Problem number 21. Each face of a die is colored with one of two colors. How many different two color dice can be made? Now on each dice there are six different sides which we can label one through six and on each of those sides we can pick one of two colors. Let's say red and blue. Now that means we have two to the sixth power different combinations of colors that we can put on the different faces. That's 2 times 2 times 2 times 2 times 2 times 2, 2 to the 6th power. That is how many different color combinations we have. But the problem is asking for two color dice and that means that we can't have one dice that's entirely red or one dice that's entirely blue. But those are included in the 2 to the 6th power combinations since all red or all blue is a possible combination. So we have 2 to the 6th is equal to 64 and we're gonna subtract those two options where we have a totally red and totally blue die and that leaves us with 62 possibilities. So our final answer is B. Problem number 22. A train consists of an engine and five cars. 1, 2, 3, 4, and 5. How many ways can the cars be arranged so that car 1 is always closer to the engine than car 2? I'm gonna go ahead and list out all the math first and then explain where it's coming from. And there's also a visual aid here. Go ahead and imagine that these are the five positions that the cars can be in where the further left you are the closer you are to the engine. Imagine the engine is right over here. So the leftmost is the closest. Now what we're gonna do is we're gonna break this up into cases. We're first going to consider when car 1 is in the first position and then the second and third and fourth and we don't care about the fifth position because then the second car will definitely be closer than the first car so we can just ignore that. Now looking at the case where position where car 1 is in position 1 we know for sure that the second car or that the car 2 will definitely be farther away than car 1 since car 1 is just the closest car out of all of them. That means that we can organize these remaining four cars however we want. So we have four choices for the second car and once we've used one of those choices we will have three more for the for the third car. Then we'll have two for the second car and one for the for the last car. So it'll be four times three times two times one possibilities for these four cars. So when one is first we have 24 possibilities. What if one is second? In this case we're not allowed to have car 2 in the first spot. It can only be in the third fourth or fifth in order to stay with the condition of the problem. So we have three choices for car 1. We can do 3, 4 or 5. Then we will have three choices for the third position because we now can place car 2 here and whatever two cars we didn't put over in position 1. So we have three choices here, three choices here and then two choices here since we used one up and then one choice here since we used one up. So that'll be 3 times 3 times 2 times 1 which equals 18. If 1 is in the third position then we can't have car 2 in the first or second position. So we have three choices here, two choices here and then we have whatever car we didn't use from 3, 4 and 5 and we also have car 2. So that's going to be two choices again. So 3 times 2 times 2 and then times 1. And then if 1 is in the fourth position then we will have, we can only have one choice for this last car and for these first three cars we have three choices times two choices times one choice times 1 which is all equal to 6. So we now have considered all the combinations when 1 is in the first position, second position, third position, fourth position and fifth position. Now we can add up all of these numbers of combinations and the total will list all the possible arrangements such that car 1 is closer to the engine than car 2. So taking the sum we get 24 plus 18 plus 12 plus 6 is equal to 60. So our final answer is B. Problem number 23. X, Y and Z are positive real numbers. Let X be greater than or equal to Y be greater than or equal to Z and X plus Y plus Z equals 20. Which of the statements below is always true? So our strategy here is going to be to try finding counterexamples for each of the statements and if we can't find a counterexample then we're going to try thinking about why that statement might always be true sort of logically and if we come up with a counterexample for each of the statements then that's easy we just know that the final answer is E. So let's go ahead and start looking for counterexamples. Answer choice A. X times Y is less than 99. Well here we can let X equal 10 and let Y equal 9.99 and you multiply those and you get 99.9 and we know that there is going to be a positive Z that we can add it's going to be 0.01 if we want to add up to 20. So we can eliminate answer choice A. B. X times Y is greater than 1. Well in this case we can let X equal 19.99 and Y equal a very very tiny number 0.0099 and because this is such a tiny number it's basically it's basically dividing 19.99 by a multiple of 1,000 we know that this is going to be less than 1. So if you want to be extra safe you can go ahead and add more nines to the 19.9999 and more zeros here to make it an even tinier number but we can definitely make this number less than 1 and we can always add a point 0.000000 however many zeros with a 1 at the end for Z and that's still going to be a positive number so that's not going to be an issue so B definitely doesn't hold. X times Y does not equal 25. Well here we could choose 18 and 25 over 18 and in particular we know that 25 over 18 is less than 36 over 18 which would be equal to 2 and to get 36 over 18 we would need to add 11 over 18 which is less than 25 over 18 so we see that we can let X equal 18 let Y equal 25 over 18 which is definitely less than X and then Z will be 11 over 18 and 11 is less than 25 so nothing is being violated here and we have a counterexample for C and for D we have X times Y does not equal 75. Well this counterexample is fairly straightforward it's just going to be 10 times 7.5 where we can let Z equal 2.5 so we have a counterexample for A, B, C, and D which means our final answer must be E. Problem number 24 numbers 1, 2, and 3 are written on the circumference of a circle. Afterwards the sum of each pair of neighboring numbers is written between them. In this way six numbers are obtained 1, 3, 2, 5, 3, and 4. This operation of writing the sums of two neighboring numbers is repeated three more times resulting in 48 numbers on the circle. What is the sum of these numbers? So it would take a very long time to actually perform this operation and I don't think that would be very practical so instead we're going to look for a pattern. Now here's what the problem starts off talking about 1, 2, and 3 on a circle. Notice by the way that the sum of these three numbers is 6. 1 plus 2 plus 3 equals 6. Now if we perform the operation once we fill in the numbers like this because 1 plus 2 equals 3, 2 plus 3 equals 5, 3 plus 1 equals 4. Now we have 6, 1, 2, and 3, plus 4 which is 10, plus 3 which is 13, plus 5 which gives us 18. So at first we had 6 as our sum now we have 18 as our sum which is 3 times greater. In particular the way that we can look at it is like this. We could rewrite 3 as 1 plus 2 so it's like 1 and 2 appear a second time here and we can think of 5 as 2 plus 3 so it's like 2 and 3 appear a second time here so 2 appears an extra time here and 2 appears an extra time here if we rewrite it this way and then 4 is equal to 1 plus 3 so 1 appears another time here it appears once in 1 plus 3 and it appears once in 1 plus 2 and then 3 appears in 1 plus 3 and in 2 plus 3 so it's like when we're adding these numbers each of the old numbers gets repeated two more times or in other words it's multiplied by 3. You're adding two more copies of the number so you have three copies of the number total. So in fact what this does to the sum of the numbers and this is a pattern you could check by performing the operation one more time but it actually takes the sum of the numbers and it multiplies it by 3 since all of the individual numbers are effectively multiplied by 3 this way so looking at our sum we're performing the operation a total of four times as given in the problem because it's repeated it's done once and then repeated three more times so we have our original sum which is 1 plus 2 plus 3 which equals 6 and we multiply it by 3 four times 3 to the fourth is equal to 81 and 6 times 81 is equal 486 so our final answer is C. Problem number 25 a square with sides of 10 centimeters length is rolled without sliding along the line until point P again lies on the line. What is the length of the curve that point P has made? We can divide up this curve that point P has made into three arcs one arc that travels up along along in this direction and then it'll be up in the top left corner of the square when the square rotates again it will make another rotation which will be further up like an extra bump and then it'll be on the top right portion of the square and then after one more rotation it'll move down back to the line so the arc on the left and the arc on the right are going to be exactly equal to each other they are both arcs of a circle of radius 10 centimeters as you can see it being rotate you can imagine it if you can't visualize it go ahead and take a square sheet of paper mark one of the points and try rotating it in the way that's described in the problem you'll see that it's rotating around a circle of radius 10 centimeters so the question now is what is the what is the length of the top arc because we can figure out the lengths of the two outer arcs the they're both just reliant on formulas from the from the circle of radius 10 centimeters but this upper arc here is a different length in particular it comes from a quarter rotation of a circle that has radius equal to the diagonal of the square which is square root of 2 times the side length which is 10 so this circle up here that we're moving the arc along is is 10 times square root of 2 radius so here's what the math should look like we have 1 4th of it's a quarter of a circle in each of the turns it's a quarter circle here a quarter circle here and a quarter circle here and the first arc is going to be a quarter of the circumference of this circle with radius 10 the circumference of a circle is 2 pi times the radius so 2 pi times 10 and here we also have a circumference 2 pi times 10 and it's a quarter of a rotation of each of these and then the radius of this middle arc is 10 times square root of 2 so it's 2 pi times the radius and we add all these together and since each of them are just a quarter of the circle that they're rotating around it's 1 4th times this quantity and if we expand this all out if we multiply 1 4th by this and by this we should get 5 pi here and 5 pi here which gives us 10 pi and then 2 times 10 gives us 20 and 1 4th of that is 5 so here we have 5 pi square root of 2 so multiplying it all out and then adding we get 10 pi plus 5 pi square root of 2 and thus our final answer is C now if it's confusing why this central arc length is along a circle that has radius 10 square root of 2 again I suggest you go ahead and make a physical model and actually mark one of the corners and try rotating it that should help clarify why it's along a circle with the radius of the diagonal of the square but if you can't visualize it in your head then that's really the only way that you can that you could help imagine that actually work with a physical model of it and that's true for anything you can't visualize if you have difficulty visualizing any of these sorts of problems just try making it in real life and seeing how it works and that should help clear up any issues so again our final answer is C problem number 26 working on a certain math kangaroo problem Barbara noticed that the following conclusions are correct if answer A is true then answer B is also true if answer C is not true then answer B is also not true if answer B is not true then neither D nor E is true which answer is correct if there is only one correct answer now we're going to use process of elimination here the most obvious one we can eliminate is answer choice a since if answer choice a is true then answer choice B is also true but we're only allowed to have one true answer so let's eliminate A. Now, what about answer choice, let's say, B? What can we say about this? Well, looking at statement number two, if answer C is not true, then answer B is also not true. Now, imagine this, and this is a little difficult to visualize at first, so really try thinking about this logic. Imagine that answer B is true. This means that answer choice C must also be true, and here's why. Because if B is true and C is not true, then we have a logical contradiction, because we have if C is not true, then B is also not true, but we just assumed that B is true. So it's sort of like you can flip around these two statements and write down their negative forms, so instead of saying not true, you say true, and instead of saying not true, it's true, and it turns out that that statement is also true. So if C is not true, then B is not true is exactly the same as saying if B is true, then C is true. In fact, if you ever take a course in higher level logic, this is sort of one of the most useful things to know in writing down rigorous mathematical proofs. This is called contrapositive, when you flip around the two statements and you write down their negatives, and that the contrapositive should always be logically equivalent to the original implication that is being stated, where an implication is if this, then that. So that's going a little deeper than we need to for the sake of this problem, but if you want to really understand why this sort of logic holds, just look up what contrapositive means in terms of mathematical logic. It's some very interesting stuff if you like higher level math, and it has nothing to do with numbers or anything. It's just pure rigorous logic, and a lot of very important mathematical facts are proven by sort of saying, oh, if A then B, well, I don't know how to prove that, but let's assume if not B, and then try showing that that means not A. So that's called a proof by contrapositive. Now going back to the problem, if C is not true, then B is not true. Well, that means if B is true, then C is true. But that's basically the exact same as statement one, where we've said, oh, if A is true, then B is true, but we're only allowed to have one true answer. So we can eliminate answer choice B. Now, again, we could sort of use this contrapositive approach using statement three. If either D or E is true, then B is going to be true. So we can't have D or E either, because statement three tells us that that would mean that B is also true. So those answers will not work either, again, with this contrapositive approach. By process of elimination, we are left with only answer choice C. So that is our final answer. Problem number 27. The numbers 257 and 338 have the property that when their digits are put in reverse order, they create larger numbers, 752 and 833. How many three digit numbers have this property? So let's start by eliminating some kinds of numbers that definitely don't have this property. In particular, we can't use numbers that have zeros as their last digit, as their rightmost digit, and we can't use symmetrical numbers. So like 121 or 333, for instance, numbers that are perfectly symmetrical to each other. Now, how many numbers have a zero at the end of them when considering all of the three digit numbers? Well, going from, let's say, 100 to 199, we have 100, 110, 120, 130, all the way up to 190. That's a total of 10 numbers that have that property. Since we have sort of nine sets of 100, we have 100 to 199, 200 to 299, 300 to 399, and so on. There's nine blocks of numbers like that, and each of them have ten such numbers. We have a total of 90 numbers which end in a zero, which means that they definitely don't satisfy the property. Similarly, there are 10 symmetrical numbers in each block of 100. For instance, in the 100 to 199 block, there is 101, 111, 121, 131, and so on. So again, that's going to be 10 numbers per block times nine blocks, which means that there's 90 numbers which are symmetrical. So there's a total of 900 three digit numbers going from 100 to 999. There's a total of 900 of those numbers, and we just showed that there are 90 symmetrical numbers and 90 numbers ending in a zero, which we can eliminate. So we have 900 minus 90 plus 90, which is equal to 900 minus 180, which equals 720. Now of these 720 remaining numbers, half of them satisfy the property and half of them do not. And you can probably imagine why, since there are no other sort of special kinds of numbers which violate the property that's given in the problem. So all of the other numbers are sort of normal numbers. They don't have zeros and they don't have symmetry. So half of the remaining numbers will become bigger when you flip them, and the other half will be smaller when you flip them. In fact, you could see all those numbers as mirror images of each other. That's one way to sort of think about it. So we have 720 divided by 2, which gives us 360. So our final answer is E. Problem number 28. Points M and N are chosen on square ABCD. The square is then divided into 8 parts, and 3 of the areas are given. See the picture. What is the area of the shaded region? Now, first let's quickly note something about how this square has been divided up. Let's talk about triangle AND. Notice that the base of this triangle, AD, is the side length of the square, and the height of the triangle, going from the base to N, is also equal to the side length of the square. That means that the area, 1 half of base times height, is just equal to 1 half of the side length of the square squared. So in other words, it's exactly half the area of the square. And by the exact same logic, this triangle, CDM, is also half the area of the square, because the base is the side length, and the height is the side length. So each of these two triangles are equal in area, and they are each half the area of the square. Now let's go ahead and label all of the regions that we don't know. So we'll call S the area of the shaded region, and then we have A, B, C, and D. Now, knowing that we have half of the area covered by one triangle, we could take all the other spaces, and claim that that's the other half of the area of the triangle. And so from this, we can write two equations. One is S plus B plus D, which is this triangle, AND, is equal to all of the other areas added together, since they make up the other half. And then similarly, we have S plus A plus C, that's S, A, and C, that's this triangle, CDM, that's this triangle making up half the area, is equal to all of the others added together. So we know that S plus B plus D is equal to S plus A plus C. But we don't know how A and C relate to B and D. But the nice thing we can do here is, since S plus A plus C equals S plus B plus D, we could actually substitute in one for the other, and get an equation like S plus A plus C, from the second equation, is equal to A plus 3 plus 2 plus C plus 9, from the first equation. We could have also paired up this term from the first equation, and this term from the second equation. Or we could have even added both equations together and solved. Either way would have been fine. So once we solve for this equation, we get S is equal to 14, since the A and C cancel each other out, all we're left with is just adding together some constants. So S is equal to 14, and our final answer is A. Problem number 29. Number Y is defined as the sum of the digits of number X, and Z as the sum of the digits of number Y. How many natural numbers X satisfy equation X plus Y plus Z equals 60? Now here's how we can go about solving this. It really is a lot of trial and error, but it helps if you can pick a starting point. What number X should you actually sort of start looking at? And a good place to start is letting X equal the number 40 or higher. Because if you have anything less than 40, then you're going to be more than 20 away from 60. And that's really a lot, because the biggest one-digit number is 9, and 9 plus 9 is equal to 18, which is less than 20. And so just by knowing that, you can imagine it will be very difficult if your first number is more than 20 away from 60. It will be very unlikely that the other two numbers can add up to something that's greater than 20. So that's kind of a hand-wavy way to describe it, but the idea is that you're probably not going to have any X's that are farther away from 60 than 40. So start with X equals 40 and move up. And once you go ahead and check all of the numbers going from 40 to 60, you should find that there are three numbers that actually satisfy this property. Now you can go ahead and look back and try some of the numbers that are less than 40, but you'll quickly see that there's no way that any of the numbers can possibly add up to 60. It's just going to be too small. You'll see that you're getting farther and farther away from 60 the lower you go. So checking through all of the possible answers above 40, you'll get 44, 47, and 50 as good values of X. And so our final answer is D. Problem number 30. Suppose the final result of a soccer match is 5-4, the home team winning. If the home team scored first and kept the lead until the end, in how many different orders could the goals have been scored? So the general strategy here is to start by drawing this sort of a diagram. We want to list all of the different possible scores where the home team is leading, and we want to categorize these scores somehow. So each row corresponds to a different score for the home team, and each of the columns represents a different score for the away team. So we can ignore this upper half entirely since here the scores are equal or the away team is doing better. Now these are all of the possible scores where the home team is leading. Let's go ahead and start at the end. Now there's only one way to get to 5-4 following the conditions of the problem, and that's from 5-3. And there's only one path to get from 5-3 to 5-4. Now let's look at the ways that we can get to 5-3. We could get there from 4-3, in which case there's one path that will take us to 5-4. And we could get there from 5-2, in which case there's one path that takes us to 5-4. So there's one path and one path. Now what about 4-2? Notice that here we could take either this path or this path that we were just looking at. So from the point 4-2, there's actually two paths, 1 plus 1. Now here from 3-2, there's only one way to get, there's only one way to go from here. You can only go to 4-2. So there's only one path that you could take. And from path 4-2, we know that there's two paths to get to 5-4. So that means that there's two paths from 3-2 to get to 5-4. So the wording might be a little bit confusing at first, but basically the idea is you add the number of paths of all of the places that you can go to in your next step. So the next step for 4-2 will either be 4-3 or 5-2. So you add up all of the possible paths and all of the possible paths, in that case that's 1 plus 1, and that gives you 2. 3-2, you're just adding 2 and nothing else. And for instance now for 3-1, you can go ahead and add all the paths for 3-2 and all the paths for 4-1. So for instance, you're adding 2 from 3-2, and then how many paths are there from 4-1? Well, there's two paths coming from 4-2, and then there's one more path coming from 5-1. So that's actually three paths total, 2 plus 1. So 4-1 has three paths, 3-2 has two paths, and 3 plus 2 is equal to 5. So we know that 3-1 has five paths. And if this doesn't make sense, then try going over it a few times, because this is really a great shortcut as opposed to trying to count all of the paths, and you won't know if you counted one path twice, or if you counted one path three times, or maybe you missed a path. By counting backwards this way, by adding the paths of the next place that you can go to, you can sort of work your way backwards until you eventually see that there are 14 paths from 2-0, and that means that there's 14 paths from 1-0. So I think this shortcut is a good thing to learn, and really just sort of sit on it, think about it. This is one of those things that you should really just try to make sure that you have down this sort of a shortcut, because it's a lot easier than counting it out by brute force, where you might lose track of something. In general, when you have something like this, where you have a lot of counting to do, always just ask yourself if there's a quick shortcut, if there's some faster way to count everything. That's the best advice I can offer for for these sorts of problems where it looks like there's going to be a lot of tricky counting. Just look for a pattern, look for a shortcut, but in this case we see that the answer, the number of paths from 1-0 will be 14, so our final answer is D.

math problems

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logical statements

goal sequences