💡First, I need to determine how many guests Alex is inviting. Since Alex is inviting two-thirds of the number of guests Bridgette is inviting, I calculate:[frac{2}{3} times 84 = 56 text{ guests}]Next, I add the number of guests Bridgette and Alex are inviting together:[84 text{ (Bridgette's guests)} + 56 text{ (Alex's guests)} = 140 text{ guests}]The caterer makes 10 extra plates, so I add that to the total number of guests:[140 text{ guests} + 10 text{ extra plates} = 150 text{ plates}]Each plate will have 8 asparagus spears, so I multiply the number of plates by the number of asparagus spears per plate:[150 text{ plates} times 8 text{ asparagus spears} = 1200 text{ asparagus spears}]Therefore, the caterer will need (boxed{1200}) asparagus spears in all.

💡Alright, so I have this problem about a regular hexagon, ABCDEF. Point K is on side DE, and the line AK divides the area of the hexagon in a 3:1 ratio. I need to find out how K divides DE, meaning the ratio DK:KE.First, I should recall some properties of a regular hexagon. A regular hexagon can be divided into six equilateral triangles, all congruent, with each triangle having a central angle of 60 degrees. So, the area of the hexagon can be considered as six times the area of one of these equilateral triangles.Let me denote the area of the hexagon as S. If I consider each of those equilateral triangles to have an area of 1, then the total area S would be 6. That might make calculations easier.Now, the line AK divides the hexagon into two parts with areas in the ratio 3:1. So, one part has an area of 4.5 (which is 3/4 of the total area) and the other has an area of 1.5 (which is 1/4 of the total area). But I need to figure out exactly where K is located on DE to achieve this.Let me visualize the hexagon. Starting from point A, going through B, C, D, E, F, and back to A. So, DE is one of the sides. Point K is somewhere on DE. The line AK connects A to K.I think it might help to break down the hexagon into smaller regions whose areas I can calculate. Since AK is a line from A to DE, it will divide the hexagon into two regions: one that includes triangle ADE and part of the hexagon, and the other that includes the rest.Wait, maybe I should consider the areas of the triangles formed by AK. Let me think. If I draw AK, it will form triangle AKD and quadrilateral AKFE or something like that. Hmm, maybe not. Let me try to be more precise.Let me label the hexagon with coordinates to make it easier. Let's place point A at (0,0), and since it's a regular hexagon, I can assign coordinates to all other points assuming a unit side length. But maybe that's overcomplicating things.Alternatively, I can use symmetry and proportions. Since the hexagon is regular, all sides are equal, and all internal angles are 120 degrees. The side DE is opposite to side AB, right? So, DE is one of the sides, and K is somewhere along it.If I consider the line AK, it will pass through the interior of the hexagon and divide it into two regions. The area ratio is 3:1, so one region is three times larger than the other.I think the key is to find the position of K on DE such that the area on one side of AK is three times the area on the other side. Let me denote DK as x and KE as y, so that x + y = DE. Since DE is a side of the hexagon, its length is equal to the side length, say 's'. But since we're dealing with ratios, the actual length might not matter; the ratio x:y will suffice.So, I need to express the areas in terms of x and y. Let me think about the areas created by AK. The area closer to A will include triangle ADE and some other regions, while the area closer to DE will include triangle AKD and the rest.Wait, maybe I should calculate the area of triangle AKD and set up the ratio accordingly. If AK divides the hexagon into two regions with area ratio 3:1, then the area of one region is 3 parts and the other is 1 part.Assuming the total area is 6, as I considered earlier, then one region would be 4.5 and the other 1.5. But I need to figure out which side of AK corresponds to which area.Wait, actually, if AK is drawn from A to DE, the area closer to A would be smaller, right? Because AK is closer to A, so the region near A would be smaller, and the region near DE would be larger. But the problem says the ratio is 3:1. So, does that mean the larger area is 3 parts and the smaller is 1 part?Yes, that makes sense. So, the area closer to DE is 3 parts, and the area closer to A is 1 part. Therefore, the area of the region containing DE is 3/4 of the total area, and the area of the region containing A is 1/4 of the total area.So, the area of triangle AKD plus the area of quadrilateral KDEF should be 3/4 of the total area. Wait, but I need to think carefully about the regions.Alternatively, maybe it's better to consider the area on one side of AK. Let me think of the hexagon as being split into two polygons by AK: one polygon is triangle AKD, and the other is a pentagon AKFEA or something like that.Wait, no. If I draw AK from A to K on DE, the two regions would be triangle AKD and the rest of the hexagon, which is a pentagon. So, the area of triangle AKD is 1/4 of the total area, and the pentagon is 3/4.Therefore, S(AKD) = 1.5 (since total area is 6, 1/4 of that is 1.5). So, the area of triangle AKD is 1.5.Now, how can I relate this to the position of K on DE? Let me recall that in a regular hexagon, the area of triangle ADE is 2. Because each equilateral triangle has area 1, and triangle ADE is made up of two such triangles.Wait, actually, let me confirm that. If the hexagon is divided into six equilateral triangles, each with area 1, then the area of triangle ADE would be 2, since it spans two of those triangles.So, triangle ADE has area 2. Now, triangle AKD is a part of triangle ADE. So, if I can find the ratio of areas between AKD and ADE, that would give me the position of K on DE.Given that S(AKD) = 1.5 and S(ADE) = 2, the ratio of areas is 1.5:2, which simplifies to 3:4. So, the area of AKD is 3/4 of the area of ADE.Since both triangles share the same base AD, the ratio of their areas is equal to the ratio of their heights. But in this case, the base is not the same. Wait, actually, triangle AKD and triangle ADE share the same vertex A, but their bases are KD and DE respectively.Wait, no. Triangle AKD has base KD and height from A, while triangle ADE has base DE and the same height from A. So, the ratio of their areas is equal to the ratio of their bases, KD:DE.But wait, triangle AKD and triangle ADE share the same height from A to DE. Therefore, the ratio of their areas is equal to the ratio of their bases KD:DE.So, since S(AKD)/S(ADE) = 3/4, then KD/DE = 3/4. Therefore, KD = (3/4) DE.But DE is the entire side, so if KD is 3/4 of DE, then KE is 1/4 of DE. Therefore, the ratio DK:KE is 3:1.Wait, that seems straightforward, but let me double-check.If the area of AKD is 3/4 of ADE, which is 2, then AKD is 1.5. That matches the earlier calculation where AKD is 1.5, which is 1/4 of the total hexagon area. So, that seems consistent.Therefore, the ratio DK:KE is 3:1.I think that makes sense. So, the point K divides DE in the ratio 3:1, with DK being three times as long as KE.**Final Answer**The point ( K ) divides the side ( DE ) in the ratio boxed{3:1}.

💡Alright, so Johann has 80 fair coins, and each coin is flipped. If a coin lands on tails, it's flipped again up to three more times, making a total of four flips per coin. We need to find the expected number of coins that show heads after all the tosses are completed. The answer choices are 70, 72, 75, 78, and 80.First, I need to understand the problem clearly. Each coin is flipped up to four times, but only if it lands on tails initially. So, for each coin, we keep flipping it until it lands on heads or until we've flipped it four times. The goal is to find the expected number of heads after all these flips.I remember that expectation can be calculated by finding the probability of an event and then multiplying it by the number of trials. In this case, the event is getting a head on a coin, and the number of trials is 80 coins.So, I need to find the probability that a single coin shows heads after up to four flips. Once I have that probability, I can multiply it by 80 to get the expected number of heads.Let me think about how to calculate this probability. For a single coin, the probability of getting heads on the first flip is 1/2. If it doesn't get heads on the first flip (which is also a probability of 1/2), it gets flipped again. The probability of getting heads on the second flip is then 1/2, but this only happens if the first flip was tails. So, the combined probability for getting heads on the second flip is (1/2) * (1/2) = 1/4.Similarly, if the first two flips are tails, the coin is flipped a third time. The probability of getting heads on the third flip is (1/2)^3 = 1/8. And if the first three flips are tails, the coin is flipped a fourth time, with a probability of (1/2)^4 = 1/16.So, the total probability of getting at least one head in four flips is the sum of these probabilities:P(heads) = 1/2 + 1/4 + 1/8 + 1/16Let me calculate that:1/2 is 8/16,1/4 is 4/16,1/8 is 2/16,1/16 is 1/16.Adding them up: 8/16 + 4/16 + 2/16 + 1/16 = 15/16.So, the probability that a single coin shows heads after up to four flips is 15/16.Now, since there are 80 coins, the expected number of heads is 80 multiplied by 15/16.Let me compute that:80 * (15/16) = (80/16) * 15 = 5 * 15 = 75.Therefore, the expected number of coins showing heads after all the tosses is 75.Wait, let me double-check my calculations to make sure I didn't make a mistake.First, the probability of getting at least one head in four flips is indeed 15/16. That's because the only way a coin doesn't show heads in four flips is if all four flips are tails, which has a probability of (1/2)^4 = 1/16. So, the probability of getting at least one head is 1 - 1/16 = 15/16. That checks out.Then, multiplying 80 by 15/16:80 divided by 16 is 5, and 5 multiplied by 15 is 75. That seems correct.So, I think my answer is right. The expected number of heads is 75.

💡Okay, so I need to prove that any segment of length 1/n on the number line contains no more than (n+1)/2 irreducible fractions of the form p/q, where p and q are integers and 1 ≤ q ≤ n. Hmm, let me think about how to approach this.First, I should recall what an irreducible fraction is. An irreducible fraction is one where the numerator and denominator have no common divisors other than 1. So, for example, 1/2 is irreducible, but 2/4 is not because it can be simplified to 1/2.Now, the problem is about counting how many such irreducible fractions can lie within a segment of length 1/n. The segment can be anywhere on the number line, right? So, it's not fixed; it could be from a to a + 1/n for some real number a.I think it might help to consider the Farey sequence here. The Farey sequence of order n is the set of irreducible fractions between 0 and 1 with denominators ≤ n. Maybe there's a way to use properties of Farey sequences to bound the number of fractions in any interval.Wait, but the segment isn't necessarily between 0 and 1. It could be anywhere on the number line. Hmm, so maybe I need a more general approach.Let me think about the spacing between irreducible fractions. If two fractions are close to each other, their difference must be at least some minimal amount. Maybe I can use the concept of the mediant of two fractions or something related to the Farey sequence's properties.In the Farey sequence, between any two consecutive terms a/b and c/d, the determinant bc - ad = 1. This implies that the difference between them is 1/(bd). So, the minimal distance between two consecutive terms in the Farey sequence is at least 1/(n^2), since b and d are at most n.But wait, in our case, the segment has length 1/n, which is much larger than 1/n^2. So, maybe the number of fractions in such a segment can be bounded by considering how many such minimal distances fit into 1/n.But I'm not sure if that's the right way to go. Maybe I should think about the number of possible denominators. For each denominator q, there are φ(q) numerators p such that p/q is irreducible. So, the total number of irreducible fractions with denominators up to n is the sum of φ(q) from q=1 to n. But that's the total number, not the number in a specific interval.But in our case, we're looking at a specific interval of length 1/n. So, maybe I can bound the number of fractions in such an interval by considering how many fractions with each denominator can lie within it.For a fixed denominator q, the fractions p/q are spaced 1/q apart. So, in an interval of length 1/n, the maximum number of such fractions is roughly (1/n)/(1/q) = q/n. But since q ≤ n, this is at most 1. So, for each q, there can be at most one fraction p/q in the interval.Wait, that seems too restrictive. For example, if q=1, then the fractions are integers, spaced 1 apart. So, in an interval of length 1/n, which is less than 1, there can be at most one integer. Similarly, for q=2, fractions are spaced 1/2 apart, so in an interval of length 1/n, you can have at most one fraction of the form p/2.But wait, if n is small, say n=2, then 1/n=1/2. So, an interval of length 1/2 could contain two fractions like 0 and 1/2, but 0 is an integer, so p=0, q=1, and 1/2 is p=1, q=2. Both are irreducible. So, in this case, the number is 2, which is equal to (2+1)/2=1.5, but since we're dealing with integers, maybe it's 2. Hmm, so the bound might not be tight for small n.But let's think more generally. For each denominator q, the number of fractions p/q in the interval is at most 1, because the spacing is 1/q, and the interval is 1/n. Since q ≤ n, 1/q ≥ 1/n, so the interval can contain at most one such fraction.Therefore, the total number of irreducible fractions in the interval is at most the number of denominators q, which is n. But the problem states that it's at most (n+1)/2. So, my initial thought is too loose.I must be missing something. Maybe the key is to consider that for each denominator q, the number of possible numerators p such that p/q is irreducible is φ(q). But in the interval, for each q, you can have at most one p/q. So, the total number is at most the number of q's, but since we're considering irreducible fractions, maybe we can pair them up in some way.Wait, perhaps I should use the concept of Farey sequences and the mediant property. In Farey sequences, between any two consecutive terms, the mediant has a denominator larger than n. So, maybe in any interval of length 1/n, you can't have too many terms from the Farey sequence.Alternatively, maybe I can use the concept of continued fractions or something related to the distribution of fractions.Wait, another idea: for each fraction p/q in the interval, consider its denominator q. Since q ≤ n, and the interval has length 1/n, the spacing between fractions with denominator q is 1/q. So, in the interval, you can have at most one fraction for each q.But again, that would give a total of n fractions, which is more than (n+1)/2.So, perhaps the key is to consider that for each q, the number of fractions p/q in the interval is at most 1, but also, for each q, there's a corresponding q' such that q' is related to q in a way that limits the total count.Wait, maybe I can use the concept of reciprocal. For each fraction p/q, there's a corresponding fraction q/p, but that might not be directly useful.Alternatively, perhaps I can use the fact that for each q, the number of fractions p/q in the interval is at most 1, but also, for each q, there's a unique p such that p/q is in the interval. So, the total number is at most the number of q's, but we need to show it's at most (n+1)/2.Wait, maybe I can pair the denominators q and n+1 - q or something like that. Let me think.Suppose I consider all denominators q from 1 to n. For each q, there's a unique p such that p/q is in the interval. Now, if I can show that for each pair (q, n+1 - q), there's at most one fraction in the interval, then the total number would be at most (n+1)/2.But I'm not sure if that's the case. Maybe I need to think differently.Another approach: consider that any two irreducible fractions p/q and p'/q' in the interval must satisfy |p/q - p'/q'| < 1/n. Using the property of fractions, |p/q - p'/q'| = |pq' - p'q| / (qq'). Since the fractions are irreducible, the numerator |pq' - p'q| must be at least 1, right? Because if pq' - p'q = 0, then p/q = p'/q', which would mean they are the same fraction.So, |pq' - p'q| ≥ 1. Therefore, |p/q - p'/q'| ≥ 1/(qq'). But since q and q' are ≤ n, qq' ≤ n^2, so |p/q - p'/q'| ≥ 1/n^2.But the interval has length 1/n, so the number of such fractions is at most (1/n) / (1/n^2) = n. Again, that's the same as before, which is too loose.Wait, but maybe I can use a better lower bound for |pq' - p'q|. If p/q and p'/q' are distinct irreducible fractions, then |pq' - p'q| ≥ 1, but actually, it's at least the minimal difference between two distinct fractions with denominators ≤ n.I recall that in Farey sequences, the minimal difference between two consecutive terms is 1/(n(n-1)), but I'm not sure.Alternatively, maybe I can use the concept of the mediant. The mediant of two fractions a/b and c/d is (a+c)/(b+d). If b+d > n, then the mediant is not in the Farey sequence of order n. So, maybe in any interval, the number of fractions is limited by the number of mediants.But I'm not sure how to apply this directly.Wait, another idea: consider that for each denominator q, the fractions p/q are spaced 1/q apart. So, in an interval of length 1/n, the maximum number of such fractions is floor((1/n)/(1/q)) + 1, which is floor(q/n) + 1. Since q ≤ n, q/n ≤ 1, so floor(q/n) is 0 or 1. Therefore, for each q, there can be at most 1 fraction p/q in the interval.Thus, the total number of fractions is at most n, but we need to show it's at most (n+1)/2. So, perhaps we need a different approach.Wait, maybe I can use the concept of the number of Farey points in an interval. There's a theorem that states that the number of Farey points of order n in an interval of length L is at most something like 2nL + 1. But I'm not sure about the exact statement.Alternatively, maybe I can use the concept of the circle method or something from analytic number theory, but that might be too advanced.Wait, perhaps I can use the concept of the Stern-Brocot tree. Each fraction can be represented in the Stern-Brocot tree, and the number of fractions in an interval can be bounded by the number of levels in the tree. But I'm not sure.Alternatively, maybe I can use the concept of the number of solutions to |p/q - a| < 1/n, where a is the center of the interval. Then, rearranging, |p - aq| < q/n. Since p and q are integers, |p - aq| must be less than q/n. But since p is an integer, the distance from aq to the nearest integer is less than q/n. So, the number of such p for each q is at most 1, as before.But again, that gives a total of n, which is too much.Wait, maybe I can consider that for each q, the number of p such that p/q is in the interval is at most 1, but also, for each q, there's a corresponding q' such that q' = n + 1 - q, and the fractions p/q and p'/q' can't both be in the interval. So, for each pair (q, n+1 - q), there's at most one fraction in the interval. Therefore, the total number is at most (n+1)/2.But I'm not sure if that's rigorous.Wait, let's think about it more carefully. Suppose we have a fraction p/q in the interval. Then, consider the fraction (q - p)/q. Is that necessarily in the interval? Not necessarily, because the interval is of length 1/n, and the distance between p/q and (q - p)/q is |p/q - (q - p)/q| = |2p/q - 1|. So, unless 2p/q is close to 1, which would mean p is close to q/2, the distance could be large.But maybe I can use the fact that for each q, the fractions p/q and (q - p)/q are symmetric around 1/2. So, if one is in the interval, the other might not be, depending on where the interval is.But I'm not sure how to make this precise.Wait, maybe I can use the concept of the number of distinct denominators. For each q, there's at most one p such that p/q is in the interval. So, the total number of fractions is at most the number of q's, which is n. But we need to show it's at most (n+1)/2.Wait, perhaps the key is to note that for each q, there's a unique p such that p/q is in the interval, but also, for each q, there's a corresponding q' such that q' = n + 1 - q, and the fractions p/q and p'/q' can't both be in the interval unless they are the same fraction.Wait, that might not be true. For example, if q=1, then q'=n, and the fractions 0/1 and something/n could both be in the interval.But maybe if I consider that for each q ≤ (n+1)/2, there's a corresponding q' = n + 1 - q, and if both q and q' are ≤ n, then we can pair them up, and for each pair, there's at most one fraction in the interval.But I'm not sure.Wait, another idea: consider that for each q, the number of fractions p/q in the interval is at most 1, so the total number is at most n. But we need to show it's at most (n+1)/2. So, perhaps we can show that for each q, there's a unique q' such that q' = n + 1 - q, and if both q and q' are ≤ n, then the fractions p/q and p'/q' can't both be in the interval unless they are the same.But I'm not sure how to formalize this.Wait, maybe I can use the concept of the number of distinct denominators. For each q, there's at most one p such that p/q is in the interval. So, the total number of fractions is at most the number of q's, which is n. But we need to show it's at most (n+1)/2.Wait, perhaps the key is to note that for each q, there's a unique p such that p/q is in the interval, but also, for each q, there's a corresponding q' such that q' = n + 1 - q, and the fractions p/q and p'/q' can't both be in the interval unless they are the same.But I'm not sure.Wait, maybe I can think about the number of fractions with q ≤ (n+1)/2 and q > (n+1)/2 separately. For q ≤ (n+1)/2, each q can have at most one p/q in the interval. For q > (n+1)/2, since q ≤ n, then n + 1 - q < (n+1)/2. So, for each q > (n+1)/2, there's a corresponding q' = n + 1 - q < (n+1)/2. Now, if both q and q' have fractions in the interval, then p/q and p'/q' are both in the interval. But since q' < (n+1)/2, and q > (n+1)/2, maybe their fractions can't both be in the interval unless they are the same.Wait, perhaps if p/q and p'/q' are both in the interval, then |p/q - p'/q'| < 1/n. But since q' = n + 1 - q, maybe we can derive a contradiction unless p/q = p'/q'.But I'm not sure.Wait, let's try to make this precise. Suppose we have two fractions p/q and p'/q' in the interval, where q' = n + 1 - q. Then, |p/q - p'/q'| < 1/n. Let's express this as |pq' - p'q| / (qq') < 1/n. Since q' = n + 1 - q, then qq' = q(n + 1 - q). So, |pq' - p'q| < qq'/n.But qq' = q(n + 1 - q) = q(n + 1) - q^2. Since q ≤ n, qq' ≤ n(n + 1) - q^2. But I'm not sure how to proceed.Wait, maybe I can use the fact that |pq' - p'q| must be at least 1 because p/q and p'/q' are distinct irreducible fractions. So, |pq' - p'q| ≥ 1. Therefore, 1 ≤ |pq' - p'q| < qq'/n. So, qq'/n > 1, which implies qq' > n.But qq' = q(n + 1 - q) = q(n + 1) - q^2. Let's see when qq' > n.We have q(n + 1) - q^2 > n.Rearranging, q^2 - q(n + 1) + n < 0.This is a quadratic in q: q^2 - (n + 1)q + n < 0.The roots of the quadratic equation q^2 - (n + 1)q + n = 0 are q = [ (n + 1) ± sqrt((n + 1)^2 - 4n) ] / 2 = [ (n + 1) ± sqrt(n^2 + 2n + 1 - 4n) ] / 2 = [ (n + 1) ± sqrt(n^2 - 2n + 1) ] / 2 = [ (n + 1) ± (n - 1) ] / 2.So, the roots are q = [ (n + 1) + (n - 1) ] / 2 = (2n)/2 = n, and q = [ (n + 1) - (n - 1) ] / 2 = 2/2 = 1.So, the quadratic is negative between q=1 and q=n. Therefore, for q in (1, n), qq' > n.But q must be an integer between 1 and n. So, for q=2 to q=n-1, qq' > n.Therefore, for q=2 to q=n-1, we have qq' > n, so |pq' - p'q| < qq'/n > 1, which contradicts |pq' - p'q| ≥ 1. Therefore, our assumption that both p/q and p'/q' are in the interval must be false.Thus, for q=2 to q=n-1, if p/q is in the interval, then p'/q' cannot be in the interval, where q' = n + 1 - q.Therefore, for each pair (q, q'), where q=2 to q=(n+1)/2, we can have at most one fraction in the interval.Now, for q=1 and q=n, since q'=n and q'=1 respectively, which are the same as q=n and q=1. So, for q=1, q'=n, and for q=n, q'=1.So, in the interval, we can have at most one fraction with q=1 and at most one fraction with q=n. But since q=1 and q=n are paired, we can have at most one of them in the interval.Wait, but actually, for q=1, the fractions are integers, spaced 1 apart. So, in an interval of length 1/n, which is less than 1, there can be at most one integer. Similarly, for q=n, the fractions are spaced 1/n apart, so in an interval of length 1/n, there can be at most one fraction of the form p/n.But wait, if the interval is exactly of length 1/n, it could contain two fractions p/n and (p+1)/n if the interval is aligned that way. For example, if the interval is [k/n, (k+1)/n), then it contains exactly one fraction p/n. But if the interval is [k/n - ε, k/n + 1/n - ε), for small ε, it could contain two fractions: k/n and (k+1)/n.Wait, no, because the length is exactly 1/n. So, if the interval is [a, a + 1/n), then it can contain at most one fraction p/n, because the distance between consecutive fractions p/n and (p+1)/n is exactly 1/n. So, the interval can't contain both unless it's aligned exactly at p/n, in which case it would contain p/n and possibly p/n itself, but not two distinct fractions.Wait, actually, if the interval is [p/n, p/n + 1/n), then it contains p/n and (p+1)/n if p/n + 1/n > (p+1)/n, which is always true. Wait, no, because (p+1)/n = p/n + 1/n, so the interval [p/n, p/n + 1/n) would include p/n but not (p+1)/n, because (p+1)/n is the endpoint, which is not included in the interval.Wait, actually, it depends on whether the interval is open or closed. If it's closed on the left and open on the right, then [p/n, (p+1)/n) contains p/n but not (p+1)/n. Similarly, if it's [p/n - 1/n, p/n), it contains p/n but not (p-1)/n.But in any case, an interval of length 1/n can contain at most one fraction of the form p/n.Similarly, for q=1, the interval can contain at most one integer.Therefore, for q=1 and q=n, we can have at most one fraction each, but since they are paired, we can have at most one of them in the interval.Wait, but actually, the interval could contain both an integer and a fraction with denominator n, but that would require the interval to contain both k and k + 1/n, which is possible if the interval is [k, k + 1/n). But in that case, the integer k is included, and k + 1/n is the endpoint, which may or may not be included depending on the interval's definition.But in any case, the number of fractions from q=1 and q=n is at most 2, but since they are paired, we can consider them as contributing at most 1 to the total count.Wait, I'm getting confused. Let me try to summarize:- For each q from 2 to n-1, we can have at most one fraction p/q in the interval, and for each such q, there's a corresponding q' = n + 1 - q, and we can have at most one fraction from each pair (q, q').- For q=1 and q=n, we can have at most one fraction each, but since they are paired, we can have at most one of them in the interval.Therefore, the total number of fractions is at most:- 1 (for q=1 or q=n) + sum over q=2 to q=(n+1)/2 of 1.Wait, but if n is even, (n+1)/2 is not an integer. So, maybe it's better to say that the number of pairs (q, q') is floor((n-1)/2), and for each pair, we can have at most one fraction. Plus, for q=1 and q=n, we can have at most one fraction.So, total number is at most 1 + floor((n-1)/2). If n is odd, floor((n-1)/2) = (n-1)/2, so total is 1 + (n-1)/2 = (n+1)/2. If n is even, floor((n-1)/2) = (n-2)/2, so total is 1 + (n-2)/2 = (n)/2 + 1 - 1 = n/2. Wait, but (n+1)/2 is larger than n/2, so maybe the bound is (n+1)/2 regardless.Wait, let me check for n=2:- q=1: can have at most 1 fraction.- q=2: can have at most 1 fraction.Total: 2, which is equal to (2+1)/2=1.5, but since we're dealing with integers, it's 2.For n=3:- q=1: 1 fraction.- q=2: paired with q'=2 (since n+1 - 2=2), so at most 1 fraction.- q=3: paired with q'=1, which we've already counted.Total: 2, which is equal to (3+1)/2=2.For n=4:- q=1: 1 fraction.- q=2: paired with q'=3, so at most 1 fraction.- q=3: paired with q'=2, already counted.- q=4: paired with q'=1, already counted.Total: 2, which is equal to (4+1)/2=2.5, but since we're dealing with integers, it's 2.Wait, but for n=4, the bound is 2.5, but we have 2, which is less. So, maybe the bound is actually floor((n+1)/2).But the problem states (n+1)/2, which could be a non-integer, but since the number of fractions must be an integer, maybe it's the ceiling or floor.But regardless, the key idea is that for each q from 2 to n-1, we can pair them with q' = n + 1 - q, and for each pair, we can have at most one fraction in the interval. Plus, for q=1 and q=n, we can have at most one fraction.Therefore, the total number is at most 1 + floor((n-1)/2). For n odd, this is (n+1)/2. For n even, it's n/2, which is less than (n+1)/2.But the problem states (n+1)/2, so maybe we can just say it's at most (n+1)/2, since for even n, n/2 ≤ (n+1)/2.Therefore, the total number of irreducible fractions in the interval is at most (n+1)/2.So, putting it all together:- For each q from 2 to n-1, pair q with q' = n + 1 - q. For each pair, at most one fraction can be in the interval.- For q=1 and q=n, at most one fraction can be in the interval.- The total number is at most 1 + floor((n-1)/2) ≤ (n+1)/2.Therefore, any segment of length 1/n contains no more than (n+1)/2 irreducible fractions of the form p/q with 1 ≤ q ≤ n.I think that makes sense. I might have missed some edge cases, but overall, the pairing of denominators and the fact that their corresponding fractions can't both be in the interval unless they are the same seems to lead to the desired bound.

💡Okay, so I need to find the ones digit of 3 raised to the 53rd power. Hmm, that seems a bit tricky at first glance because 3 to the 53rd power is a huge number. I don't think I can calculate it directly here, so I need another approach.Maybe I can look for a pattern in the ones digits of powers of 3. Let me start by calculating the first few powers of 3 and noting their ones digits.- 3^1 is 3. So the ones digit is 3.- 3^2 is 9. The ones digit is 9.- 3^3 is 27. The ones digit here is 7.- 3^4 is 81. The ones digit is 1.- 3^5 is 243. The ones digit is 3 again.Wait a minute, I see a pattern here. The ones digits are repeating every four powers: 3, 9, 7, 1, and then back to 3. So it seems like there's a cycle of 4 for the ones digits of powers of 3.If that's the case, then to find the ones digit of 3^53, I can figure out where 53 falls in this cycle. To do that, I can divide 53 by 4 and look at the remainder.Let me do that division: 53 divided by 4. 4 times 13 is 52, so 53 minus 52 is 1. So the remainder is 1.Now, according to the cycle I observed:- If the remainder is 1, the ones digit is 3.- If the remainder is 2, the ones digit is 9.- If the remainder is 3, the ones digit is 7.- If there's no remainder (i.e., the number is divisible by 4), the ones digit is 1.Since the remainder here is 1, the ones digit of 3^53 should be 3.Let me double-check to make sure I didn't make a mistake. I'll calculate a few more powers of 3 to confirm the cycle:- 3^6 is 729. Ones digit is 9.- 3^7 is 2187. Ones digit is 7.- 3^8 is 6561. Ones digit is 1.- 3^9 is 19683. Ones digit is 3.Yep, the pattern repeats every four exponents. So my initial conclusion seems correct. Therefore, the ones digit of 3^53 is indeed 3.

💡Okay, so I have this problem here about a function y = f(x) that's even and defined on all real numbers. When x is greater than 0, the inequality f(x) + x·f'(x) < 0 holds. Then I'm given three expressions: a = 3^{0.2}·f(3^{0.2}), b = (log_{π}2)·f(log_{π}2), and c = (log_{2}(1/4))·f(log_{2}(1/4)). I need to figure out the relationship among a, b, and c.First, let me parse the problem step by step. The function f(x) is even, which means f(-x) = f(x) for all x. That's a key property. Also, for x > 0, we have f(x) + x·f'(x) < 0. Hmm, that looks like the derivative of some product. Let me think: if I have a function g(x) = x·f(x), then the derivative g'(x) would be f(x) + x·f'(x) by the product rule. So, g'(x) = f(x) + x·f'(x). And since this is less than 0 for x > 0, that means g(x) is decreasing when x is positive.So, g(x) = x·f(x) is decreasing for x > 0. Now, since f(x) is even, what can I say about g(-x)? Let's compute it: g(-x) = (-x)·f(-x) = (-x)·f(x) because f is even. So, g(-x) = -x·f(x) = -g(x). That means g(x) is an odd function. So, if g(x) is odd and decreasing for x > 0, then for x < 0, since it's odd, it should also be decreasing. Because if you have an odd function, the behavior on the negative side is a mirror image of the positive side. So, if it's decreasing on the positive side, it's decreasing on the negative side as well.Wait, actually, let me think about that again. If g(x) is odd, then g(-x) = -g(x). So, if g(x) is decreasing for x > 0, then for x < 0, g(x) is also decreasing because as x increases (from more negative to less negative), g(x) increases (since it's negative of a decreasing function). Hmm, maybe I need to be careful here.Alternatively, since g(x) is odd and decreasing for x > 0, then for x < 0, it's also decreasing because the function is symmetric about the origin. So, the function g(x) is decreasing on the entire real line. So, whether x is positive or negative, as x increases, g(x) decreases.Now, let me look at the expressions for a, b, and c.a = 3^{0.2}·f(3^{0.2}) = g(3^{0.2})b = (log_{π}2)·f(log_{π}2) = g(log_{π}2)c = (log_{2}(1/4))·f(log_{2}(1/4)) = g(log_{2}(1/4))So, I need to evaluate g at three different points: 3^{0.2}, log_{π}2, and log_{2}(1/4).First, let me compute these values numerically to understand their magnitudes.Starting with 3^{0.2}. 0.2 is 1/5, so 3^{0.2} is the fifth root of 3. Let me approximate that. 3^0.2 is approximately e^{(ln3)/5} ≈ e^{1.0986/5} ≈ e^{0.2197} ≈ 1.2457. So, 3^{0.2} ≈ 1.2457.Next, log_{π}2. That's the logarithm of 2 with base π. Since π is approximately 3.1416, log_{π}2 is less than 1 because 2 is less than π. To compute it, I can use the change of base formula: log_{π}2 = ln2 / lnπ ≈ 0.6931 / 1.1442 ≈ 0.605. So, log_{π}2 ≈ 0.605.Lastly, log_{2}(1/4). That's the logarithm of 1/4 with base 2. Since 1/4 is 2^{-2}, log_{2}(1/4) = -2.So, now I have the three points:- For a: x = 3^{0.2} ≈ 1.2457- For b: x = log_{π}2 ≈ 0.605- For c: x = log_{2}(1/4) = -2Since g(x) is decreasing on the entire real line, the larger the x, the smaller the g(x). So, the order of the x-values is: -2 < 0.605 < 1.2457. Therefore, since g is decreasing, the order of g(x) is: g(-2) > g(0.605) > g(1.2457). Which translates to c > b > a.Wait, but let me make sure. Since g(x) is decreasing, as x increases, g(x) decreases. So, the smallest x gives the largest g(x). So, x = -2 is the smallest, so g(-2) is the largest. Then x = 0.605 is next, so g(0.605) is next. Then x = 1.2457, so g(1.2457) is the smallest. So, yes, c > b > a.But let me double-check if I got the direction right. Since g is decreasing, higher x means lower g(x). So, the order from largest to smallest g(x) is from smallest x to largest x. So, indeed, c (x=-2) is the largest, then b (x≈0.605), then a (x≈1.2457).Therefore, the relationship is c > b > a, which is option A.I think that's the correct reasoning. I considered the function g(x) = x·f(x), recognized it's odd and decreasing everywhere, computed the values of x for a, b, c, ordered them, and then applied the monotonicity of g(x) to find the relationship.**Final Answer**boxed{A}