to be back at 3 o'clock. The whole trip is the three hours that it takes to go up the mountain, plus the half hour rest at the top of the mountain, plus the two and a half hours it will take to come down the mountain. Half an hour plus two and a half hours is three hours. So we have three hours plus three hours, which together is six hours. So now we know that the whole trip will take six hours. Now, let's look at a clock. In order for the group to be back at 3 o'clock, they will need to leave six hours earlier. Let's go back six hours on the clock here. One hour, two, three, four, five, six. So we see that six hours before 3 o'clock is 9 o'clock. The group needs to leave by 9 o'clock in the morning, which is answer C. Problem two. What is the value of this expression? 2 times 0 times 0 times 6 plus 2006. The most important rule to remember in this problem is that we need to do multiplication before we do the addition. Let me rewrite the problem down here. 2 times 0 times 0 times 6 plus 2006 will be equal to the value of the expression 2 times 0 times 0 times 6 plus 2006. If you remember, anything times 0 is 0. So the first part becomes 0, and then we add 2006. Adding 2006 to 0 gives us 2006. So the answer is B. This expression is equal to 2006. Problem number three. How many cubes have been removed from the first block to obtain the second block? Figure out how many cubes are in the top layer, and then subtract the number of the cubes that remain in the top layer. The figure is made out of 18 cubes, 9 in each layer. In order to have 2 cubes left on the top layer, we need to remove 7 cubes from that layer. We see that since 9 minus 2 is 7, the answer is 7. The answer is D. 7 cubes have been removed from the figure. Problem number four. Katie's birthday was yesterday. It is Thursday tomorrow. What day was Katie's birthday? Let's figure out the days of the week in order. The week starts with Sunday. We have Monday, Tuesday, Wednesday, Thursday, Friday, and Saturday. From the problem, we know that tomorrow will be Thursday. If tomorrow is Thursday, then today is the day before Thursday, which is Wednesday. So now we know that today is Wednesday. The birthday was yesterday. Yesterday is the day right before today. We need to go back one more day. And we get to Tuesday. So Katie's birthday was on Tuesday. This is answered. Problem number five. John is playing with darts. He brings back all the darts after he has thrown them. And for each time he hits the bullseye, he gains 2 additional darts. At the beginning, he has 10 darts, and at the end, 20. How many times did he hit the bullseye? Notice that John does not use up or lose any of the darts that he throws. So we know that the number of darts he gained is the same as the difference of the number of darts he had at the end, which was 20, and the darts he had at the beginning, which was 10. 20 minus 10 is 10. So John gained 10 darts. Now let's see how many times John had to hit the bullseye to gain 10 darts. He has 2 darts for each time he hits the bullseye. So 10 darts divided by 2 is 5. So John hit the bullseye 5 times and gained 10 darts. The answer is D, 5. Problem number six. A kangaroo enters the building as shown in the picture. He only passes through triangular rooms. Where does he leave the building? To solve this problem, we will need to see which rooms are shaped as triangles, that is, which ones have exactly three sides. Let's mark all the rooms that are triangles in orange. Now we see that the path that the kangaroo needs to take will go like this. There are some other triangular rooms in this building, but they either are not connected or do not lead to an exit. So the kangaroo will come out of any of these triangular rooms. But they either are not connected or do not lead to an exit. So the kangaroo will come out of exit E. The answer is E. Problem number seven. Four people can sit at a certain kind of square table. For a school party, the students put together seven such square tables in order to make one long rectangular table. How many people can sit at this long table now? First, let's get an idea of what one of the square tables looks like. It should be something like this. Here's the table, and then we have one, two, three, four seats. If we put two of the tables together, here's one. And here's another. We can have three seats on one of the end tables. And three seats on the other. So, if we put more than two of these tables together in a row, we'll have a table at the end with three seats. A table at the other end, also with three seats. And then any tables that we put in between them, we'll see two people each. Now, if we want to put seven tables together, we will have two tables at either end. And in order to make 7 tables, we will have 5 tables in between here. 1, 2, 3, 4, 5. And each one of these will seat 2 people. So if we put seven tables in one long rectangle, we can seat two times three people, those are the end tables, plus five times two people, those are the five tables in between them, two times three is six, plus five times two is ten, which means that all together we can seat 16 people. The answer is B, 16. Problem number eight. In his wallet, Stanley has one five dollar bill, one two dollar bill, and one one dollar bill. Which of the following amounts can Stan not make out of the bills that he has? A, three dollars, B, four dollars, C, six dollars, D, seven dollars, or E, eight dollars. Remember that he only has one of each kind of bill. This problem asks which amount Stanley cannot make. So we will need to try making the amounts, and we should be able to make four out of the five that are listed. First let's write down the bills that Stanley has. There's a five dollar bill, a two dollar bill, and a one dollar bill. For answer A, we need to try to make three dollars. We can easily do that by using the two dollar bill and the one dollar bill. So we can make three dollars. So answer A will not be the one that he cannot make. Answer B asks to make four dollars. Now we don't have anything that adds up to four dollars here. We would need two two dollar bills, so this is probably the correct answer because there's no way that we can add up these bills to make four dollars. Just to make sure, let's check the other answers. For C, we need to make six dollars, which we can make using the five dollar bill and the one dollar bill. Six dollars. For D, make seven dollars. We can use the five dollar bill and the two dollar bill. And then answer E asks us to make eight dollars, and that is the sum of all three bills put together. Five plus two plus one is eight, so we can make eight dollars using all the bills. So again, we see that the answer is B, four dollars, and that's the amount Stan cannot make out of the bills that he has. Problem nine. On one side of Long Street, the houses are numbered with consecutive add numbers from 1 to 19. On the other side of that street, the houses are numbered with the consecutive even numbers from 2 to 14. How many houses are there on Long Street? One way to solve this problem is to notice that because the add numbers start with 1 and the even numbers start with 2, up to a certain point, the houses will be numbered with consecutive natural numbers. Then there will be just a few more house numbers to take account of. The house number with add numbers will be numbered 1, 3, 5, and so on, all the way up to 19. The even numbers will be 2, 4, 6, and so on, up to 14. Let's try the last few add numbers to see where the overlap is. So 13, 15, 17, and 19. Now we can easily tell that the consecutive numbers go from 1 to 15. Since 14, it's right in here. And beyond that, we just have house numbers 17 and 19. The houses from 1 to 15 make up 15 houses, and then there we have two more. So 15 plus 1 plus 1 will be 17. So there are 17 houses on Long Street. This is answer C. Problem number 10. From which of the figures below was the figure to the left cut out? The figure shown here on the left was cut out from one of the rectangles shown in the answers. We need to find out what the least length and height of the rectangle is that would allow us to cut out this figure. To make things a little easier to see, I'm going to outline again the figures and the answers. So looking back at our original figure, let's see what the length and the height are. Let's find the widest parts that it goes to, and the tallest parts right there. And we can see that it's 1, 2, 3, 4, 5, 6 squares across. And 1, 2, 3, 4, 5, 6, 6 squares in height. So the rectangle that it was cut out from used to be at least 6 by 6. Let's look at the biggest one, which looks like it's figure E. It's 1, 2, 3, 4, 5, 6 squares high, squares high, and 1, 2, 3, 4, 5, 6 squares across. So rectangle E is the one that the figure was cut out of. The answer is E. Problem number 11. The picture below shows bus routes and ticket prices between six towns. What is the least amount of money to pay for tickets to get from town A to town B? To solve this problem, we will need to pick routes and just check the prices. Looking at answers helps us because if we can find a route that costs 90, we know that we have found the smallest route. Otherwise, it might be 100 or 110. We need to find the route from town A to town B. We can kind of see that the less expensive routes, like this one is 10, 20, 30, 20, 10 here get us to the next destination in a less expensive way than the longer ones. We can get to the town in the middle here from town B for 10 plus 20 plus 30, which is 60 if we go to town A. The roundabout way as opposed to going straight for 70. The same is true on the other side. We can get to this town for 20 plus 10 plus 30, which is 60. Just cheaper than going the straight way for 80. And this is actually a route that connects us from A all the way to B. If we add all these, we have 20 plus 10 plus 30 plus 20 plus 10, which is 2 plus 1 is 3, plus 3 is 6, plus 2 is 8, plus 1 is 9. It makes the cost of this trip 90, which actually is our lowest possible price of those listed. So the answer will be A, 90. Problem number 12. What is the least number we can get by arranging the six cards with numbers shown in the picture in one row, one after another? No matter how we arrange the cards, the number will have the same number of digits. When we compare numbers of the same number of digits, we start by comparing the first digit. So we need to start by making the first digit as small as possible, and then at each choice, also finding the smallest digit of those left. So let's start by looking at the first digit of each card. First digit of each card. In this case, that's 3, 4, here's 5, 7, 6, and 2. Obviously, the smallest digit that we have here is 2, so we'll use the card with the 2. That's gone. Now, let's find the next smallest digit, 3, so we need to use 3, 0, 9. And we have 4, we need to write 41, then 5, just a 5. Those left, we have 6 here, so we'll write 68, and then 7 at the end. So our number will be 2, 3, 0, 9, 4, 1, 5, 6, 8, 7. And this will be answer D, 2, 3, 0, 9, 4, 1, 5, 6, 8, 7. So the answer is D. Problem number 13. Six weights weighing 1 pound, 2 pounds, 3 pounds, 4 pounds, 5 pounds, and 6 pounds were placed into three boxes, two weights in each box. The weights in the first box weigh 9 pounds together, and those in the second box weigh 8 pounds. Which weights are in the third box? To solve this problem, find out which two weights are in the first box, and which two weights are in the second box. The remaining two weights are the weights that are in the third box. Here are the three boxes, and here are the six weights. We know that the weights in the first box weigh 9 pounds, and the weights in the second box weigh 8 pounds. We are trying to find which weights are in the third box. Now, let's find which two weights can add up to 9, and which two weights can add up to 8. There are two ways in which we can make 9 out of the numbers that we have. We can use the weights 4 and 5, or we can use the weights 3 and 6. Now, let's look at our options for making 8. We can make 8 using the weights 6 and 2, or 5 and 3. Notice that we need to use either the 5-pound weight or 3-pound weight to make 9 pounds for the first box. 5 or 3. So, we cannot use both those weights to make 8. Our only choice is to use the weights 6 and 2 to make 8 pounds, and that leaves us with 4 and 5 for making 9 pounds. Now, we are left with the weights 1 pound and 3 pounds. So, these are the ones that are in the third box. The answer is C, 3 and 1. Problem 14. Four routes are drawn between two points. Which route is the shortest? To solve this problem, notice that each of the paths is made up of the same length segments. These segments are the diagonals of the little squares, like this. Since we know that the paths are made up of segments of the same length, and these segments are clearly marked in our pictures, we can just count how many such segments there are in each answer. In A, there are 1, 2, 3, 4, 5, 6, 7, 8 segments. In B, there are 1, 2, 3, 4, 5, 6, 7, 8 also. In C, there are 1, 2, 3, 4, 5, 6, 7, 8 segments. In D, there are 1, 2, 3, 4, 5, 6, 7, 8. Because they are all 8 segments long, there is no route that is the shortest. All these routes are equal. So, the answer is E. Problem 15. Numbers are written on a number flower. Mary pulled out all the petals with numbers which give a remainder of 2 when divided by 6. What is the sum of the numbers on the petals that Mary pulled out? If a number gives a remainder of 2 when it is divided by 6, that means that 6 goes into it a certain number of times or that it's a multiple of 6 and then 2 is added to that. Let's start by writing out the multiples of 6. 1 times 6 is 6. 2 times 6 is 12. 3 times 6 is 18. 4 times 6 is 24. 5 times 6 is 30. 6 times 6 is 36. 7 times 6 is 42. 8 times 6 is 48. 9 times 6 is 54. And 10 times 6 is 60. None of our numbers go above 60, so we don't need to go higher than that. Now, let's add 2 to each of these numbers to get numbers that give a remainder of 2 when divided by 6. 6 becomes 8. 12 becomes 14. 18 becomes 20. 24 becomes 26. 30 becomes 32. 36 becomes 38. 42 becomes 44. 48 becomes 50. 54 becomes 56. And 60 becomes 62. Now it's time to see which of these numbers are actually on our flower. 8 is on the flower. 14 is not. 20 is not. 26 is not. 32 is not. 38 is right here. 44 isn't. 50 is not. 56 also isn't on the flower. And 62 isn't. And again, since our numbers are all less than 60, we don't need to check any further. So the only 2 numbers on the petals that give a remainder of 2 when divided by 6 are 8 and 38. Since the problem asks for the sum of the numbers on these petals, we need to add these 2 together. We can do 38 plus 8. 8 plus 8 is 16. And we carry the 1 and get 46 as our final answer. So the answer is 46a. Problem 16. You can move or rotate any of the shapes of the puzzles, but you cannot flip them over. Which of the shapes below does not appear in the puzzle to the right? The problem states that one of the pieces does not appear in the puzzle. That means that the other four will. Let's try looking for them in our puzzle. To make it easier to see, I am going to move our large puzzle a little lower. I have also made it bigger. It is easy to see some of the shapes in the puzzle. For example, shape D is right here in the middle. Other shapes we will need to find by analyzing the puzzle itself. I will mark the pieces of the puzzle that don't match any of the shapes gray as we come to them. For example, there is no short line, no little square, no straight lines at all. There is no puzzle like this one right above the small square that looks like a sideways W. And the one up here is similar to that. If we look in the upper left-hand corner, there is an L-shaped piece, which we do not have. Now we can look for the pieces in our answers more easily, and we can rotate them as we go to see if we can find them. I don't see piece A the way that is shown in the picture. I will copy it lower and try rotating it. First, I will rotate it a quarter of the way clockwise. We can see that a piece of this shape is found right here. So we found piece A, so that will not be the answer. Next, we can look for piece B. Again, we might try to rotate it. Here's the piece, and now let's do the rotation, again, starting with a quarter of the way around clockwise. I actually see two of this shape in our picture, here and here. Now it's time to look for piece C. I'll bring it down and rotate. I don't see a piece here without this space in between the two ends. We can try rotating again. There's no piece shaped like this shorter version of piece A, and there is no piece shaped like this one, which is what we get after another rotation. So it looks like piece C is the one that we cannot find. To make sure, let's see if we can find piece E anywhere. Here's our piece. Let's try rotating by 90 degrees. After first rotation, it's not really easy to see where this piece is, so let's rotate again, and now we can more easily see that this piece, E, can be found right here. So it's not a piece that does not appear in the puzzle. So our answer is indeed C. This is the shape that does not appear in the puzzle. Problem 17. Four crows are sitting on a fence. Their names are Dana, Hannah, Lina, and Benny. Bena sits exactly in the middle between Hannah and Lina. The distance between Hannah and Dana is the same as the distance between Lina and Benny. Bena sits four feet away from Benny. How far away is Hannah sitting from Benny? To solve this problem, we will start with what we know and draw a picture. Here's the fence. First thing that we know for sure is that Dana sits exactly in the middle between Hannah and Lina. So we can draw three crows, one in the middle, one here, and one here. We can call them Dana in the middle, and say this one is Hannah, and this one Lina. The other thing we're told is that the distance between Hannah and Dana is the same as the distance between Lina and Benny. Now we see the distance between Hannah and Dana right here. The distance between Dana and Lina then needs to be the same as the distance between Hannah and Dana, because Dana is right in the middle, which means that Benny is sitting on the other side of Lina, over here, because he is not sitting right on top of Dana. Here we can draw Benny. As far as measuring the distance between Hannah and Benny, here and here, we need to go with the other information we are given, which is that Dana is four feet away from Benny. So Dana is here, and Benny is here, and this is four feet. We already saw that all the distances between the crows are equal, which means that if there is four feet between Dana and Benny, there are two feet from Benny to Lina, and two feet from Lina to Dana. Now the third distance from Dana to Hannah will also be two feet. Now we know the three distances are each two feet. So from Hannah to Benny, it's going to be three times two feet, which is six feet. The answer is B, six feet. Problem 18, Johnny is building a house out of cards. In the picture, one-story, two-story, and three-story houses are shown. How many cards does Johnny need to build a four-story house? A good way to approach this type of problem is to see if there's a pattern in the number of cards that need to be added to make each extra story. When we look at the first figure, the one-story house, we see that only one, two, two cards are used. In the two-story house, there are two cards on top. There's one card separating the stories, and one, two, three, four cards underneath. So this one uses two plus five cards. In the three-story building, the two cards on top, the one, two, three, four, five cards underneath, and then added to that are two horizontal cards, one, two, and then one, two, three, four, five, six more cards. So this one uses two plus five plus eight cards. If we're looking for a pattern, we see we started with two, then to that, we added five, and to the last one, we added eight. We see that these numbers are increasing by three. So this suggests that for a four-story house, we use two plus five plus eight cards, which is this figure, plus eight plus three, or 11 cards. Now adding these all together, two plus eight makes 10, plus five makes 15, and 15 plus 11 makes 26. Our answer should be 26, which is easy. Let's check our work by adding 11 cards to the three-story building to make a four-story building. Let's clear a space. Here is the three-story building again. Let's draw in a bottom story to make it a four-story building. One, two, three, four, five, six, seven, eight, nine, 10, 11. So there's 11 cards, and just to double-check, let's count. We have 11, so we just drew in 12, 13, 14, 15, 16, 17, 18, 19, 20, 21, 22, 23, 24, 25, 26. So checking our work confirms that the answer is 26. Problem 19. The structure shown in the picture is made by gluing together the sides of 10 cubes. Newton painted the entire structure, including the bottom. How many faces of the cubes did he paint? Since the structure is made up of cubes, all of which are positioned the same way, one way of solving this problem is to see how many faces of cubes we can see from each of the six sides. Let's start with the sides that we can see easily from the picture as it is shown. Looking from the front, we can see one, two, three, four, five, six sides of the cubes. The tops are also visible in this perspective. We see one, two, three, four, five, six sides here as well. And then we have the sides that we can see from the left, one, two, three, four, five, six. When we think about the sides that we can't see, we can recognize that since this figure is made up of 10 cubes, it will be solid all the way through it. From the right, from this side, the figure would look something like this. The furthest column has three cubes, the one that's a little closer is made up of two, and then there is one in the row that's closest to us. This means that there are six cubes that we can see and therefore six sides that are painted when you look at the figure from the right. When we look at the figure from behind, the layout is going to be similar. We just need to rotate the cube in our mind a little bit and it looks something like this. Again, we have six cubes. Now when we look at the figure from the bottom, the pattern is also three cubes in the row furthest from us, two in the row that is closer to us, and one in the row that is closest, which is six. So in total, we have six views and in each one of them, we can see six faces of cubes painted, which means that there are 36 faces of cubes that have been painted all together. The answer is D, 36. Problem 20, Irina, Anne, Kate, Olga, and Elena live in the same two-story house. Two of the girls live on the first floor, three of them live on the second floor. Olga lives on a different floor than Kate and Elena. Anne lives on a different floor than Irina and Kate. Who lives on the first floor? When we look at the information given, we see that in the two statements, Kate's name is repeated twice. Kate is listed in the first statement, which says Olga lives on a different floor than Kate and Elena, and she is listed in the second statement, Anne lives on a different floor than Irina and Kate. Since there are only two floors, and there are three girls on one of them and two on the other one, we see that Kate, Elena, and Irina live on the same floor. This would be the second floor because three of them live there. So the remaining two girls who are Olga and Anne live on the first floor, so the answer is E, Anne and Olga. In the expression 2002, blank, 2003, blank, 2004, blank, 2005, blank, 2006, either plus or minus can be written in the place of each blank, which result is impossible. The answers given, 1998, 2001, 2002, 2004, and 2006, differ from the first term, 2002, by at most four. So we will not be adding any thousands to the problem, which means that for each of the other terms that we add, we will need to subtract a term so that the extra thousands cancel out. So no matter how we arrange the pluses and minuses, we will need to use two pluses and two minuses, so there are no extra thousands added to the result. Now we can start figuring out which results are possible. Let's list our terms, 2002, 2003, 2004, 2005, and 2006. To make things easier, let's rewrite this by getting rid of the thousands since we know that they will cancel out. So instead of 2003, we'll just have three, so 2004, we'll just have four, 2005, we'll have five, and instead of 2006, we will have six, and we will start with 2002. We know that we are using two pluses and two minuses, so that does not give us too many combinations that we can't try all of them. So let's first start putting the two pluses first, and then putting in the two minuses. 2002 plus three makes 2005, plus four more makes 2009, minus five makes 2004, minus six makes 1998. So that's answer A, so we can make answer A. Let's erase and try a different combination. Let's do plus, minus, plus, minus. 2002 plus three is 2005, minus four is 2001, plus five is 2006, minus six is 2000. 2000 is not one of the answers listed, so I will just write it down here, and let's go ahead and erase and try the next combination. This time, let's do minus, plus, minus, plus. 2002 minus three is 1999, plus four gives us 2003, minus five makes it 1998, and plus six makes 2004. So that's answer D, so that will not be one that we cannot make. If we try minus, minus, plus, plus, we have 2,002 minus three is 1,999. Minus four gives us 1,195. Plus five gives us 2,000. And plus six gives us 2,006. So this is answer E. That will not be one that we cannot get. Next combination to try could be plus, minus, minus, plus. 2002 plus 3 is 2005, minus 4 is 2001, minus 5 is 1996, and plus 6 makes it 2002. Just answer C, so it's not an impossible result. And it seems that the only combination we have not tried yet is minus, plus, plus, minus. 2002 minus 3 is 1999, plus 4 will make it 2003, plus 5 will make it 2008, and minus 6 will make it 2002. So 2002 again. So the only result that we have not gotten is B, 2001. So the result that's impossible to get is 2001. The answer is B. Problem 22. One year in March, there were five Mondays. Which day of the week below could not appear in this month also five times? Choose our Saturday, Sunday, Tuesday, Wednesday, or Thursday. March has 31 days. That means that there are four full weeks, which is four times seven or 28 days, and three other days. It's these three days that can appear five times. We know that Monday is one of the days that appears five times, so we know that there are five Mondays. Let's mark them here like this. We are left with two more days that can also appear five times. But if you think about a calendar, all the days need to be consecutive. So, for example, we could start with Saturday, Sunday, and Monday. Have those appear three times and fill up the calendar. Like this. Or, obviously, we could start with Sunday and have Sunday and Tuesday appear. Also, five times, let's start with Monday and have Tuesday and Wednesday be the days that appear five times. However, there's no way that we can have Thursday appear five times because we cannot have gaps. We can't have a Thursday here and then a Sunday or a Thursday here, a Wednesday, and no Tuesday. So, Thursday is the only day it cannot appear five times of those. The answer is Thursday. Problem 23. In each of the nine cells of the square, we need to write one of the digits 1, 2, or 3. We need to do this in such a way that each of the digits 1, 2, and 3 will be written in each horizontal row and each vertical column. If we start with one in the upper left cell, in how many different ways can the square be filled? Each of the three numbers needs to appear once in each row and in each column. Since we're already given the number 1 in the upper left hand circle, we only have two choices as to what we can do in that top row. We can either put the 2 first or the 3 first and then the other number will go after it. Likewise, we have two choices as to what to do in the first column. Either put the 2 on top or 2 on the bottom and 3 in the other spot. That means that we can start with four different choices. What we then need to do is to figure out whether all four of those choices work and whether putting in the 2s and 3s in a given way in the top row and left column gives us any choices as to how we can arrange other numbers in the remaining part of the square. Here are four blank squares for us to start with. We'll start by filling out the first square. We will make the top row 1, 2, 3 and the first column also 1, 2, 3. Now let's see how we can arrange the other numbers in this square and if there is more than one choice of doing it in the top row and the leftmost column are filled in. Looking at the second row, we already have the 2, so we need a 1 and a 3. Since there's a 3 in this right column, we cannot put the 3 for the row there, so we'll put the 1 here and we will put the 3 in the middle and that helps us figure out that in the last row we need to have the 1 first and then the 2. So there were no other choices or other ways of doing it because we could only put the 1 and the 3 this way. Now let's try leaving the row the same way we had it 1, 2, 3 and trying 1, 3, 2 in the column. For the second row, we have the 3 that's given. We need to put in the 2 and the 1. There's already a 2 in the second column, so the 2 will have to go here, which means the 1 goes here, which is the only way we can do this. Now the only way to put the 1 and the 3 in the last row is to put the 1 in the last column and the 3 in the middle column here. So again, there was only one choice. If we try 1, 3, 2 for the row and 1, 3, 2 for the column, looking at the second row, we have a 3, so we need to place the 1 and the 2. We cannot place the 2 in the last column because we have it there, so the 1 here and the 2 here, and the 2 here, which means in the last row, we will need to have 1 in the middle column and the 3 at the end. Again, there was only one way of doing this. And finally, let's try 1, 3, 2 for the top row and for the first column. Let's try 1, 2, 3. In the second row, we need to find a place for the 1 and the 3. We cannot put the 3 in the middle, so we'll put the 1 here and then 3 at the end. And we know that in the last row, we need to put 2 in the middle spot and 1 at the end. So there were four different ways of arranging the numbers in the top row and in the leftmost column, and each one of those only gave us one way of arranging the remaining numbers. So there are exactly four ways that the square can be filled if we put the 1 in the left top square. The answer is C, 4. Problem 24. The weights in the figure are in balance. The same shapes have the same weight. The weight of each circular shape is 30 ounces. What is the weight of the square shape indicated with the question mark? If the shapes are in balance, that means that the weights are equal. We are given the weight of the circular shape down here, and we're told the same shapes are the same weight. So we already know that the other circle is also going to weigh 30. Now this is the first place where they are balanced, because this piece holding them up is held here, so the right half and the left half of this part of the mobile need to weigh the same. Now what we're looking for is weight on the other side of the mobile. So we will need to figure out the weight of this whole half and then break it down on the other side. So we already know that this part weighs 60 because 30 plus 30 is 60, and so this other part, which is balanced of it, will also weigh 60, which means that together these two weigh 120 ounces. So if this half weighs 120 ounces, then the left half will also weigh 120. Notice that this plank here is holding up three pieces, because the one in the middle here, which could be any weight and still leave it balanced, is the same as the weight on the far left of both parts. All three of these parts will have to have the same weight. So we need to divide this 120 by 3. 12 divided by 3 is 4, so each one of these parts is going to be 40 ounces. 40 ounces here, 40 ounces here, and 40 ounces here. Now we see that the two squares both together weigh 40 ounces, so each one of them will be 20 ounces. The answer is B, 20.

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