💡First, determine how many seashells Vail had by subtracting 5 from Aiguo's total of 20.Next, calculate Stefan's seashells by adding 16 to Vail's total.Finally, sum the seashells of Stefan, Vail, and Aiguo to find the total number of seashells collected by all three children.

💡Okay, so I have this problem here about three skew lines in space, and I need to figure out how many lines can intersect all three of them. The options are 0, 1, more than 1 but finite, or infinite. Hmm, let me think.First, I remember that skew lines are lines that don't lie in the same plane and don't intersect. So, if I have three lines a, b, and c, each pair is skew. That means none of them intersect each other, and they aren't parallel either because parallel lines would lie in the same plane.Now, the question is about lines that intersect all three of these skew lines. I need to visualize this. Maybe I can think of each line as being in a different direction and position in space. Since they're skew, they don't interfere with each other in terms of intersection or being parallel.I think about the concept of transversals. A transversal is a line that intersects two or more lines. So, in this case, we're looking for transversals that intersect all three skew lines. I wonder how many such transversals exist.I recall that in three-dimensional space, given two skew lines, there is exactly one line that intersects both and is perpendicular to both. But here, we have three lines, so it's more complicated.Maybe I can approach this by considering two lines first. Let's say I have lines a and b. Since they're skew, there's a unique common perpendicular line between them. But that line doesn't necessarily intersect line c. So, that might not help directly.Alternatively, maybe I can think about the set of all lines that intersect a and b. Since a and b are skew, the set of lines intersecting both forms a ruled surface, specifically a hyperboloid of one sheet. I think that's right. So, this surface is doubly ruled, meaning it can be generated in two different ways by moving a line in space.Now, if I have a third line c, which is skew to both a and b, how does it interact with this hyperboloid? The line c might intersect the hyperboloid in some points, and each intersection point would correspond to a line that intersects a, b, and c.But how many such points are there? If the line c intersects the hyperboloid, it can intersect it in two points, one point, or none. But since c is skew to both a and b, I think it must intersect the hyperboloid in two points. Therefore, there would be two lines that intersect all three skew lines.Wait, but the answer choices don't include two. They have 0, 1, more than 1 but finite, or infinite. So, maybe I'm missing something here.Let me think again. Maybe the number of lines that intersect all three skew lines isn't just two. Perhaps it's more. I remember something about the number of common transversals for three skew lines. I think it can be up to three or maybe even infinite.Alternatively, maybe it's always infinite. If I imagine moving a line in space such that it intersects all three skew lines, perhaps there are infinitely many ways to do this.Wait, but how? If I fix two lines, a and b, the set of lines intersecting both is a hyperboloid. Then, for each point on line c, there might be a line passing through that point and intersecting both a and b. But since c is skew, it doesn't lie on the hyperboloid, so maybe only a finite number of points on c lie on the hyperboloid.But earlier, I thought it might be two points. So, maybe two lines? But again, the answer choices don't have two.Wait, maybe I'm confusing something. Let me try a different approach. Let's consider the geometry of the situation.If I have three skew lines, I can try to construct lines that intersect all three. Let's say I pick a point on line a, then try to draw a line from there that intersects both b and c. But since b and c are skew, the line from a point on a to b might not intersect c, and vice versa.Alternatively, maybe I can use the concept of the regulus. A regulus is a set of lines that are all skew to each other and lie on a hyperboloid. If three skew lines lie on a regulus, then there is a corresponding set of lines (the other ruling) that intersect all three. So, if a, b, and c lie on a hyperboloid, then the other ruling consists of lines that intersect all three.But wait, are three skew lines always lying on a hyperboloid? I think so, yes. Any three skew lines lie on a unique hyperboloid of one sheet. Therefore, the other ruling of the hyperboloid consists of lines that intersect all three. Since a hyperboloid has infinitely many lines in each ruling, that would mean there are infinitely many lines intersecting all three skew lines.So, that would mean the answer is infinite, which is option D.But let me verify this. If three skew lines lie on a hyperboloid, then yes, the other ruling provides infinitely many transversals. So, that makes sense.Alternatively, I can think about the degrees of freedom. In three-dimensional space, a line is determined by four parameters. To intersect three skew lines, we have three conditions. So, the number of solutions should be infinite, as the degrees of freedom exceed the number of conditions.Wait, actually, each intersection condition imposes two constraints (since a line intersecting another line must satisfy two equations). So, three lines would impose six constraints, but a line in space has four degrees of freedom. That suggests that the system is overdetermined, but in reality, because the lines are skew, the constraints are not all independent.Hmm, maybe that's not the right way to think about it. Instead, considering the hyperboloid approach seems more concrete.So, putting it all together, since three skew lines lie on a hyperboloid, and the other ruling of the hyperboloid provides infinitely many lines intersecting all three, the answer should be infinite, which is option D.**Final Answer**boxed{D}

💡First, I need to determine the total number of women on the flight. Since 65% of the 180 passengers are women, I can calculate this by multiplying 180 by 65%.Next, I need to find out how many of these women are sitting in first class. Given that 15% of the women are in first class, I will multiply the number of women by 15%.Finally, I will round the result to the nearest whole number to determine the exact number of women in first class.

💡First, I need to determine the total number of US states, which is 50.Paul has license plates from 40 different states.To find out what percentage of the total US states Paul has, I divide the number of states he has plates from by the total number of states and then multiply by 100.Next, I calculate the amount Paul earns by multiplying the percentage points by the amount his parents give him per percentage point, which is 2.Finally, I present the calculated amount Paul earns from his parents.

💡First, I need to evaluate the expression (-6 cdot 3 - (-8 cdot -2) + (-7 cdot -5) - 10).I'll start by performing the multiplications according to the order of operations.1. Multiply (-6) by (3): [ -6 cdot 3 = -18 ]2. Multiply (-8) by (-2). Since a negative times a negative is a positive: [ -8 cdot -2 = 16 ]3. Multiply (-7) by (-5). Again, a negative times a negative is a positive: [ -7 cdot -5 = 35 ]Now, substitute these results back into the original expression:[-18 - 16 + 35 - 10]Next, I'll perform the additions and subtractions from left to right.4. Subtract (16) from (-18): [ -18 - 16 = -34 ]5. Add (35) to (-34): [ -34 + 35 = 1 ]6. Finally, subtract (10) from (1): [ 1 - 10 = -9 ]So, the final result of the expression is (-9).

💡Alright, so I have this problem where I need to prove that there's a way to color the numbers from 1 to 1987 using four colors such that no arithmetic progression of 10 terms is all the same color. Hmm, okay. Let me try to break this down.First, I know that an arithmetic progression (AP) is a sequence of numbers where each term after the first is obtained by adding a constant difference. So, for example, 1, 3, 5, 7,... is an AP with a common difference of 2. In this problem, we're dealing with APs that have exactly 10 terms.The set M is {1, 2, 3, ..., 1987}, and we need to color each number in this set with one of four colors. The goal is to ensure that no such 10-term AP is monochromatic, meaning all 10 terms aren't the same color.I remember something about Ramsey Theory, which deals with conditions under which order must appear. Maybe this is related? But I'm not entirely sure. Maybe I should think about it in terms of probability or combinatorics.Let me consider how many possible 10-term APs exist in M. Each AP is determined by its starting term and its common difference. So, for a starting term k, the common difference d must satisfy k + 9d ≤ 1987. That means d can be at most floor((1987 - k)/9). So, the number of possible APs would be the sum over all possible k of the number of possible d for each k. That is, A = sum_{k=1}^{1978} floor((1987 - k)/9). Hmm, that seems a bit complicated, but maybe I can approximate it.If I approximate floor((1987 - k)/9) as (1987 - k)/9, then A ≈ sum_{k=1}^{1978} (1987 - k)/9. This simplifies to (1/9) * sum_{k=1}^{1978} (1987 - k). The sum from k=1 to 1978 of (1987 - k) is the same as the sum from j=9 to 1986 of j, right? Because when k=1, 1987 - 1 = 1986, and when k=1978, 1987 - 1978 = 9. So, it's the sum of integers from 9 to 1986.The sum of integers from 1 to n is n(n+1)/2, so the sum from 9 to 1986 is sum_{j=1}^{1986} j - sum_{j=1}^{8} j. That would be (1986*1987)/2 - (8*9)/2 = (1986*1987)/2 - 36.So, A ≈ (1/9) * [(1986*1987)/2 - 36]. Let me compute that:First, 1986*1987. Let me compute 1986*2000 = 3,972,000. Then subtract 1986*13 = 25,818. So, 3,972,000 - 25,818 = 3,946,182. Then divide by 2: 3,946,182 / 2 = 1,973,091. Subtract 36: 1,973,091 - 36 = 1,973,055. Then divide by 9: 1,973,055 / 9 ≈ 219,228.33.So, A is approximately 219,228.33. But since we approximated floor((1987 - k)/9) as (1987 - k)/9, the actual number of APs is slightly less. So, A < 219,228.33.Now, 4^9 is 262,144. So, A is less than 4^9. That seems important. Why? Because if I consider the number of colorings where a particular AP is monochromatic, it's 4 * (number of APs) * (number of ways to color the rest). But since the number of APs is less than 4^9, maybe we can use the probabilistic method or something like that.Wait, the total number of colorings is 4^1987. The number of colorings where at least one AP is monochromatic is less than 4*A*4^{1987 - 10} = 4*A*4^{1977}. Since A < 4^9, this becomes 4*4^9*4^{1977} = 4^{1987 + 1}. Wait, that can't be right because 4^{1987 + 1} is larger than 4^{1987}, which is the total number of colorings. That suggests that my approach is flawed.Maybe I need to think differently. Perhaps instead of counting all possible APs, I can use a specific coloring strategy. Maybe coloring based on residues modulo some number. For example, if I color each number based on its residue modulo 4, then any AP with difference not divisible by 4 will have all residues, so they won't be monochromatic. But if the difference is divisible by 4, then the entire AP will have the same residue, hence the same color. That's bad because we could have a monochromatic AP.So, maybe mod 4 isn't enough. What if I use a higher modulus? Let's see. If I color based on residues modulo 5, then APs with difference not divisible by 5 will have multiple colors. But APs with difference divisible by 5 will still be monochromatic. Hmm, same problem.Wait, maybe I need to use multiple moduli. Like, color based on residues modulo several numbers. For example, using the Chinese Remainder Theorem, if I have a coloring based on residues modulo 4 and 5, then each number is assigned a pair (a, b) where a is mod 4 and b is mod 5. But that would give me 20 colors, which is more than 4. Hmm, not helpful.Alternatively, maybe I can use a coloring that's periodic with a certain period. If the period is chosen such that any AP of length 10 can't stay within a single color. For example, if the period is 10, then each color repeats every 10 numbers. But an AP with difference 1 would cycle through all colors, so it wouldn't be monochromatic. But an AP with difference 10 would stay within the same color. That's bad again.Wait, maybe if I choose the period to be larger than 10? If the period is, say, 11, then an AP with difference 1 would cycle through 10 different colors, but since the period is 11, it might not complete a full cycle. Hmm, not sure.Alternatively, maybe I can use a coloring based on the highest power of 2 dividing each number. For example, color each number based on the exponent of 2 in its prime factorization. But I'm not sure how that would prevent APs from being monochromatic.Wait, maybe I should think about the problem differently. Since we're dealing with 10-term APs, maybe we can use a coloring that ensures that in any AP of length 10, at least two terms have different colors. How can we ensure that?Perhaps by ensuring that the coloring is such that within any 10 consecutive numbers, there are at least two different colors. But that's too restrictive because APs can have larger differences.Wait, another idea: use a coloring based on the remainder when divided by 10. So, color each number based on n mod 10. Then, any AP with difference not divisible by 10 will have terms with different remainders, hence different colors. But APs with difference divisible by 10 will have the same remainder, hence same color. So, again, that's bad because we could have monochromatic APs.Hmm, maybe instead of using a single modulus, I can combine multiple moduli in a way that any AP of length 10 must hit at least two different colors. For example, using a coloring based on multiple residues such that any AP of length 10 can't stay within a single color class.Wait, I remember something called the van der Waerden theorem, which states that for any given coloring, there exists a monochromatic AP of a certain length. But in this case, we want the opposite: to ensure that no monochromatic AP of length 10 exists. So, maybe we need a coloring that avoids van der Waerden's theorem.But I'm not sure how to apply that directly. Maybe instead, I can use a coloring that's based on a higher-dimensional structure. For example, color each number based on its coordinates in a 4-dimensional grid, ensuring that any AP in one dimension can't stay within a single color.Wait, that might be too abstract. Let me try a different approach. Maybe use a greedy algorithm: color each number one by one, ensuring that no AP of length 10 is monochromatic. But that seems complicated because coloring one number affects many potential APs.Alternatively, maybe use a probabilistic method: show that the probability of a random coloring having a monochromatic AP is less than 1, hence such a coloring exists. That might work.So, let's consider a random coloring where each number is colored independently with one of four colors. The probability that a specific AP is monochromatic is 4 * (1/4)^10 = 4^{-9}. Since there are A APs, the expected number of monochromatic APs is A * 4^{-9}.If we can show that this expectation is less than 1, then there exists a coloring with no monochromatic APs. So, we need A * 4^{-9} < 1.Earlier, I approximated A ≈ 219,228.33. Let's compute 219,228.33 * 4^{-9}.4^9 = 262,144. So, 219,228.33 / 262,144 ≈ 0.836. That's less than 1. So, the expected number of monochromatic APs is less than 1, which implies that there exists at least one coloring with no monochromatic APs.Wait, but I approximated A as 219,228.33, but the actual A is slightly less because I used floor((1987 - k)/9). So, the actual A is less than 219,228.33, meaning that A * 4^{-9} is less than 0.836, which is still less than 1. Therefore, the expectation is less than 1, so such a coloring exists.Okay, that seems to work. So, by the probabilistic method, there exists a 4-coloring of M such that no 10-term AP is monochromatic.But wait, the problem asks to prove that such a coloring exists, not just to show that it's likely. So, maybe I need to construct such a coloring explicitly. Hmm, that's trickier.Alternatively, maybe I can use the Lovász local lemma, which can sometimes show the existence of objects with certain properties without constructing them. The Lovász local lemma states that if each "bad event" (like having a monochromatic AP) is not too dependent on others, then the probability that none occur is positive.In this case, each AP is a bad event, and each AP shares numbers with many other APs. But maybe the dependencies are limited enough that the local lemma applies.Let me recall the symmetric version of the Lovász local lemma: if each bad event is independent of all but D other bad events, and if e * p * (D + 1) ≤ 1, where p is the probability of each bad event, then the probability that none of the bad events occur is positive.Here, p = 4^{-9}, and D is the maximum number of APs that share a common number. How many APs share a common number? For a given number n, how many APs of length 10 include n?Well, an AP is determined by its starting term and difference. For a given n, it can be the k-th term of an AP. So, for each possible difference d, n can be expressed as a + (k-1)d, where a is the starting term. So, for each d, a = n - (k-1)d. Since the AP must have 10 terms, a must satisfy a + 9d ≤ 1987.So, for each d, the number of APs that include n is roughly the number of possible k such that a ≥ 1. That is, for each d, k can range from 1 to floor((n - 1)/d) + 1, but ensuring that a + 9d ≤ 1987.This seems complicated, but maybe we can bound D. For each n, the number of APs containing n is roughly O(n), since for each d, there's at most one AP with difference d containing n. But actually, for each d, there's at most one AP with difference d that includes n as the k-th term. So, the number of APs containing n is roughly the number of possible d such that d divides (n - a) for some a.Wait, maybe it's better to think that for each n, the number of APs containing n is roughly O(n), because for each possible difference d, there's at most one AP with difference d that includes n. But actually, for each d, there are multiple APs that include n, depending on where n is in the AP.Wait, no. For a fixed d, n can be in multiple positions in different APs. For example, n can be the first term, second term, etc., as long as the AP stays within M.So, for each d, the number of APs containing n is roughly the number of positions k such that n = a + (k-1)d, and a ≥ 1, a + 9d ≤ 1987.So, for each d, the number of such k is roughly floor((1987 - n)/d) + 1. But this varies with d.Alternatively, maybe the total number of APs containing n is roughly O(n^2), but that seems too high.Wait, actually, for each d, the number of APs containing n is roughly 1, because for each d, there's at most one AP with difference d that includes n. Because if you fix d, then n can be expressed as a + (k-1)d for some a and k, but a is determined once k is chosen.Wait, no. For a fixed d, n can be in multiple APs with difference d, depending on where it is in the AP. For example, n can be the first term, second term, etc., as long as the AP doesn't exceed M.So, for each d, the number of APs containing n is roughly the number of possible starting positions a such that a ≤ n and a + 9d ≤ 1987. So, a can range from max(1, n - 9d) to n. Therefore, the number of APs with difference d containing n is roughly min(n, 1987 - 9d + 1).This is getting too detailed. Maybe I should instead use an upper bound. For each n, the number of APs containing n is at most 1987, since each AP is determined by its starting term and difference, and n can be in at most one AP for each difference d.But that's probably an overestimate. Maybe a better bound is that each n is in at most 1987 APs, but that's still too high.Wait, actually, for each n, the number of APs containing n is roughly O(n), because for each possible difference d, there's at most one AP with difference d that includes n. So, the number of APs containing n is roughly the number of possible d, which is up to n - 1.But n can be up to 1987, so D is roughly 1987. Then, using the Lovász local lemma, we need e * p * (D + 1) ≤ 1.We have p = 4^{-9} ≈ 1.52587890625e-6, and D ≈ 1987. So, e * p * (D + 1) ≈ 2.718 * 1.52587890625e-6 * 1988 ≈ 2.718 * 1.52587890625e-6 * 2000 ≈ 2.718 * 3.0517578125e-3 ≈ 0.00832, which is less than 1. Therefore, the Lovász local lemma applies, and there exists a coloring with no monochromatic APs.Wait, but I approximated D as 1987, but actually, for each n, the number of APs containing n is much less. For example, for n=1000, the number of APs containing 1000 is roughly the number of possible differences d such that there's an AP with difference d containing 1000. For each d, there's at most one AP with difference d containing 1000, so the number of APs containing 1000 is roughly the number of possible d such that 1000 - 9d ≥ 1, which is d ≤ (1000 - 1)/9 ≈ 111. So, D is roughly 111 for n=1000. Similarly, for smaller n, D is smaller.So, maybe D is roughly O(n) for each n, but in reality, it's much less. So, perhaps D is around 1000 on average, but even then, e * p * D ≈ 2.718 * 1.52587890625e-6 * 1000 ≈ 4.13e-3, which is still less than 1. Therefore, the Lovász local lemma still applies.Therefore, there exists a 4-coloring of M such that no 10-term AP is monochromatic.But wait, the problem asks to prove that such a coloring exists, not just to show that it's likely. So, maybe I need to construct such a coloring explicitly. Hmm, that's trickier.Alternatively, maybe I can use a specific coloring strategy. For example, color each number based on its residue modulo 4, but shift the coloring periodically to avoid long monochromatic APs. Or use a more sophisticated coloring based on multiple residues.Wait, another idea: use a coloring based on the binary representation of the numbers. For example, color each number based on the parity of the number of 1s in its binary representation. But I'm not sure if that would prevent monochromatic APs.Alternatively, maybe use a coloring based on the highest power of 2 dividing each number. For example, color each number based on the exponent of 2 in its prime factorization. But again, I'm not sure.Wait, maybe I can use a coloring based on the remainder when divided by 10, but with a twist. For example, color each number based on (n mod 10) mod 4. That way, any AP with difference not divisible by 10 will have terms with different colors. But APs with difference divisible by 10 will still be monochromatic. Hmm, same problem.Wait, maybe instead of mod 10, use a larger modulus. For example, mod 20. Then, color each number based on (n mod 20) mod 4. But then APs with difference divisible by 20 will still be monochromatic.Hmm, this seems like a dead end. Maybe I need to use a different approach.Wait, going back to the probabilistic method, I showed that the expected number of monochromatic APs is less than 1, so such a coloring exists. That should be sufficient for the proof, even if it doesn't construct the coloring explicitly.So, to summarize:1. The total number of colorings is 4^1987.2. The number of 10-term APs in M is A ≈ 219,228.3. The probability that a random coloring has a monochromatic AP is A * 4^{-9} ≈ 0.836, which is less than 1.4. Therefore, there exists at least one coloring with no monochromatic APs.Hence, such a 4-coloring exists.But wait, I think I made a mistake in step 3. The probability that a specific AP is monochromatic is 4 * (1/4)^10 = 4^{-9}. The expected number of monochromatic APs is A * 4^{-9} ≈ 219,228 * 4^{-9} ≈ 0.836, which is less than 1. Therefore, the probability that there are no monochromatic APs is positive, so such a coloring exists.Yes, that makes sense. So, the proof is complete using the probabilistic method.