Problem number one states, we put 2, 0, 1, and 3 into an adding machine as shown. What is the result in the box with the question mark? So first, let's start out with the leftmost branch. We'll start out by adding these two yellow boxes. We know they add together because their sum should be in the box below. So the sum of 2 and 0 is going to be equal to 2. Then let's move on to the right branch. In this addition, we have 1 plus 3. The sum of 1 and 3 is 4. So we'll put a 4 in that box. Then next, we have to do one more addition from these two branches. And the sum of 2 and 4 is 6. So really, all we did here was just add all the original inputs and get the answer. So the question asks us, what is the result in the box with the question mark? The answer is 6, letter E. Problem number two states, Nathalie wants to build the same cube as Diana had at figure one. However, Nathalie ran out of small cubes and built only a part of the cube, as you can see in figure two. How many small cubes must be added to figure two to form figure one? Okay, so let's just try to reconstruct figure one from figure two by adding cubes. And let's keep a counter of how many cubes we add. So here we'll have the counter. Right now it's at zero because we have not added any cubes yet. Then let's add a cube into the back shelf. Right here will represent all the cubes that we add in yellow. So here we've added one cube. Now let's add two more cubes at the very bottom to make the whole plane even. And we've added two, so now our counter is at three. Now let's add two more to the left to make it into a one-by-two section that we need to fill. In order to do this we need to add two more cubes, so now we're up to five cubes. And then finally let's just add the additional two cubes in order to make the three-by-three cube. And we did this by only using seven cubes. So the question asked us, how many small cubes must be added to figure two to form figure one? The answer is seven cubes. Letter C. Problem number 3 states, find the distance which Mara covers to get to her friend Bhoonika. First, the problem tells us that an eighth of the distance between Mara and Bhoonika is 100 meters. But let's figure out what each of these different segments has in length relative to how big the length between Mara and Bhoonika are. If that doesn't make sense, I'll show you an example. So, for example, we know that Mara is at 0 out of 8, right? And then, the first one-eighth is one-eighth along. So the distance between Mara and that one-eighth is one-eighth of the total distance from Mara to Bhoonika. Then the next distance to the right, between one-eighth and one-fourth, if we subtract one-fourth from one-eighth, we find a common denominator, we get another one-eighth. Then for the next distance, between one-fourth and one-half, one-half minus one-fourth is one-fourth, so the distance there is one-fourth of the distance between Mara and Bhoonika. Then we know that Bhoonika is the distance away from Mara and Bhoonika, so if we subtract one-half from one, we'll get the distance between one-half and Bhoonika, which is one-half. So now we know all the relative distances between all of the points. And we know that one-eighth is equal to 100 meters. This means that the one-eighth to the right is also 100 meters, and one-fourth is twice as big as 100 meters, so that must be 200 meters, and one-half is twice as big as one-fourth, so that must be 400 meters. So in total, the distance between Mara and Bhoonika is 800 meters, because we do 100 plus 100 plus 200 plus 400 added up equals 800 meters. So the question asked us, find the distance which Mara covers to get to her friend Bhoonika. The answer is 800 meters. Letter C. Problem number four states, Nick is learning to drive. He knows how to turn right, but cannot turn left. What is the smallest number of turns he must make in order to get from A to B, starting in the direction of the arrow? OK, so we know that Nick can only make right turns, and we have to get to point B. So Nick is heading upwards, and we know that he has to turn right. So he has two options. He can either take the first right, or he can take the second right. Really doesn't matter which one he takes, because ultimately he will end up on the road, on the long road, all the way on the right, going downwards. He has to be going downwards, because he has to turn right onto that road. Then he's going to be heading down, and I guess he can turn right again here, but that doesn't really make any sense, because then he'd just be going in a circle. So I would advise against Nick doing that, if he wants to make as little right turns as possible. Next, Nick could take a right here. If he takes a right here, then it makes no sense for him to make another right where it says A, because then he would just be going again in a circle, and that's probably not advised. So he could continue on going down, and then take a right. If he does this, he then has some options. But either way, whichever way he goes, he will end up at a place that he was already at, which is not very efficient, since we want to turn right as little as possible. So turning right at that intersection is not advised. He should probably go all the way down, and take a right there. He probably shouldn't take the first right, because that will set him back to where we were before, so he should take the second right. If he takes the second right, he will arrive at his destination point B. So we've kind of tested all the possible combinations, so this has to be the best one. In order to do this, he only has to turn right four times. So the question asked us, what is the smallest number of turns he must make in order to get from A to B, starting in the direction of the arrow? The answer is four turns, letter B. Problem number five states the sum of the ages of Anne, Bob, and Chris is 31 years. What will the sum of their ages be in three years? So we'll represent A as Anne's age, B as Bob's age, and C as Chris's age. We know if we take the sum of all of those, we should have an answer of 31. So now let's try to figure out what all of their ages will be in three years. Well, in three years, they'll be three years older. So Anne's age will be A plus 3, Bob's age will be B plus 3, and Chris's age will be C plus 3. So now if we take the sum of all of their ages, we should take A plus 3 plus B plus 3 plus C plus 3. That'll equal x, which is what we're looking for, the answer. So now we can drop the parentheses and combine like terms to get A plus B plus C plus 9 equals x. So now it kind of seems that we're stuck because what's the value of A plus B plus C? Well, the problem tells us. As we can see in the line above, A plus B plus C is equal to 31. So we can substitute A plus B plus C for 31 to get 31 plus 9 equals x. 31 plus 9 is 40, so x must equal 40. So the question asked us, what will the sum of their ages be in three years? The answer is 40, letter E. Problem number six states, what digit must be placed in all three boxes of blank blank times blank equals 176 in order to make the multiplication work? Okay, so on the left are all the possible answers to the multiple choice question. So let's try plugging in some of these into the equation. Of course, this would be quite time consuming because it's quite hard to compute something like 66 times 6. Instead of this, let's say that the second digit in the two digit number, the first one in the product is zero. This will be easier to compute, and it'll still give us an answer close to what the answer would be if that zero was another box. We can then see if the answer is close. If it is, we can compute the exact value. So first, let's start out with the largest number, 9. So let's plug in 9 into the box. We would have 90 times 9. 9 times 9 is 81, and then we have an extra zero, so the answer would be 810. Nowhere near 176, so 9 cannot be the answer. Next, let's try 8. If we plug in 8, we would have 80 times 8. 80 times 8 will equal 640. 640 is not close to 176 either, so that doesn't work. Next, let's try 7. 7, we would have 70 times 7, which would be 490. 490 is not close to 176, so that one would also not work. Next, let's try 6. 6 would be 60 times 6, which would give us a value of 360, which is again not close to 176, so let's continue. Well, we've gotten rid of 6, 7, 8, and 9, and we only have 4 left, so 4 is probably going to be the answer, but let's just make sure. So let's try 4. It would be 40 times 4, which would be 160, which is quite close to 176. In order to confirm this, let's put in the other 4 and actually do the actual multiplication. So 44 times 4 would be 160, plus 16, which would be 176. So this one works. So the question asked us, what digit must be placed in all three boxes of blank blank times blank equals 176 in order to make the multiplication work? The answer is 4, letter B. Problem number 7 states, Michael has to take a pill every 15 minutes. He took the first pill at 11.05. What time did he take the fourth pill? So here we have a timeline of him taking the first pill, the second pill, the third pill, and the fourth pill. We know that he took the first pill at 11.05. So let's try to figure out at what time he took all the next pills. So when he took the second pill, 15 minutes had passed from 11.05, so 15 minutes plus 11.05 is 11.20. Then he took, after 15 minutes, he took the third pill, so 11.20 plus 15 is 11.35. Then for the fourth pill, we would have to add another 15 minutes, so 11.35 plus 15 would be 11.50. So he took the fourth pill at 11.50. So the question asked us, what time did he take the fourth pill? The answer is 11.50, letter B. Problem number eight states, by drawing two circles, Mike obtained a figure which consists of three regions. See picture. At most, how many regions could he obtain by drawing two squares? So here we have two squares, and probably the first thing you think he could do is intersect them like this. And there we go, we have three regions just like he had it. But actually there's a better way of making regions. If we rotate one of the squares and then overlap them like this, we can see that we make a lot of regions. Specifically in this case, we make nine regions. This is the most you can make by intersecting two squares. So the question asked us, at most how many regions could he obtain by drawing two squares? The answer is nine. Letter E. Problem number 9 states, the number 36 has the property that it is divisible by the digit in the 1's position, because 36 is divisible by 6. The number 38 does not have this property. How many numbers between 20 and 30 have this property? So here are all the numbers written out from 20 to 30. And let's start out with 20. So in order for 20 to have this property, it has to be divisible by 0. So now the question is, is 20 divisible by 0? Well, if we divide 20 by 0, we get an undefined answer. As a matter of fact, if we divide any number by 0, it's undefined. So 20 is definitely not divisible by 0. So this one doesn't work. Next, let's move on to 21. The question, is 21 divisible by 1? Well, all numbers that are whole are divisible by 1, so yes, 21 is divisible by 1. Next, let's move on to 22. Is 22 divisible by 2? In order to be divisible by 2, the number has to be even, and 22 is even, so 22 is in fact divisible by 2. So 22 works. Is 23 divisible by 3 is the next question. 23 is prime, meaning it is only divisible by 1 and 23, so it is definitely not divisible by 3. Next, is 24 divisible by 4? 4 goes into 24 six times, and it is one of its factors, so yes, 4 goes into 24. It works. Next, 25. Is 25 divisible by 5? If a number ends in a 5 or a 0, then that number is divisible by 5. So yes, 25 is definitely divisible by 5. Next, is 26 divisible by 6? The factors of 26 are going to be 1, 2, 13, and 26. 6 is not in those numbers, so 26 is not divisible by 6. Next, let's move on to 27. Is 27 divisible by 7? I know 28 is divisible by 7, which means that one number less cannot be divisible by 7, so 27 is not divisible by 7. Next, is 28 divisible by 8? The factors of 28 are going to be 1, 2, 4, 7, 14, and 28, none of which are 8, so 28 is not divisible by 8. Next, let's move on to 29. Is 29 divisible by 9? 29 is prime, so no number is divisible by it, so 9 is definitely not divisible by it. So that one doesn't work. And then finally, is 30 divisible by 0? And finally, is 30 divisible by 0? I don't know any numbers that are divisible by 0, so 30 is definitely not one of them. So this leaves us with four numbers, 21, 22, 24, and 25. So the question asked us, how many numbers between 20 and 30 have this property? The answer is four of them do. Letter C. Problem number 10 states, Ann has a lot of pieces like the one shown on the top of the picture. She tries to put as many as possible in the 4x5 rectangle as shown. The pieces may not overlap each other. What is the largest possible number of pieces Ann can put in the rectangle? In order to make it more clear where we put the pieces, let's actually color the pieces blue. And now, let's try to pack these pieces as tightly as we can to make room for as many as possible. So let's put the first piece like this, and then the second piece like this, and then the third one like this. This very efficiently covers up the top section of the rectangle. Now let's put down a fourth piece. And actually, it doesn't matter where we put down this fourth piece, it is actually impossible to put down another fifth piece. We will always have some segment of four squares that we cannot quite fill up with this shape. So we've used four pieces to do this. So the question asked us, what is the largest possible number of pieces Ann can put in the rectangle? The answer is four pieces, letter C. Problem number 11 states, which of the following pieces covers the largest number of dots in the table? So first, let's start out with the piece in the top left. When we put it on the grid, we can see that it covers up two dots. But this is inaccurate, because we can move the shape up or down. In this case, regardless of whether or not we move it up or down, it will cover up four spots. We'll have to keep this in mind in the future. So this shape covers up four dots. This next one seems to cover up also four dots. If we move it down though, we can see that it also covers four dots. So this one covers at maximum four dots. Next, for this one, this one seems to cover up one dot. But then when we move it down, it actually ends up covering five dots. For this next one, we can move it in very many ways. We'll start out here, in this case it covers four dots. And if we move it anywhere, it covers four dots, because it's a three by three grid. So there we go, four dots. And this next one, in this orientation, it covers up three dots. If we move it down, it also covers three dots. So this one can only cover three dots. So the best shape is this one. At most, this one can cover up five dots. So the question asked us, which of the following pieces covers the largest number of dots in the table? The answer is this shape, letter C. Problem number 12 states, Mary shades various shapes on square sheets of paper as shown. How many of these shapes have the same perimeter as the sheet of paper itself? So in order to solve this problem, what we need to do is check whether the lines that are drawn inside of the square match up with the parts of the perimeter of the square that are not touched by the shape. So first, let's take a look at this L shape. We can see that it touches the perimeter of the square along all of this red line, so we don't have to worry about any of that section. Next, let's just worry about the square that we have left. So we can see that these two yellow lines are exactly the same in length, which means that they make up the perimeter, so that right side is good. Then we also have this green line. These two green lines also match up with the same length on the figure as the perimeter of the square, so this one also works. Next, let's take a look at the H piece. Again, let's disregard everywhere where the lines overlap the perimeter. And here we can see that these two lines are the same, so they cancel out with the perimeter, and these two are as well. But what about these lines? These lines don't cancel out with anything, so we can't do anything with them. So the perimeters don't match up, and this one cannot work. Now let's take a look at this one. Let's take a look at where the edges of the shape match up with the perimeter. And then let's do the same analysis as last time. We can see that these two lengths are the same. But just like in the last example, we have these two side lengths, which aren't accounted for at all, so this one also doesn't work. Next, let's take a look at this S shape. For the S shape, let's again mark all the areas where the shape's exterior matches up with the perimeter. And then let's start up matching sides. So these two green ones match up. These two yellow ones match up. These two pink ones. And finally, these two blue ones. So the S shape works. Next for the T. Match up all the sides that match up with the perimeter. And then these sides work. These sides work. And these sides work. And these sides also work. So the T works. Now for the cross. Match up all the side lengths with the perimeters. And then I won't go through every single one of them, because that would be very long. But here they are, and you can see that they all match up. So the cross also works. So we've had two shapes that don't work, and four shapes that work. So the question asked us, How many of these shapes have the same perimeter as the sheet of paper itself? The answer is four of them do. Letter C. Problem number 13 states, Anne rides her bicycle throughout the afternoon with a constant speed. She sees her watch at the beginning and at the end of the ride with the following result. Which picture shows the position of the minute hand when Anne finishes one-third of the ride? So first let's read the clocks to see when she started and when she stopped. The hour hand is in between the 1 and the 2, so it must be 1.30 on the left. And on the right, the hour hand is between the 3 and the 4, so it must be 3.30. Now let's figure out what the steps are in the middle. So we know that she rode her bike from 3.30 to 1.30, which is two hours. If we divide this two hours into three segments, we get 0.66666 hours. 0.6666 hours is another way of saying 40 minutes. So we know that each of these segments are spaced out by 40 minutes. So for the first one, 1.30 plus 40 minutes would be 2.10, 2.10 plus 40 minutes would be 2.50, and 2.50 plus 40 minutes would be 3.30. So it all checks out. This means that the clock with the question mark's minute hand should be at the 10 minute place. So the question asked us, Which picture shows the position of the minute hand when Anne finishes one-third of the ride? And the answer is this one, letter D. Problem number 14 states, Matthew is catching fish. If he had caught three times as many fish as he actually did, he would have 12 more than he does. How many fish did he catch? So first, let's start out with some variables. Let's say that f is going to represent the number of fish caught by Matthew. This means that 3f should represent the number of fish, the three times as many fish that he caught. Now let's set up an equation. The problem tells us that if he had caught three times as many fish as he actually did, he would have 12 more fish than he does. Mathematically represented, this would be 3f equals 12 plus f. The 3f is the part where it says if he had caught three times as many fish as he actually did. And the 12 plus f is he would have 12 more fish than he does. So now that we have our equation, we have to solve for f. So first, let's subtract f from both sides. We have 2f equals 12. And then we divide both sides by 2 to get f equals 6, which means that Matthew caught six fish. So the question asked us, how many fish did he catch? The answer is 6. Letter b. Problem number 15 states, John made a building of cubes standing on a 4x4 grid. The diagram shows the number of cubes standing on each cell. When John looks from the front, what does he see? So first let's start out with the left column. We know that John is going to be looking from the front, so his left is also going to be our left. So in the leftmost column, the largest number we have is 4. Which means that on the leftmost side, he should see a tower of 4 blocks tall. Then let's move over one column to the right. In this column, the largest number is 3. So John would see a stack of 3 cubes next to the one of 4 cubes. And then to the right of that, the largest number here is also 3. So John would see another stack of 3 cubes next to the stack of 3 cubes that we put down previously. And then finally, in the last column, the largest number we have is 2. This is the rightmost column, and on the rightmost side, he would see 2 cubes stacked. So this is the shape that John would see. So the question asked us, when John looks from the front, what does he see? The answer is this shape, letter E. Problem number 16 states, In an election, each of the five candidates got a different number of votes. The candidates received 36 votes in total. The winner got 12 votes. The candidate in last place got 4 votes. How many votes did the candidate in second place get? So here we have some variables to help us out. We'll let A represent the number of votes first place got, B the number of votes second place got, and on and on and on, C, D, and E. We know they're all different values because the first thing the problem told us is that each of the candidates got a different number of votes. So now let's take a look at what type of information the problem gives us. The first thing they say is that the candidates received 36 votes in total. This means that the sum of all of the votes must be equal to 36. Next the problem tells us the winner got 12 votes, and that last place got 4 votes. That means that A equals 12, and that E equals 4. We also know that if A got first place, B got second place, C got third place, D got fourth place, and E got fifth place, that A must be greater than B, which must be greater than C, which must be greater than D, which must be greater than E. So now with this information, let's try to figure out what B is. First let's substitute in 12 for A and 4 for E, and then combine like terms and move them all to one side, and we get that B plus C plus D must be equal to 20. And then if we plug these numbers into our inequality, we get that B must be less than 12, D must be greater than 4, and all the information that we already knew. So now let's try to figure out what B, C, and D are. First let's just say that D is equal to 5. This will mean that B plus C plus 5 equals 20. This fits our constraints, so we're doing fine right now. For C, let's just say that C equals 6, and then in order for the sum to be equal to 20, B must be equal to 9. So here we go, we have one possible way of doing it, and we get that B equals 9. Let's try a different one. Let's keep D as 5, but let's say that C is 7. If we get this, then B would be equal to 8. 8 plus 7 plus 5 fits all of our constraints, and it adds up to 20, so that's good too. But we have different numbers for B, which means that B can either be 8 or 9. So the question asked us, how many votes do the candidates in second place get? The answer is 8 or 9. Letter B. Problem number 17 states, from a wooden cube with a side of 3 cm, we cut out in the corner a little cube with a side of 1 cm. See picture. What is the number of faces of the solid after cutting out such a small cube at each corner of the big cube? So first, let's take a look at how many additional faces cutting out this cube creates. We can see here that it creates an additional 3 faces. So let's write that down, that each corner adds 3 faces. We know that a cube is made up of 8 vertices, so a cube has 8 corners. And we know that just a standard cube has 6 faces. So if we were to count up all the faces, if all of these chunks were removed from the corner, we would first have the initial 6 faces of the cube, and then we would have an additional 3 for each corner of the cube. So 8 times 3 additional faces. 8 times 3 is 24, 6 plus 24 is 30. That means that this shape would have 30 faces. So the question asked us, what is the number of faces of the solid after cutting out such a small cube at each corner of the big cube? The answer is 30. Letter D. Problem number 18 states, find the number of pairs of two-digit natural numbers whose difference is equal to 50. So for example, we need to find two numbers a minus b such that they're equal to 50. But a and b are both whole numbers, and they have to have two digits. So one such example would be 60 minus 10 equals 50. Now, in order to get another combination, we can just add one to both sides of the subtraction. These numbers will then cancel out and still give us 50. So for example, we can do 61 minus 11, and that also equals 50. We can keep on doing this, on and on and on, until we get to 99 minus 49 equals 50. In order to do this, we went from 60 to 99, which is 40 numbers. Or you could additionally also look at it as we went from 10 to 49, which is also 40 numbers. So there we go. There are 40 possible ways to make 50 of a difference of two natural two-digit numbers. So the question asked us, find the number of pairs of two-digit natural numbers whose difference is equal to 50. The answer is 40, letter A. Problem number 19 states. The final of the local hockey championship was a match full of goals. There were 6 goals in the first half, and the guest team was leading after the first half. After the home team scored 3 goals in the second half, the home team won the game. How many goals did the home team score all together? So first let's let A represent the number of goals scored by the home team in the first half, and B the number of goals scored by the guest team in the first half. We know that in total 6 goals were scored in the first half, so we know that A plus B must be equal to 6. And we also know that after that, the guest team was leading, so we know that B must have been greater than A, because the guest team must have scored more than the home team. And we also know that after the home team scored 3 goals, A plus 3, that the home team won the game, which means that A plus 3 must be greater than B. So first let's simplify the inequality. We can say that A plus 3 is greater than B, which is greater than A. This means that B is sandwiched between A and A plus 3. The only numbers that B can be is either A plus 2 or A plus 1. So let's plug these numbers into our equation and see if they work. First let's say that B is going to be equal to A plus 1. So let's substitute that in. So we have A plus A plus 1 equals 6. Let's combine like terms, so we'll have 2A plus 1 equals 6, subtract 1. We have 2A equals 5. And if we divide both sides by 2 to solve for A, we can see a problem, that A is equal to 2.5. I've never seen a hockey game where a team has scored half of a point, so this is impossible. B cannot be equal to A plus 1. B must be equal to A plus 2. Let's verify this. Let's plug in A plus 2 into our equation, of A plus B equals 6, to get A plus A plus 2 equals 6. Combine like terms and subtract 2 from both sides, we get 2A equals 4. Dividing both sides by 2 to solve for A, we get that A equals 2. That's a much more reasonable score. Which means that the home team scored two goals in the first half, and then the problem tells us that in the second half, they scored three goals, so in total, the home team scored five goals in the hockey championship. So the question asked us, how many goals did the home team score altogether? The answer is 5. Letter C. Problem number 20 states, In the squares of the 4x4 board, numbers are written such that the numbers in adjacent squares differ by 1. Numbers 3 and 9 appear in the table. Number 3 is in the top left corner as shown. How many different numbers appear in the table? So first let's try to figure out what appears next to the 3. And we have several options. We can either say that both of the squares adjacent to the 3 are a 4, they could both be a 2, or one of them could be a 2 and one of them could be a 4. In the example all the way on the left, the square in the corner that's next to the 4 and the 4 could either be a 3 or a 5. In the middle example, next to the 2 and the 2, the number could either be a 1 or a 3. And on the example on the right, the number could only be a 3. The problem tells us that numbers 3 and 9 appear in the table, which means that we want to get these numbers as high as possible. In the middle example, we could either go down or stay the same at 3, so that's probably not advised. And on the right, we could stay the same as 3, so that's not advised either. But on the left, we could go up all the way to 5, which each diagonal stride, so we should definitely go with the left example. Obviously, since we want the numbers to be as high as possible, we'll say that that number is a 5. Next, let's fill in all the 5s diagonally. And then all the unknown cells that are adjacent to the 5s we'll say are 6s. And all the cells adjacent to the 6s, which we don't know, will be 7s. And next to those will be 8s. And finally, there will be a 9. So this fits the rule that a 3 and a 9 appear in the table. So the question asked us how many different numbers appear in the table, and we can see that there are 6 different numbers that appear. 3, 4, 5, 6, 7, 8, and 9. So the question asked us, how many different numbers appear in the table? The answer is 7. Letter D. Problem number 21 states, Aaron, Burn, and Carl always lie. Each of them owns one stone, either a red stone or a green stone. Aaron says, my stone is the same color as Burn's stone. Burn says, my stone is the same color as Carl's stone. Carl says, exactly two of us own red stones. Which of the following statements is true? So first, let's take a look at all the possible combinations of stone ownership. A represents Aaron, B represents Burn, and C represents Carl. Now let's go through their statements and figure out which of these combinations is not possible. First, Aaron says, my stone is the same color as Burn's stone. The problem tells us that all of them are liars, so this is not true. So let's toss out all the combinations where the color of A is the same as the color of B. There are four such combinations. And now let's move on to Burn's statement. Burn says, my stone is the same color as Carl's stone. Again, Burn is a liar, so this is a false statement. So let's toss out all the combinations where Burn and Carl have the same colored stones. There are two of these. And lastly, Carl says, exactly two of us own red stones. This is again a lie, so we can toss out the left combination. And we're only left with one combination. Aaron having a green stone, Burn having a red stone, and Carl having a green stone. So out of all of the statements, only one of them is true. That Aaron's stone is green. So the question asked us, which of the following statements is true? The answer is that Aaron's stone is green. Letter A Problem number 22 states, 66 cats signed up for the contest Miss Cat 2013. After the first round, 21 were eliminated because they failed to catch mice. 27 cats out of those that remained in the contest had stripes and 32 of them had one black ear. All the striped cats with one black ear got to the final. What is the minimum number of finalists? So first we know that we had 66 cats at the beginning of the Miss Cat 2013 contest. Then the problem tells us that 21 of them were eliminated. 66 minus 21 is 45. So we have 45 cats remaining. Out of those cats that are remaining, we know that 27 of them had stripes and 32 of them had one black ear. 27 plus 32 is 59, which is greater than 45. This means that at least some of the cats had to have both stripes and at least one black ear. But how many? Well if we take the difference of the total number of those that had stripes and one black ear, and of all the cats, we'll see how many had to have at minimum these two traits. 59 minus 45 is 14. So at least 14 of them had to have both traits. Of course there could have been more because some of them could have had neither of the traits, but there could not have been any less. So the question asks us, what is the minimum number of finalists? The answer is 14. Letter D. Problem number 23 states, there are four buttons in a row as shown below. Two of them show happy faces and two of them show sad faces. If we press on a face, its expression turns to the opposite. A happy face turns into a sad face. In addition to this, the adjacent buttons also change their expressions to the opposite. What is the least number of times you need to press the buttons in order to get all happy faces? In this problem, there is no better strategy than just trying things out and seeing if they work. First, let's try and press the second button. This turns that button sad and the other ones around it happy. Let's press the button to the right of that one. That would turn that button sad, the button to the right sad, and the button to the left happy. Finally, let's just press the button all the way on the right. This will turn all the remaining sad faces into happy faces. We were able to do this using only three button presses. As a matter of fact, it doesn't even matter in what order we press them. If we pressed any of the three right buttons once, we would have gotten all happy faces. Of course, the proof that this is the best and most efficient way to do this is quite complicated. We will not be going into this, but this is the most efficient way. And you just have to figure this out by trying different things. So the question asked us, what is the least number of times you need to press the button in order to get all happy faces? The answer is three times. Letter B. Problem number 24 states, 40 boys and 28 girls stand in a circle hand in hand, all facing inwards. Exactly 18 boys give their right hand to a girl. How many boys give their left hand to a girl? So here we have a different example with less boys and less girls, so we can kind of analyze the situation. First, let's take a look at the girl all the way on the right. Let's see what's going on with her. To her right, we have a boy who's holding his left hand with hers. And to her left, we have a boy who's holding his right hand with hers. So this girl is causing both a left hand from a boy and a right hand from a boy to touch her. Next, let's take a look at this girl. It's the same story with the left hand from the boy, but from the right hand from the boy, it doesn't touch directly her, but it touches with the girl next to her. If we form a chain, we can see that each chain has at least one boy touching a chain of girls with his left hand and one boy touching a chain of girls with his right hand. Next, let's take a look at this chain of girls. We can observe the exact same thing, a boy touching them with his left hand and with his right hand. So we see that every single time, whenever a girl is touched by a boy with his left hand, then she is also touched with a boy with his right hand. This means that there are exactly the same number of boys touching a girl with their left hand as there are with their right hand. So if the problem tells us that exactly 18 boys give their right hand to a girl, this means that exactly 18 boys must give their left hand to a girl as well. So the question asks us, how many boys give their left hand to a girl? And the answer is 18. Letter A. Problem number 25 states, a 2x2x2 cube is to be constructed using 4 white and 4 black unit cubes. How many different cubes can be constructed in this way? Two cubes are not different if one can be obtained by rotating the other. So first let's take a look at a split view, let's take a look at the cube from the bottom and from the top. So here is one possible combination, that the whole bottom cube is black and the whole top cube is white. Another combination would be a checkerboard pattern. Another combination would be a checkerboard pattern but only from one angle. For a fourth combination what we can do is we can make the bottom have one white piece and the top have one black piece. And then in order to find different combinations we can just rotate this black piece around the top. And this will give us all different combinations that cannot be achieved by rotating the cube. In total we have achieved 7 combinations and this is all that we can get. So the question asked us, how many different cubes can be constructed in this way? The answer is 7, letter D. Problem number 26 states, how many 3 digit numbers possess the following property? After subtracting 297 from such a number we get a 3 digit number consisting of the same digits in the reverse order. So where do we start? Well first let's figure out what's going on here. So we're going to have some number and we're going to subtract 297 from it and that should equal another number with a in the reverse order of digits. So if we put in 500 we should expect 005 out. I guess that was a bad example. If we put in 502 in we should get 205 out. So let's try to find one number. First let's just put in a completely random number of 517. 517 minus 297 is 220, which is not at all equal to 715. If we lower the number by 1 to 516 we get 219, which is not equal to 615. But the number, the goal target, 615, is lower. It used to be 715, but now it went down to 615. So let's keep on lowering 516 down more until we get really close to the answer. So every single time it goes down by 100, the target number. So let's go down 4 more to get down to the 200s. So let's do 512 minus 297 and that equals 215. We found our first number. So let's mark that one down. And then if you notice, if you add 10 to 512 you get 522 minus 297 is equal to 225. So every single time that you add 10 this property still works. So that works for all of the 500s. But then once we move on to 602 it doesn't work. We get 305 instead of 206. But if we change it to 603 it works. So we also have to change the ones place every 100 numbers that we go. And so this ends up working for all the 400s through the 900s. It doesn't work for the 300s because the largest number in the 300s would be 390. If we subtract 297 we get 93, which is not a 3 digit number. In total here we have 6 columns and 10 rows, so we have 60 numbers that fit this property. So the question asked us, how many 3 digit numbers possesses the following property? After subtracting 297 from such a number we get a 3 digit number consisting of the same digits in reverse order. The answer is 60, letter D. Problem number 27 states, when Matthew and Martin found their old model railway, Matthew quickly made a perfect circle from eight identical track parts. Martin starts to make another track with two of these pieces as shown in the picture. He wants to use as few pieces as possible to make a close track. How many pieces does his track consist of? So in order to solve this problem, let's try to make the ends meet. First, let's try going up and around to the right. So let's add this piece to turn towards the other piece, and keep on adding pieces, until we get to about here. Because now it's kind of hard to get to the other part, so let's start on the other end instead. We'll add this piece and work our way around counterclockwise. And then we can see from this point that we can just put in like an S section to get back to the starting point. Like this. So it works. We did this using only 12 track pieces, and this is the minimum we need to make a close track. So the question asked us, how many pieces does this track consist of? The answer is 12. Letter B. Problem number 28 states, there were 2013 inhabitants on an island. Some of them were knights and others were liars. The knights always tell the truth and the liars always lie. Every day, one of the inhabitants says, after my departure, the number of knights on the island will equal the number of liars, and then left the island. After 2013 days, there was nobody on the island. How many liars were there initially? So in order to start this problem, let's start from the back. Let's start from the last person that left the island. So when they left, they said that after my departure, the number of knights on the island will equal the number of liars. And they would be telling the truth, because there would be nobody left after they left. So they must be a knight, because they're telling the truth. So then the second to last person came and also said that after his departure, the number of knights on the island will be equal to the number of liars. And he is obviously lying, since after he left, there were zero liars and one knight. Then the third to last person that left said that after my departure, the number of knights on the island will be equal to the number of liars. And they were telling the truth, because there would be one liar and one knight left. So we can see a sort of alternating pattern here. One knight, one liar. And this would continue on and on and on, all the way back to the first person. But we have an odd number of people, so the question is, was the last person a knight, or was the last person a liar? Well, it's quite clear that after the first person left, there was an equal number of liars as truth-tellers, since there was an even number of people. So that last person must have been a knight. This means that there were 1007 knights and 1006 liars. In total, this makes 3013 people, so that checks out. So the question asked us, how many liars were there initially? The answer is 1006. Letter B. Problem number 29 states, Starting with a list of three numbers, the change-sum procedure creates a new list by replacing each number by the sum of the other two. For example, from 3-4-6, change-sum gives 10-9-7, and a new change-sum leads to 16-17-19. If we begin with the list 20-1-3, what is the maximum difference between the two numbers of the list after 2013 consecutive change-sums? So here we have our list, 20-1-3, and let's do a change-sum. So for the first number it would be 1 plus 3, so 4. Then it would be 20 plus 3, so 23, and then 1 plus 20, so 21. So we have 4, 23, 21. Let's continue on to the next one, 23 plus 21, 44, 4 plus 21, 25, 4 plus 23, 27. And then let's just do one last one for good measure, 25 plus 27, 52, 44 plus 27, 71, and 44 plus 25, 69. So now let's figure out what the maximum difference is between any two numbers in the list. Starting off with the first one, we have 20-1, which is 19. And then for the next one we have 23-4, which is also 19. Then we have 44-25, which is also 19. And lastly we have 71-52, which is also 19. So it turns out that the maximum difference stays constant throughout the pattern, which means that after 2013 iterations, the difference must also be 19. So the question asked us, if we begin with the list 20-1-3, what is the maximum difference between two numbers of the list after 2013 consecutive change-sums? The answer is 19. Letter D. Problem number 30 states, Alice forms four identical numbered cubes using the net shown. She then glues them together to form a 2x2x1 block as shown. Only faces with identical numbers are glued together. Alice then finds the total of all the numbers on the outside surface of the block. What is the largest total that Alice can get? So first, for the faces to glue onto the middle, let's try to pick the smallest numbers. Because if we pick the smallest numbers, this means that the largest numbers will be on the outside, and that will result in the largest sum for Alice. So ideally we would use 1 and 2, but 1 and 2 are not adjacent as you can see on the net. The next best alternative would be 1 and 3, and luckily 1 and 3 are adjacent on the net. This means that 1 and 3 will go onto the inside of all the cubes, and that means that on the outside we will have 2, 4, 5, and 6. This means that on the outside we will have four pairs of 6, 5, 4, and 2. So let's figure out what the total sum of that would be. So 6 plus 5 plus 2 plus 4 is going to be 17, and we will have four of those sums, so 4 times 17 is going to be 68. So we'll have a total sum on the outside of 68. So the question asked us, what is the largest total that Alice can get? The answer is 68, letter B.

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