Problem number one states, the average of four numbers is nine. What is the fourth number if three of the numbers are five, nine, and twelve? So first let's go over how to find the arithmetic mean of a list of numbers. So let's say we have some numbers a1, a2, a3 all the way up through a sub n. In order to find the arithmetic mean we would have to add them all up and then divide by the number of numbers that there are. So that would be n. So if we have four numbers we would do a sub 1 plus a sub 2 plus a sub 3 plus a sub 4 all over 4. And we know that this value is going to be equal to the average. And we know three of the numbers that we're taking the average of. We know that they are five, nine, and twelve. And we'll call the other number b. And then we also know what the average is going to be equal to. We know that that is nine. So now we have one unknown variable in our equation so we can solve. First let's multiply both sides by four to get that it equals thirty-six. Now let's combine like terms on the left side to get twenty-six plus b equals thirty-six. And finally subtracting twenty-six from both sides we get that b equals ten. So the missing number in our average was ten. So the question asked us, what is the fourth number if three of the numbers are five, nine, and twelve? The answer is ten, letter d. Problem number two states, which of the following numbers is the closest to the result of seventeen times 0.3 times 20.6 divided by 999? So the problem asks us for a solution that's close. So in other words we can sort of approximate all these numbers and not use them exactly. So for example dividing something by 999 isn't all that easy. But dividing something by a thousand is a little bit easier. So we can change this 999 to a thousand. Same thing with this seventeen, seventeen is kind of hard to multiply by and doesn't have many factors, it's actually prime. But we can change that to twenty. 0.3 we can change to 0.25. And then 20.6 we can round down to twenty. So now our new equation is going to be twenty times 0.25 or one-fourth times twenty over a thousand. So what does this equal? Well first let's do twenty times twenty. So that's going to be 400. 400 times 0.25 is going to be 100. And then we're left with a hundred over a thousand. Dividing the numerator and the denominator by a hundred we get one-tenth or 0.1. So our answer is going to be closest to 0.1. The exact value for the quotient is something like 0.1029, so we're pretty close with our estimate of 0.1. And this helps with the fact that we overestimated the numbers almost as much as we underestimated them. So that'll give us a better estimate. So the question asked us, which of the following numbers is the closest to the result of 17 times 0.3 times 20.6 over 999? The answer is 0.1. Letter B. Problem number 3 states, On a test consisting of 30 questions, Ruth had 50% more right answers than she had wrong answers. Each answer was either right or wrong. How many correct answers did Ruth have, assuming she answered all questions? So first let's let the variable c represent the number of correct answers she got, and the variable w the number of wrong answers she got. So we know from the problem that the test consisted of 30 questions. So we know that c plus w is equal to 30, since she either got the answer correct or wrong. Next the problem told us that Ruth had 50% more right answers than she had wrong answers. So mathematically that can be represented as c is equal to 1.5 times w. So she has 1.5 times more correct answers than wrong answers, which is that 50% more. So now we have two equations with two unknown variables so we can solve. Let's substitute 1.5w for c, and we get 1.5w plus w is equal to 30. Combining like terms on the left side we get 2.5w is equal to 30. Now dividing both sides by 2.5 we get that w is equal to 12. So we know that she got 12 wrong answers. But the problem asks us for how many correct answers she got. So now we can plug in this w equals 12 into our original equation to get c plus 12 is equal to 30. Subtracting 12 from both sides we get the number of correct answers that Ruth had was 18. So the question asked us, how many correct answers did Ruth have assuming she answered all questions? The answer is 18. Letter d. Problem number four states. In a coordinate system, four of the following points are the vertices of a square, which is not a vertex of the square. So here we have our five points, and we'll just label them with the letter that the answer is. So point A will be minus one three, point B will be zero minus four, and on and on and on. So let's say that we have two points, P and Q, and we'll say that P's coordinates are x1, y1, and Q's coordinates are x2, y2. If we wanted to find the distance between these two points, we would take the square root of x2 minus x1 squared plus y2 minus y1 squared, and this would get us the distance between point P and point Q. So now let's take a look at an example. Let's find the distance between point A and point B. So plugging in the relevant numbers, and then we get the square root of one squared plus seven squared, or the square root of 50. So the distance between point A and point B is the square root of 50. So now let's do this for points A and C, and we get the distance between A and C is root 17. Between A and D, we get the distance is the square root of eight, and between A and E, we get that the distance is square root 41. So here we have the distances from A to every other point. Now if we take a look at a square, and we say that this corner is A, and we know that it has two points it is connected to, we'll call them G and H. And we know that the length from A to G has to be equal to the length of A of H, since all the side lengths of a square are the same. That comes from the definition of the square. So if A is part of the square, that means the distance from it to two other points must be the same. However, all the distances are different, so A cannot be part of the square. You could have also graphed it, and this is how it would have looked, and it's pretty clear here that C, D, E, and B all form a square, while A is alone on the outside. So the question asked us, which point is not a vertex of this square? The answer is (-1, 3). Letter A. Problem number five states, when the positive integer x is divided by 6, the remainder is 3. What is the remainder when 3x is divided by 6? So first we know we have some integer x. And we know that then when we divide it by 6, we get a remainder of 3. In other words, x minus 3 is divisible by 6. So we can say that x consists of some part that's divisible by 6 plus 3. So we can represent this as 6k plus 3, where k is some integer. Now if we were to divide this number by 6, obviously 6k would be divisible by 6 and we would get k, and then we would just have a remainder of 3. The question doesn't ask about x, it asks about 3x. So let's multiply both sides by 3. And then now let's use the distributive property. And we get 3x is equal to 18k plus 9. Let's break the 9 up into 6 plus 3. And now let's factor out a 6 from the two left terms. To get 3x is equal to 6 quantity 3k plus 1 plus 3. Now if we see if we divide 3x by 6, we can see that obviously 6 times 3k plus 1 would be divisible by 6, since it is a multiple of 6, but in addition to that we have plus 3. So the remainder would be 3. So the question asked us, what is the remainder when 3x is divided by 6? The answer is 3. Letter B. Problem number six states, how many weeks are equivalent to 2016 hours? So we're starting out with 2016 hours, and we want to end up with weeks. So let's do some unit conversions. First let's convert it to days, so we know that we have 24 hours in one day. We put hours in the denominator, so the hours cancel out. Now that we have days, we can convert it to weeks. So we have seven days in a week. So we put seven days in the denominator for them to cancel out with the days in the numerator, and then we would be left with weeks. So in order to convert hours to weeks, we need to divide by 24, and then divide by 7. So this is what we would get. So in the denominator, 24 times 7 is going to be 168. So the number of weeks is going to be 2016 divided by 168, which is equal to 12. So 2016 hours is 12 weeks. So the question asked us, how many weeks are equivalent to 2016 hours? The answer is 12. Letter D. Problem number 7 states, Little Lucas invented his own way to write down negative numbers before he learned the usual way with the minus sign in front. Counting backwards, he'd write 3, 2, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, on and on and on. What is the result of 0, 0, 0 plus 0, 0, 0, 0 in his notation? So rather than figuring out how addition works in his notation, let's simply convert it back to our usual negative form and then convert it back into his form. So to start off with, here we have a conversion table of our values with the minus sign in front to his values with all the 0's. So we know that the sum that he wants to figure out consists of 3 0's plus 4 0's. So if we look up at our table, 3 0's is minus 2 and 4 0's is minus 3. So really he's doing minus 2 plus minus 3, which is minus 5. And we can see that the solution of minus 5 is 6 0's. So the answer would be 6 0's. His method of counting is very similar to base 1 or tallying. So you could add up the numbers of 0, but then you would have to also subtract 1 because the base value of 0 already has a 0. So if you want to do it that way, you could have also done it that way. So we have 6 0's. So the question asked us, what is the result of 0, 0, 0 plus 0, 0, 0, 0 in his notation? The answer is 0, 0, 0, 0, 0, 0, letter C. Problem number 8 states, I have some strange dice. The faces show the numbers 1 to 6 as usual, except that the odd numbers are negative. Minus 1, minus 3, minus 5, in the place of 1, 3, 5. If I throw two such dice, which of these sums cannot be achieved? So first, let's take a look at our dice. One dice consists of minus 1, 2, minus 3, 4, minus 5, and 6. And the other dice is exactly the same. And these are the values 3, 4, 5, 7, and 8, which we are trying to achieve by summing up the individual values of the two dice. So first, let's try and make 3. 3 is possible if one dice rolls 4 and the other one rolls minus 1. 4 minus 1 is 3. Next, for 4, we could do dice 1 with 2 and dice 2 also with 2 to get 2 plus 2 equals 4. For 5, you could do 6 and minus 1, because 6 minus 1 is equal to 5. And then trying to do 7, it's going to be quite hard because it's not possible. And then we already know the answer. We know it's 7. Let's just try and do 8. And 8 is possible if we do 2 and 6. So 2 plus 6 is 8, so that one works. So the only one that doesn't work is 7, so that's our answer. So the question asked us, if I throw two such dice, which of these sums cannot be achieved? The answer is 7, letter D. Problem number 9 states, what is the necessary number of transportations of two adjacent letters in order to change, step by step, the word V-E-L-O into the word L-O-V-E? So we know that first we start out with a VLO and we want to end up with LOVE. And we can only swap two adjacent characters. So to start out with, we have three options. Either we can swap the left two, the right two, or the middle two. If we swap the left two, we'll realize it doesn't really make much sense, because this puts the E all the way to the left side. In order to put the E all the way to the right side, where it belongs in the word LOVE, we would need to do four flips. If we flip the right two, we would have a similar problem, because now we need to put the letter L to the left side, and that would also take four flips. There would be even more flips involved to turn it into LOVE, because we're just focusing on the first and last characters here. So swapping the left two and swapping the right two clearly isn't the right move. So instead, let's swap the middle two. At this point, swapping the middle two again would just reverse what we did, so there's no point. And swapping the left two and swapping the right two, we'll actually get the L and the E in the right spot. So doing that, we get this. And swapping on the right side, we get this. And now only two letters are out of place, the V and the O. So it'll take one more swap to flip those around to get the word LOVE. And it took four transportations to turn VLO into LOVE. So the question asked us, what is the necessary number of transportations of two adjacent letters in order to change the word VLO into the word LOVE? The answer is 4. Letter B. Problem number 10 states, Sven wrote five different one digit positive integers on a blackboard. He discovered that no sum of any two numbers is equal to 10. Which of the following numbers did Sven definitely write on the blackboard? So we know that he wrote down five one digit numbers. So we'll call these numbers a1, a2, a3, a4, and a5. And we know that their possible values are the numbers 1 through 9 inclusive. And we know that any number plus any other number cannot be equal to 10. So we can write this list. We can say that a1 is going to be greater than or equal to 1, a2 is going to be greater than that, a3 than that, a4 than that, a5 than that, which is going to be less than or equal to 9. Now what we can do is we can take 10 minus all of these sides and we would get this. Since we're effectively multiplying this side by a negative, we would have to flip all the signs as well. And this is what we would get. This is very similar to our initial statement. As a matter of fact, it's exactly the same. But now what we have is we have two lists of numbers. This a1, a2, a3, a4, a5, and then 10 minus all of those terms. And since we only have 10 entries in the list, but only nine possible values, that means that two of these values must repeat. So in other words, ai must be equal to 10 minus aj. Adding aj to both sides, we get ai plus aj is equal to 10. But that's impossible because earlier we realized that ai plus aj cannot be equal to 10. So what's going on here? Well, ai plus aj cannot be equal to 10 because of a restriction they gave us at the beginning. They told us that i cannot be equal to j. So effectively, if you have the number 2, you cannot add it to itself. That does not count. But this ai plus aj is equal to 10 lifts this restriction. So we can say that i equals j. So in this case, ai plus ai would be equal to 10. So ai would be equal to 5, which means that one of our values must be 5. The other four would either be 1, 2, 3, 4, or 6, 7, 8, 9. So the question asks us, which of the following numbers did Sven definitely write on the board? The answer is 5. Letter e. Problem number 11 states, let a plus 5 equal b squared minus 1 equals c squared plus 3 equals d minus 4. Which of the numbers a, b, c, d is the largest? So first we know that all the numbers are equal to one another, so we can say that all of these values equal the same number x. So first let's consider the bottom two equations. c squared plus 3 equals d minus 4, and let's set them equal to one another. Let's add 4 to both sides to get d in terms of c, and we get that d is equal to c squared plus 7. Let's get all of the variables in terms of c. So now let's go, let's set c squared plus 3 equal to b squared minus 1. So now let's add 1 to both sides, and we get that b squared is equal to c squared plus 4. Now let's do this with the last equation as well. So a plus 5 is going to be equal to c squared plus 3. Subtract 5 from both sides, and we get that c squared minus 2 is equal to a. So what does this tell us? Well from these three equations, we know that a is less than b squared, which is less than d. How do we know this? Well because obviously c squared minus 2 is less than c squared plus 4, which is less than c squared plus 7. But now here's the question, where does c fit into this? Right now d is looking like the greatest one, but let's make sure of that. So if c is between minus 2 and positive 2, c would be greater than c squared minus 2 and less than c squared plus 4. But if c was less than negative 2 or greater than 2, then c would be less than c squared minus 2. So regardless of what the value of c is, we know it occurs on the left side, so it's going to be less than d. But we have one more problem, what about this b squared value? Well since the square root of any number greater than 1 is going to be less than that number, we don't have to worry if b is greater than 1. And if b is less than 1, the greatest the number is going to be is 1. And on c squared, we have a plus 7, which is 3 more than the c squared plus 4. So again, we don't have to worry. So here it's clear that d is the greatest number. So the question asked us, which of the numbers a, b, c, d is the largest? The answer is d, the letter d. Problem number 12 states the three by three table is divided into nine unit squares and two circles are inscribed in two of them. See picture. What is the distance between the two circles? So here we have our shape and we know that each of the squares is a unit square, which means that side lengths are one. So we can mark all of those here. And the question is asking us what's the distance between the two circles? So that would be this length. Rather than finding this length, let's first try and figure out what this length is. So in order to do that, we can make a right triangle like this. In order to find the length of the green side, we can use Pythagoras's theorem, a squared plus b squared equals c squared, where a and b are the side lengths of the right triangle and c is the hypotenuse. In this case, we have an isosceles right triangle. So we actually know that a and b are equal to one another. So we can say that b is equal to a and just have it as two a squared is equal to c squared. So now let's find the length of a, we know that the length of a is going to be two radii of the circle less than three. And we know what the radius of the circle is, it's half of the side length of the square, which is one half. So if we plug in one half for our, we get three minus two times one half, the two and the one half cancel out to make one. And so the side length is going to be two. So what does that make the hypotenuse? Well, we can plug that into our equation, two a squared is equal to the hypotenuse squared. And we get that the hypotenuse squared is equal to eight, taking the square root of both sides, we get the hypotenuse is root eight. Of course, we can break that up into two squared times two, and then factor out a two to get the hypotenuse is two root two. The problem wasn't asking about the lengths between the centers of the two circles, but rather about the length between the two circles. Luckily, we can figure this out quite easily. We know that the distance between the two circles centers is equal to the distance between the two circles plus two times the radius. And we know all of these values except for one, so we can solve. We know that the radius is one half, and we know that the distance between the two circles centers is two root two. So plugging all that in, we get that c minus one is equal to d, or two root two minus one is equal to the distance between the two circles. So that's our answer. So the question asked us, what is the distance between the two circles? The answer is two root two minus one, letter a. Problem number 13 states, in a tennis knockout tournament, six of the results of the quarterfinals, the semifinals, and the finals were not necessarily in this order. Bella beat Anne, Celine beat Dona, Gina beat Holly, Gina beat Celine, Celine beat Bella, and Emma beat Farah. Which result is missing? So in a knockout tournament, once somebody loses, they are out of the tournament for good. So here we have our competitors, we have eight of them. And first we're told that Bella beats Anne. That means that Anne would be out of the tournament. Next we're told that Celine beat Dona, so Dona would be out of the tournament. Next Gina beat Holly, so Holly would be out of the tournament. Next Gina beat Celine, so Celine would be out of the tournament. Then we know that Celine beat Bella, which doesn't really make sense since Celine is already out of the tournament, but it said it's not in the same order so it's okay. But that means that Bella would be out of the tournament. And then we're told that Emma beat Farah, which means that the only two people that are left are Emma and Gina. And here are our possible answers. So Gina could not have beat Bella since Bella already lost one game, and you can only lose one game in this knockout style of tournament, and then you'll be out. Celine beat Anna could not happen because both Celina and Anna have already lost. Emma beat Celine could not have happened because Celine had already lost. And Bella beat Holly could not have happened because both Bella and Holly have already lost. However, Gina beat Emma is possible because neither Gina or Emma have lost yet. So the question asks us, which result is missing? The answer is Gina beat Emma. Letter E Problem number 14 states what percentage of the area of the triangle and the figure is shaded. So here we have our figure and if we consider the equilateral triangle on the outside we can see that all of its side lengths are 5. And if we consider this tiny equilateral triangle on each of its vertices we can see that its side lengths are all 1. So now what does this tell us? Well we know that the ratio of the side lengths is 1 to 5. So therefore the ratio of the areas would be 1 to 5 squared or 1 to 25. That means that the yellow triangle has an area 25 times greater than the red triangle. And in this case units don't really matter. Well we were told that they would be percentages so let's say that the yellow triangle has an area of a hundred percent. This means that the red triangle would have 1 25th of the area so 4%. So now let's plug in our values back into the original drawing and we know that the yellow triangle has an area of a hundred percent and that each of the small red triangles have an area of 4%. And we know we have three of them so we would subtract three areas of 4% each for a total of 12%. So we would have a hundred minus twelve. So in total the shaded area has an area of 88% of the original yellow triangle. So the question asked us what percentage of the area of the triangle in the figure is shaded? The answer is 88%. Letter C. Problem number 15 states, Jill is making a magic multiplication square using the numbers 1, 2, 4, 5, 10, 20, 25, 50, and 100. The products of the numbers in each row, in each column, and in the two diagonals must all be the same. In the figure, you can see how she has started. Which number should Jill place in the cell with the question mark? So these are the numbers that we can put into our cells, and we have nine of them. So first let's figure out what the product of each of the rows, the columns, and the diagonals is going to be. In order to do this, we can multiply all of these numbers together, and then take the cube root to figure out what the value will be in each row and each column. And since it's the same for the diagonals, it'll be in that as well. So first let's multiply all of these numbers together. And we can see that a lot of these products actually make up 100, so this multiplication is not that hard. We have four factors of 100, and then one of 10, which gives us a value of 10 to the 9th. So now let's take the cube root of that, so it would be 1 3rd times 9, so we would get 10 to the 3rd. 10 to the 3rd is 1000. So this tells us that each row and each column must have a product of 1000. So here we are back with the square, and let's call this value y. So we know that 20 times 1 times y has to equal 1000. So dividing 20 by both sides, we get that y equals 50. So now let's move on and try and figure out what the question mark is going to be. Let's call the question mark x. Let's call the value in this cell a. So now taking a look at the column with 50x and a, we know that 50 times x times a is going to be equal to 1000. Dividing 50x by both sides, we get a is equal to 1000 over 50x, or that a is equal to 20 over x. So now we can substitute that in for a. Now if we plug in the same column back in, all the x's will cancel out, and we'll just get that 1000 equals 1000. So we need to make a row, column, or diagonal out of different rows, columns, and diagonals. And we'll see that later. So let's call the value in this cell b, and let's use the diagonal of 20b and 20 over x. So 20 times b times 20 over x is equal to 1000. And then let's divide by 400 over x on both sides to get 1000x over 400, or 5x over 2. So the middle cell value is going to have a value of 5x over 2. Next let's try and figure out what the bottom one is going to be. So in this case we'll use the column 1, 5x over 2, and c. So 1 times 5x over 2 times c is going to be equal to 1000. Divide both sides by 5x over 2 to get 2000 over 5x, or that c is equal to 400 over x. And if we do one more in the bottom left corner, then we'll be able to represent the diagonal. So let's do that. So we'll call that one d. And let's do the bottom row. So we're going to have d times 400 over x times 20 over x is going to be equal to 1000, or 8000 over x squared. So then we're going to have d is equal to 1000x squared over 800, or d is equal to x squared divided by 8. So now we can take a look at the relationship on the diagonal. Multiplying all the numbers on the diagonal, we get x squared over 8 times 5x over 2 times 50 is equal to 1000, which gives us 250x cubed over 16 is equal to 1000. Dividing by 250 over 16, we get x cubed is equal to 64. Taking the cube root of 64, we get that x equals 4. So now we've solved the mystery number, we know that it is 4. And this is how the whole square would look. So the question asked us, which number should Jill place in the cell with the question mark? The answer is 4. Letter B. Problem number 16 states, Jack wants to hold 6 circular pipes each with a diameter of 2 cm together by a rubber band. Now considering other possibilities, he decided between the following two options as shown in the picture. Which of the following is true about the lengths of the rubber bands? So here we have a combination of the pipes with the rubber bands around them. And let's independently consider the straight sides and the curved sides of the rubber band. So here we know that there is a constant radius of the pipe, we'll call that r. And we can see here that this value is also going to be r, this value is going to be r, and this value, the diameter of the circle, is going to be 2r. That means that that bottom red section is going to have a length of 4r, r being the radius of the pipe. We can do the same thing up here to realize that that length is also going to be 4r. And again over here that this length is going to be 2r, and over here that this length is also going to be 2r. So now we figure out all the lengths of the straight sides, let's figure out the lengths of the curved sides. And here we can see that the yellow segments on the left side add up to an arc of 180 degrees, and the yellow segments on the right also add up to an arc of 180 degrees. That means that in total this yellow section makes a length of the circumference of one of the pipes, which would be 2pi r. So in total the length of the rubber band is 12r plus 2pi r. Now let's consider the other one. Using the same trick that we used last time we can figure out the lengths of the straight sides. So in this case each of them will be 4r. And again realizing that the sum of each of these yellow arcs is going to add up to a full circle, we can see that the length of the arcs summed up is going to be that of the circumference of one of the pipes, which is going to be 2pi r. So that means that both of the lengths of the rubber bands are going to be 12r plus 2pi r, so they're going to be the same. So the question asked us, which of the following is true about the lengths of the rubber bands? The answer is both have the same length. Letter. E. Problem number 17 states, each of eight identical envelopes contains one of the following numbers. 1, 2, 4, 8, 16, 32, 64, 128. Eve chooses a few envelopes randomly. Allie takes the rest. Both sum up their numbers. Eve's sum is 31 more than Allie's. How many envelopes did Eve take? So first, let's get some variables going. So we'll say that E is going to be Eve's sum, and A is going to be Allie's sum. And we know that Eve's sum is 31 greater than Allie's sum. Adding capital E to both sides here, we would get 2E is equal to A plus E plus 31. And we actually know one of these values. We know what A plus E is. Since each of them picked one of the envelopes, that means that A plus E is going to be the sum of all the numbers in the envelopes. So 1 plus 2 plus 4 plus 8 plus 16 plus 32 plus 64 plus 128, which equals 255. So we can substitute that in to get 2E is equal to 255 plus 31. So 2E is going to be equal to 286. And dividing both sides by 2, we get that E is equal to 143. So now we know Eve's sum. But let's figure out how she could have made this sum. So these are our numbers that we have. 1, 2, 4, 8, 16, 32, 64, and 128. In this case, it's going to take 128, 8, 4, 2, and 1 to make them. So she should have picked 5 envelopes in order to make 143. So the question asked us, how many envelopes did Eve take? The answer is 5. Letter D. Problem number 18 states, Peter wants to color the cells of a 3x3 square in such a way that each of the rows, the columns, and both diagonals have 3 cells of 3 different colors. What is the least number of colors Peter could use? So here we have our 3x3 square, and we need to start out somewhere, so let's start out with the diagonal. We'll give it 3 different colors. We'll have blue, green, and white. They have to be 3 different colors because the same color cannot appear in a diagonal, so they must be different. Next let's consider the other diagonal. If the square in the bottom left corner is white, then it will have the same color in the bottom row. If it's blue, it will have the same color in that column, and if it's green, it will have the same color on that diagonal. So we cannot do this with 3 colors, so we need another one. We'll make this one yellow. Next let's consider the one in the top right corner. It can't be blue, green, or white for the same reasons, and it also can't be yellow because yellow already exists on that diagonal, so it needs to be a fifth color. We'll say pink. Next let's consider the empty square on the right. It cannot be green, pink, or white. It doesn't exist on a diagonal, so we don't have to consider those, but that means that it can either be blue or yellow. Next considering this one, it can't be blue, green, or pink, so it'll be yellow or white. Next for this one, it can't be blue, yellow, green, or blue or yellow, which we already have so that doesn't matter, so it's either going to be pink or white. And now the one on the bottom can't be yellow, white, green, or yellow or white, which we already have so again it doesn't matter, so it can either be blue or pink. So this is our final square that consists of 5 colors. The boxes that have 2 colors can be either of those colors and it'll still work. For example, for the left and the top one we can have white, and for the bottom one and the right one we can have blue, and it'll still work. Any combination of those works, which means that there are 16 possibilities, but that doesn't matter, what matters is that we had to use 5 colors at the very least. So the question asked us, what is the least number of colors Peter could use? The answer is 5. Letter C. Problem number 19 states, the picture on the left shows a cube with four marked angles. What is the sum of these angles? So first, let's consider this angle. This angle is one of the angles of a square, the square all the way on the back that the cube consists of. All the angles in a square are 90 degrees, so we know that this angle is right. If we look at this angle, we know the two lines that the angle consists of are perpendicular to one another, so we know that this angle is also right, so 90 degrees. Same story with this one, the two lines that consist the angle are perpendicular to one another, so we know that the value of this angle is also 90 degrees. Next, let's take a look at this angle. The lines that consist of this angle are perpendicular, so it's not right, but let's figure out what it is. So let's consider this triangle, which consists of that angle, and we can see that each of the side lengths of this triangle are the diagonals of a square of the cube, and we know that all of the squares on a cube are the same, therefore we know that their diagonals are the same, which means that each of these lengths is the same. So we have an equilateral triangle right here, and if we have an equilateral triangle, that means that each of the angles are 60 degrees. So we have three angles that are 90 degrees, and one that is 60 degrees, so let's add all of them up. So 90 plus 90 plus 90 plus 60, so we have 270 plus 60, which gives us a sum of 330 degrees. So the question asked us, what is the sum of these angles? The answer is 330 degrees. Letter. B. Problem number 20 states. There are 2016 kangaroos. Each of them is either grey or red, and at least one of them is grey and at least one is red. For every kangaroo k, we compute the fraction of the number of kangaroos of the other color divided by the number of kangaroos of the same color as k, including k. Find the sum of the fractions of all 2016 kangaroos. So first let's take a look at the smaller example to make sure we know what the problem is asking. So here we'll only have three kangaroos. One of them will be grey and two of them will be red. So first we'll say that this kangaroo is kangaroo k. So we take the fraction of all of the other colored kangaroos over all of the same colored kangaroos. So in this case there are two kangaroos of another color, and there's one of the same color, kangaroo k. So we get a fraction of 2 over 1, or 2. If a red kangaroo is k, then the fraction is flipped, since there's only one kangaroo of another color and there are two kangaroos of the red color. And it would be the same thing for this other kangaroo who is also red. And the question asks us to find the sum of all of these fractions. So in this case we would have 2 over 1 plus 1 half plus 1 half, which would be equal to 3. So enough with our simple example, let's move on to the actual problem. We'll say that r represents the number of red kangaroos and g represents the number of grey kangaroos. So if k is a red kangaroo, then the quotient would be g over r. The number of kangaroos of a different color divided by the number of kangaroos of the same color. And if k is grey, then the quotient would be r over g. So now we have to add up all of these. So since we have r red kangaroos, the g over r fraction would occur r times. And since we have g grey kangaroos, the r over g fraction would occur g times. Here we can see that we can cancel out some numbers in the numerator and denominator. And we're just left with g and r. So let's add those two up. So now the question is, what is the number of grey kangaroos plus the number of red kangaroos? Well that would be the total sum of kangaroos. And we know that there are 2016 kangaroos. So g plus r is going to be 2016, which is our answer. So the question asks us, Find the sum of the fractions of all 2016 kangaroos. The answer is 2016. Letter A Problem number 21 states, a plant wound itself exactly 5 times around a pole with a height of 1 meter and a circumference of 15 cm as shown in the picture. As it climbed, its height increased at a constant rate. What is the length of the plant? So we know the pole that the plant wrapped itself around is 1 meter tall, and we know the circumference of the pole that it wrapped around is 15 cm. We know that there are 5 equal segments of the pole that it wrapped around, so we know that each of these segments is 1 fifth of a meter or 20 cm. Now we can strip the bark off the pole at one of those 20 cm segments to get this piece of bark. The left side of this piece of bark and the right side of this piece of bark would be together when it would be wrapped around the pole. This is a flattened version. And we know this piece of bark is 20 cm tall and 15 cm wide. 15 cm is the circumference of the circle. So this piece of bark would form a hollow cylinder, and we know that the plant climbs over a constant distance over this piece of bark. So now let's find the length of the plant over this piece of bark, and we can use Pythagoras' theorem, a squared plus b squared equals c squared, since we have a right triangle. For a and b, we can plug in 15 and 20, and we have a 3-4-5 triangle, so we know that c is going to be equal to 25 cm. So now we know the length of 1 fifth of the plant, but we have 5 identical segments, each are 25 cm, so the total length will be 5 times 25 cm, or 125 cm, which is 1.25 m. So the question asked us, what is the length of the plant? The answer is 1.25 m, letter c. Problem number 22 states, what is the largest possible remainder that can be obtained when a two-digit number is divided by the sum of its digits? So we have a two-digit number, we'll say AB, and we need to divide it by the sum of its two digits, so by A plus B. And this division is going to yield with some remainder, which we have to maximize. So let's say that we're dividing C by D. In this case, we could represent C as being equal to K times D plus R. So D would divide C K times with the remainder of R. So let's convert AB into this form. So we would have AB is equal to K times A plus B plus R. And again, here what we want to do is maximize R. But we have a restriction on R. R has to be less than A plus B. Because if R was greater than A plus B, well, that wouldn't make any sense, since then we would have K plus 1 and then a different remainder, or K plus some integer with a greater remainder. So in other words, the remainder can't be greater than the number that we're dividing by, because then the remainder would be divisible by it and we would have a different remainder. So it must be less than A plus B. So if we want the highest chance of having a high remainder, we need to maximize A plus B. So let's start out with the highest digits of A and B we can have, so 99. So 99 divided by 9 plus 9 will give us 5 with a remainder of 9. We know we've reached the highest number possible when we have a remainder that is 1 less than A plus B, because that would be the maximum possible remainder. We don't have it yet, since 9 is smaller than 18. So let's keep going to the second highest number, 98. So 98 divided by 9 plus 8 will give us 5, remainder of 13. So that's closer, but still not there. The next one is actually 89. 89 divided by 8 plus 9 will give us 5, remainder of 4. So we still don't have that condition, so let's keep going. So 97 divided by 9 plus 7 will give us 6, remainder of 1. So still nothing. Next highest one would be 88 divided by 8 plus 8, which would give us 5, remainder of 8. So that's still not good. And then we get 79 divided by 7 plus 9, which gives us 4, remainder of 15. And with a remainder of 15, that's the highest remainder that we can get. Because remember that the remainder has to be less than A plus B. And we're going from the maximum value of A plus B, remember we started at 9 plus 9, and now we're going down to 7 plus 9, So our value of R can only get smaller from here. So 15 is the maximum possible value. So the question asked us, what is the largest possible remainder that can be obtained when a two digit number is divided by the sum of its digits? The answer is 15. Letter C. Problem number 23 states A 5 by 5 square is divided into 25 cells. Initially all its cells are white, as shown on the left. Neighboring cells are those that share a common edge. In each move, two neighboring cells have their colors changed to the opposite colors. White cells become black, and black ones become white. What is the minimum number of moves required in order to obtain the chest-like coloring shown on the right? So on the right board with the checkerboard pattern, we have 12 dark squares. So at the very least, we need to flip each of these once. So that would mean that we would have 12 moves. But then obviously that would flip some of the white cells to black. But if we arrange our moves into these L-shaped patterns, what we can do is we can turn two of them black and leave the one white in only two moves. So for example we could flip these two, and then we could flip these two in the next move. So now we just need to make sure that we can fit these L-shapes all throughout our checkerboard pattern. And it seems that we can. So we've proven that we can do it in 12 moves, but let's just take a look at how this might look. So let's take the original L and do that here, and let's do the next L to the right of that. And each L takes two moves. So in total it took us 12 moves to get the checkerboard pattern, and this is the least number of moves we can do. So the question asked us, what is the minimum number of moves required in order to obtain the chest-like coloring shown on the right? The answer is 12. Letter. B. Problem number 24 states it takes four hours for a motorboat to travel downstream from x to y. To return upstream from y to x it takes the motorboat six hours. How many hours would it take a wooden log to be carried from x to y by the current assuming it is unhindered by any obstacles? So here we have our river and here on the left side we have point x and on the right side we have point y and the river goes downstream from x to y and we'll say the distance between point x and point y is d and we know that the water is flowing and we know that every single point in this river is flowing at some rate. We'll call that w that'll be the speed of the water going from x to y. So now let's say that we have a log that starts at x. How long will it take for the log to reach point y? Well it'll be carried at the speed of the water and taking the speed of the object multiplied by the time that it takes to travel will give us the distance that it traveled so w times t sub l is equal to d. t sub l is the time it took the log to get from point x to point y. So rearranging this we can solve for t sub l. So t sub l is going to be the distance between point x y divided by the speed of the water w. So next let's take a look at what a motorboat would do if it was going from point x to point y. Well it would travel at a speed of the motorboat's speed plus the speed of the water and we would add those together since the motorboat is going downstream and then we would multiply it by the amount of time it took the motorboat to go downstream and that would equal the distance between x and y. And we know the amount of time that it took for it to get downstream we know that is four hours. So we know m plus w times four is equal to d. Now going upstream the only thing that would change would be the speed of the boat. Now the current would be working against the boat so we would say the speed of the boat minus the speed of the current and that would be multiplied by t sub u which is the amount of time it takes for the boat to go upstream from y to x and that would again equal the distance between point x and point y. And we know that this would take six hours so we would have m minus w times six is equal to d. So now we have our three equations so let's try and solve for t sub l. So first let's do the distributive property on the two motorboat equations to get 4m plus 4w is equal to d and 6m minus 6w is equal to d. So now let's try and solve by elimination so let's multiply the left side by three and the right side by two to get 12m plus 12w is equal to 3d and 12m minus 12w is equal to 2d. So now let's add them up together and we get 24m is equal to 5d. Now we can solve for m and we can get m is equal to 5d over 24. So now we can plug this into either of our equations but let's plug it into the one on the left and we would get 5d over 24 plus w times 4 is equal to d. So again using the distributive property we would get 5d over 6 plus 4w is equal to d. Subtracting 5 6d from both sides we get 4w is equal to d over 6. Now multiplying by 6 and dividing by w we get 24 is equal to d over w and you might see that the time it takes to log to float down the river is also equal to d over w which means that time is 24 hours. So the question asked us how many hours would it take a wooden log to be carried from x to y by the current assuming it is unhindered by any obstacles? The answer is 24. Letter e Problem number 25 states, In the Kangaroo Republic, each month consists of 40 days numbered 1 to 40. Any day whose number is divisible by 6 is a holiday, and any day whose number is a prime is also a holiday. How many times in a month does a single working day occur between two holidays? So here we have a calendar of the Kangaroo Republic, and each month has 40 days. And what we're told is that every single day that's divisible by 6 is a holiday, so we can mark all of those. That would be 6, 12, 18, 24, 30, and 36. And next they tell us that every single prime number is also a holiday, so we can mark all of those as well. So that would be 2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31, and 37. Now the only working day that's sandwiched between two holidays is the fourth. All the other working days are either sandwiched between two working days or a working day and a holiday. So the answer is that there is only one day. So the question asks us, how many times in a month does a single working day occur between two holidays? The answer is only once. Letter A Problem number 26 states, two of the altitudes of a triangle are 10 centimeters and 11 centimeters, which of the following cannot be the length of the third altitude? So let's say that we have this shape, and it has three sides, so perhaps it's a triangle. Well obviously we see that A and C would not line up together, so it can't be a triangle. And mathematically speaking in this case, A plus C has to be greater than B. If it's equal than B, we would just have a straight line. So the sum of any two sides of a triangle must be greater than the third side. So here we have a triangle, and let's take a look at one of its altitudes. The altitude of a triangle starts at the point of the triangle and then runs perpendicular to one of the sides. So here we would have the other two altitudes. And we know the values of two of them. We know that one of them is 11 centimeters and the other one is 10 centimeters. Now we know that the area of a triangle is one half base times height. And if we use the altitude as the height and the side length as the base, we can get an expression for the base in terms of the area and the height. We would get two times the area over the height equals the base. So for the magenta side, its length would be twice the area of the triangle divided by 11. And we can do that for the remaining two sides. We would get 2A over 10 and 2A over C. So now we have the three side lengths of the triangle. Now let's put into effect those relations to make sure that it's a real triangle. So the sum of any two sides has to be greater than the third side. And we can see that this equation right here is going to be true for any value of C. Since 2A over 10 has to be greater than 2A over 11 regardless of the value of C. So really we only have two equations we need to match up. Here it's pretty clear that we can divide 2A by all sides. And we would get 1 tenth plus 1 eleventh is greater than 1 C. And on the other side, 1 eleventh plus 1 over C is greater than 1 tenth. So adding up a tenth and an eleventh requires a common denominator. So we'll do 110. So then 11 plus 10 is going to be 21 over 110 is greater than 1 over C. On the right side, let's subtract 1 eleventh from both sides. Find a common denominator and we get that 1 C is greater than 1 over 110. Now combining these two equations this is what we get. And now we can take the reciprocal of both sides. And when we take the reciprocal we need to flip the signs. So we get 110 over 21 is less than C which is less than 110. So C cannot be less than about 5.2 and has to be less than 110. And these were our answers. And the only one that doesn't fit this range is C equals 5. So the third altitude cannot be 5 centimeters. So the question asked us, which of the following cannot be the length of the third altitude? The answer is 5 centimeters. Letter A Problem number 27 states, Jacob wrote down four consecutive positive integers. He then calculated the four possible totals made by taking three of the integers at a time. None of these totals was a prime. What is the smallest integer Jacob could have written? So we know that he wrote down four consecutive integers, we'll say s, t, u, and v. And since they're consecutive, we know that t is equal to s plus 1, u is equal to s plus 2, and v is equal to s plus 3. So we can substitute those in. So he chose s, s plus 1, s plus 2, and s plus 3. And now he takes the sum of three of these. So there are four possibilities. He could have summed up the first three, the first two and the last one, the first one and the last two, or just the last three. So these are the sums that we would get, 3s plus 3, 3s plus 4, 3s plus 5, and 3s plus 6. And for the first sum and the last sum, you can see that we can factor out a 3 actually. So we can do 3 quantity s plus 1, and 3 quantity s plus 2. So the four sums are 3 quantity s plus 1, 3s plus 4, 3s plus 5, 3 quantity s plus 2. And we know that none of these are going to be prime. Well, for the first one and the last one, it doesn't matter what the value of s is going to be. They're never going to be prime regardless of the value of s. Why? Because it's three times a number. And three times any number other than 1 is going to be not prime. But what if s is 0? s can't be 0, because the problem told us that Jacob only chose positive integers. So we don't even have to consider the first one and the last one, because we know for sure that they're not going to be prime. So we're only left with 3s plus 4 and 3s plus 5. We need to make sure that these two are not prime. And we need to find the minimum value of s such that this occurs. And here are possible values of s. These come from the answers. So first let's try 12. If s is equal to 12, then we would have 40 and 41. 41 is prime, so 12 doesn't work. Next let's try 10. We would get 34 and 35, neither of which are prime, so 10 works. But let's just make sure that there are no lower numbers that also work. So trying out 7, 7 also works. So in this case 7 would be the answer. But let's keep going. Maybe 6 and 3 will also work. 6 doesn't work since 19 is prime, and 3 also doesn't work since 11 is prime. So the lowest number in which these sums are not prime is 7. So the question asked us, what is the smallest integer Jacob could have written? The answer is 7. Letter C. Problem number 28 states, Four sportsmen and sportswomen, a skier, a speed skater, a hockey player, and a snowboarder had dinner at a round table. The skier sat at Andrea's left. The speed skater sat opposite Ben. Ava and Phillip sat next to each other. A woman sat at the hockey player's left hand. What sport does Ava participate in? So here we have our round table, and we'll say that there are four positions around the table. And so the first statement tells us that the skier sat at Andrea's left. So having nowhere to start with, let's just assume that Andrea sat at one, so the skier would be at four. The next statement tells us that the speed skater sat opposite Ben. Well this gives us two possibilities for where Ben could have sat. Obviously Ben couldn't sit where Andrea sat, because Andrea sat there. And Ben couldn't sit at position two, since on the other side of position two sat the skier. And we know that the speed skater sat opposite Ben. So those are the two possibilities for where Ben could have been seated. So let's keep on going. Next, it tells us that Ava and Phillip sat next to each other. So if Ben sat at position three, it would have been impossible for Ava and Phillip to sit next to each other, since they would have been at spots two and four, which aren't next to each other. So that means that Ben sat at position four, and Ava and Phillip sat at position two and three in some order. So now backtracking a little bit, the problem tells us that the speed skater sat opposite Ben, which means that the person at position two is the speed skater, since now we know that Ben sat at position four. Next, the problem tells us a woman sat at the hockey player's left hand. So there are two spots remaining for the hockey player, either at position one or at position three. Since we know that a woman sat to the left of the hockey player, and we know that Ben is not a woman, that means that we know that the person at position one is not a hockey player, it is the person at position three. And now we know who sits at position two. We know that it's a woman, and between Phillip and Ava, Ava is the woman. So, Ava sits at position two, and Phillip sits at position three. We have one more sport, which is snowboarding, which goes to Andrea at position one. So now we've figured out where everybody's sitting, and everybody's sport. And the question asked us for Ava's sport, and Ava is a speed skater. So the question asked us, which sport does Ava participate in? The answer is speed skating. Letter A Problem number 29 states, dates can be written in the form D-D-M-M-Y-Y-Y-Y. For example, today's date is 17-03-2016. A date is called surprising if all eight digits in its written form are different. In which month will the next surprising date occur? So here we have our date that we start off with, which the problem says is today. And this date is actually not surprising since it has two copies of 0 and two copies of 1. So we know that our date will take the form A-B-dot-C-D-dot-E-F-G-H. And we know if the date is surprising, A is not equal to B, which is not equal to C, which is not equal to D, which is not equal to E, which is not equal to F, which is not equal to G, which is not equal to H. So let's check if we have any valid months if our year starts with 2-0. So first, November will never be valid because November consists of two 1's. And none of the other months will work because either all of them have a 0 or they have a 2. So the year cannot start out with 2-0. Let's move on to the next possible one, which would be 2-1. For 2-1, January, February, October, and December would be tossed out in addition to November, but we already know November will never work. So now we only have these seven months. So now let's take a look at the days of the month. And none of the days are possible. Well, why is this? Well, because the 2 and the 1 are taken up by the year, and the 0 is taken up by the month. Because no matter which month we choose, it must have a 0 in it. And every single day of the month contains either a 0, a 1, or a 2. So our beginning assumption that the year starts off with 2-1 is not valid. So let's move on to the next possible one, the year starting off with 2-2, but that doesn't make any sense since the 2's would repeat. So let's move on to the next one, 2-3. So for 2-3, all of these months would be canceled out, February, March, December, and again, November, but we already know that one's not going to work. So we're only left with these eight. And let's take a look at the days of the month. And a couple of them are canceled out, but not all of them. So this one's possible. So next, let's move on to figuring out what g is. If we assume that g is 0, well, then that would cut out every single one of our months, since every single one of our remaining months contains a 0. So g cannot be 0. So let's try 1 instead. This would cut out January and October. However, this would cut out all of our remaining days of the month. So it can't be 1. It can't be 2 or 3, since we already have those. So let's try 4 next. This would only cut out April. And there would still be five days of the month that could work. So next, let's try and figure out what h is. We're trying to find the first date at which this occurs. So let's try the lowest valid number. We know 0 and 1 won't work, because that'll cancel out all of our months or all of our days. So let's try h equals 5. In this case, that would cancel out May and the 15th. So, so far, we're looking good. We're seeing that this date might occur in the year 2345. So now let's try and look through the earliest possible dates. So the earliest date would occur in January. So let's assume it's January. January wouldn't work, since that would cross out every single one of our days of the month. So instead, let's try June. June would only cross out the 16th, which leaves us with the 17th, the 18th, and the 19th for CD. And obviously, we want the most recent date, so we would pick the 17th. So June 17, 2345 would be the first surprising date since March 6, 2016. And the month is June. So the question asked us, in what month will the next surprising date occur? The answer is June, letter B. Problem number 30 states, at a conference the 2016 participants are registered with the numbers P1 to P2016. Each participant from P1 to P2015 shook hands with exactly the same number of participants as their registration number. How many hands did the 2016th participant shake? So first let's start out by putting all of these people into a large number of lists. So for the first list we'll put everybody from P1 to P2015 with P2016. Then in the second list we'll put everybody from P2 to P2014 and we'll also put in P2016. For list 3 we'll put in P3 through P2013 and P2016. So for list N we'll put in PN through P2016-N in addition to P2016. And we would have 1008 lists with the last list having the person 1008 and 2016. So first let's analyze what would happen with this first list. We know this list has 2016 people and let's take a look at the handshakes of person 2015. 2015 has to shake hands with 2015 people. So let's say that he shakes hands with people 1 through 2014. That means that person 2015 would shake hands with 2014 people, which isn't enough. He needs to shake one more person's hand. So let's say that he also shakes the hand of 2016. So 2016 so far has one handshake. Both person 1 and person 2015 have already reached their limit for handshakes, so we'll take them out in the next list. In list 2, this one only has 2014 people. So now let's consider who person 2014 shakes. So person 2014 would shake the hands of people 2 through 2013, which would result in him shaking the hands of 2012 people. But remember that 2015 also shook his hand, so really he's already shaking hands with 2013 people. So in order to get his number up to 2014, he should also shake his hand with 2016. So now 2016 has two handshakes. 2014 has his 2014 handshakes. Person 2 has person 2014's handshake and 2015's handshake, so we can drop 2 and 2014 in our third list. So our third list only has 2012 people. Again focusing on the second to last person, 2013, they would shake their hands with people 3 through 2012. So person 2013 would shake hands with 2010 people, but they already shook hands with person 2014 and person 2015, so they already shook hands with 2012 people. They need to shake one more person's hand, so they would shake it with person 2016. So currently at list 3, person 2016 has shook hands with 3 people. Continuing on this pattern, you would see that once we reach list 1008, person 2016 would have already had 1007 handshakes, and we only have 2 people in the list. Person 1008 has absolutely nobody to shake hands with, but he's already shook hands with 1007 other people, the people 1009 through 2015. So we only need to shake hands with one more person, and that would be person 2016. So that's him also satisfied, and that's everybody satisfied with person 2016 shaking 1008 people's hands. So the question asked us, how many hands did the 2016th participant shake? The answer is 1008. Letter D.

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