🧠:Okay, so I have this problem here where I need to find the number of ways to choose three integers a, b, c from the given list such that a < b < c and their product abc is a multiple of 12. The list provided is: 1, 5, 8, 21, 22, 27, 30, 33, 37, 39, 46, 50. Let me start by understanding the problem thoroughly.First, the main goal is to count all possible triplets (a, b, c) where each is from the list, in increasing order, and their product is divisible by 12. To approach this, I should recall that a number is divisible by 12 if it has at least two factors of 2 (since 4 = 2²) and at least one factor of 3. So, the product abc must have at least two 2s and one 3 in its prime factorization.Alternatively, thinking in terms of divisibility, the product must be divisible by 4 and by 3. So, maybe I can break down the problem into two parts: ensuring divisibility by 4 and divisibility by 3, then use inclusion-exclusion or something similar.But perhaps a better approach is to categorize the numbers in the list based on their prime factors. Let me first list out all the numbers and note their factors related to 2 and 3.Given list:1. 1: factors are none (1 is a unit)2. 5: prime, factors 53. 8: 2³4. 21: 3 × 75. 22: 2 × 116. 27: 3³7. 30: 2 × 3 × 58. 33: 3 × 119. 37: prime, 3710. 39: 3 × 1311. 46: 2 × 2312. 50: 2 × 5²Now, let's categorize each number based on their divisibility by 2 and 3.Divisibility by 2:- Numbers divisible by 2: 8, 22, 30, 46, 50- Numbers not divisible by 2: 1, 5, 21, 27, 33, 37, 39Divisibility by 3:- Numbers divisible by 3: 21, 27, 30, 33, 39- Numbers not divisible by 3: 1, 5, 8, 22, 37, 46, 50Now, for the product to be divisible by 12, the product must have at least two factors of 2 and at least one factor of 3.So, the triplet must include:1. At least two numbers divisible by 2 (to get 2²), or one number divisible by 4 (since 4 contributes two 2s). But 8 is divisible by 8, which is 2³, so even better.2. At least one number divisible by 3.Alternatively, if a triplet has:- One number divisible by 4 (like 8) and one number divisible by 2 (since 4 contributes two 2s and another 2 would make it three, but maybe just needing two 2s, so 4 and any even number), and at least one number divisible by 3.Or, two numbers divisible by 2 (even numbers) and one number divisible by 3. Because two even numbers can contribute at least two factors of 2 (depending on their exact factors), but need to check if two even numbers can give at least 2².Wait, if a number is divisible by 2 but not by 4, it contributes one factor of 2. If a number is divisible by 4, it contributes at least two factors. So, to get at least two factors of 2 in total, the triplet must either have:- At least one number divisible by 4 and at least one other even number (divisible by 2), or- Two numbers divisible by 2 (each contributing at least one factor of 2, but if both are only divisible by 2, not 4, then total factors of 2 would be 2, which is sufficient). Wait, but if we have two numbers divisible by 2, even if both are only divisible by 2, then together they contribute two factors of 2, which is enough for divisibility by 4. However, that's not exactly correct. Wait, if you have two numbers each with one factor of 2, then the product has 2*2=4, which is 2². So actually, two even numbers, each divisible by 2 but not necessarily 4, would suffice for the 2² part. However, if you have one number divisible by 4 (which is 2²) and another even number (divisible by 2), that also gives 2² * 2 = 2³, which is more than enough. Similarly, a number divisible by 8 (which is 2³) alone can provide the needed two factors of 2, but actually, even more. But even so, since 8 is 2³, if we include 8, then even if the other numbers are odd, the total number of 2s would be 3, which is sufficient. So, the key is that the product must have at least two 2s and at least one 3.Therefore, there are two cases for the factors of 2:Case 1: The triplet includes at least two even numbers (regardless of their exact power of 2). Because even numbers contribute at least one 2 each, so two evens would give at least two 2s. However, wait, if both are only divisible by 2 (not 4), then 2*2=4, which is 2², which is okay. If one is divisible by 4 and another by 2, then 2²*2=2³, which is still okay. If there's a number divisible by 8 (like 8 itself), then even one such number gives 2³, so if you have 8 and another even number, that's more than enough.Case 2: The triplet includes a single even number that's divisible by 4 (i.e., contributes at least two 2s). Wait, but in that case, if we have only one even number which is divisible by 4, then the total factors of 2 would be at least two, so that's sufficient. However, is that correct? Let's see: if the triplet has only one even number, which is divisible by 4 (like 8), then that contributes 2², which is sufficient for the 2² requirement, and the other two numbers can be odd. Then, as long as one of the three numbers is divisible by 3, the product is divisible by 12. Alternatively, if the triplet has one number divisible by 4 and another even number (divisible by 2), then even if the third number is odd, the total factors of 2 would be 2² * 2 = 2³, but the requirement is just 2².Therefore, the two cases for satisfying the 2² requirement are:1. At least two even numbers (each contributing at least one 2), regardless of whether any are divisible by 4.2. Exactly one even number, but that even number is divisible by 4 (so it contributes two 2s).Wait, but actually, if you have one even number divisible by 4, that gives two 2s, which is sufficient. So, the cases can be broken down into:- Either two or three even numbers (since with three even numbers, you have more 2s, but it's still okay), or one even number that's divisible by 4.But wait, if you have one even number divisible by 4, that's sufficient for the 2² part. So, combining both possibilities, the total number of triplets satisfying the 2² condition is:Number of triplets with at least two even numbers + number of triplets with exactly one even number which is divisible by 4.But we have to be careful not to double-count. Wait, no, because in the first case (at least two even numbers), even if one of them is divisible by 4, it's still counted there, and the second case (exactly one even number divisible by 4) is separate. So they are mutually exclusive.Therefore, total ways for the 2² requirement:Triplets with ≥2 even numbers + triplets with exactly 1 even number (which is divisible by 4).Now, let's compute these.First, let's note the even numbers in the list: 8, 22, 30, 46, 50. So there are 5 even numbers. Among these, numbers divisible by 4 are: 8 (since 8 is 2³), and 50 is 2 × 5², so not divisible by 4. 22 is 2 × 11, 30 is 2 × 3 × 5, 46 is 2 × 23. So only 8 is divisible by 4. So, the numbers divisible by 4 are just [8].Therefore, the second case (exactly one even number, which is divisible by 4) would be choosing 8 and two odd numbers. But the odd numbers are 1, 5, 21, 27, 33, 37, 39. So 7 odd numbers.But then, we need to also ensure that the product is divisible by 3. So, even if we have the 2² from 8, we need at least one multiple of 3 in the triplet.Therefore, when calculating the second case (one even number divisible by 4, which is 8, and two odd numbers), we need to subtract the cases where neither of the two odd numbers is divisible by 3.Similarly, for the first case (triplets with ≥2 even numbers), we need to ensure that at least one of the numbers (either in the even numbers or the remaining numbers) is divisible by 3.So, this problem requires combining the two conditions: sufficient factors of 2 and at least one factor of 3.Therefore, perhaps a better approach is to compute all possible triplets and subtract those that don't meet the 12 divisibility. But since the list has 12 numbers, the total number of triplets is C(12,3) = 220. Then, subtract the triplets that are not divisible by 12. But maybe that's more complicated because you have to consider both factors of 2 and 3.Alternatively, break it down into cases based on the number of even numbers and the presence of multiples of 3.But let's try the constructive approach: count all triplets that have either:1. At least two even numbers and at least one multiple of 3.2. Exactly one even number (which is 8, the only one divisible by 4) and at least one multiple of 3 among the three numbers.Then, sum these two cases.But we need to ensure that in case 1, the triplet can have two or three even numbers. So let's compute each part.First, let's handle case 1: triplets with at least two even numbers and at least one multiple of 3.Number of such triplets = total triplets with at least two even numbers - triplets with at least two even numbers and no multiples of 3.Similarly, for case 2: triplets with exactly one even number (which is 8) and at least one multiple of 3.So, let's compute each step.First, let's find the number of triplets with at least two even numbers.Number of even numbers: 5 (8,22,30,46,50)Number of odd numbers: 7 (1,5,21,27,33,37,39)Triplets with at least two even numbers are:C(5,2)*C(7,1) + C(5,3)= 10*7 + 10 = 70 + 10 = 80.So 80 triplets with at least two even numbers.But among these, some may not have any multiples of 3. So we need to subtract those.So, how many triplets with at least two even numbers have no multiples of 3?First, note that among the even numbers, 30 is divisible by 3 (since 30 = 2×3×5). The other even numbers: 8,22,46,50 are not divisible by 3.So, in the even numbers, only 30 is divisible by 3. The rest are not.Therefore, to have a triplet with at least two even numbers and no multiples of 3, we need to choose all even numbers from the non-3-divisible evens and the third number (if it's a triplet with two evens and one odd) must also be a non-3-divisible odd.So, let's compute that.First, the even numbers not divisible by 3: 8,22,46,50. So 4 even numbers.The odd numbers not divisible by 3: 1,5,37. So 3 odd numbers (since 21,27,33,39 are divisible by 3).Therefore, triplets with at least two even numbers and no multiples of 3:= [C(4,2)*C(3,1) + C(4,3)]= (6*3) + 4 = 18 + 4 = 22.Therefore, the number of triplets with at least two even numbers and at least one multiple of 3 is 80 - 22 = 58.Now, moving to case 2: triplets with exactly one even number (which must be 8, since it's the only even number divisible by 4) and at least one multiple of 3.Number of such triplets: we choose 8 and two odd numbers, with at least one of those two odd numbers being divisible by 3.Total number of triplets with 8 and two odd numbers: C(7,2) = 21.Number of triplets with 8 and two odd numbers not divisible by 3: C(3,2) = 3 (since there are 3 odd numbers not divisible by 3: 1,5,37).Therefore, the number of triplets with 8 and two odd numbers where at least one is divisible by 3 is 21 - 3 = 18.Therefore, total valid triplets are case1 + case2 = 58 + 18 = 76.Wait, but let me verify if this is correct.First, case1: 58 triplets with at least two even numbers and at least one multiple of 3.Case2: 18 triplets with exactly one even number (8) and two odd numbers, at least one of which is a multiple of 3.Total: 58 + 18 = 76.But let's cross-check with another method.Alternatively, compute the total number of triplets that satisfy the divisibility by 12 by considering the necessary conditions.The product must be divisible by 12, so:- At least two factors of 2 and at least one factor of 3.Therefore, possible scenarios:1. Two even numbers and one multiple of 3.But need to ensure that in these two even numbers, they contribute at least two factors of 2. Wait, but if you have two even numbers, each contributing at least one 2, then total 2s are two. However, if one of them is divisible by 4 (i.e., contributes two 2s), then even one such number would suffice. But in the case where we have two even numbers, regardless of their divisibility by 4, the total 2s are at least two. Because even if both are only divisible by 2, 2*2=4. So the total 2s would be two. If one is divisible by 4, then total 2s would be three. So either way, the 2² requirement is met.But then, additionally, we need at least one multiple of 3 in the triplet. So, in this case, triplets with two even numbers and one multiple of 3 (could be among the two even numbers or the third number).Wait, but 30 is the only even number divisible by 3. So, if in the two even numbers, one of them is 30, then that contributes both a 2 and a 3. So in that case, even if the third number is not a multiple of 3, it's still okay. Similarly, if the two even numbers are not 30, then the third number must be a multiple of 3.But this complicates the counting. Alternatively, perhaps the initial approach is better.Wait, in case1, we considered triplets with at least two even numbers and at least one multiple of 3. This includes all triplets where:- There are two even numbers and one multiple of 3 (could be the third number or one of the two even numbers if it's 30).- There are three even numbers, and at least one of them is 30, or the third number (but wait, if there are three even numbers, all three are even, but 30 is the only even number divisible by 3. So, if three even numbers include 30, then the product has a multiple of 3. If three even numbers are 8,22,46,50 (excluding 30), then the triplet would have three even numbers but no multiples of 3, which would not satisfy the 3 condition.Therefore, in the triplets with three even numbers, only those that include 30 would satisfy the multiple of 3 condition. So perhaps my initial calculation overcounts?Wait, no. In the initial calculation for case1, we calculated all triplets with at least two even numbers (including three even numbers) and subtracted those with at least two even numbers and no multiples of 3. So that should correctly account for the overlap. Let me verify.Total triplets with at least two even numbers: 80.Triplets with at least two even numbers and no multiples of 3: 22.So 80 - 22 = 58, which should be the number of triplets with at least two even numbers and at least one multiple of 3. This includes:- All triplets with two even numbers where at least one is 30 or the third number is a multiple of 3.- All triplets with three even numbers where at least one is 30.Therefore, this should be correct.Similarly, case2: triplets with exactly one even number (8) and two odd numbers, at least one of which is a multiple of 3. That's 18.Total 58 + 18 = 76.But let's cross-verify by another method. Let's compute the total number of valid triplets by considering all possibilities where the product is divisible by 12.Another approach: the total number of triplets is C(12,3) = 220.Now, subtract the triplets that are not divisible by 12. These are triplets that either:1. Don't have enough factors of 2 (i.e., fewer than two factors of 2), or2. Don't have a factor of 3.But using inclusion-exclusion: the number of triplets not divisible by 12 is equal to (number of triplets with insufficient 2s) + (number of triplets with insufficient 3s) - (number of triplets with both insufficient 2s and insufficient 3s).First, compute number of triplets with insufficient 2s: these are triplets with zero or one even number, but even here, if the triplet has one even number divisible by 4, then it's sufficient. Wait, so actually, the triplets with insufficient 2s are those that have:- Zero even numbers, or- One even number not divisible by 4.Because if you have one even number divisible by 4, then it contributes two 2s, which is sufficient.Therefore, the number of triplets with insufficient 2s is:- Triplets with zero even numbers: C(7,3) = 35.- Triplets with one even number not divisible by 4: the even numbers not divisible by 4 are 22,30,46,50 (since 8 is divisible by 4). Wait, 30 is 2 × 3 × 5, so divisible by 2 but not 4. Similarly, 22 is 2 × 11, 46 is 2 × 23, 50 is 2 × 5². So all except 8 are divisible by 2 but not 4. Wait, no: 8 is divisible by 4, the others (22,30,46,50) are divisible by 2 but not 4. So there are 4 even numbers not divisible by 4 (22,30,46,50) and 1 even number divisible by 4 (8).Therefore, triplets with one even number not divisible by 4 and two odd numbers: C(4,1)*C(7,2) = 4 * 21 = 84.But wait, the odd numbers are 7, but we need to subtract the ones that are multiples of 3? No, in this case, we are just counting the triplets with insufficient 2s, regardless of the 3s.Therefore, insufficient 2s: 35 (zero even) + 84 (one even not divisible by 4) = 119.Now, number of triplets with insufficient 3s: these are triplets with no multiples of 3. The multiples of 3 in the list are 21,27,30,33,39. So 5 numbers. Therefore, non-multiples of 3 are 12 - 5 = 7 numbers: 1,5,8,22,37,46,50.So triplets with no multiples of 3: C(7,3) = 35.Now, the intersection: triplets with both insufficient 2s and insufficient 3s. These are triplets that have no multiples of 3 and also insufficient 2s (either zero even or one even not divisible by 4).First, the non-multiples of 3 are 1,5,8,22,37,46,50.Among these, the even numbers are 8,22,46,50. Of these, 8 is divisible by 4, the others (22,46,50) are divisible by 2 but not 4.So, in the non-multiples of 3, the even numbers not divisible by 4 are 22,46,50 (three numbers), and the even number divisible by 4 is 8. The other numbers (1,5,37) are odd.Therefore, triplets with insufficient 2s and no multiples of 3 are:- Triplets with zero even numbers from the non-3-multiples: C(3,3) = 1 (since the odd non-3s are 1,5,37).- Triplets with one even number not divisible by 4 (from non-3-multiples: 22,46,50) and two odd numbers (from non-3-multiples: 1,5,37). So C(3,1)*C(3,2) = 3*3=9.So total intersection: 1 + 9 = 10.Therefore, by inclusion-exclusion, triplets not divisible by 12 = 119 + 35 - 10 = 144.Therefore, triplets divisible by 12 = total triplets - 144 = 220 - 144 = 76.This matches our previous result of 76. So this cross-verification confirms that the answer is 76.But to make sure, let's re-express the steps.First approach (constructive):Case1: At least two even numbers and at least one multiple of 3: 58Case2: Exactly one even number (8) and at least one multiple of 3 in the triplet: 18Total: 58 + 18 = 76.Second approach (subtractive):Total triplets: 220Triplets not divisible by 12: 144 (computed as 119 + 35 -10)220 - 144 = 76.Therefore, the answer is 76.But let me check once more to ensure no miscalculations.First, total triplets: C(12,3)=220. Correct.Triplets with insufficient 2s: 35 (zero even) + 84 (one even not divisible by 4) = 119. Correct.Triplets with insufficient 3s: C(7,3)=35. Correct.Intersection: triplets with insufficient 2s and insufficient 3s: 10. Correct.Thus, 119 +35 -10=144. 220-144=76. Correct.Therefore, the final answer is 76.Final Answerboxed{76}

✅:To determine the number of ways to choose integers (a, b, c) with (a < b < c) from the list (1, 5, 8, 21, 22, 27, 30, 33, 37, 39, 46, 50) such that the product (abc) is a multiple of 12, we need to ensure the product has at least two factors of 2 and one factor of 3. Key Steps:1. Categorize Numbers: - Even numbers (divisible by 2): 8, 22, 30, 46, 50 (5 numbers) - Numbers divisible by 3: 21, 27, 30, 33, 39 (5 numbers) - Numbers divisible by 4: 8 (1 number)2. Case Analysis: - Case 1: Triplets with at least two even numbers and at least one multiple of 3. - Total triplets with at least two even numbers: ( binom{5}{2} binom{7}{1} + binom{5}{3} = 70 + 10 = 80 ) - Subtract triplets with no multiples of 3: ( binom{4}{2} binom{3}{1} + binom{4}{3} = 18 + 4 = 22 ) - Valid triplets: ( 80 - 22 = 58 ) - Case 2: Triplets with exactly one even number (8) and at least one multiple of 3. - Total triplets with 8 and two odd numbers: ( binom{7}{2} = 21 ) - Subtract triplets with no multiples of 3: ( binom{3}{2} = 3 ) - Valid triplets: ( 21 - 3 = 18 )3. Total Valid Triplets: - Sum of valid triplets from both cases: ( 58 + 18 = 76 )4. Verification using Inclusion-Exclusion: - Total triplets: ( binom{12}{3} = 220 ) - Triplets not divisible by 12: ( 119 + 35 - 10 = 144 ) - Valid triplets: ( 220 - 144 = 76 ) Final Answer:[boxed{76}]

🔑:To determine the number of ways to choose integers (a), (b), and (c) from the given list so that (a < b < c) and (abc) is a multiple of 12, we need to analyze the divisibility properties of the integers in the list. The list of integers is:[ 1, 5, 8, 21, 22, 27, 30, 33, 37, 39, 46, 50 ]A product to be a multiple of 12 must include at least one factor of 3 and two factors of 2. Therefore, let's categorize the given numbers based on their divisibility by 2 and 3:- (A): Numbers not divisible by 2 or 3: [ { 1, 5, 37 } ]- (B): Numbers divisible by 2, but not by 4 or 3: [ { 22, 46, 50 } ]- (C): Number divisible by 4 but not by 3: [ { 8 } ]- (D): Numbers divisible by 3 but not by 2: [ { 21, 27, 33, 39 } ]- (E): Number divisible by 2 and 3 (i.e., divisible by 6): [ { 30 } ]We consider two main cases: when 30 is chosen and when it is not chosen. Case 1: 30 is chosenWhen 30 is chosen, (a, b,) or (c = 30), only one of the remaining chosen numbers needs to include an additional factor of 2 because 30 already provides one factor of both 2 and 3.# Sub-case 1.1: Both remaining numbers are evenChoose 2 of the 4 remaining even numbers from (B) and (C):[{ 22, 46, 50, 8 }]The number of ways to choose 2 out of 4:[binom{4}{2} = 6]# Sub-case 1.2: One remaining number is even, and the other is oddChoose 1 even number from (B) or (C):[{ 22, 46, 50, 8 }]Number of ways to choose:[4]The remaining chosen number must be odd; it can be from (A) or (D):[{ 1, 5, 37, 21, 27, 33, 39 }]Number of ways to choose:[7]Total for this sub-case:[4 times 7 = 28]Combining both sub-cases of Case 1:[6 + 28 = 34] Case 2: 30 is not chosenOne or two of the chosen numbers must be from (D) to ensure the product has at least one factor of 3.# Sub-case 2.1: Two of the chosen numbers are from (D)Choose 2 out of 4 from (D):[binom{4}{2} = 6]The remaining number must be a multiple of 4 to provide two factors of 2. Only choice:[8]Total for this sub-case:[6]# Sub-case 2.2: One of the chosen numbers is from (D)Choose 1 from (D):[4 text{ ways}]We need to choose the remaining two numbers:1. One multiple of 4 and another even number from (B):[ (8) ] and [ { 22, 46, 50 } ]Number of ways to choose the even number from (B):[3]Total for this sub-case:[4 times 3 = 12]2. One multiple of 4 and an odd number from (A):[ (8) ] and [ { 1, 5, 37 } ]Number of ways to choose the odd number from (A):[3]Total for this sub-case:[4 times 3 = 12]3. Both non-4 multiples even numbers from (B):[binom{3}{2} = 3]Total for this sub-case:[4 times 3 = 12]Combining all sub-cases of Case 2:[6 + 12 + 12 + 12 = 42] Total Number of Ways:Adding the results from both main cases:[34 + 42 = 76]Conclusion:[boxed{76}]

🧠:Okay, so I have this geometry problem here. Let me try to figure it out step by step. The problem is about point A(1,2) and a line that passes through (5, -2) and intersects the parabola y² = 4x at two other points B and C. We need to determine the nature of triangle ABC—whether it's acute, obtuse, right-angled, or indeterminate. The options are A, B, C, D accordingly.First, I need to visualize this. There's a parabola y² = 4x, which I know is a right-opening parabola with vertex at the origin. The point A is (1,2), which is on the parabola because if x=1, y²=4*1=4, so y=±2. So point A is actually on the parabola. Wait, but the line passes through (5, -2) and intersects the parabola at B and C. So the line passes through (5, -2) and cuts the parabola at two points B and C. But since point A is on the parabola, could it be one of the intersection points? The problem says "two other points B and C," so A is not one of them. So the line intersects the parabola at B and C, which are different from A. Then, triangle ABC is formed by these three points.So, to determine the nature of triangle ABC, we need to check the angles. The options are about the angles: right, acute, obtuse, or indeterminate. So, maybe we need to compute the lengths of the sides and then check the angles using the dot product or the Pythagorean theorem.Let me outline the steps:1. Find the equation of the line passing through (5, -2). But wait, the problem says "a line passing through (5, -2)" intersects the parabola at B and C. However, there are infinitely many lines passing through (5, -2). Each such line would intersect the parabola at two points B and C. But how do we determine the specific line? Wait, maybe the problem is saying that for any such line, triangle ABC has a certain nature? Or is there a specific line? Wait, the problem is stated as: "a line passing through (5, -2) intersects the parabola y² = 4x at two other points B and C". Then, given this, determine the nature of triangle ABC. Wait, but the line is arbitrary? Or is there a specific line? The problem doesn't specify, which is confusing. Wait, maybe there is a typo or something missing. Wait, no, let me read again.Original problem: "Given point A(1,2) and a line passing through (5,-2) intersects the parabola y² = 4x at two other points B and C. Determine the nature of triangle △ABC." So the line is passing through (5, -2) and intersects the parabola at B and C, which are two other points besides A? Wait, but A is (1,2), which is on the parabola. So if the line passes through (5, -2) and intersects the parabola at B and C, which are two points different from A. Then, triangle ABC is formed by these three points. But the line intersects the parabola at B and C, so A is not on the line? Because if the line passes through (5, -2) and intersects the parabola at B and C, unless A is also on the line, which would mean three intersection points, but a line and a parabola can intersect at maximum two points. So since the problem states "two other points B and C," that implies that A is not on the line. So the line passes through (5, -2) and intersects the parabola at B and C, which are distinct from A. Therefore, the three points A, B, C form a triangle. Then, the question is about the nature of this triangle.But here's the confusion: how can the line passing through (5, -2) be arbitrary? If the line is arbitrary, then the nature of triangle ABC could vary depending on the line. But the problem is presented as a single question with options, so perhaps there is a unique answer regardless of the line, or there's a specific line intended. Wait, maybe there's a misinterpretation here. Let me check the problem again.Wait, the problem says: "a line passing through (5,-2) intersects the parabola y² = 4x at two other points B and C". The wording is ambiguous. Is the line passing through (5, -2) and point A(1,2)? But then that would mean the line connects (5, -2) and (1,2), but then intersecting the parabola at two other points? Wait, but (1,2) is already on the parabola. So if the line passes through (5, -2) and (1,2), then it would intersect the parabola at A(1,2) and another point. So that would be only one other point. But the problem says "two other points B and C". Therefore, the line cannot pass through A. Hence, the line passes through (5, -2) but not through A(1,2), and intersects the parabola at two distinct points B and C. Then, triangle ABC is formed by A, B, C. Since A is not on the line, the three points are not colinear, so they form a triangle.But the problem is to determine the nature of triangle ABC. The options are acute, obtuse, right-angled, or indeterminate. Since the line is arbitrary (any line passing through (5, -2) and intersecting the parabola at B and C), perhaps the answer is always the same, regardless of the line. So maybe no matter which line we choose through (5, -2), the triangle ABC is always right-angled or something else. So the answer would be one of the options regardless of the line. Therefore, we need to check if for any line passing through (5, -2) intersecting the parabola at B and C, triangle ABC has a fixed nature.Alternatively, maybe there's a specific line intended here, but the problem doesn't specify. That seems unlikely. The problem must be designed so that regardless of the line, the triangle is always a certain type.Alternatively, maybe there's a parametrization approach where we can parametrize the line passing through (5, -2) with a certain slope, find points B and C, then compute the vectors AB, AC, BC, and check the angles.So let's try to approach this step by step.First, let's parametrize the line passing through (5, -2). Let's suppose the line has a slope m. Then, the equation of the line is:y - (-2) = m(x - 5)Which simplifies to y = m(x - 5) - 2Alternatively, since the line passes through (5, -2), we can write parametric equations.But perhaps substituting into the parabola equation is the way to go. The parabola is y² = 4x.So if the line is y = m(x - 5) - 2, substitute this into the parabola equation:[m(x - 5) - 2]^2 = 4xExpanding the left-hand side:[m(x - 5) - 2]^2 = m²(x - 5)² - 4m(x - 5) + 4Therefore:m²(x² -10x +25) -4m(x -5) +4 = 4xExpanding further:m²x² -10m²x +25m² -4mx +20m +4 =4xBring all terms to one side:m²x² -10m²x +25m² -4mx +20m +4 -4x =0Combine like terms:m²x² + (-10m² -4m -4)x +25m² +20m +4 =0This is a quadratic equation in x. Since the line intersects the parabola at two points B and C, the solutions to this quadratic will give the x-coordinates of B and C. Let's denote these roots as x₁ and x₂.But we know that point A is (1,2). Let me check if x=1 is a root of this quadratic equation. If we substitute x=1 into the quadratic:m²(1)² + (-10m² -4m -4)(1) +25m² +20m +4 = m² -10m² -4m -4 +25m² +20m +4Simplify:(m² -10m² +25m²) + (-4m +20m) + (-4 +4) =16m² +16m +0=16m² +16mFor x=1 to be a root, this must equal zero:16m² +16m =0 →16m(m +1)=0 →m=0 or m=-1But the problem states that the line intersects the parabola at two other points B and C, meaning that point A is not on the line. Therefore, x=1 should not be a root. Therefore, 16m² +16m ≠0 →m(m +1) ≠0 →m≠0 and m≠-1Therefore, lines with slopes m=0 or m=-1 would pass through point A(1,2), but since the problem states the line intersects the parabola at two other points B and C, we must exclude these slopes. Therefore, m≠0 and m≠-1.But for other slopes, the line does not pass through A, so B and C are different from A. Therefore, for any m≠0 and m≠-1, we have two intersection points B and C. Then, triangle ABC is formed by points A(1,2), B(x₁,y₁), and C(x₂,y₂). We need to analyze the angles of this triangle.To determine the nature of the triangle, we can compute the lengths of the sides and then use the Law of Cosines or check the dot products between vectors to see if any angle is 90°, greater than 90°, or all less than 90°.Alternatively, maybe there's a geometric property or algebraic identity that can be leveraged here.Given that the problem is likely designed to have a consistent answer regardless of the line (as otherwise the answer would be indeterminate), perhaps triangle ABC is always right-angled.Let me try to verify this.First, let's find the coordinates of points B and C in terms of m. From the quadratic equation above:Quadratic in x: m²x² + (-10m² -4m -4)x +25m² +20m +4 =0Let me denote the quadratic as Ax² + Bx + C =0, where:A = m²B = -10m² -4m -4C =25m² +20m +4The roots x₁ and x₂ can be found using quadratic formula:x = [10m² +4m +4 ± sqrt(( -10m² -4m -4 )² -4*m²*(25m² +20m +4 ))]/(2m²)But this seems complicated. Alternatively, perhaps we can use Vieta's formulas:x₁ + x₂ = -B/A = (10m² +4m +4)/m² =10 + (4m +4)/m²x₁*x₂ = C/A = (25m² +20m +4)/m² =25 + (20m +4)/m²Similarly, the corresponding y-coordinates for B and C would be y₁ = m(x₁ -5) -2 and y₂ = m(x₂ -5) -2Now, points B(x₁, y₁) and C(x₂, y₂), and point A(1,2). We can compute vectors AB and AC, then check the angle at A, or compute all sides and check the angles.But this might be tedious. Alternatively, perhaps there's a reflection property or parametric approach.Alternatively, consider parametric equations of the parabola. The parabola y²=4x can be parametrized as (t², 2t). Let me confirm:If x = t², then y² =4x =4t² → y=±2t. So parametrization can be (t², 2t) for parameter t. Similarly, another parametrization is (a², 2a).Therefore, points B and C can be represented as (t₁², 2t₁) and (t₂², 2t₂). Then, the line passing through (5, -2) and these two points.But the line passes through (5, -2), so the equation of the line passing through (5, -2) and (t², 2t) can be found. But since the line passes through both (t₁², 2t₁) and (t₂², 2t₂), the slope between these two points should be the same as the slope between (5, -2) and either of them.Alternatively, using the parametrization approach, let's consider points B(t₁², 2t₁) and C(t₂², 2t₂) on the parabola. The line joining B and C passes through (5, -2). Therefore, the point (5, -2) lies on the line BC. So, we can write the equation of line BC and substitute (5, -2) into it.The equation of line BC can be written using the two-point form. The slope of BC is (2t₂ - 2t₁)/(t₂² - t₁²) = 2(t₂ - t₁)/[(t₂ - t₁)(t₂ + t₁)] = 2/(t₁ + t₂)Therefore, the slope is 2/(t₁ + t₂). The equation of line BC is:y - 2t₁ = [2/(t₁ + t₂)](x - t₁²)Since point (5, -2) lies on this line, substituting x=5, y=-2:-2 -2t₁ = [2/(t₁ + t₂)](5 - t₁²)Multiply both sides by (t₁ + t₂):(-2 -2t₁)(t₁ + t₂) = 2(5 - t₁²)Expand left side:-2(t₁ + t₂) -2t₁(t₁ + t₂) = -2t₁ -2t₂ -2t₁² -2t₁t₂Right side: 10 - 2t₁²Set them equal:-2t₁ -2t₂ -2t₁² -2t₁t₂ =10 -2t₁²Simplify both sides by adding 2t₁²:-2t₁ -2t₂ -2t₁t₂ =10Divide both sides by -2:t₁ + t₂ + t₁t₂ = -5So we get an equation relating t₁ and t₂: t₁ + t₂ + t₁t₂ = -5This is a key equation. Let's note this.Our goal is to find the nature of triangle ABC, where A is (1,2), which corresponds to parameter t such that 2t =2 → t=1. So point A is (1², 2*1) = (1,2). Therefore, in the parametrization, point A corresponds to t=1.Therefore, points B and C are parametrized by t₁ and t₂, with t₁ + t₂ + t₁t₂ = -5.We need to analyze triangle ABC with vertices at A(1,2), B(t₁², 2t₁), C(t₂², 2t₂), where t₁ + t₂ + t₁t₂ = -5.To check if triangle ABC is right-angled, we can compute the vectors AB, AC, BC and check if any pair is perpendicular.Let's compute the coordinates:A(1,2), B(t₁², 2t₁), C(t₂², 2t₂)Vector AB = B - A = (t₁² -1, 2t₁ -2)Vector AC = C - A = (t₂² -1, 2t₂ -2)Vector BC = C - B = (t₂² - t₁², 2t₂ - 2t₁)To check for a right angle at A, vectors AB and AC should be perpendicular. For a right angle at B, vectors BA and BC should be perpendicular. Similarly for angle at C.Alternatively, compute the distances squared and use the Pythagorean theorem.But maybe using dot products is more straightforward.First, check if angle at A is right:AB · AC = (t₁² -1)(t₂² -1) + (2t₁ -2)(2t₂ -2)If this dot product is zero, then angle at A is right.Similarly, check angle at B:BA · BC = ( - (t₁² -1) )*(t₂² - t₁²) + ( - (2t₁ -2) )*(2t₂ - 2t₁ )And angle at C:CA · CB = ( - (t₂² -1) )*(t₁² - t₂²) + ( - (2t₂ -2) )*(2t₁ - 2t₂ )But this might get complicated. Let's see if we can find a relationship using the condition t₁ + t₂ + t₁t₂ = -5.Let me denote s = t₁ + t₂ and p = t₁t₂. Then, the equation becomes s + p = -5.Our variables are s and p with s + p = -5.Also, for the parametrization, we can express other terms in terms of s and p.But let's see if we can compute the dot product AB · AC.First, compute AB · AC:= (t₁² -1)(t₂² -1) + (2t₁ -2)(2t₂ -2)Expand both terms:First term: t₁²t₂² - t₁² - t₂² +1Second term: 4t₁t₂ -4t₁ -4t₂ +4So total:t₁²t₂² - t₁² - t₂² +1 +4t₁t₂ -4t₁ -4t₂ +4Combine like terms:t₁²t₂² + (-t₁² - t₂²) +4t₁t₂ + (-4t₁ -4t₂) + (1 +4)= t₁²t₂² - t₁² - t₂² +4t₁t₂ -4t₁ -4t₂ +5This seems complex, but maybe we can express in terms of s and p.Recall that s = t₁ + t₂, p = t₁t₂, and s + p = -5.Also, t₁² + t₂² = s² - 2pSimilarly, t₁²t₂² = p²And t₁ + t₂ = sTherefore, substitute into the expression:t₁²t₂² - (t₁² + t₂²) +4t₁t₂ -4(t₁ + t₂) +5= p² - (s² - 2p) +4p -4s +5= p² - s² +2p +4p -4s +5= p² - s² +6p -4s +5But since s + p = -5, we can express s = -5 - pSubstitute s = -5 - p into the expression:p² - (-5 - p)^2 +6p -4(-5 - p) +5First compute (-5 - p)^2 =25 +10p +p²Therefore:p² - (25 +10p +p²) +6p +20 +4p +5Simplify term by term:p² -25 -10p -p² +6p +20 +4p +5The p² terms cancel:-25 -10p +6p +4p +20 +5Combine like terms:-25 +20 +5 + (-10p +6p +4p)= (-25 +25) + (0p)=0 +0=0Wow! So the dot product AB · AC =0.Therefore, the vectors AB and AC are perpendicular, meaning that angle at A is a right angle. Therefore, triangle ABC is a right-angled triangle at A. Therefore, the answer is C. Right-angled triangle.But wait, let's confirm this result. The calculation led us to AB · AC =0, implying that angle at A is 90 degrees. So regardless of the values of t₁ and t₂ (as long as they satisfy the line passing through (5, -2)), the angle at A is right. Therefore, triangle ABC is always right-angled at A.Therefore, the correct answer is C. Right-angled triangle.This seems surprising but the algebra checks out. By parametrizing the points and using the condition from the line passing through (5, -2), we found that the dot product AB · AC equals zero. Hence, angle at A is right, making triangle ABC a right-angled triangle.Final Answerboxed{C}

✅:Given point (A(1,2)) and a line passing through ((5, -2)) that intersects the parabola (y^2 = 4x) at two other points (B) and (C), we need to determine the nature of triangle (triangle ABC).1. Equation of the Line: The line passing through ((5, -2)) can be written as (y = m(x - 5) - 2), where (m) is the slope.2. Intersection with Parabola: Substituting (y = m(x - 5) - 2) into (y^2 = 4x) gives: [ [m(x - 5) - 2]^2 = 4x ] Expanding and simplifying this equation results in a quadratic equation in (x): [ m^2x^2 + (-10m^2 - 4m - 4)x + 25m^2 + 20m + 4 = 0 ]3. Vieta's Formulas: Using Vieta's formulas, we find the sum and product of the roots (x_1) and (x_2) (coordinates of (B) and (C)): [ x_1 + x_2 = frac{10m^2 + 4m + 4}{m^2}, quad x_1x_2 = frac{25m^2 + 20m + 4}{m^2} ]4. Parametrization: Parametrizing the parabola (y^2 = 4x) as ((t^2, 2t)), points (B) and (C) are ((t_1^2, 2t_1)) and ((t_2^2, 2t_2)). The line through these points passing through ((5, -2)) gives the condition: [ t_1 + t_2 + t_1t_2 = -5 ]5. Dot Product Calculation: Calculating the dot product of vectors (overrightarrow{AB}) and (overrightarrow{AC}): [ overrightarrow{AB} cdot overrightarrow{AC} = (t_1^2 - 1)(t_2^2 - 1) + (2t_1 - 2)(2t_2 - 2) ] Simplifying using (s = t_1 + t_2) and (p = t_1t_2) with (s + p = -5), we find: [ overrightarrow{AB} cdot overrightarrow{AC} = 0 ] This implies the vectors are perpendicular, hence the angle at (A) is a right angle.Thus, triangle (triangle ABC) is a right-angled triangle.[boxed{C}]

🔑:To determine the type of triangle ( triangle ABC ), we need to carefully examine the given points and equations.1. Identify points on the parabola: - From the given problem, we know the line passes through point ( (5, -2) ) and intersects the parabola ( y^2 = 4x ) at other points ( B ) and ( C ). Points ( B ) and ( C ) will have coordinates of the form ( B(t^2, 2t) ) and ( C(s^2, 2s) ).2. Equation of the line passing through ( B ) and ( C ): - The general form of a line passing through two points ( (x_1, y_1) ) and ( (x_2, y_2) ) can be written as: [ begin{vmatrix} x & y & 1 x_1 & y_1 & 1 x_2 & y_2 & 1 end{vmatrix} = 0 ] - Using this determinant form, the equation of line ( BC ) passing through ( B(t^2, 2t) ) and ( C(s^2, 2s) ) simplifies to: [ 2x - (s+t)y + 2st = 0 ]3. The line passing through ((5, -2)): - Since the line passes through the point ( (5, -2) ), substituting these coordinates into the line equation yields: [ 2(5) - (s+t)(-2) + 2st = 0 implies 10 + 2(s+t) + 2st = 0 implies 2(s+t + st) = -10 implies s+t+st = -5 ]4. Properties ensuring a right angle: - To determine if the triangle is a right triangle, we need to check the perpendicularity of the sides. This involves looking at the slopes of specific lines. - The slopes of lines ( AB ) and ( AC ) can be computed from coordinates: [ k_{AB} = frac{2t - 2}{t^2 - 1}, quad k_{AC} = frac{2s - 2}{s^2 - 1} ]5. Calculate product of slopes: - The ( k_{AB} cdot k_{AC} ) will give information if the product is (-1): [ k_{AB} cdot k_{AC} = frac{2t - 2}{t^2 - 1} cdot frac{2s - 2}{s^2 - 1} = frac{4}{(s+1)(t+1)} ] - Given ( (s+1)(t+1) = -4 ), we substitute back in: [ k_{AB} cdot k_{AC} = frac{4}{-4} = -1 ] - Since ( k_{AB} cdot k_{AC} = -1 ), it verifies that the angle between these lines is ( 90^circ ), conforming a right angle. Conclusion:Since ( triangle ABC ) has a right angle, ( triangle ABC ) is a right-angled triangle.[boxed{text{C. Right-angled triangle}}]

🧠:Okay, let's try to figure out how many pictures the collective will paint on September 27. Hmm, let's start by understanding the problem step by step.First, the collective has different groups of artists. There are six artists who paint one picture every two days. Then there are eight artists who paint one picture every three days. The rest don't paint at all. From September 22 to September 26, they painted a total of 30 pictures. We need to find out how many they'll paint on September 27.Wait, let's clarify the dates. September 22 to September 26 inclusive—that's five days, right? Because 22, 23, 24, 25, 26. So five days total. And then September 27 is the next day after the 26th, so that's the sixth day. But we need to find the number of pictures on just September 27.So, the key here is to figure out the rate at which each artist paints and then sum up their contributions on the 27th. But first, maybe we need to figure out how many artists there are in total? Wait, the problem doesn't specify the total number of artists in the collective. It just mentions six, eight, and the rest. But maybe we don't need the total number because the rest don't paint. Hmm, maybe we can proceed without knowing the total?Wait, but let's check. The problem says "the rest never paint pictures," so they contribute zero. So, the total number of artists is 6 + 8 + (rest). But since the rest don't matter for painting, maybe the total number isn't necessary. However, the problem mentions that from September 22-26 (5 days), they painted 30 pictures. So perhaps we need to figure out how many pictures each group paints per day on average, and then use the 30 to confirm the numbers? Wait, but maybe there's more to it because the artists don't paint every day. They have intervals.So, the six artists paint every two days. That means each of them paints 1 picture every 2 days. Similarly, the eight artists paint 1 picture every 3 days. So, their painting schedules are periodic. To find the total number of pictures painted over the five days, we need to know how many times each artist painted during those days. Then, using the total of 30, maybe we can confirm that the numbers fit? Wait, but if we already know the number of artists in each group, why do we need the total? Hmm, perhaps the problem is that the days from September 22 to 26 could include different numbers of painting days depending on their schedule cycles. For example, if an artist paints every two days, whether they paint on a particular day depends on when they started.Wait, this is a problem. The problem doesn't specify when each artist started their schedule. For example, an artist painting every two days could have started on September 22, or maybe September 21, which would affect whether they paint on the 22nd, 24th, 26th, etc. Similarly for the ones painting every three days. Without knowing their starting days, how can we determine how many pictures each artist painted during the period?But the problem states that from September 22 to 26, they painted 30 pictures. Maybe the problem assumes that each artist starts painting on the first day of the period, i.e., September 22? Or perhaps the problem expects us to consider the average rate per day regardless of the starting day. Hmm, this is a bit ambiguous. Let me think.If we consider the average rate, for the six artists who paint every two days, each can paint 0.5 pictures per day on average. Similarly, the eight artists paint 1/3 pictures per day each. Then over five days, the total would be 6*(0.5)*5 + 8*(1/3)*5. Let's compute that:6 artists * 0.5 pictures/day = 3 pictures/day8 artists * 1/3 pictures/day ≈ 2.6667 pictures/dayTotal per day ≈ 3 + 2.6667 ≈ 5.6667 pictures/dayOver five days, that's 5.6667 * 5 ≈ 28.3333 pictures. But the problem states they painted 30 pictures. Hmm, so this discrepancy suggests that the average rate approach might not be accurate here because the actual number depends on the specific days they paint. Therefore, the starting day must matter. So maybe the problem assumes that each artist paints on their scheduled days starting from the first day of the period, i.e., September 22.Alternatively, maybe the artists could have started their schedules before September 22, so their painting days could be any days depending on their cycle. However, without knowing their starting days, we can't compute the exact number of paintings in the given period. Therefore, the problem must have a standard approach, perhaps assuming that the artists paint on every nth day starting from the first day of the period.Wait, but let's think differently. Maybe the question is designed such that on each day in the period September 22-26, the number of paintings per day is constant? But given that artists have cycles of 2 or 3 days, the number of paintings per day might not be constant. For example, artists painting every two days would paint on days 1, 3, 5, etc., if the period starts on day 1 (September 22). Similarly, artists painting every three days would paint on days 1, 4, 7, etc. So over the five days, the every-two-days artists would paint on days 1, 3, 5 (September 22, 24, 26) and the every-three-days artists would paint on days 1, 4 (September 22 and 25). Then, the total paintings from September 22-26 would be:For the six artists: 3 days (22,24,26) each painting 1 picture, so 6*3 = 18 pictures.For the eight artists: 2 days (22,25), so 8*2 = 16 pictures.Total pictures: 18 + 16 = 34. But the problem states 30 pictures. Hmm, this is more than 30, so this approach must be wrong.Alternatively, maybe the artists don't all start on September 22. Maybe their schedules are offset. For example, some artists in the every-two-days group could start on September 23, so their painting days would be 23, 25, etc. Similarly, some in the every-three-days group could start on 23 or 24. But without knowing the distribution, how can we determine?This seems complicated. Maybe the problem is intended to be solved by considering the average rate, despite the discrepancy in the total. Wait, but when I calculated the average rate earlier, I got approximately 28.33 pictures over five days, but the actual total is 30. So maybe there's an assumption that the artists are distributed in such a way that their painting days maximize or minimize the total? Or perhaps there's a specific distribution.Alternatively, maybe the key is to realize that the total number of paintings from September 22-26 is 30, and we need to use that information to figure out how many paintings each group contributed, and then use that to determine the number on September 27.Wait, let's think of it as a modular arithmetic problem. Each artist has a cycle: 2 days or 3 days. So, depending on their starting day, their painting days repeat every 2 or 3 days. If we can figure out how many times each artist painted during the five-day period, given their cycle, we can set up equations.But since we don't know their starting days, perhaps the problem expects us to consider the number of painting days each artist could have in the five-day period, regardless of their starting day. For example, an artist with a 2-day cycle can paint either 2 or 3 times in five days, depending on whether their first painting day is day 1 or day 2. Similarly, an artist with a 3-day cycle can paint 1 or 2 times in five days.But if that's the case, then the total number of paintings could vary. However, the problem states that the total is 30. So perhaps we need to find possible numbers of paintings for each group that sum to 30, then see which one allows us to compute September 27's paintings.Alternatively, maybe there's a standard way to compute the number of paintings in a period given cyclic schedules. Let me recall that in such problems, sometimes the least common multiple (LCM) of the cycles is considered to find a repeating pattern. The LCM of 2 and 3 is 6, so every 6 days the pattern repeats. But our period is only 5 days, so maybe not directly applicable.Wait, perhaps another approach: For each artist painting every 2 days, in 5 days, the maximum number of paintings they can do is 3 (if they paint on days 1, 3, 5) and the minimum is 2 (if they start on day 2, painting on days 2, 4). Similarly, for artists painting every 3 days, in 5 days, they can paint 2 times (days 1, 4) or 1 time (starting on day 2: days 2, 5; but wait, starting on day 2, painting every 3 days would be days 2, 5, which is two times in five days. Starting on day 3: days 3, 6 (but day 6 is outside the period). So actually, depending on the starting day, an artist with a 3-day cycle can paint 1 or 2 times in five days.But given that we have multiple artists in each group, maybe their starting days are spread out such that on average, the number of paintings per artist is the same. For example, for the six artists painting every two days: if their starting days are evenly distributed, then over the five-day period, each artist would have painted either 2 or 3 times. Similarly for the eight artists painting every three days.But how to model this? Let's suppose that for the six artists with a 2-day cycle, half of them start on day 1 and half on day 2. Then three artists would paint on days 1,3,5 and three on days 2,4. So total paintings from this group would be 3*3 + 3*2 = 9 + 6 = 15.Similarly, for the eight artists with a 3-day cycle, if their starting days are spread over days 1,2,3, then maybe some start on day 1, some on day 2, some on day 3. Let's see: starting day 1: paintings on 1,4; starting day 2: paintings on 2,5; starting day 3: paintings on 3 (day 6 is outside). So if we have artists starting on days 1, 2, and 3, each group would contribute 2, 2, and 1 paintings respectively in five days. If the eight artists are distributed as 3 on day 1, 3 on day 2, and 2 on day 3, then total paintings would be 3*2 + 3*2 + 2*1 = 6 + 6 + 2 = 14.So total paintings from both groups would be 15 + 14 = 29, which is close to 30 but not exact. Hmm, maybe a different distribution.Alternatively, maybe all eight artists in the 3-day cycle started on day 1. Then they would paint on days 1 and 4, contributing 2 paintings each, so 8*2=16. Then the six artists with a 2-day cycle: if all started on day 1, they paint on 1,3,5: 3 paintings each, so 6*3=18. Total 18+16=34, which is higher than 30.Alternatively, if the six artists are split such that some start on day 1 and others on day 2. For example, 3 start on day 1 (paint 3 times) and 3 start on day 2 (paint 2 times), total 15. For the eight artists, if they start on day 1, they paint 2 times each (16). Total 15+16=31. Still not 30.Alternatively, maybe some of the 3-day cycle artists started on day 3. For example, if 4 start on day 1 (paint 2 each: 8), 2 start on day 2 (paint 2 each: 4), and 2 start on day 3 (paint 1 each: 2). Total 8+4+2=14. Then with the six artists split as 3 and 3: 15. Total 15+14=29. Still not 30.Alternatively, maybe some of the 2-day cycle artists started on day 2. Let's say 4 start on day 1 (paint 3 each: 12) and 2 start on day 2 (paint 2 each: 4). Total 16. Then eight artists: suppose 5 start on day 1 (paint 2 each: 10), 3 start on day 3 (paint 1 each: 3). Total 13. Total paintings: 16 +13=29. Still not 30.This trial and error isn't working. Maybe there's another way. Let's denote:Let x be the number of 2-day cycle artists who started on day 1 (painting on days 1,3,5) => 3 paintings each.Let y be the number of 2-day cycle artists who started on day 2 (painting on days 2,4) => 2 paintings each.Similarly, for the 3-day cycle artists:Let a be the number starting on day 1 (painting on 1,4) => 2 paintings each.Let b be the number starting on day 2 (painting on 2,5) => 2 paintings each.Let c be the number starting on day 3 (painting on 3) =>1 painting each.We know that:For the 2-day cycle artists: x + y = 6Their total paintings: 3x + 2yFor the 3-day cycle artists: a + b + c =8Their total paintings: 2a + 2b +1cTotal paintings: 3x +2y +2a +2b +c =30But also, x + y =6, so y=6 -xAnd a + b +c=8, so c=8 -a -bSubstitute into total:3x +2(6 -x) +2a +2b + (8 -a -b) =30Simplify:3x +12 -2x +2a +2b +8 -a -b =30(3x -2x) + (2a -a) + (2b -b) +12 +8=30x +a +b +20=30Thus, x +a +b =10But we also have from c=8 -a -b, so c=8 - (a +b). But we don't have more equations. However, since x is the number of 2-day artists starting on day1, and a,b,c are numbers of 3-day artists starting on days1,2,3.But x can be from 0 to6, and a from0 to8, b from0 to8, but with a +b <=8 since c=8 -a -b >=0.So x +a +b=10. We need integer solutions where x<=6, a<=8, b<=8, a +b<=8.So x=10 -a -bBut x<=6, so 10 -a -b <=6 => a +b >=4Also, a +b <=8 (since c=8 -a -b >=0)So a +b is between4 and8, inclusive.Possible combinations:If a +b=4, then x=6. Then c=8 -4=4.Check if possible: x=6, which is allowed (since there are6 artists in the 2-day group). Then a and b must sum to4. For example, a=0, b=4, c=4. Then total paintings:3x +2y +2a +2b +cBut x=6, y=6 -x=0So 3*6 +2*0 +2*0 +2*4 +4=18 +0 +0 +8 +4=30. Yes, that works.Similarly, other combinations where a +b=4:a=1, b=3, c=4. Then total paintings:3*6 +0 +2*1 +2*3 +4=18 +0 +2 +6 +4=30. Yes.Similarly, a=2, b=2, etc.So possible. Alternatively, a +b=5, x=5:Then c=8 -5=3.x=5, y=1.Total paintings:3*5 +2*1 +2a +2b +3.Which is15 +2 +2*5 +3=15+2+10+3=30. Wait, 2a +2b=2*(a +b)=2*5=10. So yes, total 15+2+10+3=30.Similarly, a +b=5, x=5. So this is another solution.Similarly, a +b=6, x=4:c=8 -6=2x=4, y=2Total paintings:3*4 +2*2 +2*6 +2=12 +4 +12 +2=30. Yes.Similarly, a +b=7, x=3:c=1x=3, y=3Total paintings:3*3 +2*3 +2*7 +1=9 +6 +14 +1=30. Yes.a +b=8, x=2:c=0x=2, y=4Total paintings:3*2 +2*4 +2*8 +0=6 +8 +16=30. Yes.So there are multiple possible distributions of starting days that lead to the total 30 paintings. Therefore, the problem doesn't have a unique solution unless there's additional constraints. But the question is asking for the number of pictures on September 27. Depending on the starting days of the artists, the number could vary. But the problem must have a unique answer, so likely there's an assumption that all artists started their cycles on September 22 (day1). Let's check that.If all 6 two-day artists started on day1 (Sep22), they paint on 22,24,26 (days1,3,5). So each paints 3 pictures, total 6*3=18.If all 8 three-day artists started on day1, they paint on 22,25 (days1,4). So each paints 2 pictures, total 8*2=16.Total paintings:18+16=34. But problem states 30. So this can't be.Alternatively, maybe some artists started earlier. Wait, perhaps the artists have been painting before September22, and their schedules are ongoing. For example, a two-day artist who painted on September21 would paint on 23,25,27 etc. during the period.But the problem states the period from September22 to26. So maybe we need to consider their painting days within this interval, regardless of when they started. In that case, an artist painting every two days would paint on days that are even or odd within the period. Wait, but the exact days depend on their cycle.Alternatively, perhaps the problem is designed such that September27 is the next day after the period, and we need to calculate based on the same cycles. If we can find how many artists are painting on September27, which is the 6th day if we count September22 as day1. Wait, September22 is day1, 23 day2, 24 day3, 25 day4, 26 day5, 27 day6.So for two-day artists: if they painted on day1,3,5, then they will paint again on day7, which is outside. If they started on day2, they paint on day2,4,6. So on day6 (Sep27), those who started on day2 would paint.Similarly, three-day artists: those who started on day1 would paint on day1,4,7; day4 is in the period, day7 is outside. Those who started on day2 paint on day2,5,8. Day5 is in the period. Those who started on day3 paint on day3,6,9. Day6 is Sep27. Those who started on day4 paint on day4,7. Etc.But again, without knowing how many artists started on which days, we can't determine exactly. However, the problem must have a unique answer, so perhaps there's a standard approach here.Wait, let's think differently. Suppose that the total number of paintings from September22-26 is 30. We need to find how many paintings occur on September27. Let's denote:Let’s define the number of paintings on any day as the sum of contributions from both groups.For the two-day artists: Each artist paints every two days, so their contribution on day t (where t=1 for Sep22) is 6/(2) = 3 per day on average, but actually, depending on the day's parity. Similarly, three-day artists contribute 8/3 ≈2.6667 per day on average. However, the actual number per day depends on the day's position modulo 2 or 3.But since the period Sep22-26 is 5 days, and the next day is Sep27 (day6), perhaps we can find a pattern.Wait, over a 6-day period (days1-6), the two-day artists will paint on days1,3,5 (if they started on day1) or days2,4,6 (if started on day2). Each artist paints 3 times in 6 days. Similarly, three-day artists will paint twice in 6 days: days1,4 or days2,5 or days3,6.But our period is days1-5. So in days1-5, two-day artists could have painted 3 times (if started on day1) or 2 times (started on day2). Three-day artists could have painted 2 times (started day1 or2) or1 time (started day3).But given that the total over five days is30, we need to find how many paintings each group contributed. Let’s denote:Let’s suppose that x of the two-day artists started on day1, and (6 -x) started on day2.Each two-day artist starting on day1 paints on days1,3,5: 3 paintings.Each starting on day2 paints on days2,4: 2 paintings.Similarly, for three-day artists: let’s say a started on day1, b on day2, c on day3, d on day4, e on day5, f on day6. Wait, but since they paint every three days, starting on day4 would be equivalent to starting on day1 (since day4 = day1 +3). Similarly, starting on day5 is day2 +3, etc. So actually, the starting days modulo3 define their schedule.Therefore, for three-day artists, the possible starting days modulo3 are 1,2,3. So we can group them into three categories:- Starting on day1: paint on days1,4,7,...- Starting on day2: paint on days2,5,8,...- Starting on day3: paint on days3,6,9,...But since our period is days1-5 (Sep22-26), starting on day3 would mean painting on day3 and6 (Sep27). So in the five-day period, they paint only on day3.Similarly, starting on day1: days1,4.Starting on day2: days2,5.So for the three-day artists, let’s say:a started on day1: paint on1,4: 2 paintings each.b started on day2: paint on2,5: 2 paintings each.c started on day3: paint on3:1 painting each.With a +b +c=8.Similarly, for the two-day artists:x started on day1:3 paintings each.y started on day2:2 paintings each.With x + y=6.Total paintings:3x +2y +2a +2b +1c=30.But we also have:x + y=6a +b +c=8Let’s substitute y=6 -x and c=8 -a -b into the total paintings equation:3x +2(6 -x) +2a +2b + (8 -a -b)=30Simplify:3x +12 -2x +2a +2b +8 -a -b=30(3x -2x) + (2a -a) + (2b -b) +20=30x +a +b +20=30Thus:x +a +b=10But since x ≤6 and a +b ≤8 (because c=8 -a -b ≥0), the maximum x +a +b can be is6 +8=14, but we have x +a +b=10.So possible solutions:We need to find integers x, a, b such that x ≤6, a +b ≤8, and x +a +b=10.Possible combinations:1. x=6, then a +b=4. Then c=8 -4=4.Check if possible: Yes. Then total paintings from two-day artists:3*6 +2*0=18. From three-day artists:2a +2b +c=2*4 +4=8 +4=12. Total 18+12=30. Correct.2. x=5, a +b=5. c=3.Total two-day:15 +2=17. Three-day:2*5 +3=10 +3=13. Total 17+13=30.3. x=4, a +b=6. c=2.Two-day:12 +4=16. Three-day:12 +2=14. Total 30.4. x=3, a +b=7. c=1.Two-day:9 +6=15. Three-day:14 +1=15. Total30.5. x=2, a +b=8. c=0.Two-day:6 +8=14. Three-day:16 +0=16. Total30.So all these are possible. Each case gives a different distribution of starting days, which would affect the number of paintings on September27 (day6).Now, September27 is day6. Let's see which artists paint on day6.For two-day artists:Those who started on day2 paint on days2,4,6,... So on day6, the two-day artists who started on day2 will paint.Number of such artists is y=6 -x.For three-day artists:Those who started on day3 paint on days3,6,9,... So on day6, the three-day artists who started on day3 will paint.Number of such artists is c=8 -a -b.Therefore, total paintings on day6:Two-day artists: y=6 -x.Three-day artists: c=8 -a -b.Total paintings on day6: (6 -x) + (8 -a -b).But from earlier, x +a +b=10.Therefore, 6 -x +8 -a -b= (6 +8) - (x +a +b)=14 -10=4.So regardless of the values of x, a, b, the total paintings on day6 (September27) is4.Wow, that's clever! Because x +a +b=10, so substituting into the expression gives 14 -10=4. Therefore, the number of paintings on September27 is4.So the answer is4.

✅:To determine the number of paintings the collective will produce on September 27, we analyze the artists' schedules and their contributions during September 22–26. Key Steps:1. Group Contributions: - 6 artists (2-day cycle): Each paints 1 picture every 2 days. - 8 artists (3-day cycle): Each paints 1 picture every 3 days. - Total paintings from September 22–26 (5 days): 30.2. Modeling Painting Days: - 2-day artists: Depending on their starting day (September 22 or 23), they paint 3 or 2 times in 5 days. - 3-day artists: Depending on their starting day (September 22, 23, or 24), they paint 2, 2, or 1 time in 5 days.3. Equation Setup: - Let (x) = 2-day artists starting September 22 (paint 3 times). - Let (y) = 2-day artists starting September 23 (paint 2 times). Thus, (x + y = 6). - Let (a, b, c) = 3-day artists starting September 22, 23, 24, respectively. Thus, (a + b + c = 8). - Total paintings: (3x + 2y + 2a + 2b + c = 30).4. Solving Constraints: - Substitute (y = 6 - x) and (c = 8 - a - b) into the total paintings equation: [ x + a + b = 10 ] - This constraint holds for all valid distributions of starting days.5. September 27 (Day 6) Calculation: - 2-day artists starting September 23: Paint on day 6 (September 27). Number = (y = 6 - x). - 3-day artists starting September 24: Paint on day 6. Number = (c = 8 - a - b). - Total paintings on September 27: [ (6 - x) + (8 - a - b) = 14 - (x + a + b) = 14 - 10 = 4 ]Final Answer: The collective will paint (boxed{4}) pictures on September 27.

🔑:1. Identify the interval duration and number of paintings per interval: - The interval from September 22 to September 27 inclusive spans 6 days. - Six artists paint one painting every two days. - Eight artists paint one painting every three days.2. Determine individual contributions: - An artist who paints one painting every two days will paint: [ leftlfloor frac{6}{2} rightrfloor = 3 text{ paintings in this span of days} ] - An artist who paints one painting every three days will paint: [ leftlfloor frac{6}{3} rightrfloor = 2 text{ paintings in this span of days} ]3. Calculate the total number of paintings by all artists from September 22 to September 27: - There are 6 artists who paint every two days: [ 6 times 3 = 18 text{ paintings} ] - There are 8 artists who paint every three days: [ 8 times 2 = 16 text{ paintings} ] - Adding these together gives: [ 18 + 16 = 34 text{ paintings} ]4. Confirm the number of paintings from September 22 to September 26: - It is given that they painted a total of 30 paintings from September 22 to September 26.5. Calculate the number of paintings produced on September 27: - Total paintings for 6 days: (34) - Paintings from September 22 to 26: (30) - Paintings for September 27: [ 34 - 30 = 4 text{ paintings} ] Conclusion:[boxed{4}]

🧠:Okay, so I need to find the distribution of Z = |X - Y| where X and Y are independent uniform variables on [0,1]. Then show the density is f_Z(z) = 2(1 - z) for z in [0,1], and finally that the expectation E[Z] is 1/3. Let me start by recalling how to handle distributions of functions of random variables, especially absolute differences.First, since X and Y are independent and uniform on [0,1], their joint distribution is uniform on the unit square [0,1] x [0,1]. The variable Z is the absolute difference between X and Y, so we can visualize this as the distance from the line X = Y in the square. To find the distribution function F_Z(z) = P(Z ≤ z) = P(|X - Y| ≤ z), which is the probability that (X,Y) lies within a band around the line X = Y with width 2z. But wait, actually, since it's absolute difference, |X - Y| ≤ z corresponds to the region between the lines Y = X - z and Y = X + z. However, since X and Y are both between 0 and 1, these lines will be truncated at the edges of the square.So, for z between 0 and 1, the region where |X - Y| ≤ z is a band around the diagonal of the square. To find the area of this region, which will be equal to F_Z(z) because the joint distribution is uniform. Then the density f_Z(z) is the derivative of F_Z(z).Let me draw a mental picture. When z is 0, the area is zero (just the line). As z increases, the band widens. For z ≥ 1, the entire square is covered, so F_Z(z) = 1. But since z is in [0,1], we focus on 0 ≤ z ≤ 1.The area where |X - Y| ≤ z can be calculated by subtracting the area where |X - Y| > z from the total area (which is 1). So F_Z(z) = 1 - P(|X - Y| > z). Therefore, maybe it's easier to compute P(|X - Y| > z) first.So, P(|X - Y| > z) is the probability that either X - Y > z or Y - X > z. These are two regions in the square: above the line Y = X - z and below the line Y = X + z. Each of these regions is a triangle (or a trapezoid?) depending on z. Let me check for different values of z.If z is between 0 and 1, then the line Y = X + z starts at (0, z) and ends at (1 - z, 1). Similarly, the line Y = X - z starts at (z, 0) and ends at (1, 1 - z). So the regions where Y > X + z and Y < X - z are two triangles each with base (1 - z) and height (1 - z). Wait, is that correct?Let's compute the area where Y > X + z. This is the area above the line Y = X + z. The intersection points of this line with the square are (0, z) and (1 - z, 1). So this region is a triangle with vertices at (0, z), (1 - z, 1), and (1,1). Wait, no. From (0, z) to (1 - z, 1), and then the right boundary is x = 1, but y can't exceed 1. So the triangle is formed by (0, z), (1 - z, 1), and (0, 1). Hmm, maybe not. Let me think.Alternatively, the area above Y = X + z can be calculated as the integral over x from 0 to 1 - z of the length of y from X + z to 1. Because for x from 0 to 1 - z, the line Y = X + z is below Y = 1. For x from 1 - z to 1, the line Y = X + z would exceed 1, but since Y cannot exceed 1, the upper limit is 1. Wait, actually, when x > 1 - z, Y = X + z would be greater than 1, so the region Y > X + z is empty for x > 1 - z. Therefore, the area is the integral from x = 0 to x = 1 - z of (1 - (x + z)) dx.Similarly, the area below Y = X - z is the same by symmetry, since X and Y are symmetric. So total area where |X - Y| > z is twice the area above Y = X + z.Therefore, let's compute the area above Y = X + z:Integral from x = 0 to x = 1 - z of [1 - (x + z)] dx= Integral from 0 to 1 - z of (1 - x - z) dx= Integral from 0 to 1 - z of (1 - z - x) dx= (1 - z)x - (1/2)x² evaluated from 0 to 1 - z= (1 - z)(1 - z) - (1/2)(1 - z)^2= (1 - z)^2 - (1/2)(1 - z)^2= (1/2)(1 - z)^2Therefore, the area above Y = X + z is (1/2)(1 - z)^2. Similarly, the area below Y = X - z is also (1/2)(1 - z)^2. Therefore, total area where |X - Y| > z is 2*(1/2)(1 - z)^2 = (1 - z)^2.Therefore, P(|X - Y| > z) = (1 - z)^2. Hence, F_Z(z) = 1 - (1 - z)^2.Wait, but hold on. Let me verify that. If z is 0, then F_Z(0) = 1 - 1 = 0, which is correct since the probability that |X - Y| ≤ 0 is zero (only the line X=Y). For z = 1, F_Z(1) = 1 - 0 = 1, which is correct. For z between 0 and 1, F_Z(z) = 1 - (1 - z)^2. Therefore, the density f_Z(z) is the derivative of F_Z(z) with respect to z.So f_Z(z) = d/dz [1 - (1 - z)^2] = 2(1 - z). That's exactly what we needed to prove. Therefore, the density is 2(1 - z) for z in [0,1].But wait, let me check another way. Maybe using geometry. The area where |X - Y| ≤ z is the area between the lines Y = X + z and Y = X - z. Since the total area is 1, subtracting the two triangles we computed earlier gives 1 - (1 - z)^2. Therefore, F_Z(z) = 1 - (1 - z)^2, so derivative is 2(1 - z). Correct.Alternatively, another approach is to compute the CDF directly. For Z = |X - Y|, F_Z(z) = P(|X - Y| ≤ z). Let's fix z and compute the probability over the unit square.The set of points (X,Y) where |X - Y| ≤ z is the region between Y = X - z and Y = X + z. To compute this area, we can integrate over X and find the range of Y for each X. Let's consider two cases: when X is between 0 and z, and when X is between z and 1 - z, and when X is between 1 - z and 1? Wait, maybe splitting into different intervals.Alternatively, for a given X, Y must be in [X - z, X + z], but since Y is in [0,1], we have to adjust the limits.So for X from 0 to z, Y must be from 0 to X + z (since X - z would be negative, which is below 0). For X from z to 1 - z, Y can range from X - z to X + z. For X from 1 - z to 1, Y can range from X - z to 1 (since X + z would exceed 1). Therefore, we can split the integral into three parts.So integrating over X from 0 to z:For each X in [0, z], Y ranges from 0 to X + z. The length of Y is X + z.Then, integrating over X from z to 1 - z:Y ranges from X - z to X + z. The length is 2z.Finally, integrating over X from 1 - z to 1:Y ranges from X - z to 1. The length is 1 - (X - z) = 1 - X + z.Therefore, total area is:Integral_{0}^{z} (X + z) dX + Integral_{z}^{1 - z} 2z dX + Integral_{1 - z}^{1} (1 - X + z) dXCompute each integral:First integral:∫₀^z (X + z) dX = [0.5X² + zX] from 0 to z = 0.5z² + z² = 1.5z²Second integral:∫_{z}^{1 - z} 2z dX = 2z*(1 - z - z) = 2z*(1 - 2z)Wait, hold on. The integral of 2z from X = z to X = 1 - z is 2z*(1 - z - z) = 2z*(1 - 2z). Wait, but if z > 0.5, then 1 - z < z, which would make the upper limit less than the lower limit. But since z is in [0,1], when z < 0.5, 1 - z > z, so the integral is valid. However, when z ≥ 0.5, the interval [z, 1 - z] becomes invalid because z ≥ 1 - z when z ≥ 0.5. Therefore, we actually need to handle z ≤ 0.5 and z ≥ 0.5 separately?Wait a second, this complicates things. Maybe my initial split is only valid when z ≤ 0.5. Because if z > 0.5, then 1 - z < z, so the middle integral would have lower limit greater than upper limit, which doesn't make sense. Therefore, actually, we need to consider two cases: z ≤ 0.5 and z > 0.5.But in our original problem, z is in [0,1], so perhaps the CDF expression is different for z ≤ 0.5 and z > 0.5. But wait, earlier we derived F_Z(z) = 1 - (1 - z)^2, which is valid for z in [0,1]. Let's check this.Wait, if z = 0.5, then F_Z(0.5) = 1 - (1 - 0.5)^2 = 1 - 0.25 = 0.75. Let's compute it via integrating:For z = 0.5, split into three integrals:First integral from 0 to 0.5:∫₀^0.5 (X + 0.5) dX = [0.5X² + 0.5X] from 0 to 0.5 = 0.5*(0.25) + 0.5*(0.5) = 0.125 + 0.25 = 0.375Second integral from 0.5 to 0.5 (since 1 - z = 0.5), which is zero.Third integral from 0.5 to 1:∫_{0.5}^1 (1 - X + 0.5) dX = ∫_{0.5}^1 (1.5 - X) dX = [1.5X - 0.5X²] from 0.5 to 1 = (1.5*1 - 0.5*1) - (1.5*0.5 - 0.5*(0.25)) = (1.5 - 0.5) - (0.75 - 0.125) = 1 - 0.625 = 0.375Total area = 0.375 + 0 + 0.375 = 0.75, which matches F_Z(0.5) = 0.75. So that works.But if z > 0.5, say z = 0.6. Then 1 - z = 0.4. So the integral splits would be:First integral from 0 to 0.6:Wait, but if z = 0.6, then in the first integral, X from 0 to z = 0.6, but the middle integral from X = z = 0.6 to X = 1 - z = 0.4, which is impossible because 0.6 > 0.4. Therefore, in reality, when z > 0.5, the middle integral vanishes, and the entire area is covered by the first and third integrals.Wait, maybe when z > 0.5, the lines Y = X + z and Y = X - z intersect the square such that the region between them is actually covered by overlapping parts. Let me try computing F_Z(z) when z > 0.5.Take z = 0.6. Then the lines Y = X + 0.6 and Y = X - 0.6. The line Y = X + 0.6 starts at (0, 0.6) and ends at (0.4, 1), since when X = 0.4, Y = 1. Similarly, Y = X - 0.6 starts at (0.6, 0) and ends at (1, 0.4). So the area between these lines is a hexagon? Or perhaps two trapezoids?Wait, actually, when z > 0.5, the region |X - Y| ≤ z becomes overlapping near the corners. Let's visualize this. For z = 0.6, the area between Y = X - 0.6 and Y = X + 0.6 would cover the central band, but due to the high z, this band overlaps with the edges. Wait, but actually, when z is larger, the band is wider, but since the square is bounded, the overlapping occurs.Wait, perhaps the area is still calculated as 1 - 2*(1/2)(1 - z)^2 even when z > 0.5. Wait, let's test z = 0.6. Then 1 - (1 - z)^2 = 1 - (0.4)^2 = 1 - 0.16 = 0.84. Let's compute it by integration.For z = 0.6, compute the area where |X - Y| ≤ 0.6. Let's split the integral:When X is from 0 to 1 - z = 0.4, Y ranges from X - z (but since X < 1 - z = 0.4 and z = 0.6, X - z is negative, so Y starts at 0) to X + z (which is up to 0.4 + 0.6 = 1). Wait, no. Wait, maybe for X from 0 to 1, regardless of z, we can adjust the Y ranges.Wait, perhaps it's better to handle it as follows:For all X in [0,1], the valid Y range is [max(0, X - z), min(1, X + z)].Therefore, the length of Y for each X is min(1, X + z) - max(0, X - z).So integrating over X from 0 to 1:F_Z(z) = ∫₀^1 [min(1, X + z) - max(0, X - z)] dXThis integral can be split into intervals based on where max(0, X - z) and min(1, X + z) change their expressions.Case 1: X - z < 0 and X + z < 1. This is when X < z and X < 1 - z. But if z > 0.5, then 1 - z < z, so X < 1 - z is stricter. Therefore, for z > 0.5, X < 1 - z is the condition. But for z < 0.5, X < z.Wait, this is getting complicated. Let me consider two cases: z ≤ 0.5 and z > 0.5.Case 1: z ≤ 0.5.Then, for X in [0, z], Y ranges from 0 to X + z.For X in [z, 1 - z], Y ranges from X - z to X + z.For X in [1 - z, 1], Y ranges from X - z to 1.Case 2: z > 0.5.Then, 1 - z < z. So:For X in [0, 1 - z], Y ranges from 0 to X + z (since X + z ≤ (1 - z) + z = 1).For X in [1 - z, z], Y ranges from 0 to 1 (since X - z < 0 and X + z > 1).Wait, but if z > 0.5, then 1 - z < z. So when X is in [1 - z, z], X is between 1 - z and z. But since z > 0.5, this interval is only valid if 1 - z < z, which is true, but for example, if z = 0.6, then 1 - z = 0.4, so X from 0.4 to 0.6. In this interval, X - z is negative (since X ≤ z), so Y starts at 0, and X + z exceeds 1 when X > 1 - z. Wait, when X > 1 - z, X + z > (1 - z) + z = 1, so Y ends at 1.Therefore, for X in [1 - z, z], Y ranges from 0 to 1. Hence, the length is 1.For X in [z, 1], Y ranges from X - z to 1.Therefore, in case z > 0.5, the integral splits into three parts:1. X from 0 to 1 - z: Y from 0 to X + z. Length = X + z.2. X from 1 - z to z: Y from 0 to 1. Length = 1.3. X from z to 1: Y from X - z to 1. Length = 1 - (X - z) = 1 - X + z.Therefore, total area is:∫₀^{1 - z} (X + z) dX + ∫_{1 - z}^z 1 dX + ∫_{z}^1 (1 - X + z) dXCompute each integral:First integral:∫₀^{1 - z} (X + z) dX = [0.5X² + zX] from 0 to 1 - z = 0.5(1 - z)^2 + z(1 - z) = 0.5(1 - 2z + z²) + z - z² = 0.5 - z + 0.5z² + z - z² = 0.5 - 0.5z²Second integral:∫_{1 - z}^z 1 dX = z - (1 - z) = 2z - 1Third integral:∫_{z}^1 (1 - X + z) dX = ∫_{z}^1 (1 + z - X) dX = [(1 + z)X - 0.5X²] from z to 1 = (1 + z)(1) - 0.5(1)^2 - [(1 + z)z - 0.5z²] = (1 + z - 0.5) - (z + z² - 0.5z²) = (0.5 + z) - (z + 0.5z²) = 0.5 - 0.5z²Adding all three parts:0.5 - 0.5z² + 2z - 1 + 0.5 - 0.5z² = (0.5 + 0.5 - 1) + (2z) + (-0.5z² - 0.5z²) = 0 + 2z - z²So total area when z > 0.5 is 2z - z².But wait, for z > 0.5, F_Z(z) = 2z - z². Let's check at z = 0.6:F_Z(0.6) = 2*0.6 - (0.6)^2 = 1.2 - 0.36 = 0.84. Which matches 1 - (1 - z)^2 = 1 - (0.4)^2 = 0.84.So indeed, whether z ≤ 0.5 or z > 0.5, F_Z(z) = 1 - (1 - z)^2. Because when z ≤ 0.5, the integral calculation gives 1 - (1 - z)^2, and when z > 0.5, the integral gives 2z - z², which can be rewritten as 1 - (1 - z)^2:Expand 1 - (1 - z)^2 = 1 - 1 + 2z - z² = 2z - z². So yes, same expression.Therefore, regardless of z in [0,1], F_Z(z) = 1 - (1 - z)^2. Therefore, the density is the derivative:f_Z(z) = d/dz [1 - (1 - z)^2] = 2(1 - z). But wait, derivative of -(1 - z)^2 is -2(1 - z)(-1) = 2(1 - z). Correct. Therefore, f_Z(z) = 2(1 - z) for z ∈ [0,1]. That's exactly what we needed to prove.Now, for the expectation E[Z]. Since we have the density f_Z(z) = 2(1 - z), we can compute E[Z] as the integral from 0 to 1 of z * f_Z(z) dz.So E[Z] = ∫₀¹ z * 2(1 - z) dz = 2 ∫₀¹ z(1 - z) dz = 2 ∫₀¹ (z - z²) dz = 2 [0.5z² - (1/3)z³] from 0 to 1 = 2 [0.5 - 1/3 - 0] = 2 [ (3/6 - 2/6) ] = 2 [1/6] = 2/6 = 1/3.Hence, E[Z] = 1/3 as required.Alternatively, recalling that for a non-negative random variable, E[Z] = ∫₀^∞ P(Z > z) dz. Since Z is between 0 and 1, we can write E[Z] = ∫₀¹ P(Z > z) dz. Earlier, we found that P(Z > z) = (1 - z)^2. Therefore:E[Z] = ∫₀¹ (1 - z)^2 dz = ∫₀¹ (1 - 2z + z²) dz = [z - z² + (1/3)z³] from 0 to 1 = (1 - 1 + 1/3) - 0 = 1/3. Same result.Therefore, both methods confirm that E[Z] = 1/3.Let me recap to ensure there are no mistakes. The key steps were:1. Recognize that Z = |X - Y| and X, Y ~ Uniform[0,1] independent.2. The joint distribution is uniform over [0,1]^2.3. Compute P(|X - Y| ≤ z) by calculating the area between Y = X ± z.4. Found that this area is 1 - (1 - z)^2, leading to F_Z(z) = 1 - (1 - z)^2.5. Differentiated F_Z(z) to get f_Z(z) = 2(1 - z).6. Computed E[Z] by integrating z * f_Z(z) over [0,1], resulting in 1/3.All steps check out, and the alternative method using the expectation formula for non-negative variables also confirms the result. I think this is solid.

✅:To determine the distribution of ( Z = |X - Y| ) where ( X ) and ( Y ) are independent uniform random variables on ([0,1]), we analyze the joint distribution over the unit square ([0,1] times [0,1]).Step 1: Cumulative Distribution Function (CDF) of ( Z )The CDF ( F_Z(z) = P(|X - Y| leq z) ) corresponds to the area between the lines ( Y = X + z ) and ( Y = X - z ) within the unit square. This area is calculated as:[F_Z(z) = 1 - P(|X - Y| > z) = 1 - 2 cdot frac{1}{2}(1 - z)^2 = 1 - (1 - z)^2.]Here, ( P(|X - Y| > z) ) is the combined area of two triangles outside the band ( |X - Y| leq z ), each with area ( frac{1}{2}(1 - z)^2 ).Step 2: Probability Density Function (PDF) of ( Z )Differentiating the CDF with respect to ( z ):[f_Z(z) = frac{d}{dz} F_Z(z) = frac{d}{dz} left[1 - (1 - z)^2right] = 2(1 - z).]Thus, the PDF is ( f_Z(z) = 2(1 - z) ) for ( z in [0,1] ).Step 3: Expected Value ( E[Z] )Using the PDF, compute the expectation:[E[Z] = int_0^1 z cdot f_Z(z) , dz = int_0^1 2z(1 - z) , dz = 2 int_0^1 (z - z^2) , dz.]Evaluating the integral:[2 left[ frac{1}{2}z^2 - frac{1}{3}z^3 right]_0^1 = 2 left( frac{1}{2} - frac{1}{3} right) = 2 cdot frac{1}{6} = frac{1}{3}.]ConclusionThe density function of ( Z ) is ( f_Z(z) = 2(1 - z) ) for ( z in [0,1] ), and the expected value is ( E[Z] = frac{1}{3} ).Final AnswerThe density of ( Z ) is ( f_Z(z) = 2(1 - z) ) for ( z in [0,1] ), and thus ( E[Z] = boxed{dfrac{1}{3}} ).

🔑:1. Let ( X ) and ( Y ) be independent random variables uniformly distributed on ([0, 1]). We want to evaluate the distribution of ( Z = |X - Y| ).2. Note that the joint density function of (X) and (Y) is ( f(x, y) = 1 ) for ( (x, y) in [0, 1] times [0, 1] ).3. Define ( U = X - Y ). Since ( Z = |X - Y| ), we need to first find the distribution of ( U ) and use it to find the distribution of ( Z ).4. The support of ( U ) is ([-1, 1]), because if ( X in [0, 1] ) and ( Y in [0, 1] ), then ( X - Y ) ranges from (-1) to 1.5. We calculate the density function ( f_U(u) ) of ( U ): For ( 0 leq u leq 1 ): [ f_U(u) = int_{-infty}^infty f(u + y, y) , dy = int_0^{1-u} dy = 1 - u ] Here, we integrate from ( y = 0 ) to ( y = 1 - u ) because ( f(u + y, y) neq 0 ) only when ( 0 leq y leq 1 - u ) and ( 0 leq u + y leq 1 ). For ( -1 leq u < 0 ): [ f_U(u) = int_{-infty}^infty f(u + y, y) , dy = int_{-u}^1 dy = 1 + u ] Here, again we integrate within feasible bounds of ( y ).6. Now that we have ( f_U(u) ) for the range ([-1, 1]): [ f_U(u) = begin{cases} 1 - u & text{if } 0 leq u leq 1 1 + u & text{if } -1 leq u < 0 end{cases} ]7. Since ( Z = |U| ), the density function ( f_Z(z) ) can be obtained by combining ( f_U(z) ) and ( f_U(-z) ): [ f_Z(z) = f_U(z) + f_U(-z) ] For ( 0 leq z leq 1 ): [ f_Z(z) = (1 - z) + (1 - (-z)) = (1 - z) + (1 + z) = 2(1 - z) ]8. Thus, the density function ( f_Z(z) ) is confirmed as: [ f_Z(z) = 2(1 - z), quad 0 leq z leq 1 ]Conclusion:[boxed{f_Z(z) = 2(1 - z), quad z in [0, 1]}]Now, to find the expected value ( E[Z] ):9. Using the density function ( f_Z(z) ), we calculate: [ E[Z] = int_0^1 z cdot f_Z(z) , dz = int_0^1 z cdot 2(1 - z) , dz ]10. Solve the integral: [ E[Z] = 2 int_0^1 z(1 - z) , dz = 2 int_0^1 (z - z^2) , dz ]11. Compute the integrals separately: [ int_0^1 z , dz = left[frac{z^2}{2}right]_0^1 = frac{1}{2} - 0 = frac{1}{2} ] [ int_0^1 z^2 , dz = left[frac{z^3}{3}right]_0^1 = frac{1}{3} - 0 = frac{1}{3} ]12. Combining the results: [ E[Z] = 2 left( frac{1}{2} - frac{1}{3} right) = 2 left( frac{3}{6} - frac{2}{6} right) = 2 cdot frac{1}{6} = frac{1}{3} ]Conclusion:[boxed{E[Z] = frac{1}{3}}]

🧠:Okay, so I have this problem here: I need to factorize the polynomial ( x^2 - 2x - n ) into a product of two linear polynomials with integer coefficients. Then, determine how many natural numbers ( n ) (not greater than 100) satisfy this condition. The options are (A) 8, (B) 9, (C) 10, (D) 11. Alright, let's start breaking this down.First, factorizing a quadratic polynomial into two linear polynomials with integer coefficients. The general form of such a quadratic is ( x^2 + bx + c ), which factors into ( (x + m)(x + k) ) where ( m + k = b ) and ( mk = c ). But in our case, the quadratic is ( x^2 - 2x - n ). So comparing that to the standard form, here ( b = -2 ) and ( c = -n ).Wait, but if we factorize ( x^2 - 2x - n ), it should be written as ( (x + a)(x + d) ), right? Let me check. Expanding ( (x + a)(x + d) ) gives ( x^2 + (a + d)x + ad ). But our quadratic is ( x^2 - 2x - n ). So matching coefficients:1. The coefficient of ( x ) is ( a + d = -2 )2. The constant term is ( ad = -n )So, we have two equations:1. ( a + d = -2 )2. ( a cdot d = -n )Our goal is to find all natural numbers ( n leq 100 ) such that there exist integers ( a ) and ( d ) satisfying these two equations.Wait, but since ( n ) is a natural number, ( -n ) is negative. Therefore, ( ad = -n ) must be negative. So the product of ( a ) and ( d ) is negative. That implies that one of ( a ) or ( d ) is positive and the other is negative.Additionally, ( a + d = -2 ). So the sum of these two numbers is -2. Let me think about how to approach this.Let me denote the two numbers as ( a ) and ( d ). Since their sum is -2 and their product is -n. Let's solve for ( a ) and ( d ).Let me set ( a = p ) and ( d = -2 - p ), since their sum is -2. Then, their product is ( p cdot (-2 - p) = -n ).So:( -2p - p^2 = -n )Multiply both sides by -1:( 2p + p^2 = n )Therefore:( n = p^2 + 2p )So, ( n = p(p + 2) )Since ( n ) must be a natural number not greater than 100, we can think of this as ( n = p(p + 2) leq 100 ), where ( p ) is an integer. But wait, ( p ) can be positive or negative? Let's see.But in the original equations, ( a ) and ( d ) are integers, so ( p ) can be any integer. However, ( n ) is a natural number, so ( n ) must be positive. Therefore, ( p(p + 2) ) must be positive. So when is ( p(p + 2) > 0 )?The product of ( p ) and ( p + 2 ) is positive when both factors are positive or both are negative.Case 1: Both ( p > 0 ) and ( p + 2 > 0 ). This is true when ( p > 0 ).Case 2: Both ( p < 0 ) and ( p + 2 < 0 ). This is true when ( p < -2 ).So, ( p ) can be any integer greater than 0 or less than -2. However, let's check if both cases give distinct values of ( n ).If ( p ) is positive, then ( n = p(p + 2) ).If ( p ) is negative, say ( p = -k ) where ( k > 2 ), then ( n = (-k)(-k + 2) = (-k)(- (k - 2)) = k(k - 2) ).Wait, let's substitute ( p = -k ), then ( p + 2 = -k + 2 ). So ( n = (-k)(-k + 2) = k(k - 2) ). So for ( p < -2 ), ( k > 2 ), and ( n = k(k - 2) ).Therefore, the values of ( n ) when ( p ) is negative are ( k(k - 2) ) where ( k > 2 ). Let's see if these values overlap with the positive ( p ) case.For positive ( p ), ( n = p(p + 2) ). For negative ( p ), ( n = k(k - 2) ), where ( k = -p ).So for example, if ( p = 1 ), ( n = 1*3 = 3 ). If ( p = -3 ), then ( k = 3 ), so ( n = 3*1 = 3 ). So actually, ( n ) values repeat for positive and negative ( p ). Therefore, each ( n ) can be generated by both a positive and a negative ( p ), except possibly when ( p = -1 ) or ( p = 0 ), but those are excluded because when ( p = -1 ), ( k = 1 ), then ( n = 1*(-1) = -1 ), which is not natural, and ( p = 0 ) gives ( n = 0*2 = 0 ), which is not a natural number.Therefore, to avoid double-counting, we can consider only positive ( p ) and generate all ( n = p(p + 2) leq 100 ), then also include ( n = k(k - 2) ) where ( k > 2 ), but since these are the same as the positive ones, we need to check if they produce the same ( n ).Wait, actually, for a given ( n ), there are two possible factorizations: one with positive ( p ) and one with negative ( p ). However, since ( n ) must be positive, the negative ( p ) case would generate the same ( n ) as some positive ( p ). Let me test with a specific example.Take ( p = 2 ): ( n = 2*4 = 8 ). For ( p = -4 ): ( k = 4 ), ( n = 4*(4 - 2) = 4*2 = 8 ). So indeed, same ( n ). Similarly, ( p = 3 ): ( 3*5 = 15 ); ( p = -5 ): ( 5*3 = 15 ). Therefore, each positive ( p ) corresponds to a negative ( p' = - (p + 2) ), because:If ( n = p(p + 2) ), then taking ( k = p + 2 ), then ( n = k(k - 2) ). So, for each positive ( p ), there is a corresponding ( k = p + 2 ), and ( n = k(k - 2) ). Hence, the same ( n ) is generated by both a positive ( p ) and a negative ( p ). Therefore, the set of possible ( n ) is exactly the numbers of the form ( p(p + 2) ) where ( p ) is a positive integer, and each such ( n ) is unique.Wait, but ( p(p + 2) = (p + 1)^2 - 1 ). Hmm, interesting. So perhaps there's another way to look at it. For example, if ( n + 1 ) must be a perfect square. Wait, ( n = (p + 1)^2 - 1 ). So ( n + 1 = (p + 1)^2 ). Therefore, ( n + 1 ) must be a perfect square. Therefore, for each ( n ) such that ( n + 1 ) is a perfect square, then ( n ) can be expressed as ( p(p + 2) ).Wait, let's check. Let me compute ( n + 1 ):If ( n = p(p + 2) ), then ( n + 1 = p^2 + 2p + 1 = (p + 1)^2 ). So yes, ( n + 1 ) is a perfect square. Therefore, all such ( n ) are one less than a perfect square.But wait, is that correct? Let me verify with the earlier examples:When ( p = 1 ), ( n = 3 ), ( n + 1 = 4 = 2^2 ). Correct.When ( p = 2 ), ( n = 8 ), ( n + 1 = 9 = 3^2 ). Correct.When ( p = 3 ), ( n = 15 ), ( n + 1 = 16 = 4^2 ). Correct.Similarly, when ( p = 4 ), ( n = 24 ), ( n + 1 = 25 = 5^2 ). So yes, indeed, ( n + 1 ) is a perfect square.Therefore, the problem reduces to finding all natural numbers ( n leq 100 ) such that ( n + 1 ) is a perfect square.Therefore, the number of such ( n ) is equal to the number of perfect squares greater than 1 (since ( n geq 1 )) and less than or equal to 101 (since ( n + 1 leq 101 )).So, let's find all integers ( k ) such that ( k^2 leq 101 ).The largest integer ( k ) where ( k^2 leq 101 ) is ( k = 10 ), since ( 10^2 = 100 leq 101 ), and ( 11^2 = 121 > 101 ).Therefore, ( k ) can be 2, 3, ..., 10. Because when ( k = 1 ), ( n + 1 = 1 ), so ( n = 0 ), which is not a natural number. Therefore, starting from ( k = 2 ):k = 2: n = 4 - 1 = 3k = 3: n = 9 - 1 = 8k = 4: 16 -1 =15k=5:25-1=24k=6:36-1=35k=7:49-1=48k=8:64-1=63k=9:81-1=80k=10:100-1=99So the values of n are 3,8,15,24,35,48,63,80,99. That's 9 values. So option B is 9. But wait, let me check if these are all less than or equal to 100.Yes, 99 is the largest one. So 9 values. But hold on, in the problem statement, they are asking for natural numbers n not greater than 100. So n can be 1,2,...,100. But in our case, the n's generated are 3,8,15,24,35,48,63,80,99. That's 9 numbers. So the answer is 9? Which is option B.But wait, let's double-check. Because maybe there's an n=0? But n must be a natural number, so depending on the definition, sometimes natural numbers start at 1, sometimes at 0. But since the problem says "natural numbers n not greater than 100", and typically in math problems, natural numbers start at 1. So n=0 is excluded. So the k starts at 2, giving n=3 as the first. Then up to k=10, giving n=99. So that's k from 2 to 10, inclusive. So that's 9 values. So answer is 9, which is option B.But wait, let me check this another way. Suppose we model the original quadratic equation: ( x^2 - 2x - n ). To factor this into two linear polynomials with integer coefficients, we need to find integers a and d such that:( x^2 - 2x - n = (x + a)(x + d) )Expanding the right-hand side: ( x^2 + (a + d)x + ad ). Comparing coefficients:1. Coefficient of x: ( a + d = -2 )2. Constant term: ( ad = -n )So we can set up the equations:1. ( a + d = -2 )2. ( a cdot d = -n )We can solve for a and d. Let me consider a as an integer variable, then d = -2 - a. Then substituting into the second equation:( a(-2 - a) = -n )Which simplifies to:( -2a - a^2 = -n )Multiply both sides by -1:( a^2 + 2a = n )Therefore, ( n = a^2 + 2a )So n must be equal to ( a(a + 2) ). Since n is a natural number, ( a(a + 2) ) must be positive. As we saw earlier, this happens when a > 0 or a < -2.But since n must be the same for both a and d, which are integers, and a can be positive or negative, but n is positive. Therefore, for each positive a, n is ( a(a + 2) ), and for each a < -2, n is ( a(a + 2) ). However, if we take a negative a, say a = -k where k is a positive integer greater than 2, then:n = (-k)(-k + 2) = (-k)(- (k - 2)) = k(k - 2)So for a = -k, n = k(k - 2). So when k = 3, n = 3*1 = 3; k = 4, n=4*2=8; k=5, n=5*3=15; etc. Wait, this is the same sequence as when a is positive. For example, a=1: n=1*3=3; a=2:2*4=8; a=3:3*5=15; which is the same as when k=3,4,5,... in the negative a case. Therefore, these are overlapping values. Therefore, the distinct values of n are given by either a positive a or a negative a, but they generate the same n.Therefore, the number of distinct n is equal to the number of positive integers a such that ( a(a + 2) leq 100 ). Let's solve for a.Find all positive integers a where ( a(a + 2) leq 100 ).Let's solve the inequality:( a^2 + 2a - 100 leq 0 )Solving the quadratic equation ( a^2 + 2a - 100 = 0 ):Using quadratic formula:( a = frac{ -2 pm sqrt{4 + 400} }{ 2 } = frac{ -2 pm sqrt{404} }{ 2 } )Compute sqrt(404). Since 20^2 = 400, so sqrt(404) ≈ 20.09975.Therefore, the positive root is:( frac{ -2 + 20.09975 }{ 2 } ≈ frac{18.09975}{2} ≈ 9.049875 )Therefore, the inequality ( a(a + 2) leq 100 ) holds for positive real numbers a ≤ 9.049875. Since a must be a positive integer, the maximum a is 9.So let's check a=9:9*11 = 99 ≤ 100, okay.a=10: 10*12 = 120 > 100, too big.Therefore, a can be from 1 to 9, inclusive. Therefore, there are 9 possible values of a, leading to 9 values of n: 3,8,15,24,35,48,63,80,99. Which matches our earlier result.But hold on, wait. If a can be from 1 to 9, inclusive, then n would be 3,8,15,24,35,48,63,80,99. That's 9 numbers. Therefore, the answer should be 9, which is option B.But the options given are (A) 8, (B) 9, (C) 10, (D) 11. Hmm. Wait, but let's check again.Wait, maybe I missed something here. Let's list all the n for a from 1 to 9:a=1: 1*3=3a=2:2*4=8a=3:3*5=15a=4:4*6=24a=5:5*7=35a=6:6*8=48a=7:7*9=63a=8:8*10=80a=9:9*11=99Yes, that's 9 values. So 9 natural numbers n ≤100. Therefore, answer is B)9.But let me check once more to be absolutely sure. Maybe there's another case where a and d are both integers but not covered by this approach?Wait, the original problem says "factorize into a product of two linear polynomials with integer coefficients". So, the factorization must have integer coefficients. Therefore, the roots of the quadratic equation ( x^2 -2x -n =0 ) must be integers. Because if it factors into linear terms with integer coefficients, then the roots are integers.Therefore, the discriminant must be a perfect square. The discriminant D is:( D = (-2)^2 - 4(1)(-n) = 4 + 4n = 4(n + 1) )Therefore, for the roots to be integers, the discriminant must be a perfect square. So, 4(n + 1) must be a perfect square, which implies that n + 1 must be a perfect square, because 4 is already a square. Let me explain:If 4(n +1) is a perfect square, then 4(n +1) = k^2 for some integer k. Then, n +1 = (k/2)^2. Since n +1 must be an integer (as n is natural), then k/2 must be an integer, so k must be even. Let k = 2m, then n +1 = (2m/2)^2 = m^2. Therefore, n +1 = m^2. Therefore, n = m^2 -1.Therefore, n must be one less than a perfect square. Therefore, n = m^2 -1, where m is an integer ≥2 (since n ≥1). Then, m^2 -1 ≤100 ⇒ m^2 ≤101 ⇒ m ≤10 (since 10^2=100 ≤101, 11^2=121>101). Therefore, m can be 2 through 10 inclusive. That gives m=2:3, m=3:8, m=4:15,...,m=10:99. Exactly as before. So 9 values of m, leading to 9 values of n. So answer is 9. Option B.But the options are given as (A)8, (B)9, (C)10, (D)11. Therefore, the correct answer is B)9.Wait a second, but when m=1, n=1^2 -1=0, which is excluded. So m starts at 2. Therefore, m=2 to m=10, inclusive. That's 9 values. Therefore, answer is 9. So why the confusion?Alternatively, maybe there's another approach where n can be written as (x - a)(x - b), but let's check:The quadratic is ( x^2 -2x -n ). Suppose it factors into (x - a)(x - b). Then expanding gives:( x^2 - (a + b)x + ab ). Comparing coefficients:- Coefficient of x: - (a + b) = -2 ⇒ a + b = 2- Constant term: ab = -nTherefore, same as before but with signs flipped. So in this case, a + b = 2 and ab = -n. So the product ab is negative, so one of a or b is positive and the other is negative. Let a = p and b = 2 - p. Then, ab = p(2 - p) = -n. Therefore:( 2p - p^2 = -n )Multiply by -1:( p^2 - 2p = n )So, n = p(p - 2). Now, since n must be positive, p(p - 2) >0. Therefore, p >2 or p <0.Case1: p >2. Then n = p(p -2). For example, p=3:3*1=3, p=4:4*2=8, p=5:5*3=15, etc.Case2: p <0. Let p = -k where k >0. Then n = (-k)(-k -2) = k(k +2). For example, p=-1: n=1*3=3, p=-2:2*4=8, etc.Wait, so here, in this case, n can be written as k(k +2) for k positive integers (when p negative), and as p(p -2) when p >2. However, notice that when p=3, n=3*1=3; when k=1 (p=-1), n=1*3=3. Similarly, p=4, n=4*2=8; k=2 (p=-2), n=2*4=8. So again, these generate the same values. Therefore, regardless of the approach, n is of the form m(m +2) or m(m -2), but they end up being the same sequence. Therefore, the number of n's is the same as before.But in this case, solving for p >2:Find all integers p >2 such that n = p(p -2) ≤100.Similarly, solving p(p -2) ≤100.For p=3:3*1=3p=4:4*2=8p=5:5*3=15...Find maximum p where p(p-2) ≤100.Solving p^2 -2p -100 ≤0.Quadratic equation p^2 -2p -100=0:Solutions p = [2 ± sqrt(4 + 400)] / 2 = [2 ± sqrt(404)] / 2 ≈ [2 ± 20.09975]/2.Positive solution: (2 + 20.09975)/2 ≈ 11.0498.Thus, p can be up to 11. But p must be integer greater than 2.Check p=11:11*9=99 ≤100p=12:12*10=120>100Therefore, p=3 to p=11 inclusive. That's 9 values. So again, 9 n's. Same result.So regardless of the approach, it's 9.Therefore, the answer must be B)9.But why do the options include up to 11? Maybe I made a mistake in considering something. Wait, let's list all possible n:From m=2 to m=10:m=2: 4 -1=3m=3:9 -1=8m=4:16-1=15m=5:25-1=24m=6:36-1=35m=7:49-1=48m=8:64-1=63m=9:81-1=80m=10:100-1=99That's 9 numbers. All are ≤100. So 9 values. So answer is B)9.Therefore, the correct answer is option B)9.Final Answerboxed{B}

✅:To factorize the polynomial ( x^2 - 2x - n ) into a product of two linear polynomials with integer coefficients, we need to find integers ( a ) and ( d ) such that:1. ( a + d = -2 )2. ( a cdot d = -n )We can express ( n ) as ( n = p(p + 2) ) where ( p ) is a positive integer. This is derived from solving the equations ( a + d = -2 ) and ( a cdot d = -n ), leading to ( n = p(p + 2) ) when considering positive integers ( p ).The values of ( n ) must be natural numbers not greater than 100. This translates to finding the number of positive integers ( p ) such that ( p(p + 2) leq 100 ).Solving the quadratic inequality ( p^2 + 2p leq 100 ):1. The quadratic equation ( p^2 + 2p - 100 = 0 ) gives the roots using the quadratic formula: [ p = frac{-2 pm sqrt{4 + 400}}{2} = frac{-2 pm sqrt{404}}{2} ] The positive root is approximately ( 9.05 ), so the maximum integer ( p ) is 9.Listing the values of ( n ) for ( p ) from 1 to 9:- ( p = 1 ): ( n = 1 cdot 3 = 3 )- ( p = 2 ): ( n = 2 cdot 4 = 8 )- ( p = 3 ): ( n = 3 cdot 5 = 15 )- ( p = 4 ): ( n = 4 cdot 6 = 24 )- ( p = 5 ): ( n = 5 cdot 7 = 35 )- ( p = 6 ): ( n = 6 cdot 8 = 48 )- ( p = 7 ): ( n = 7 cdot 9 = 63 )- ( p = 8 ): ( n = 8 cdot 10 = 80 )- ( p = 9 ): ( n = 9 cdot 11 = 99 )These values are all natural numbers not greater than 100, resulting in 9 valid values of ( n ).Thus, the number of natural numbers ( n ) not greater than 100 is (boxed{B}).

🔑:Given the quadratic polynomial (x^2 - 2x - n), we want to express it as the product of two linear polynomials with integer coefficients. 1. Assume Factorization: Let (x^2 - 2x - n = (x - a)(x - b)). 2. Find Relations Between Coefficients: From the factorization, we get two equations from the coefficients: - Sum of the roots: (a + b = 2) - Product of the roots: (ab = -n)3. Assume Values for (a) and (b): Since (a) and (b) are integers and (ab = -n), one must be positive and the other must be negative (as (n) is a natural number and hence positive). Assume (a > 0) and (b < 0).4. Express (b) in Terms of (a): From (a + b = 2), we get (b = 2 - a).5. Express (n) in Terms of (a): Using the product (ab = -n): [ n = -ab = -a(2 - a) = a^2 - 2a ]6. Simplify (n): Rewrite (a) as (a - 1 = c), where (c) is a non-negative integer ((c geq 0)). [ n = a(a - 2) = (a - 1)^2 - 1 = c^2 - 1 ]7. Identify Possible Values of (n): The expression (n = c^2 - 1) suggests that (n) is obtained from subtracting 1 from a perfect square. We need (n) to be a natural number and less than or equal to 100. Thus, (c^2) is a perfect square less than or equal to 101. The perfect squares ( leq 100) are: [ 1, 4, 9, 16, 25, 36, 49, 64, 81, 100 ]8. Calculate Corresponding Values of (n): Subtract 1 from each perfect square: [ 1 - 1 = 0, quad 4 - 1 = 3, quad 9 - 1 = 8, quad 16 - 1 = 15, quad 25 - 1 = 24, quad 36 - 1 = 35, quad 49 - 1 = 48, quad 64 - 1 = 63, quad 81 - 1 = 80, quad 100 - 1 = 99 ] The possible values of (n) that are natural numbers are: [ 3, 8, 15, 24, 35, 48, 63, 80, 99 ]9. Count the Number of Valid Values: There are (9) valid values for (n). Conclusion:The number of values of (n) is ( boxed{9} ).

🧠:Okay, so I need to find the minimum value of k, which is the sum of the sizes of two non-empty finite sets S and T, given that the symmetric difference between S and T is 1. Let me start by recalling what symmetric difference means. The symmetric difference A Δ B is defined as (A B) ∪ (B A). So, it's the set of elements that are in either A or B but not in both. The problem states that |S Δ T| = 1, which means there's exactly one element that is in either S or T but not in both. Now, the goal is to minimize k = |S| + |T|. Since both S and T are non-empty, they each have at least one element. Let me think about how the symmetric difference being 1 affects the sizes of S and T. Let me denote the symmetric difference as having exactly one element. That element is either in S but not in T, or in T but not in S. Let's break it down:Case 1: The element is in S but not in T. Then, all other elements in S must also be in T, and all elements in T must also be in S except for this one element. Wait, no. If the symmetric difference is exactly one element, then S and T can differ by only that one element. So, the intersection of S and T would be all the elements they have in common, and then one of them has an extra element not in the other.Let me formalize this. Suppose that S and T share n elements. Then, if |S Δ T| = 1, either S has one more element than T, or T has one more element than S. So, either |S| = n + 1 and |T| = n, or |T| = n + 1 and |S| = n. Therefore, the symmetric difference is that one extra element. But wait, is that the only possibility? Suppose S and T have some elements not in each other, but in such a way that the total number of differing elements is 1. For example, S could have an element not in T, and T could have an element not in S, but if those two elements are the same, then that would not contribute to the symmetric difference. Wait, no. The symmetric difference is the union of S T and T S. So, if S has an element not in T and T has an element not in S, then the symmetric difference would have two elements. Unless those elements are the same, which they can't be because they are different sets. Wait, no, elements are distinct. So if S has an element a not in T, and T has an element b not in S, then S Δ T = {a, b}, which has size 2. Therefore, to have |S Δ T| = 1, there must be exactly one element in either S T or T S, but not both. That is, either S has an extra element that T doesn't have, and T has no extra elements, or vice versa. Therefore, the only way to have symmetric difference of size 1 is if one set is a subset of the other and their sizes differ by 1. For example, if S is a subset of T and |T| = |S| + 1, then the symmetric difference would be the element in T that's not in S, so |S Δ T| = 1. Similarly, if T is a subset of S and |S| = |T| + 1, then again the symmetric difference is 1. So, in that case, the sum k = |S| + |T| would be |S| + (|S| + 1) = 2|S| + 1. Alternatively, if it's T that's the larger set, then similarly k = |T| + (|T| + 1) = 2|T| + 1. But since we are to find the minimum k, we need to minimize 2n + 1 where n is the size of the smaller set. Since both sets must be non-empty, the smallest possible n is 1. If n = 1, then k = 2*1 + 1 = 3. Let me check if this is possible.Suppose S = {a}, and T = {a, b}. Then S is a subset of T, the symmetric difference is {b}, so |S Δ T| = 1. Then |S| + |T| = 1 + 2 = 3. Similarly, if S = {a, b} and T = {a}, then the symmetric difference is {b}, sum is 2 + 1 = 3. Alternatively, is there a way to have smaller k? Let's see. If both sets have size 1, then S and T are both singletons. If they are the same set, then the symmetric difference is empty, which is not allowed. If they are different, then the symmetric difference would be 2 elements, since S T is the element in S not in T, and T S is the element in T not in S. So, for example, S = {a}, T = {b}, then |S Δ T| = |{a, b}| = 2, which is too big. So two singleton sets can't work. If one set has size 1 and the other has size 2, but they are not subsets? For example, S = {a}, T = {b, c}. Then S Δ T = {a, b, c}, which has size 3. That's even worse. If S has size 2 and T has size 2, but they share one element. Let's say S = {a, b}, T = {a, c}. Then S Δ T = {b, c}, size 2. Still not 1. So that's not helpful. If S has size 2 and T has size 3, with S being a subset of T, then S Δ T is {c}, so size 1. Then k = 2 + 3 = 5. But we already have a case where k = 3. So the minimal k seems to be 3. Wait, but maybe there's another configuration where the symmetric difference is 1 but the sets are not subsets. Let me think. Suppose S and T have some overlapping elements, and each has one unique element, but somehow the unique elements are the same. Wait, that's impossible because sets can't have duplicate elements. So, for instance, if S has elements {a, b}, and T has elements {a, c}, then the symmetric difference is {b, c}. If I have S and T such that one of the differences is empty. For example, S = {a, b}, T = {a}. Then S Δ T = {b}, size 1. That works. Similarly, S = {a}, T = {a, b}, same result. So in that case, the sum is 1 + 2 = 3. Alternatively, if S has 3 elements and T has 2 elements, but T is a subset of S. Then S Δ T is the element in S not in T. So sum is 3 + 2 = 5, which is higher. So the minimal case is when one set has 1 element, the other has 2. Wait, but the problem says both sets are non-empty. So the minimal sizes possible would be 1 and 2. But let me check if there are any other possibilities. Suppose S and T have more elements, but arranged in such a way that the symmetric difference is 1. For example, S = {a, b, c}, T = {a, b, d}. Then S Δ T = {c, d}, which has size 2. Not helpful. If S = {a, b, c}, T = {a, b}, then S Δ T = {c}, which is size 1. Then the sum is 3 + 2 = 5. But again, 5 is larger than 3. So that's worse. Alternatively, if S = {a}, T = {a, b}, sum 3. If S = {a, b, c, d}, T = {a, b, c}, sum 4 + 3 = 7. So, clearly, the minimal sum is 3. Wait, but let me confirm that there are no other configurations where the symmetric difference is 1 with smaller sum. Suppose S and T are such that they share some elements, but one set has an extra element. The minimal case would be when the shared part is as small as possible. Wait, actually, if the shared part is smaller, then the total sum would be larger. For example, if S and T share 0 elements. Then, if they share 0 elements, then S Δ T = S ∪ T. So |S Δ T| = |S| + |T|. But we need this to be 1. But since both S and T are non-empty, the minimum |S| + |T| would be 2, but |S Δ T| would be 2. Which is not 1. Therefore, the shared part cannot be zero. If they share one element. Suppose S = {a}, T = {a}. Then S Δ T = empty set, which is not allowed. If S = {a}, T = {a, b}, then S Δ T = {b}, size 1. Sum is 1 + 2 = 3. If they share one element, but S has 2 elements and T has 2 elements. For example, S = {a, b}, T = {a, c}. Then symmetric difference is {b, c}, size 2. Not helpful. Alternatively, S = {a, b}, T = {a}. Then symmetric difference {b}, sum 2 + 1 = 3. Wait, so the key is that one set is a subset of the other, and their sizes differ by 1. Then the symmetric difference is 1. Therefore, the minimal sum is achieved when the smaller set has size 1 and the larger has size 2. So 1 + 2 = 3. Therefore, the minimal k is 3. Let me check if there are any other edge cases. For example, if both sets have size 1. Then if they are the same, symmetric difference is 0. If they are different, symmetric difference is 2. So that's not allowed. So the minimal possible sum is indeed 3. Another way to think about it is using the formula for the size of the symmetric difference. Recall that |A Δ B| = |A| + |B| - 2|A ∩ B|. We have |A Δ B| = 1. So 1 = |A| + |B| - 2|A ∩ B|. Let me denote |A| = a, |B| = b, |A ∩ B| = c. Then 1 = a + b - 2c. We need to minimize a + b. So, rearrange the equation: a + b = 1 + 2c. Since c is the size of the intersection, which is at least 0 and at most the minimum of a and b. Since both a and b are at least 1 (non-empty sets), c is at least 1? Wait, no. If A and B have no elements in common, then c = 0, but then a + b = 1 + 0 = 1, but a and b are each at least 1, so a + b would be at least 2. But |A Δ B| would be a + b in that case, so |A Δ B| = 2. But we need |A Δ B| = 1. Therefore, c cannot be 0. Therefore, c must be at least 1. So c ≥ 1. Then, a + b = 1 + 2c. To minimize a + b, we need to minimize c. The minimal c is 1. Then a + b = 1 + 2*1 = 3. Therefore, the minimal sum is 3. So this confirms the earlier result. To check if this is achievable, set c = 1. Then a + b = 3. Since c = |A ∩ B| = 1, then the intersection has one element. Therefore, the sets A and B each have at least that one element. Suppose A has one more element and B doesn't, or vice versa. For example, A = {x, y}, B = {x}. Then |A| = 2, |B| = 1, intersection is {x}, so c = 1. Then |A Δ B| = |{y}| = 1, which satisfies the condition. Sum is 2 + 1 = 3. Alternatively, A = {x}, B = {x, y}. Same result. Therefore, the minimal k is indeed 3. Is there a case where c is greater than 1 but a + b is still 3? Let's see. If c = 1, then a + b = 3. If c = 2, then a + b = 1 + 4 = 5. So a + b increases as c increases. Therefore, the minimal sum is when c is minimal. Hence, 3 is the minimal value. Therefore, after analyzing different cases and using the formula for symmetric difference size, it's clear that the minimum value of k is 3.

✅:The problem requires finding the minimum value of ( k = |S| + |T| ) given that ( |S Delta T| = 1 ) for non-empty finite sets ( S ) and ( T ).Key Steps:1. Understanding Symmetric Difference: ( S Delta T = (S setminus T) cup (T setminus S) ). The condition ( |S Delta T| = 1 ) implies exactly one element differs between ( S ) and ( T ).2. Set Relationship: For the symmetric difference to have one element, one set must be a subset of the other with their sizes differing by 1. For example, ( S subset T ) and ( |T| = |S| + 1 ), or vice versa.3. Minimizing ( k = |S| + |T| ): Let ( |S| = n ) and ( |T| = n + 1 ). Then ( k = n + (n + 1) = 2n + 1 ). The smallest ( n ) for non-empty sets is 1, leading to ( k = 2(1) + 1 = 3 ).4. Verification: - If ( S = {a} ) and ( T = {a, b} ), then ( S Delta T = {b} ) and ( |S| + |T| = 1 + 2 = 3 ). - Using the formula ( |S Delta T| = |S| + |T| - 2|S cap T| ), set ( |S Delta T| = 1 ). Solving ( 1 = |S| + |T| - 2|S cap T| ) confirms the minimum ( k = 3 ) when ( |S cap T| = 1 ).Conclusion: The minimum value of ( k ) is 3, achieved when one set has 1 element and the other has 2 elements, with the smaller set being a subset of the larger one.(boxed{3})

🔑:Let us define the expressions and constraints given in the problem:1. The symmetric difference of sets (A) and (B) is defined as: [ A Delta B = (A setminus B) cup (B setminus A) ] 2. We have two non-empty finite sets ( S ) and ( T ) such that: [ |S Delta T| = 1 ]Let's analyze the conditions given for ( S ) and ( T ):1. If ( |S Delta T| = 1 ), this implies that either: - ( |S setminus T| = 1 ) and ( |T setminus S| = 0 ), or - ( |S setminus T| = 0 ) and ( |T setminus S| = 1 ). Case Analysis# Case 1: ( |S setminus T| = 1 ) and ( |T setminus S| = 0 )2. This would mean ( T ) is a subset of ( S ), and ( S ) has exactly one more element than ( T ). Hence, ( T ) is a proper subset of ( S ). Let's express the cardinalities: [ |S| > |T| quad text{and} quad |T| geq 1 ]3. Therefore, we can derive: [ k = |S| + |T| ]4. Given ( |S| = |T| + 1 ), we have: [ k = (|T| + 1) + |T| = 2|T| + 1 ]5. Since ( |T| geq 1 ): [ k geq 2 cdot 1 + 1 = 3 ]Thus, ( k ) is minimized when ( |T| = 1 ). Example Construction for Minimum ValueConsider a concrete example:Let:[S = {1, 2}, quad T = {2}]1. We can verify: [ S Delta T = (S setminus T) cup (T setminus S) = {1} cup emptyset = {1} ]2. Therefore: [ |S Delta T| = 1 ]3. Moreover: [ k = |S| + |T| = 2 + 1 = 3 ] Conclusion:The smallest possible value of ( k = |S| + |T| ) where ( |S Delta T| = 1 ) is ( 3 ).[boxed{3}]