Problem number one states, what is the sum of the last two digits of the product 1 times 2 times 3 times 4 times 5 times 4 times 3 times 2 times 1? So let's take a look at this product. First let's realize that we can drop the 2 ones since multiplying a number by 1 does not change its value. Next let's break this number up into two smaller numbers and then we'll just multiply those numbers at the end. So we'll break it up into 2 times 3 times 4 times 5 and 4 times 3 times 2. For the left one we get a product of 120 and for the right one we get a value of 24. Now we just have to multiply these and we're going to use long multiplication. So 0 times 4 or 4 times 0 is going to be 0, 4 times 2 is going to be 8, and since we really only care about the last two digits we can kind of stop here but let's just keep going for completion's sake, 4 times 1 is going to be 4, and now since we're multiplying from the tens place we'll shift over one so 2 times 0 is going to be 0 but that's going to be shifted, 2 times 2 is going to be 4, and 2 times 1 is going to be 2. Adding those numbers up we get a total of 2,880. So the problem asks us for the sum of the last two digits which are 8 and 0, and 8 plus 0 is simply 8. So the question asked us, what is the sum of the last two digits of the product 1 times 2 times 3 times 4 times 5 times 4 times 3 times 2 times 1? The answer is 8, letter D. Problem number two states, an ant walked every day on a straight horizontal path from A to B, which are five meters apart. One day, humans placed two strange obstacles of height one meter each on the path. Now the ant walks along or above the same straight line, except that it has to climb up and down vertically over both the obstacles, as shown in the picture. How long is the ant's path now? So first let's take an aerial view of these boxes. We can see that the ant still walks from A to B in a straight line, meaning that whenever the ant encounters a box, it walks vertically upwards, and the problem even tells us that. So we can split up the problem into the spots where the ant walks horizontally, the red lines, and where the ant walks vertically, the yellow lines. We know the sum of the values of the red lines has to be equal to five meters, because the problem told us that the distance from point A to B is five meters, and that the ant takes the same path, only when it encounters a box, it has to go either up or down it. Now we just need to know what the height of the boxes are, and luckily the problem tells us that both of them are one meter tall. And we have four one meter tall, so it would be four meters. So the ant travels five meters horizontally and four meters vertically, for a total of nine meters. So the question asked us, how long is the ant's path now? The answer is nine meters. Letter B. Problem number three states, Rini marked two points A and B as accurately as possible on the number line. Which of the points PQRST on the number line best represents their product AB? So first let's take a look at the restrictions on values A and B. We know that A and B are both less than 1, and then we know that A is less than B. So we say A is less than B, which is less than 1. Now let's take a look at the fact that if we take a number, we'll call it X, and multiply it by another number, we'll call it Y, and Y is less than 1, then the product XY is less than X. If you don't believe me, we can substitute Y for 0.9. And we can clearly see that X times 0.9 would be less than X. So in this case, let's replace our numbers with A and B. So X will be A and Y will be B. So A times B is less than A, since B is less than 1. So therefore, we know that the value has to be less than A, so it cannot be R, S, or T. So now we're only left with P and Q. So now let's try and find the minimum value for A and B. Let's assume that they have the minimum value possible, which in this case would be 3 fourths. We know that they actually can't be 3 fourths, since we know that the number is greater than 3 fourths, and we know that B is even greater than A, but this will give us an even lower minimum, so it's okay. So let's multiply them together. A times B would give us 9 sixteenths, which is slightly more than a half and lies right here. Since this value, even less than the minimum value, is greater than P, therefore, there is no possible way that the product AB could be P. That means the product AB must be Q. So the question asked us, which of the points PQRST on the number line best represents their product AB? The answer is Q, letter B. Problem number four states the pie chart shows how the students at my school get to school. Approximately twice as many go by bike as use public transportation and roughly the same number come by car as walk. The rest use a moped. What percentage use a moped? So here we have our pie chart and the problem tells us that roughly the same number come by car as walk. The only two numbers that are really close to one another are 11 and 12 percent. That means that these are our two values. One of them represents the percentage of students that walk and the other represents the percentage of students that come by car. The other statement that we're given is that approximately twice as many go by bike as use public transportation. The only value that is roughly twice that of another is 47. 47 is roughly twice that of 24. That means that 47 is the percentage of people that use a bike and 24 is the percentage they use public transportation. That means the six percent that we have left represents the percentage of kids that come to school by moped. So the question asked us what percentage use a moped? The answer is six percent. Letter A. Problem number five states, the sum of five three-digit numbers is 2,664, as shown on the board. What is the value of a plus b plus c plus d plus e? So first let's take each of these numbers and convert their digits into their value form. So for example for the first number, we would have 100 times a plus 10 times b plus c. Since a is in the 100 spot, b is in the 10 spot, and c is in the 1 spot. If we do that for all of them, so for the next one it would be 100b plus 10c plus d, for the third one it would be 100c plus 10d plus e, and on and on and on for all of them. If we then summed up all of these values, we should get 2,664. We can see that we have many numbers here that we can factor out. For example, we can factor out 100 from 100a, 100b, 100c, 100d, and 100e. As a matter of fact, each of the terms has each of the variables. So each of them can be represented as 100 times a plus b plus c plus d plus e, 10 plus a plus b plus c plus d plus e, and 1 times a plus b plus c plus d plus e. Now let's combine all of our like terms to get 111 times a plus b plus c plus d plus e equals 2,664. Let's divide both sides by 111, and we get a plus b plus c plus d plus e equals 24, which is the answer we're looking for. So the question asked us, what is the value of a plus b plus c plus d plus e? The answer is 24. Letter C. Problem number six states, what is the value of 1,010 squared plus 2,020 squared plus 3,030 squared divided by 2,020? So let's take our quotient here. And we can see that all the terms in the numerator and the denominator are multiples of 1,010. So let's try and transform all of these into some form of 10,010. So let's try and transform all of these into some form of 1,010. So for example, at the top with 20,20 we can factor out a 2 to get 2 times 1,010 squared. And for 30,30 we can factor out a 3. And in the denominator we can factor out a 2. Now in order to take these values out of the parentheses, we have to square them. So we would have 1,010 squared plus 4 times 1,010 squared plus 9 times 1,010 squared divided by 2 times 1,010. Combining like terms in the numerator, we get 14 times 1,010 squared divided by 2 times 1,010. 14 over 2 is just 7, so those cancel out. And we can cancel out one of the 1,010s from the numerator with the one in the denominator to get 7 times 1,010. 7 times 1,010 is 7,070. So the question asked us, what is the value of 1,010 squared plus 2,020 squared plus 3,030 squared divided by 2,020? The answer is 7,070. Letter E. Problem number 7 states, let a, b, and c be integers satisfying 1 is less than or equal to a, which is less than or equal to b, which is less than or equal to c, and the product a, b, c equals 1,000,000. What is the largest possible value of b? So to solve this problem, let's work backwards from the answers. First, let's take a look at the restrictions on a, b, and c. We know that 1 has to be less than or equal to a, which has to be less than or equal to b, which has to be less than or equal to c, and we know the product a, b, c has to be equal to a million. So now let's start out at the largest possible value for e, since we're trying to find the largest number of b. So if b is equal to 2,000, that means the minimum value for c would also be 2,000, and the minimum value for a would be 1. That means the minimum product, if b is 2,000, a, b, c would be 4,000,000, which is about 3,000,000 too high. So that means that if b was 2,000, we could not make a million. So now let's move on to the second highest, 1,000. If b is 1,000, the lowest value for c would also be 1,000, and the lowest value for a would be 1. If we multiply them together, a, b, c would be 1,000,000, which is perfect. This means that d is the answer, and 1,000 is the highest possible value for b. It can't be c, b, or a, because although there's certainly some way to make a million, b would be much lower than 1,000. So we know the answer is 1,000. So the question asked us, what is the largest possible value of b? The answer is 1,000. Letter d. Problem number eight states, if D dogs weigh K kilograms and E elephants weigh as much as M dogs, how many kilograms does one elephant weigh? So first let's represent the weight of an elephant and the weight of a dog. We'll say that each elephant weighs lowercase e and each dog weighs lowercase d. So the first thing the problem tells us is that D dogs weigh K kilograms. That means that capital D, the number of dogs, times lowercase d, the weight of each of the dogs, is equal to uppercase K, the weight of the D dogs. Next, the problem tells us that E elephants weigh as much as M dogs. So if we have E elephants and each of them weighs lowercase e, that would be a total mass of capital E times lowercase e, and that's gonna be equal to the weight of M dogs, capital M being the number of dogs and d being the weight of each dog. So equals uppercase M times lowercase d. So these are our two equations. Now we just have to solve for e. In order to do that, let's put both of the equations in terms of lowercase d. So in the first equation, let's divide both sides by uppercase d to get lowercase d equals uppercase K over uppercase d. And in the bottom one, let's divide both sides by uppercase M to get lowercase d is equal to uppercase E times lowercase e over uppercase M. Now we're able to set these two equations equal to one another. And now we just have to solve for lowercase e. We can do that by multiplying both sides by uppercase M and dividing both sides by uppercase e. So we get lowercase e, the weight of each elephant is equal to uppercase K times uppercase M over uppercase d times uppercase e. So the question asked us, how many kilograms does one elephant weigh? The answer is KM over DE, letter D. Problem number nine states, there are two dice. Each one has two red faces, two blue faces, and two white faces. If we roll both dice together, what is the probability that both show the same color? So if we roll one of the dice, we know there's a one in six chance of it turning out on each face. We know two of the faces are red, two more are blue, and the last two are white. Since there are two of each face, there's a one-third chance of each of the colors showing up. A one-third chance of red, a one-third chance of blue, and a one-third chance of white. So let's say that's on roll one. Then on roll two, there's another one-third chance of it being each color. But together, in order to roll two reds in a row, it would only be a one in nine chance. In order to roll a red and a blue, it would also be a one in nine chance. In order to roll a red and then a white, it would also be a one in nine chance. So all of the ones on roll two have a one in nine chance. Out of all of these rolls, only one from each of the initial rolls would have the same color value as the previous roll. This means that we have three chances, if we roll both dice together, that they will both show the same color. So one out of nine plus one out of nine plus one out of nine would be equal to one-third. So there's a one in three chance of this happening. So the question asked us, what is the probability that both show the same color? The answer is one out of three. Letter E. Problem number 10 states, which of the following numbers is not divisible by 3 for any integer of n? So first let's create a table of all the values of n and then all the values of each of these expressions. So first let's take a look if n equals 0. These are the values that we would get. So out of these, 0 is divisible by 3, so it cannot be n squared or n times n plus 1. If n is equal to 1, we get a 6 here, which is divisible by 3. So it cannot be 5n plus 1. If we get n equals 2, we get values of 11 and 7 for the remaining two expressions, which neither of them are divisible by 3. So let's keep going to n equals 3, 17 and 26, also not divisible by 3. When we get to 4, we get 23 and 63. And 63 is divisible by 3, which means that the expression is not n cubed minus 2. That means it must be 6n minus 1. But let's prove this. So we know 6 can be broken up into 3 times 2, so we would have 3 times 2 times n minus 1. Since we know n is an integer, we know that any number times 3 is going to be divisible by 3. So we know that 3 times 2 times n has to be divisible by 3. And if we subtract 1 from this number, well, then we know that it can never be divisible by 3. Because the number after that is always divisible by 3, and numbers that are divisible by 3 only come every three numbers. So it is impossible for the number before that to also be divisible by 3. So 6n minus 1 can never be divisible by 3. So the question asked us, which of the following numbers is not divisible by 3 for any integer n? The answer is 6n minus 1. Letter D. Problem number 11 states a blue rectangle and a red rectangle overlapping. The figure shows four different such cases. We denote by B the area of the part of the blue rectangle that is not common to the two rectangles and we denote by R the area of the red rectangle that is not common to the two rectangles. Which of the following statements is true about the quantity B minus R? So the problem told us that the variable B represents the blue area and the variable R represents the red area. Let's introduce a new variable here we'll call it A and so be the area that both of the triangles overlap, the white area. Next let's talk about X. We'll let X be the area of the blue rectangle so that'll be the blue plus the white area and Y the area of the red rectangle so the red area plus the white area. So now let's create some math expressions. For example the area of the blue rectangle the blue plus the white is equal to just that the the area of the blue section plus the area of the overlapping section X equals A plus B. We can do the same thing for Y only instead of B we'll use R since that represents the red rectangle so Y equals A plus R. So now with these two equations let's subtract them from one another so X minus A is equal to A plus B minus A plus R. Let's distribute that negative over to get A plus B minus A minus R and we can see that the A's cancel out and now we're just left with X minus Y equals B minus R. X and Y represent the area of the blue rectangle and the red rectangle and we know that they are constant. If X and Y are constant that means that B and R must also be constant which means that the difference of their areas does not change. So the question asked us which of the following statements is true about the quantity B minus R? The answer is the quantity B minus R is the same in all cases. Letter E. Problem number 12 states, five coins are lying on a table with the head side up. At each step you must turn over exactly three of the coins. What is the least number of steps required to have all the coins lying with the tail side up? So this is where we start with five coins all head side up. We're able to flip three of them per turn and it doesn't matter which ones we flip. So let's just flip the first three. So now we have three tails and two heads. At this point we could flip all of those three tails back to heads and get back to where we started, but that's not productive so let's not do that. Instead we have two choices. We can either flip one of the coins that we already flipped or we can flip two of the coins that we already flipped. If we flip one coin that we already flipped we will have four tails and one head and if we flip two of the coins that we already flipped we will have three heads and two tails. If we have three heads and two tails we can just flip all the ones that are heads to tails and we'll be left with five tails. So this took us only three flips to get. So the question asked us, what is the least number of steps required to have all the coins lying with the tail side up? The answer is three. Letter B. Problem number 13 states four identical boxes are glued together to make the shape shown in the picture. One liter of paint is needed to paint the outside of one such box. How many liters of paint are needed to paint the outside of the glued construction? So first let's take one of these boxes and then label all of its sides. We'll label them A, B, and C. Each of these blocks has two sides of A, two sides of B, and two sides of C. So now in our glued shape let's label all of the A's. So we have three on top and then on the exact other side on the bottom we would have three more for a total of six A's. Next let's label all the B's. So these two right here are obvious, and then we have two half B's right there and there, and then we also have some B's hiding in the back. We have three of them right there, there, and there. So in total we also have six B's. Next let's label all the C's. So these faces are obvious, these four C's, and then we have two more in the back. So in total we also have six C's. We have six A's, six B's, and six C's. That means we have two total blocks of surface area. In other words our shape has the same surface area as three individual boxes. And since we know one box takes one liter of paint to paint, that means that we need three liters in order to paint this shape. So the question asked us, how many liters of paint are needed to paint the outside of the glued construction? The answer is three. Letter B. Problem number 14 states, let a, b, and c be integers, which of the following is certainly not equal to a minus b squared plus b minus c squared plus c minus a squared? So let's start out by foiling out our expression. So a minus b quantity squared will be a squared minus 2ab plus b squared. b minus c quantity squared will be b squared minus 2bc plus c squared. And c minus a quantity squared will be c squared minus 2ac plus a squared. Here we see that we have a pair of every single square, so we can combine those into 2a squared plus 2b squared plus 2c squared minus 2ab minus 2bc minus 2ac. And all of these terms have a 2, so we can factor that out to get this. We know the expression in the parentheses has to be an integer, because we know that a, b, and c are all integers, and squaring integers, adding them together, and multiplying them by each other does not change their identity as integers. And if we multiply an integer by 2, we know that that number is going to be even, which means that we know that this expression has to yield an even integer. Out of our answers, only one of our answers is odd, and that is 1. It is impossible to get 1 from this expression, with any value of a, b, c being an integer. So the question asks us, which of the following is certainly not equal to a minus b quantity squared plus b minus c quantity squared plus c minus a quantity squared? The answer is 1, letter b. Problem number 15 states, the first two digits of a 100 digit integer are 2 and 9. How many digits does the square of this number have? So we know we have an integer that starts with 2 9 and then there are 98 digits following that. We can represent this number as a times 10 to the 99th, where a is bounded on the lower side by 2.9 and on the upper side by 3. The value would be 2.9 if all of the remaining numbers were 0 and the number would be slightly less than 3 if all of the remaining numbers were 9. So that's the whole range of numbers that we can have, a times 10 to the 99th. And why is it 10 to the 99th not 10 to the 100th? Well we can do a simple example of 11. 11 is equal to 1.1 times 10 to the first and we see that 11 has two digits but the 10 is only to the first power. So you take the number of digits and then minus 1. So let's square our number. So we would have a times 10 to the 99th times a times 10 to the 99th. By the commutative property we can also rewrite that as a times a times 10 to the 99th times 10 to the 99th. So first let's do a times a. Let's assume the lower bound first, so 2.9 times 2.9. So now let's add up our two numbers, and we get 8.41. So now let's do it for the upper bound as well. In this case, we'll just substitute it with 3, as that is close enough. 3 times 3 is obviously 9. And now that we've done a times a for both the lower bound of a and the upper bound of a, let's do the 10 to the 99th part. So 10 to the 99th times 10 to the 99th, we will not be doing long multiplication here. Simply, we can realize that both the numbers have the same base. So if we add their exponents, we will get the value that we get when we multiply them. So 10 to the 99th plus 99 is 10 to the 198th. So we know that our number is sandwiched between 8.41 times 10 to the 198th and 9 times 10 to the 198th. Both of these numbers have 199 digits. That means that a times 10 to the 99th squared also has 199 digits. So the question asked us, how many digits does the square of this number have? The answer is 199. Letter B. Problem number 16 states, Matias has placed 15 numbers on a wheel. Only one of the numbers is visible, the 10 at the top. The sum of the numbers in any seven consecutive positions on the wheel, such as the ones shaded gray, is always the same. When all 15 numbers are added, exactly how many of the numbers 75, 216, 365, and 2020 are possible totals? So first let's start out by orientating our gray section like this. So first let's actually label all of the cells. So we know the sum of the values in the gray is going to be 10 plus a plus b plus c plus d plus e plus f. And then if we shift it over one, we know that the sum of these values has to be equal to the sum of the original values. And this sum is a plus b plus c plus d plus e plus f plus g. The only difference here is the g and the 10. That means if we subtract that sum from both sides, we get that 10 equals g. So let's move our gray section over here, and we get this sum. And if we move it over by one more, we get this sum, which is very similar to what we did initially. Subtracting h plus i plus k plus m plus n plus p from both sides, we get q equals 10. So let's keep doing this around the circle. And eventually we get that all of the values are 10. So we know we have 15 circles, and each of them have a value of 10. That means when we add it up, we get a sum of 150. That means that no matter what we do, we cannot get a sum of 75, 216, 365, or 2020. The only sum we can get is 150. So the question asked us, when all 15 numbers are added, exactly how many of the numbers 75, 216, 365, and 2020 are possible totals? The answer is zero of them. Letter a. Problem number 17 states, a large square touches two other squares as shown in the diagram. The numbers in the small square represent their areas. What is the area of the largest square? So for the duration of this problem we're going to be considering this right triangle. We'll say that one of its side lengths is X and that the other one is Y. Y is quite easy to solve for. We can see at the bottom length is 3 since we know that the square labeled 9 has an area of 9. The square root of 9 is 3 which means it is a side length of 3. And the same logic we know that the segment above it is 1 since the square root of 1 is 1. So we know that Y consists of 1 plus 3 and 1 plus 3 is 4. So we know the length of line segment Y is 4. Next let's take a closer look at this small right triangle. We know that this value is 3 since the square root of 9 is 3. And we know that this value is 1. And we know that this value is 1 less than 3. So 3 minus 1 which is 2. And we know that this length is also 1. So what does this tell us? This tells us that the diagonal of our yellow right triangle has a proportion of going 1 down and then 2 over. 2 over to the left. So essentially you could say that is a slope of 1 half. So now we can see that the triangle goes down by 4. And if it goes down by 4 that means it goes twice as many over or it goes 8 over. So X must be equal to 8. So now we figured out the side lengths of our right triangle. Let's label the hypotenuse of our right triangle Z. And now we can use Pythagoras's theorem a squared plus b squared equals c squared to solve for Z. So for a we'll plug in 8. And for b we'll plug in 4. 8 squared is 64. 4 squared is 16. 64 plus 16 is 80. So we know that 80 equals z squared. And the area of the big square, the question we're trying to answer, is actually z squared. One side length times the other side length. So there we go. We have the answer. The area of the square is 80. So the question asked us, what is the area of the large square? The answer is 80. Letter B. Problem number 18 states the sequence fn is given by f sub 1 equals 1, f sub 2 equals 3, and f sub n plus 2 is equal to f sub n plus f sub n plus 1 for all n greater than or equal to 1. How many of the first 2,020 elements of the sequence are even? So here we have the beginning of our sequence. We know f sub 1 and f sub 2, we know they're 1 and 3. So let's keep going and let's see if we can establish some sort of pattern. So we get 1, 3, 4, 7, 11, and 18. And now if we realize that any even number plus an even number equals an even number, and any odd number plus another odd number equals an even number, and only when an odd number and even number are summed up do we get an odd number. We can realize that there's a pattern going on here. For f sub 3, we have odd plus odd equals even. Then for f sub 4, we have odd plus even equals odd. For f sub 5, we have even plus odd equals odd. And for f sub 6, we get odd plus odd equals even. So whenever we get an even term, the two odd terms that are sandwiching it are used up to form two more odds. And only once we have two odds in a row do we make an even, but then that even is followed by two more odds. So the pattern that we get is odd, odd, even, odd, odd, even, on and on and on. But how do we know how this series looks at n equals 2,020? So we take our number, our index, because we know that we have 2,020 terms, and then we just divide it by 3, the length of our pattern, and we get 673 remainder 1, which means that we have 673 terms that go odd, odd, even, and then we have one more term, and that term is going to be the next one in the pattern, so odd. But overall we have 673 terms of odd, odd, even, which contain one even, which means there are 673 evens in the first 2,020 elements. So the question asked us, how many of the first 2,020 elements of the sequence are even? The answer is 673. Letter A. Number 19 states, a circle and a rectangle have been drawn in such a way that the circle touches two of the sides of the rectangle and passes through one of its vertices. The distances of two vertices of the rectangle from one of the points where the circle touches the rectangle and five are five and four as shown. What is the area of the rectangle? So first let's take a look at the radius of the circle. We'll call it R. We can see that this is equal to five because the five below goes from one edge of the rectangle, which is also the edge of the circle, to the center point of the circle. So therefore, that length is the radius. Next, let's consider this radius. We obviously know its value is five since all the radii of a circle are the same, and we know that this length is four. So we can see we have a 3-4-5 triangle right here, so we know that this length is three. If you don't believe me, you can do Pythagoras' theorem and see that three squared plus four squared equals five squared. So next, if we consider this line segment, it is the same length as the radius of the circle, which we know to be five. So now we have the base of our rectangle and the height of our rectangle. So in order to solve, we just have to multiply it. So area equals base times height, and the base is 5 plus 4 or 9, and the height is 3 plus 5 or 8. So the area is equal to 9 times 8, which is equal to 72. So the question asked us, what is the area of the rectangle? The answer is 72. Letter C. Problem number 20 states, 3 cuboids are arranged to make a larger cuboid as in the figure. The width of one of them is 6, and the areas of some of their faces are 14, 21, 16, and 30 as shown. What is the area of the face with the question mark? So first let's label all of the important edges, and if they have the same color they have the same value. So the only edge length that we're given is the green edge length, which we know is 6. But now we know that the green edge length times the blue edge length is equal to 21. So now if we substitute 6 in for green, and divide both sides by 6, we get the blue edge length is 3 and a half. Now that we know the blue edge length, we can solve for the yellow edge length. Since we know the blue edge length times the yellow edge length is equal to 14. Substituting 3 and a half for blue, and then dividing by 3 and a half on both sides, we get the yellow edge length is equal to 4. Knowing the yellow edge length, now we can solve for the pink edge length. Since we know the yellow edge length times the pink edge length is equal to 16. Substituting 4 in for yellow, and then dividing by 4 on both sides, we get the pink edge length is 4. Now we can solve for the red edge length. The red edge length times the pink edge length is equal to 30. Substituting 4 in for pink, and dividing both sides by 4, we get the red edge length is 7 and a half. Now we can solve for the orange edge length by realizing the red edge length is the sum of the blue edge length plus the orange edge length. So 7 and a half is equal to 3 and a half plus the orange edge length. Subtracting 3 and a half from both sides, we get that the orange edge length is equal to 4. Now we can finally solve for the area of the face with the question mark on it. We know its value is equal to the green edge length times the orange edge length, and we know both the orange edge length, 4, and the green edge length, 6. So 4 times 6 is 24. So the face with the question mark has a value of 24. So the question asked us, what is the area of the face with the question mark? The answer is 24. Letter B. Problem number 21 states the figure shows a section of the parabola with the equation y equals ax squared plus bx plus c. Which of the following numbers is positive? So let's consider each of the coefficients of the parabola a, b, and c independently and see what they do. And let's try and see if they're going to be positive or negative. So first let's take a look at this a value. If a is positive that means that the parabola will be opening upwards. However if it's negative that means the parabola will be opening up downwards. In our example we know that the parabola opens upwards so we know that a is greater than zero. Next let's take a look at the coefficient c. c determines the y-intercept of the parabola. So if it's greater than zero the parabola crosses the y-axis above zero. If it's equal to zero it crosses it at zero. And if it's below zero it crosses it below zero. In our example our parabola crosses the y-axis below zero so we know that c is less than zero. So now let's consider b. So in order to analyze the coefficient b we need to notice that h, the x-coordinate of our vertex, is equal to minus b over 2a. And in this case we know that the x-coordinate of the vertex is greater than zero so we know that h is going to be greater than zero. So now let's try and figure out whether or not b is going to be greater than or less than zero using this. So we know that h is going to be positive and we know that a is going to be positive. So in order for positive equals minus b over 2 positive to be true b must be negative. So that means that b must be less than zero. So now let's analyze each of our terms. For c it cannot be greater than zero because c is less than zero. For b plus c both of both b and c are less than zero so their sum cannot be greater than zero. a times c is a positive number times a negative number which is a negative number which is less than zero so that doesn't work. For b times c both of those numbers are less than zero so therefore the negatives would cancel out and we would get a positive number so d works. For completion sake let's do e. a is greater than zero and b is less than zero so multiplying them together we would get a number that is less than zero so our answer is bc. So the question asks us which of the following numbers is positive? The answer is bc letter d. Problem number 22 states, on a square grid paper a little kangaroo draws a line passing through the lower left corner P of the grid and colors in three triangles as shown. Which of the following could be the ratio of the areas of the triangles? So first let's take a look at this tiny little right angle here, and then we can see that we have this bigger right angle here which is similar to the first. And then we have an even bigger one like this one which is similar to the first two. We know they're similar since two of their angles are the same, the one at point P and then the 90 degree angle. We know that since they're similar their edge lengths are proportional to each other. So we know that the bottom edge length for each of them is 1, 2, and 3. If we say the height of the tiny right triangle is H, then the height of the middle right triangle will be 2H, and the height of the big right triangle will be 3H. Now we know all the yellow heights because we can count the grid cells, so we know there are 1, 2, and 3. And now we know the heights of the small triangles. So for the small one we have 1 minus H, for the middle one we have 2 minus 2H, and for the big one we have 3 minus 3H. We can factor out a 2 and a 3 out of the last two, and I guess we can factor out a 1 out of the first one as well. And now we can see the ratio of their heights is 1 to 2 to 3, but the question asks about the ratio of their areas. In order to figure that out we need to notice that the base of these triangles is proportional to their height. We know the base is proportional to the height because we know all of these triangles are similar, and therefore their edge lengths are proportional. So in order to find the ratios of the areas we simply have to multiply the base by the height, so in this case just square the number, and the proportions of the area will be 1 to 4 to 9. So the question asked us, which of the following could be the ratio of the areas of the triangles? And I don't see it here so it would be none of the above is accurate. Letter E. Problem number 23 states the length of one of the sides of a rectangular garden is increased by 20 percent and the length of the other side is increased by 50 percent. The new garden is a square as shown in the diagram. The shaded area between the diagonal of the square garden and the diagonal of the original rectangular garden is 30 meters. What was the area of the original garden? So first let's say that the width of the original rectangular garden is x and the height is y. Now we know the width of the garden was expanded by 20 percent so we know that this length is 1.2 x and the height of the garden was expanded by 50 percent so we know that this length is 1.5 y. So let's find the area of the shaded region in terms of x and y. So here we'll take the area of this triangle which is one half base times height. The base is 1.2 x and the height is 1.5 y and that simplifies for an area of 0.9 x y. Now let's find the area of this triangle so we can subtract it from the first one and then get the shaded region. The area is equal to one half base height and the base is x and the height is y so that simplifies down to 0.5 x y. So now in order to find the green region we have to subtract the yellow region from the red region so we get that that region is equal to 0.9 x y minus 0.2 x y. 0.9 x y minus 0.5 x y which is 0.4 x y. So now we were told that the area of the shaded region is 30 meters squared so we can set that equal to 0.4 x y. 0.4 x y is 2 fifths so now we can multiply both sides by 5 and divide by 2 to get that the original rectangular garden had an area of 75 meters squared. So the question asked us what was the area of the original rectangular garden? The answer is 75 meters squared. Letter d. Problem number 24 states a large integer n is divisible by all except two of the integers from 2 to 11. Which of the following pairs of integers could be these exceptions? So here we have all our numbers 1 2 3 4 5 6 7 8 9 10 11 and we know that there is some number that is divisible by all of these numbers except for a pair that is in our answers. So first we can just cut out the 1 because we know that all numbers are divisible by 1 and now we can break up all of the remaining numbers into their prime factorizations which will play an important role in analyzing this problem. So first let's take a look at 2 and 3. So we know the number cannot be divisible by 2 and 3 but it does have to be divisible by 6 but that doesn't make much sense since 6 has to be divisible by 2 and 3. So the number cannot be divisible by 2 and 3 exclusively since 6 is a multiple of 2 and 3. So next let's move on to 4 and 5. So we know our number cannot be a multiple of 4 or a multiple of 5 so we cannot have two 2's but we know it has to be divisible by 8 and 8 has three 2's so that would make it divisible by 4. And in addition we also have 10 which also has a factor of 5 so that would make it impossible to not be divisible by 5. So B is also not possible. Next let's move on to C for 6 and 7. We know for C the number cannot be divisible by 6 or 7 however 6 is just 2 times 3 and we know the number has to be divisible by both 2 and 3 and 2 and 3 appear in 9 and 10. Next let's take a look at 10 and 11. So the number cannot be divisible by 10 or 11 but both the number has to be divisible by 5 and has to be divisible by 8 which has a 2 in it so will be divisible by 10. It can't be divisible by 11 but it is divisible by 10 so E doesn't work either. So now we know the answer is 7 and 8 but let's just verify. So for 7 and 8 it can't be divisible by 7 or 8 however if it's divisible by both 2 and 4 then we would get 2 to the third power which would make it divisible by 8. So something's not right here. That's because the problem said that these integers could be the exceptions not they had to be the exceptions. So this one for example actually does work because if it's divisible by 4 then it doesn't have to be divisible by 2 because 4 would automatically make it divisible by 2 and if we know it's also divisible by 3 then doesn't have to be divisible by 6. If we know it's divisible by 5 then we don't have to make it divisible by 10 and if we make it divisible by 9 we don't have to make it divisible by 3 and then we'll just make it divisible by 11. So now if we take our product which is 4, 5, 9, and 11 we can see that this number is multiple of 1, 2, 3, 4, 5, 6, 9, 10, and 11 but not of 7 or 8. So this one works. So the question asked us which of the following pairs of integers could be these exceptions? The answer is 7 and 8. Letter D. Problem number 25 states, In the morning, the ice cream shop offers 16 flavors. Anna wants to choose a two-flavor ice cream cone. In the evening, several flavors are sold out, and Bella wants to choose a three-flavor ice cream cone from the flavors left. Both Anna and Bella can choose from the same number of possible combinations. How many flavors were sold out? So first let's take a look at Anna's choices. We know that she has 16 ice creams to choose from. Out of these, she has to choose one, which gives her only 15 options left. So that means that in total she has 16 times 15 combinations of ice creams to choose. This isn't completely accurate because the order doesn't matter. It doesn't matter if she first picks vanilla and then chocolate, or chocolate and then vanilla. So we have to divide the total number of combinations by 2. So the total number of combinations that she can make is 240 divided by 2, or 120. So now let's take a look at Bella. We don't know how many flavors of ice cream she has, so we'll just say that she has N flavors to choose from. So after she chooses N, then there's N minus 1 flavors left. So she picks one more, and now there are N minus 2 flavors left. That doesn't mean that the flavors are going away, like they're getting sold out. That just means that she can't pick the same one twice. So then after she picks the next one, the total number of combinations she could have made is N times N minus 1 times N minus 2. Now we have the same problem as last time, where this is order dependent. In order to make it order independent, let's say that we have 3 ice creams, vanilla, chocolate, and strawberry. There are 6 possible ways to arrange them, so we have to divide our number by 6. And for larger numbers, you would realize this is the number of choices that you had factorial. So now let's set this value equal to 120, since the problem tells us that they have the same number of possible combinations. Multiply both sides by 6. And if we foiled all of this out, we would get a nasty cubic, and we would be forced to probably be doing synthetic division. So rather than doing that, let's work backwards from our answers and see if we can figure it out that way. But the problem is that N doesn't represent our answer. N represents the flavors that Bella saw initially. And the problem asks us for how many flavors were sold out when Bella went to get ice cream. So that would be equal to 16 minus N. And we'll say A here for answer. So now solving for N, we get that N equals 16 minus A. So every instance of N that we have, let's just replace that with 16 minus A. And we get this. So 16 minus A times 15 minus A times 14 minus A equals 720. So now let's just plug in all of our answer choices. First let's try 2. So we would get 14 times 13 times 12. And that is a little high. That does not equal 720. If we try 3, that's also high. It doesn't work. If we try 4, it also doesn't work. If we try 5, it also doesn't work. And finally when we try 6, it works. 720 equals 720. So the question asked us, how many flavors were sold out? The answer is 6. Letter E. Problem number 26 states, Tony has 71 marbles at his disposal in a box. He is allowed to take out exactly 30 marbles from the box or to return exactly 18 marbles to it. Tony is allowed to apply each operation as many times as he wishes. What is the smallest number of marbles that can remain in the box? So first we know that Tony starts out with 71 marbles. And we know that he can take away 18 marbles as many times as he wants. So we'll represent that as n. So the total amount of marbles after he takes 18 away n times is going to be 71 minus 18m. And then he can also put back 30 marbles as many times as he wants. And we'll say that the number of times that he does that is m. So ultimately the number of marbles in the box is going to be equal to 71 minus 18n plus 30m. Where n represents the number of times that he returns the marbles. And m is the number of times that he takes 30 marbles. So we'll say that this value is going to be equal to c. And we have some restrictions on this value. For example, it has to be greater than zero. He can't take away more marbles from the box than there are marbles. And we know that n and m have to be integers. For example, he can't take away 1 tenth of 18 marbles from the box or 1 twentieth of 30 marbles from the box. That doesn't make any sense. And the problem asks us to figure out what is the minimum number of marbles that he can have left. So we need to minimize his value c. So in order to do this let's factor out a 6 from the 18n and 30m. To get 71 plus 6 quantity negative 3n plus 5m equals c. Next let's subtract c from both sides and then subtract the 6 times minus 3n plus 5m and factor out the negative. And so now we can see that we need to maximize the value 71 minus c, the number of marbles that have been taken away. So then we know that the 71 minus c must be a multiple of 6. So let's try and find the highest multiple of 6 that's close to 71. And the closest one is 66, 6 times 11. And let's just make sure that making 11 is possible with our values of n and m. And certainly making 11 from multiples of negative 3 and positive 5 is possible. For example we could have n being equal to 8 and m being equal to 7 minus 24 plus positive 35 is equal to 11. So now we know that 71 minus c is going to be equal to 66. So let's solve for c. And we get that c equals 5. So we can get 5 marbles remaining. If 8 times we return 18 marbles and 7 times we take 30 marbles. Obviously we can't take the 30 marbles 7 times because then we would run out of marbles. But doing a combination of each will yield us 5 marbles left in the box. So the question asked us, what is the smallest number of marbles that can remain in the box? The answer is 5. Letter C. Problem number 27 states, Vajra took a square piece of paper with a side of one and folded two of its sides to the diagonal as shown in the diagram to make a quadrilateral. What is the area of this quadrilateral? So first let's draw a square surrounding our quadrilateral and now let's name some important points. We'll say A, B, C, D, E, F. So this square represents the initial sheet of paper that was folded and we know the sheet of paper is square and had a height of one so therefore it had a width of one as well. Next let's draw in the diagonals of our quadrilateral or of our kite really. So it would be a line going from A, C and one going from B, D. And when folding we can see that what she did was she took point F and put it in towards point G and she took point E and put it in towards point G. So that's what she did during folding. And we can see line segments A, F and A, E were there before the paper is folded and then after they were folded they both converged onto A, G. So therefore they have the same length. A, G has the same length as A, F and A, E which is one. Awesome. So now let's consider this line segment. Its length is the hypotenuse of a right triangle that's made up of line segments A, E, E, C and then itself the hypotenuse A, C. So using Pythagoras's theorem A squared plus B squared plus C squared we can see that line segment A, C has a length of square root 2. So now let's consider the quadrilateral A, E, C, F and realize that its area is going to be one half times the lengths of the two diagonals. So A, C and B, D. Luckily we already know one of the diagonals. We know that A, C is square root 2. We just figured that out so we can plug that in. And next we can see that angle G, C, D is 45 degrees since we know that F, C, E is a right triangle since we have a square and line segment A, C is the diagonal of this square. So we know that angle G, C, D is half of B, C, D which means that line segment G, D is half that of line segment B, D. So we can instead of having it in terms of B, D have it in terms of G, D. So the area of our quadrilateral is now square root 2 over 2 times 2 times the length of G, D. So now let's figure out the length of G, D. If we consider C, G we can see that it has the same length as G, D because triangle G, C, D is a 45-45-90 triangle which means that the two sides which oppose the 45 degrees have the same value. So the length of G, C is equal to the length of C, D. Next let's consider line segment A, C and we can see that A, C is the sum of G, C and A, G and we know both A, G and A, C. We know that A, G is 1 and that A, C is the square root of 2, the diagonal of the square. So now we can solve for G, C and figure out that it's equal to the square root of 2 minus 1. Now we can plug that in and figure out what G, D is, exactly the same thing as G, C, square root of 2 minus 1. Now we can plug that into our AECF area equation and now all what's left to do is to simplify. So we can cancel out the twos, we could have done that earlier as well. Now we have root 2 quantity root 2 minus 1, distribute the root 2, we get 2 minus root 2. So the question asked us, what is the area of this quadrilateral? The answer is root 2 over 2, letter A. Problem number 28 states, a certain iceberg has the shape of a cube. Exactly 90 percent of its volume is hidden below the surface of the water. Three edges of the cube are partially visible over the water. The visible parts of these edges are 24 meters, 25 meters, and 27 meters in length. How long is an edge of the cube? So here we have our cube partially submerged in the water, and we'll say that the white part of the cube is not submerged, and the blue part is submerged. So we know that 90 percent of the volume is underwater, and that means that 10 percent is above water. And now we know the three lengths of above water edges. They are 24, 25, 27 meters, and they form an irregular tetrahedron. And if we can find the volume of this, then we can find the volume of the whole cube, and then find the side length of the cube. So let's find the volume of this irregular tetrahedron. In order to do that, the volume of an irregular tetrahedron is one-third the area of the base triangle times the height. So we'll say this is the base triangle, and we'll say that this is the height. Now the height has to be perpendicular to the area of the base triangle, and it might not look like it here, but we are working on a cube, so that right there is a right angle, in addition to this one. Since both of those are right angles, we can actually really easily figure out what the area of the base of the triangle is. It's just one-half base times height. The base is 27 meters, and the height is 25 meters, so we get an area of 337.5 meters squared. So now we can plug our values into our volume equation, and we get a volume of 2,700 meters cubed for the part of the iceberg that is above the water. So in this case, Va will denote the volume of the iceberg above the water, and Vu will denote the volume under the water. So we know the cube is 90 percent underwater and 10 percent above water. So Vu equals 9 times Va, which is 2,700. So the volume underwater is 24,300. So now we can add up our two volumes to get the total volume of the iceberg, and we get 24,300 plus 2,700, which is 27,000. So that's the total volume of our iceberg. In order to find the side length of it, we just need to take the cube root of it, and luckily 27,000 is a really nice number. If you know that three cubed is 27, then you can realize that the side length is 30 meters. No need to do cube roots. So the question asked us, how long is an edge of the cube? The answer is 30 meters. Letter A. Problem number 29 states, there are n different prime numbers, p sub 1 to p sub n, written from left to right in the bottom row of the table shown. The product of two numbers next to each other in the same row is written in the box directly above them. The number k equals p1 to the alpha 1 power, p2 to the alpha 2 power, on and on and on through pn to the alpha n power is written in the box in the top row. In a table where alpha sub 2 is equal to 8, how many numbers are divisible by the number p sub 4? Okay so here we have our table, and we know the number at the top is going to be equal to p sub 1 to the power of alpha sub 1, times p sub 2 to the power of alpha sub 2, times p sub 3 to the power of alpha sub 3, on and on and on. And we know that alpha sub 2 is 8. So let's try and figure out how many rows our pyramid has. So with two rows, we can see that we only get a k value of p2 to the first power. If we add another row though, we can see we get p2 to the second power. And above the second row, we don't show any of the other terms. So like there's a p1 and a p3 term there, but we're just concerned about the p2 term for now. If we add four rows, then the k term would have p2 to the third power, and with five rows we would have p2 to the fourth power. So we can see that it increases linearly. With two rows we had one, with three rows we had two, with five rows we had four. So if we know our alpha sub 2 value is 8, that means that we know that we have nine rows in our pyramid. So our pyramid looks like this, with k at the top, and then p4 at the bottom. The question asks us to find, in the table, how many of the numbers were divisible by p4? So since we know they're all prime, we know that only numbers that have p4 in their product will be divisible by p4. So one that we know for sure is we know that p4 is going to be divisible by p4 exactly once. And then we also know the two numbers above that are going to be divisible by p4, since we know they contain a product containing p4. And we can do that for the numbers above that, and we can do that on and on and on until we get to the very top, k. And we have four diagonals that are six long of these numbers, so in total we have 24 blocks that are divisible by p4. So the question asked us, how many numbers are divisible by the number p sub 4? The answer is 24. Letter C. Problem number 30 states, Adam and Brit try to find out which of the following figures is Carl's favorite. Adam knows that Carl has told Brit its shape. Brit knows that Carl has told Adam its color. When the following conversation takes place, Adam, I don't know Carl's favorite figure and I know that Brit doesn't know it either. Brit, at first I didn't know Carl's favorite figure, but now I do. Adam, now I know it too. Which figure is Carl's favorite? So this question almost seems impossible at first. So first let's analyze what each of them know. So we know that Adam knows the color and Brit knows the shape. So in Adam's mind, we know that he has one of these columns in mind, which could be Carl's favorite. And for Brit, these are the things that he has in mind. So first Adam says, I don't know Carl's favorite figure, but I know that Brit doesn't know it either. So what does this mean? Well, this actually tells us that the shape cannot be white. Because if the shape was white, then Adam couldn't say for certain that Brit didn't know what the shape was. If the shape was white, then Adam would not be sure if Brit knew. For example, Brit might know that it's a hexagon, or he could know it's another shape. So Adam wouldn't know. So therefore, none of the white shapes are possible. So now we're only left with these combinations. Then Brit goes on to say, at first I didn't know Carl's favorite figure, but now I do. If Brit knew that they were circles, then he still wouldn't be sure. So that means that the shape is not a circle. And since Adam knows all of this information that we know, he also knows that it is not a circle. And then finally, Adam says, now I know it too. This means that the shape cannot be black. So it cannot be the black star or the black square. So it must be the gray triangle. So the question asks us, which figure is Carl's favorite? The answer is the gray triangle, letter C. .

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