💡First, I recognize that there are some students in the class, and Taqeesha was absent on the day the test was given. This means that the number of students who took the test initially is one less than the total number of students in the class.Given that the average score of the 16 students who took the test is 77, I can calculate the total score for these students by multiplying the average by the number of students: 77 multiplied by 16 equals 1232.The next day, Taqeesha took the test, and the new average score for all the students became 78. To find the total number of students in the class, I need to determine how many students there are in total, including Taqeesha.Let’s denote the total number of students as ( N ). Since 16 students took the test initially, the total number of students is ( N = 16 + 1 = 17 ).To verify, if there are 17 students in total, the new average score of 78 would mean the total score for all students is 78 multiplied by 17, which equals 1326.The difference between the total score with Taqeesha and without her is 1326 minus 1232, which equals 94. This means Taqeesha scored 94 on the test.Finally, since the number of students who took the test initially was 16 and adding Taqeesha makes the total number of students 17, the total number of students in the class is 17.

💡Okay, so I have this probability problem here about two people, Person A and Person B, who are shooting balloons. Person A has a shooting accuracy of 0.7, which means they have a 70% chance of hitting a balloon each time they shoot. Person B has a lower accuracy of 0.4, so they have a 40% chance each time. Both of them are going to shoot twice. The problem has two parts. The first part asks for the probability that Person A hits exactly 1 balloon and Person B hits exactly 2 balloons. The second part is a bit more complex: it asks for the probability that the number of balloons hit by Person A and Person B is equal. That means both could hit 0, 1, or 2 balloons each, but the counts have to match.Starting with part (I). I need to find the probability that Person A hits 1 balloon and Person B hits 2 balloons. Since each shot is independent, I can model this using the binomial probability formula. The binomial formula is used when there are a fixed number of independent trials, each with two possible outcomes (success or failure), and the probability of success is constant.For Person A, who has a 0.7 chance of hitting each balloon, the probability of hitting exactly 1 out of 2 shots is calculated as follows:The binomial probability formula is:[ P(k) = C(n, k) times p^k times (1-p)^{n-k} ]where ( C(n, k) ) is the combination of n things taken k at a time, ( p ) is the probability of success, and ( n ) is the number of trials.So for Person A, n = 2, k = 1, p = 0.7:[ P(A = 1) = C(2, 1) times 0.7^1 times (1 - 0.7)^{2 - 1} ]Calculating the combination:[ C(2, 1) = 2 ]So,[ P(A = 1) = 2 times 0.7 times 0.3 = 2 times 0.21 = 0.42 ]Now, for Person B, who has a 0.4 chance of hitting each balloon, the probability of hitting exactly 2 out of 2 shots is:[ P(B = 2) = C(2, 2) times 0.4^2 times (1 - 0.4)^{2 - 2} ]Calculating the combination:[ C(2, 2) = 1 ]So,[ P(B = 2) = 1 times 0.16 times 1 = 0.16 ]Since the events are independent, the combined probability that both Person A hits 1 balloon and Person B hits 2 balloons is the product of their individual probabilities:[ P(A = 1 text{ and } B = 2) = P(A = 1) times P(B = 2) = 0.42 times 0.16 ]Calculating that:[ 0.42 times 0.16 = 0.0672 ]So, the probability for part (I) is 0.0672.Moving on to part (II), which is a bit trickier. I need to find the probability that the number of balloons hit by Person A and Person B is equal. That means I need to consider all scenarios where both hit 0, both hit 1, or both hit 2 balloons. Then, I'll sum those probabilities.First, let's outline the possible cases:1. Both hit 0 balloons.2. Both hit 1 balloon.3. Both hit 2 balloons.I'll calculate each probability separately and then add them up.Starting with both hitting 0 balloons.For Person A, hitting 0 balloons out of 2 shots:[ P(A = 0) = C(2, 0) times 0.7^0 times (1 - 0.7)^{2 - 0} ]Calculating the combination:[ C(2, 0) = 1 ]So,[ P(A = 0) = 1 times 1 times 0.3^2 = 1 times 1 times 0.09 = 0.09 ]For Person B, hitting 0 balloons out of 2 shots:[ P(B = 0) = C(2, 0) times 0.4^0 times (1 - 0.4)^{2 - 0} ]Calculating the combination:[ C(2, 0) = 1 ]So,[ P(B = 0) = 1 times 1 times 0.6^2 = 1 times 1 times 0.36 = 0.36 ]The combined probability for both hitting 0 balloons is:[ P(A = 0 text{ and } B = 0) = P(A = 0) times P(B = 0) = 0.09 times 0.36 = 0.0324 ]Next, both hitting 1 balloon.We already calculated ( P(A = 1) ) earlier as 0.42. Let's do the same for Person B.For Person B, hitting exactly 1 balloon:[ P(B = 1) = C(2, 1) times 0.4^1 times (1 - 0.4)^{2 - 1} ]Calculating the combination:[ C(2, 1) = 2 ]So,[ P(B = 1) = 2 times 0.4 times 0.6 = 2 times 0.24 = 0.48 ]The combined probability for both hitting 1 balloon is:[ P(A = 1 text{ and } B = 1) = P(A = 1) times P(B = 1) = 0.42 times 0.48 ]Calculating that:[ 0.42 times 0.48 = 0.2016 ]Lastly, both hitting 2 balloons.For Person A, hitting 2 balloons:[ P(A = 2) = C(2, 2) times 0.7^2 times (1 - 0.7)^{2 - 2} ]Calculating the combination:[ C(2, 2) = 1 ]So,[ P(A = 2) = 1 times 0.49 times 1 = 0.49 ]For Person B, hitting 2 balloons:[ P(B = 2) = C(2, 2) times 0.4^2 times (1 - 0.4)^{2 - 2} ]Calculating the combination:[ C(2, 2) = 1 ]So,[ P(B = 2) = 1 times 0.16 times 1 = 0.16 ]The combined probability for both hitting 2 balloons is:[ P(A = 2 text{ and } B = 2) = P(A = 2) times P(B = 2) = 0.49 times 0.16 = 0.0784 ]Now, adding up all three probabilities:[ P(text{Equal hits}) = P(A = 0 text{ and } B = 0) + P(A = 1 text{ and } B = 1) + P(A = 2 text{ and } B = 2) ][ P(text{Equal hits}) = 0.0324 + 0.2016 + 0.0784 ]Calculating the sum:[ 0.0324 + 0.2016 = 0.234 ][ 0.234 + 0.0784 = 0.3124 ]So, the probability that both Person A and Person B hit an equal number of balloons is 0.3124.Wait, let me double-check my calculations to make sure I didn't make any mistakes.For part (I):- Person A hitting 1: 2 * 0.7 * 0.3 = 0.42- Person B hitting 2: 1 * 0.4^2 = 0.16- Multiply: 0.42 * 0.16 = 0.0672That seems correct.For part (II):- Both hitting 0: 0.09 * 0.36 = 0.0324- Both hitting 1: 0.42 * 0.48 = 0.2016- Both hitting 2: 0.49 * 0.16 = 0.0784Adding them up: 0.0324 + 0.2016 = 0.234; 0.234 + 0.0784 = 0.3124That also seems correct.I think I did everything right. I used the binomial formula for each case, calculated the individual probabilities, multiplied them since the events are independent, and then summed them up for the equal hits case.Just to be thorough, let me verify the binomial coefficients and exponents.For Person A:- Hitting 0: C(2,0)=1, 0.7^0=1, 0.3^2=0.09- Hitting 1: C(2,1)=2, 0.7^1=0.7, 0.3^1=0.3, so 2*0.7*0.3=0.42- Hitting 2: C(2,2)=1, 0.7^2=0.49, 0.3^0=1, so 0.49For Person B:- Hitting 0: C(2,0)=1, 0.4^0=1, 0.6^2=0.36- Hitting 1: C(2,1)=2, 0.4^1=0.4, 0.6^1=0.6, so 2*0.4*0.6=0.48- Hitting 2: C(2,2)=1, 0.4^2=0.16, 0.6^0=1, so 0.16All these look correct. So, the calculations for each probability seem accurate.Therefore, I'm confident that my answers are correct.**Final Answer**(I) The probability is boxed{0.0672}.(II) The probability is boxed{0.3124}.

💡Okay, so I have this problem about sets and complements. Let me try to understand it step by step. The sets are M and N. M is defined as all real numbers x where x is greater than 2. So, M = {x | x > 2}. That means M includes numbers like 3, 4, 5, and so on, but not 2 itself.Then, N is defined as all real numbers x where x is greater than 1 and less than or equal to 3. So, N = {x | 1 < x ≤ 3}. This set includes numbers like 1.5, 2, 2.5, and 3, but not 1 or anything less than or equal to 1.The question is asking me to find the set N ∩ (¬_R M), where ¬_R M denotes the complement of set M in the real numbers. I think the complement of M would be all real numbers that are not in M. Since M is {x | x > 2}, the complement should be all real numbers x where x ≤ 2. So, ¬_R M = {x | x ≤ 2}.Now, I need to find the intersection of N and ¬_R M. The intersection of two sets is the set of elements that are common to both sets. So, N ∩ (¬_R M) would be all real numbers that are in both N and ¬_R M.Let me write down what I have:- N = {x | 1 < x ≤ 3}- ¬_R M = {x | x ≤ 2}So, N ∩ (¬_R M) = {x | 1 < x ≤ 3 and x ≤ 2}. Combining these conditions, x has to be greater than 1 and less than or equal to 2. Therefore, N ∩ (¬_R M) = {x | 1 < x ≤ 2}.Looking at the options provided:A: (1,2]B: [-2,2]C: (1,2)D: [2,3]The set {x | 1 < x ≤ 2} corresponds to option A: (1,2]. So, I think the correct answer is A.Wait, let me double-check. The complement of M is everything less than or equal to 2, and N is from just above 1 to 3. So, their overlap should be from just above 1 to 2, including 2. So, yes, (1,2] makes sense. Option A is correct.

💡Okay, so I have this problem where there are four excellent students: A, B, C, and D. They've been recommended for admission to three universities: Peking University, Tsinghua University, and Fudan University. The condition is that each university must receive at least one student. I need to figure out the total number of different admission schemes possible.Alright, let's break this down. First, I need to understand what exactly is being asked. It's about assigning four students to three universities with each university getting at least one student. So, it's a problem of distributing four distinct objects (students) into three distinct boxes (universities) with no empty boxes.I remember that in combinatorics, when we want to distribute objects into boxes with certain conditions, we can use the principle of inclusion-exclusion or Stirling numbers of the second kind. But since the universities are distinct, maybe I should think in terms of functions or assignments.Let me recall: If we have n distinct objects and k distinct boxes, the number of ways to distribute the objects into the boxes is k^n. But this includes cases where some boxes might be empty. Since we need each university to have at least one student, we need to subtract the cases where one or more universities are empty.So, the formula for the number of onto functions from a set of size n to a set of size k is given by the inclusion-exclusion principle:Number of onto functions = k^n - C(k,1)*(k-1)^n + C(k,2)*(k-2)^n - ... + (-1)^k*C(k,k)*(k-k)^nIn this case, n = 4 (students) and k = 3 (universities). Plugging in the values:Number of onto functions = 3^4 - C(3,1)*2^4 + C(3,2)*1^4 - C(3,3)*0^4Calculating each term:3^4 = 81C(3,1)*2^4 = 3*16 = 48C(3,2)*1^4 = 3*1 = 3C(3,3)*0^4 = 1*0 = 0So, putting it all together:Number of onto functions = 81 - 48 + 3 - 0 = 36Wait, that seems straightforward, but I want to make sure I'm not missing anything. Maybe I should think about it another way to verify.Another approach is to consider the possible ways to partition the four students into three non-empty groups and then assign each group to a university. This involves using Stirling numbers of the second kind, which count the number of ways to partition a set of n objects into k non-empty subsets.The formula for Stirling numbers of the second kind, S(n,k), is:S(n,k) = S(n-1,k-1) + k*S(n-1,k)But I might not need to compute it recursively. Instead, I can use the formula:S(n,k) = (1/k!)*sum_{i=0 to k} (-1)^i * C(k,i) * (k-i)^nWhich is similar to the inclusion-exclusion principle we used earlier.So, for n=4 and k=3:S(4,3) = (1/3!)*[C(3,0)*3^4 - C(3,1)*2^4 + C(3,2)*1^4 - C(3,3)*0^4]Calculating:C(3,0)*3^4 = 1*81 = 81C(3,1)*2^4 = 3*16 = 48C(3,2)*1^4 = 3*1 = 3C(3,3)*0^4 = 1*0 = 0So,S(4,3) = (1/6)*(81 - 48 + 3 - 0) = (1/6)*(36) = 6So, there are 6 ways to partition the four students into three non-empty groups. But since the universities are distinct, we need to assign each group to a specific university. There are 3! ways to assign the groups to the universities.Therefore, the total number of admission schemes is S(4,3)*3! = 6*6 = 36Hmm, that matches the earlier result. So, both methods give me 36.But just to be thorough, maybe I should consider enumerating the possibilities to ensure that 36 is indeed the correct number.Let's think about how the students can be distributed. Since there are four students and three universities, one university will have two students, and the other two universities will have one student each.First, choose which university gets two students. There are 3 choices for this.Then, choose which two students go to that university. There are C(4,2) = 6 ways to choose the pair.Once the pair is chosen, the remaining two students each go to one of the remaining two universities. There are 2! = 2 ways to assign them.So, the total number of ways is 3 (choices for the university with two students) * 6 (choices for the pair) * 2 (assignments for the remaining students) = 3*6*2 = 36Again, I get 36. This seems consistent across different methods: inclusion-exclusion, Stirling numbers, and direct enumeration.I think I'm confident now that the total number of different admission schemes is 36.

💡Okay, so I have this problem where I need to find all positive integers n less than 10799 such that the sum from k equals 1 to n of 1 divided by (sin k times sin(k+1)) equals m times (sin n divided by sin squared 1), where m is either 1 or 2. Hmm, that seems a bit complex, but let me try to break it down step by step.First, I remember that sometimes when dealing with sums involving trigonometric functions, there are identities or telescoping series that can simplify things. Maybe I can rewrite the term 1/(sin k sin(k+1)) in a different form. Let me think... Oh, I recall that there's an identity involving the difference of cotangents. Specifically, cot A - cot B can be expressed in terms of sine functions. Let me verify that.Yes, the identity is: cot A - cot B = (sin(B - A))/(sin A sin B). So, if I set A = k and B = k+1, then B - A = 1. So, cot k - cot(k+1) = sin(1)/(sin k sin(k+1)). That means 1/(sin k sin(k+1)) = (cot k - cot(k+1))/sin 1. So, substituting this back into the sum, I get:Sum from k=1 to n of [ (cot k - cot(k+1)) / sin 1 ].Since sin 1 is a constant, I can factor it out of the sum:(1 / sin 1) * Sum from k=1 to n of (cot k - cot(k+1)).Ah, this is a telescoping series! That means most terms will cancel out. Let's write out the first few terms to see:When k=1: cot 1 - cot 2When k=2: cot 2 - cot 3When k=3: cot 3 - cot 4...When k=n: cot n - cot(n+1)So, when we add all these up, the cot 2, cot 3, ..., cot n terms cancel out, leaving us with cot 1 - cot(n+1). Therefore, the entire sum simplifies to:(1 / sin 1) * (cot 1 - cot(n+1)).So, the original equation becomes:(1 / sin 1) * (cot 1 - cot(n+1)) = m * (sin n / sin² 1).Let me write that out:(cot 1 - cot(n+1)) / sin 1 = m * sin n / sin² 1.I can multiply both sides by sin 1 to get rid of the denominator on the left:cot 1 - cot(n+1) = m * sin n / sin 1.Hmm, so now I have:cot 1 - cot(n+1) = (m / sin 1) * sin n.I need to solve for n here. Let me recall that cot x is cos x / sin x. So, let's express cot 1 and cot(n+1) in terms of sine and cosine:cot 1 = cos 1 / sin 1,cot(n+1) = cos(n+1) / sin(n+1).Substituting these into the equation:cos 1 / sin 1 - cos(n+1) / sin(n+1) = (m / sin 1) * sin n.Let me rearrange this equation:cos 1 / sin 1 - (m / sin 1) * sin n = cos(n+1) / sin(n+1).Factor out 1 / sin 1 on the left side:[cos 1 - m sin n] / sin 1 = cos(n+1) / sin(n+1).So, cross-multiplying:[cos 1 - m sin n] * sin(n+1) = sin 1 * cos(n+1).Hmm, this is getting a bit complicated. Maybe I can use some trigonometric identities to simplify this. Let me recall that sin(n+1) can be written as sin n cos 1 + cos n sin 1, and cos(n+1) can be written as cos n cos 1 - sin n sin 1.So, substituting these into the equation:[cos 1 - m sin n] * [sin n cos 1 + cos n sin 1] = sin 1 * [cos n cos 1 - sin n sin 1].Let me expand the left side:cos 1 * sin n cos 1 + cos 1 * cos n sin 1 - m sin n * sin n cos 1 - m sin n * cos n sin 1.Simplify term by term:First term: cos 1 * sin n cos 1 = sin n cos² 1.Second term: cos 1 * cos n sin 1 = sin 1 cos 1 cos n.Third term: -m sin² n cos 1.Fourth term: -m sin n cos n sin 1.So, the left side becomes:sin n cos² 1 + sin 1 cos 1 cos n - m sin² n cos 1 - m sin n cos n sin 1.The right side is:sin 1 cos n cos 1 - sin² 1 sin n.So, let's write the equation again:sin n cos² 1 + sin 1 cos 1 cos n - m sin² n cos 1 - m sin n cos n sin 1 = sin 1 cos n cos 1 - sin² 1 sin n.Let me bring all terms to the left side:sin n cos² 1 + sin 1 cos 1 cos n - m sin² n cos 1 - m sin n cos n sin 1 - sin 1 cos n cos 1 + sin² 1 sin n = 0.Simplify term by term:sin n cos² 1: remains as is.sin 1 cos 1 cos n - sin 1 cos 1 cos n: these cancel out.- m sin² n cos 1: remains.- m sin n cos n sin 1: remains.+ sin² 1 sin n: remains.So, the equation simplifies to:sin n cos² 1 - m sin² n cos 1 - m sin n cos n sin 1 + sin² 1 sin n = 0.Let me factor out sin n from all terms:sin n [cos² 1 - m sin n cos 1 - m cos n sin 1 + sin² 1] = 0.Since n is a positive integer, sin n is not zero (unless n is a multiple of π, but n is an integer, so sin n ≠ 0). Therefore, the term in the brackets must be zero:cos² 1 - m sin n cos 1 - m cos n sin 1 + sin² 1 = 0.Let me rearrange the terms:cos² 1 + sin² 1 - m sin n cos 1 - m cos n sin 1 = 0.I know that cos² 1 + sin² 1 = 1, so this simplifies to:1 - m (sin n cos 1 + cos n sin 1) = 0.But sin n cos 1 + cos n sin 1 is equal to sin(n + 1) by the sine addition formula. So, substituting that in:1 - m sin(n + 1) = 0.Therefore:m sin(n + 1) = 1.So, sin(n + 1) = 1/m.Since m is either 1 or 2, we have two cases:Case 1: m = 1.Then, sin(n + 1) = 1.Case 2: m = 2.Then, sin(n + 1) = 1/2.Now, let's solve each case.Case 1: sin(n + 1) = 1.The solutions to sin x = 1 are x = π/2 + 2πk, where k is an integer.So, n + 1 = π/2 + 2πk.Therefore, n = π/2 + 2πk - 1.But n must be a positive integer less than 10799. So, we need to find all integers k such that n is positive and less than 10799.Similarly, for Case 2: sin(n + 1) = 1/2.The solutions to sin x = 1/2 are x = π/6 + 2πk or x = 5π/6 + 2πk, where k is an integer.So, n + 1 = π/6 + 2πk or n + 1 = 5π/6 + 2πk.Therefore, n = π/6 + 2πk - 1 or n = 5π/6 + 2πk - 1.Again, n must be a positive integer less than 10799.Now, let's compute the possible values of k for each case.First, let's note that π is approximately 3.1415926535, so 2π is approximately 6.283185307.Case 1: n = π/2 + 2πk - 1 ≈ 1.5708 + 6.283185307k - 1 ≈ 0.5708 + 6.283185307k.We need n to be a positive integer less than 10799.So, 0.5708 + 6.283185307k < 10799.Solving for k:6.283185307k < 10799 - 0.5708 ≈ 10798.4292.k < 10798.4292 / 6.283185307 ≈ 1720.000.Since k must be an integer, the maximum k is 1719.But let's check n for k=0: n ≈ 0.5708, which is less than 1, so n=0, but n must be positive, so k starts from 1.Wait, no, for k=0, n ≈ 0.5708, which is less than 1, so n=0 is not positive. So, k starts from 1.Wait, actually, n must be a positive integer, so n must be at least 1. So, let's solve for k such that n ≥ 1.0.5708 + 6.283185307k ≥ 1.6.283185307k ≥ 0.4292.k ≥ 0.4292 / 6.283185307 ≈ 0.0683.So, k starts from 1.Similarly, the maximum k is 1719, as above.So, k ranges from 1 to 1719.But wait, let's compute n for k=1:n ≈ 0.5708 + 6.283185307*1 ≈ 6.853985307.So, n ≈ 6.853985307, which is approximately 7. But n must be an integer, so n=7.Wait, but n must be exactly equal to π/2 + 2πk -1. But π is irrational, so n will never be an integer unless π/2 + 2πk -1 is integer. But since π is irrational, π/2 + 2πk is irrational, so π/2 + 2πk -1 is irrational, so n cannot be an integer. Therefore, there are no solutions for m=1.Wait, that's a problem. Because n must be an integer, but n = π/2 + 2πk -1 is irrational, so n cannot be an integer. Therefore, for m=1, there are no solutions.Hmm, that's unexpected. Let me double-check my steps.Wait, in the equation m sin(n +1) =1, for m=1, sin(n+1)=1. So, n+1= π/2 + 2πk, so n= π/2 + 2πk -1. Since π is irrational, n cannot be an integer. Therefore, indeed, there are no solutions for m=1.So, moving on to Case 2: m=2.Then, sin(n +1)=1/2.So, n +1= π/6 + 2πk or n +1=5π/6 + 2πk.Therefore, n= π/6 + 2πk -1 or n=5π/6 + 2πk -1.Again, n must be a positive integer less than 10799.Let's compute the approximate values:For n= π/6 + 2πk -1 ≈ 0.5236 + 6.283185307k -1 ≈ -0.4764 + 6.283185307k.Similarly, for n=5π/6 + 2πk -1 ≈ 2.61799 + 6.283185307k -1 ≈ 1.61799 + 6.283185307k.We need n to be positive, so for the first case:-0.4764 + 6.283185307k > 0.6.283185307k > 0.4764.k > 0.4764 / 6.283185307 ≈ 0.0758.So, k starts from 1.Similarly, for the second case:1.61799 + 6.283185307k > 0.Which is always true for k ≥0.But n must be less than 10799.Let's compute the maximum k for each case.First case: n= π/6 + 2πk -1 ≈ -0.4764 + 6.283185307k < 10799.So, 6.283185307k < 10799 + 0.4764 ≈ 10799.4764.k < 10799.4764 / 6.283185307 ≈ 1720.000.So, k can be from 1 to 1719.Similarly, for the second case: n=5π/6 + 2πk -1 ≈1.61799 + 6.283185307k <10799.So, 6.283185307k <10799 -1.61799≈10797.382.k <10797.382 /6.283185307≈1720.000.So, k can be from 0 to 1719.But n must be a positive integer, so let's check for k=0 in the second case:n≈1.61799 +0≈1.61799, which is approximately 2. So, n=2.But let's check if n=2 satisfies the original equation.Wait, let's compute n=2:Sum from k=1 to 2 of 1/(sin k sin(k+1)).Compute for k=1: 1/(sin1 sin2).Compute for k=2: 1/(sin2 sin3).So, sum=1/(sin1 sin2) +1/(sin2 sin3).We can compute this numerically to check if it equals 2 sin2 / sin²1.But let me see, since n=2 is a possible solution, but n must be an integer, and n=2 is an integer, but wait, n=2 is approximately 1.61799, which is not exactly 2. Wait, no, n=5π/6 +2πk -1 for k=0 is approximately 1.61799, which is not an integer. So, n=2 is not a solution because n must be exactly equal to 5π/6 +2πk -1, which is not an integer.Wait, but n must be an integer, so n=5π/6 +2πk -1 must be integer. But since π is irrational, 5π/6 +2πk is irrational, so n cannot be integer. Therefore, similar to the m=1 case, there are no solutions for m=2 either.Wait, that can't be right. The problem states that n is a positive integer less than 10799, and m is 1 or 2. So, perhaps I made a mistake in assuming that n must be exactly equal to π/2 +2πk -1 or similar. Maybe I need to consider that n is an integer, so n+1 must be such that sin(n+1)=1 or 1/2, but n+1 is in radians, and n is integer, so n+1 is an integer in radians. But radians are just a unit, so n+1 is an integer multiple of π/2 or π/6, etc.Wait, but n+1 is an integer, so sin(n+1) must be sin of an integer. But sin of an integer (in radians) is not necessarily 1 or 1/2. So, perhaps the equation sin(n+1)=1/m has solutions only when n+1 is such that sin(n+1)=1 or 1/2, but since n+1 is an integer, we need to find integers x=n+1 such that sin x=1 or 1/2.But sin x=1 occurs at x=π/2 +2πk, but x must be integer, so π/2 +2πk must be integer. But π is irrational, so π/2 +2πk is irrational, so x cannot be integer. Similarly, sin x=1/2 occurs at x=π/6 +2πk or 5π/6 +2πk, which are also irrational, so x cannot be integer. Therefore, there are no integer solutions for x=n+1 such that sin x=1 or 1/2. Therefore, there are no solutions for n.But that contradicts the problem statement, which asks to determine all such n. So, perhaps I made a mistake in my approach.Wait, let's go back to the equation:m sin(n +1)=1.So, sin(n +1)=1/m.But since sin(n+1) must be between -1 and 1, 1/m must be within that range. For m=1, 1/m=1, which is valid. For m=2, 1/m=0.5, which is also valid.But as we saw, sin(n+1)=1 or 0.5 requires n+1 to be specific angles, which are not integers. Therefore, there are no integer solutions for n.But the problem states that n is a positive integer less than 10799, so perhaps there are no solutions? But the problem asks to determine all such n, implying that there are solutions.Wait, maybe I made a mistake in the earlier steps. Let me go back to the equation:Sum from k=1 to n of 1/(sin k sin(k+1)) = m sin n / sin²1.We transformed it to:cot1 - cot(n+1) = m sin n / sin1.Then, we expressed cot1 and cot(n+1) in terms of sine and cosine:cos1/sin1 - cos(n+1)/sin(n+1) = m sin n / sin1.Then, we rearranged to:[cos1 - m sin n] / sin1 = cos(n+1)/sin(n+1).Cross-multiplying:[cos1 - m sin n] sin(n+1) = sin1 cos(n+1).Then, expanding sin(n+1) and cos(n+1):[cos1 - m sin n][sin n cos1 + cos n sin1] = sin1 [cos n cos1 - sin n sin1].Expanding the left side:cos1 sin n cos1 + cos1 cos n sin1 - m sin²n cos1 - m sin n cos n sin1.Which simplifies to:sin n cos²1 + sin1 cos1 cos n - m sin²n cos1 - m sin n cos n sin1.The right side is:sin1 cos n cos1 - sin²1 sin n.Bringing all terms to the left:sin n cos²1 + sin1 cos1 cos n - m sin²n cos1 - m sin n cos n sin1 - sin1 cos n cos1 + sin²1 sin n =0.Simplifying:sin n cos²1 - m sin²n cos1 - m sin n cos n sin1 + sin²1 sin n =0.Factoring out sin n:sin n [cos²1 - m sin n cos1 - m cos n sin1 + sin²1] =0.Since sin n ≠0, we have:cos²1 - m sin n cos1 - m cos n sin1 + sin²1=0.Which simplifies to:1 - m (sin n cos1 + cos n sin1)=0.So, 1 - m sin(n+1)=0.Thus, m sin(n+1)=1.So, sin(n+1)=1/m.Therefore, n+1 must satisfy sin(n+1)=1/m.But n+1 is an integer, say x=n+1.So, sin x=1/m.But x is an integer, so sin x must be 1 or 1/2, depending on m.But as we saw, sin x=1 or 1/2 requires x to be specific angles, which are not integers. Therefore, there are no integer solutions for x, hence no solutions for n.But the problem states that m is 1 or 2 and n is a positive integer less than 10799. So, perhaps the answer is that there are no such n.But that seems odd. Maybe I made a mistake in the telescoping series approach.Wait, let me double-check the telescoping sum.We had:Sum from k=1 to n of 1/(sin k sin(k+1)) = (cot1 - cot(n+1))/sin1.Yes, that seems correct.Then, setting this equal to m sin n / sin²1:(cot1 - cot(n+1))/sin1 = m sin n / sin²1.Multiplying both sides by sin1:cot1 - cot(n+1) = m sin n / sin1.Yes, that's correct.Then, expressing cot1 and cot(n+1):cos1/sin1 - cos(n+1)/sin(n+1) = m sin n / sin1.Yes.Then, rearranging:[cos1 - m sin n] / sin1 = cos(n+1)/sin(n+1).Cross-multiplying:[cos1 - m sin n] sin(n+1) = sin1 cos(n+1).Yes.Then, expanding sin(n+1) and cos(n+1):[cos1 - m sin n][sin n cos1 + cos n sin1] = sin1 [cos n cos1 - sin n sin1].Yes.Expanding left side:cos1 sin n cos1 + cos1 cos n sin1 - m sin²n cos1 - m sin n cos n sin1.Which is:sin n cos²1 + sin1 cos1 cos n - m sin²n cos1 - m sin n cos n sin1.Right side:sin1 cos n cos1 - sin²1 sin n.Bringing all terms to left:sin n cos²1 + sin1 cos1 cos n - m sin²n cos1 - m sin n cos n sin1 - sin1 cos n cos1 + sin²1 sin n =0.Simplifying:sin n cos²1 - m sin²n cos1 - m sin n cos n sin1 + sin²1 sin n =0.Factoring sin n:sin n [cos²1 - m sin n cos1 - m cos n sin1 + sin²1] =0.Since sin n ≠0, we have:cos²1 - m sin n cos1 - m cos n sin1 + sin²1=0.Which is:1 - m (sin n cos1 + cos n sin1)=0.Thus, 1 - m sin(n+1)=0.So, m sin(n+1)=1.Therefore, sin(n+1)=1/m.But n+1 is an integer, so sin(n+1)=1/m must hold for some integer n+1.But sin of an integer (in radians) can only be 1 or 1/2 if n+1 is such that sin(n+1)=1 or 1/2, but as we saw, n+1 must be π/2 +2πk or π/6 +2πk, etc., which are not integers. Therefore, there are no integer solutions for n.But the problem states that n is a positive integer less than 10799, so perhaps the answer is that there are no such n.Wait, but the problem says "determine all such n", implying that there are solutions. Maybe I made a mistake in assuming that n+1 must be such that sin(n+1)=1/m. Perhaps I need to consider that n+1 is in degrees? But the problem didn't specify, and usually, in higher mathematics, angles are in radians unless specified otherwise.Alternatively, maybe I need to consider that n is in degrees, but that would complicate things because the sum would involve degrees, but the right side is in terms of sin n, which would be in radians. That seems inconsistent.Alternatively, perhaps the problem is in degrees, so n is in degrees, and sin n is in degrees. But that's a stretch, as usually, in calculus and higher math, angles are in radians.Wait, let me check the original problem again:"Let m be equal to 1 or 2 and n<10799 be a positive integer. Determine all such n for which sum from k=1 to n of 1/(sin k sin(k+1)) = m sin n / sin²1."It doesn't specify degrees or radians, but in mathematical problems, unless specified, angles are in radians.Therefore, I think my conclusion is correct: there are no positive integers n less than 10799 satisfying the given equation for m=1 or m=2.But the problem asks to determine all such n, so perhaps the answer is that there are no solutions.Alternatively, maybe I made a mistake in the telescoping series approach. Let me try a different approach.Let me consider the sum S = sum from k=1 to n of 1/(sin k sin(k+1)).We can write 1/(sin k sin(k+1)) as (cot k - cot(k+1))/sin1, as we did before.So, S = (cot1 - cot(n+1))/sin1.Set this equal to m sin n / sin²1.Thus:(cot1 - cot(n+1))/sin1 = m sin n / sin²1.Multiply both sides by sin1:cot1 - cot(n+1) = m sin n / sin1.Express cot1 and cot(n+1) as cos1/sin1 and cos(n+1)/sin(n+1):cos1/sin1 - cos(n+1)/sin(n+1) = m sin n / sin1.Multiply both sides by sin1:cos1 - [cos(n+1)/sin(n+1)] sin1 = m sin n.So:cos1 - sin1 cos(n+1)/sin(n+1) = m sin n.Let me write this as:cos1 - sin1 cot(n+1) = m sin n.Hmm, maybe I can write this as:cos1 - m sin n = sin1 cot(n+1).So:cot(n+1) = (cos1 - m sin n)/sin1.Now, cot(n+1) is cos(n+1)/sin(n+1).So, we have:cos(n+1)/sin(n+1) = (cos1 - m sin n)/sin1.Cross-multiplying:cos(n+1) sin1 = (cos1 - m sin n) sin(n+1).Let me expand sin(n+1):sin(n+1) = sin n cos1 + cos n sin1.So, substituting:cos(n+1) sin1 = (cos1 - m sin n)(sin n cos1 + cos n sin1).Expanding the right side:cos1 sin n cos1 + cos1 cos n sin1 - m sin²n cos1 - m sin n cos n sin1.Which is:sin n cos²1 + sin1 cos1 cos n - m sin²n cos1 - m sin n cos n sin1.So, the equation becomes:cos(n+1) sin1 = sin n cos²1 + sin1 cos1 cos n - m sin²n cos1 - m sin n cos n sin1.Let me rearrange terms:cos(n+1) sin1 - sin n cos²1 - sin1 cos1 cos n + m sin²n cos1 + m sin n cos n sin1 =0.Hmm, this seems similar to what I had before. Maybe I can factor terms.Let me group terms with cos(n+1):cos(n+1) sin1.Terms with sin n:- sin n cos²1.Terms with cos n:- sin1 cos1 cos n + m sin n cos n sin1.Terms with sin²n:+ m sin²n cos1.So, let's write:cos(n+1) sin1 - sin n cos²1 + cos n (- sin1 cos1 + m sin n sin1) + m sin²n cos1 =0.Hmm, not sure if that helps. Maybe I can use the identity for cos(n+1):cos(n+1) = cos n cos1 - sin n sin1.So, substituting:[cos n cos1 - sin n sin1] sin1 - sin n cos²1 + cos n (- sin1 cos1 + m sin n sin1) + m sin²n cos1 =0.Expanding:cos n cos1 sin1 - sin n sin²1 - sin n cos²1 + cos n (- sin1 cos1 + m sin n sin1) + m sin²n cos1 =0.Simplify term by term:cos n cos1 sin1 - sin n sin²1 - sin n cos²1 - cos n sin1 cos1 + m sin n cos n sin1 + m sin²n cos1 =0.Now, let's look for cancellation:cos n cos1 sin1 - cos n sin1 cos1 =0.Similarly, - sin n sin²1 - sin n cos²1 = - sin n (sin²1 + cos²1) = - sin n.So, the equation simplifies to:- sin n + m sin n cos n sin1 + m sin²n cos1 =0.Factor out sin n:sin n [ -1 + m cos n sin1 + m sin n cos1 ] =0.Since sin n ≠0, we have:-1 + m cos n sin1 + m sin n cos1 =0.Which is:m (cos n sin1 + sin n cos1) =1.But cos n sin1 + sin n cos1 = sin(n +1).So, m sin(n +1)=1.Thus, sin(n +1)=1/m.Which is the same equation as before.Therefore, my earlier conclusion stands: sin(n +1)=1/m.But since n+1 is an integer, and sin of an integer (in radians) cannot be exactly 1 or 1/2, there are no solutions.Therefore, the answer is that there are no positive integers n less than 10799 satisfying the given equation for m=1 or m=2.But the problem asks to determine all such n, so perhaps the answer is that there are no solutions.Alternatively, maybe I need to consider that n+1 is such that sin(n+1)=1/m, but n+1 is not necessarily an integer multiple of π, but just an integer. However, sin(n+1)=1/m must hold for some integer n+1, which is not possible because sin of an integer (in radians) cannot be exactly 1 or 1/2.Therefore, the conclusion is that there are no such n.But the problem states that n is a positive integer less than 10799, so perhaps the answer is that there are no solutions.Alternatively, maybe I made a mistake in the initial steps. Let me try plugging in small values of n to see if the equation holds.For example, let n=1:Sum from k=1 to1 of 1/(sin1 sin2)=1/(sin1 sin2).Compute the right side for m=1: sin1 / sin²1=1/sin1.So, 1/(sin1 sin2) vs 1/sin1.These are equal only if 1/sin2=1, which implies sin2=1, but sin2≈0.909, not 1. So, not equal.For m=2: 2 sin1 / sin²1=2/sin1.Compare to 1/(sin1 sin2)=1/(sin1 *0.909)≈1/(0.841*0.909)≈1/0.765≈1.307.2/sin1≈2/0.841≈2.38.Not equal.So, n=1 is not a solution.Similarly, n=2:Sum from k=1 to2 of 1/(sin1 sin2) +1/(sin2 sin3).Compute numerically:1/(sin1 sin2)≈1/(0.841*0.909)≈1.307.1/(sin2 sin3)≈1/(0.909*0.141)≈1/0.128≈7.81.Total sum≈1.307+7.81≈9.117.Right side for m=1: sin2 / sin²1≈0.909 /0.708≈1.283.For m=2: 2 sin2 / sin²1≈2*0.909 /0.708≈2.566.Neither equals 9.117.So, n=2 is not a solution.Similarly, n=3:Sum≈9.117 +1/(sin3 sin4).sin3≈0.141, sin4≈-0.757.But 1/(sin3 sin4)=1/(0.141*(-0.757))≈1/(-0.106)≈-9.43.So, sum≈9.117 -9.43≈-0.313.Right side for m=1: sin3 / sin²1≈0.141 /0.708≈0.199.For m=2: 2*0.141 /0.708≈0.398.Neither equals -0.313.So, n=3 is not a solution.Similarly, n=4:Sum≈-0.313 +1/(sin4 sin5).sin4≈-0.757, sin5≈-0.959.1/(sin4 sin5)=1/((-0.757)*(-0.959))≈1/0.726≈1.377.Sum≈-0.313 +1.377≈1.064.Right side for m=1: sin4 / sin²1≈-0.757 /0.708≈-1.069.For m=2: 2*(-0.757)/0.708≈-2.138.Neither equals 1.064.So, n=4 is not a solution.Similarly, n=5:Sum≈1.064 +1/(sin5 sin6).sin5≈-0.959, sin6≈-0.279.1/(sin5 sin6)=1/((-0.959)*(-0.279))≈1/0.268≈3.73.Sum≈1.064 +3.73≈4.794.Right side for m=1: sin5 / sin²1≈-0.959 /0.708≈-1.354.For m=2: 2*(-0.959)/0.708≈-2.708.Neither equals 4.794.So, n=5 is not a solution.It seems that for small n, the sum does not equal m sin n / sin²1 for m=1 or 2.Therefore, perhaps there are no solutions for n.Thus, the answer is that there are no positive integers n less than 10799 satisfying the given equation for m=1 or m=2.

💡Okay, so I need to prove Fermat's Little Theorem, which says that if ( p ) is a prime number and ( a ) is not divisible by ( p ), then ( a^{p-1} equiv 1 pmod{p} ). Hmm, I remember hearing about this theorem in my number theory class, but I need to recall exactly how it works.First, let me understand the statement. If ( p ) is prime and ( a ) is an integer that's not a multiple of ( p ), then raising ( a ) to the power of ( p-1 ) gives a result that's one more than a multiple of ( p ). So, ( a^{p-1} - 1 ) is divisible by ( p ).I think there are a few ways to prove this. One common method is using modular arithmetic and permutations. Another approach might involve mathematical induction. I'll try to explore both methods to see which one makes more sense to me.Starting with the first method: modular arithmetic and permutations. Let's consider the set of integers ( {1, 2, 3, ldots, p-1} ). Since ( p ) is prime, none of these integers share a common factor with ( p ), which is important.Now, if I multiply each element of this set by ( a ), I get the set ( {a, 2a, 3a, ldots, (p-1)a} ). Because ( a ) is not divisible by ( p ), each of these products ( ka ) (where ( k ) ranges from 1 to ( p-1 )) will also not be divisible by ( p ). So, when I take each ( ka ) modulo ( p ), I should get a unique remainder between 1 and ( p-1 ).This means that the set ( {a, 2a, 3a, ldots, (p-1)a} ) modulo ( p ) is just a permutation of the set ( {1, 2, 3, ldots, p-1} ). Therefore, the product of the elements in both sets should be congruent modulo ( p ).So, the product of ( {a, 2a, 3a, ldots, (p-1)a} ) is ( a^{p-1} times (p-1)! ), and the product of ( {1, 2, 3, ldots, p-1} ) is ( (p-1)! ). Therefore, we have:[a^{p-1} times (p-1)! equiv (p-1)! pmod{p}]Since ( (p-1)! ) is not divisible by ( p ) (because all its factors are less than ( p ) and ( p ) is prime), we can divide both sides of the congruence by ( (p-1)! ), which gives:[a^{p-1} equiv 1 pmod{p}]Okay, that seems to work out. But let me make sure I understand why dividing by ( (p-1)! ) is valid. Since ( (p-1)! ) and ( p ) are coprime (their greatest common divisor is 1), there exists a multiplicative inverse of ( (p-1)! ) modulo ( p ). Therefore, multiplying both sides by this inverse effectively cancels out ( (p-1)! ), leaving us with ( a^{p-1} equiv 1 pmod{p} ).Now, let me try to think about another way to prove this, maybe using mathematical induction. I know that induction involves proving a base case and then showing that if the statement holds for some ( n ), it holds for ( n+1 ).Let's see. The theorem can be generalized to say that for any integer ( n ), ( n^p equiv n pmod{p} ). This is actually a slightly different statement, but it's related. If I can prove this, then setting ( n ) not divisible by ( p ) would give ( n^{p-1} equiv 1 pmod{p} ) by dividing both sides by ( n ).So, let's try induction on ( n ).**Base Case**: ( n = 1 ). Then ( 1^p = 1 ), and ( 1 equiv 1 pmod{p} ). So, the base case holds.**Inductive Step**: Assume that for some integer ( k ), ( k^p equiv k pmod{p} ). We need to show that ( (k+1)^p equiv k+1 pmod{p} ).Using the binomial theorem, we can expand ( (k+1)^p ):[(k+1)^p = sum_{i=0}^{p} binom{p}{i} k^i]Now, for ( 1 leq i leq p-1 ), the binomial coefficients ( binom{p}{i} ) are divisible by ( p ) because ( p ) is prime. This is because ( binom{p}{i} = frac{p!}{i!(p-i)!} ), and since ( p ) is prime, it doesn't divide the denominator, so ( p ) must divide the numerator, making the entire coefficient divisible by ( p ).Therefore, modulo ( p ), all the middle terms ( binom{p}{i} k^i ) for ( 1 leq i leq p-1 ) are congruent to 0. So, we have:[(k+1)^p equiv k^p + 1 pmod{p}]But by the inductive hypothesis, ( k^p equiv k pmod{p} ). Therefore:[(k+1)^p equiv k + 1 pmod{p}]This completes the inductive step. Hence, by mathematical induction, ( n^p equiv n pmod{p} ) holds for all integers ( n ). If ( n ) is not divisible by ( p ), we can divide both sides by ( n ) to get ( n^{p-1} equiv 1 pmod{p} ).Wait, does dividing by ( n ) make sense here? Since ( n ) is not divisible by ( p ), ( n ) and ( p ) are coprime, so ( n ) has a multiplicative inverse modulo ( p ). Therefore, dividing both sides by ( n ) is valid, and we indeed get ( n^{p-1} equiv 1 pmod{p} ).Okay, so both methods seem to work. The first one uses properties of permutations and modular arithmetic, while the second one uses mathematical induction and the binomial theorem. I think I understand both approaches now.But let me double-check if there are any gaps in my reasoning. In the first proof, I assumed that multiplying by ( a ) permutes the set ( {1, 2, ldots, p-1} ) modulo ( p ). Is that always true? Well, since ( a ) is not divisible by ( p ), it is invertible modulo ( p ), meaning there exists some ( b ) such that ( ab equiv 1 pmod{p} ). Therefore, multiplication by ( a ) is a bijection on the set ( {1, 2, ldots, p-1} ), which implies it's a permutation. So, that part checks out.In the second proof, I used the binomial theorem and the fact that the binomial coefficients are divisible by ( p ) for ( 1 leq i leq p-1 ). Is that correct? Yes, because ( p ) is prime, and the binomial coefficients ( binom{p}{i} ) for ( 1 leq i leq p-1 ) have ( p ) in the numerator and no factors of ( p ) in the denominator, so they must be divisible by ( p ).Another thing to consider is whether the inductive step correctly handles the transition from ( k ) to ( k+1 ). I think it does because we expanded ( (k+1)^p ) and showed that all the middle terms vanish modulo ( p ), leaving us with ( k^p + 1 ), which by the inductive hypothesis becomes ( k + 1 ).Is there any case where these proofs might fail? Well, both proofs rely on ( p ) being prime. If ( p ) were composite, these proofs wouldn't necessarily hold because, for example, the set ( {1, 2, ldots, p-1} ) wouldn't necessarily be invertible modulo ( p ), and the binomial coefficients might not be divisible by ( p ).Also, in the first proof, we crucially used that ( a ) is not divisible by ( p ) to ensure that ( a ) has an inverse modulo ( p ). If ( a ) were divisible by ( p ), then ( a equiv 0 pmod{p} ), and ( a^{p-1} equiv 0 pmod{p} ), which is not congruent to 1. So, the condition that ( a ) is not divisible by ( p ) is necessary.I think I've covered the main points. Both proofs seem solid, and they approach the problem from different angles—modular arithmetic and induction. It's reassuring that both methods lead to the same conclusion, which adds to the theorem's validity.Just to make sure, let me test the theorem with a small prime and an integer not divisible by it. Let's take ( p = 5 ) and ( a = 2 ). Then, ( 2^{5-1} = 16 ). Calculating ( 16 mod 5 ), we get ( 1 ), which matches the theorem. Another example: ( p = 7 ), ( a = 3 ). ( 3^{6} = 729 ). ( 729 mod 7 ) is ( 1 ), since ( 7 times 104 = 728 ), and ( 729 - 728 = 1 ). Yep, it works.What if ( a ) is divisible by ( p )? Let's try ( p = 5 ), ( a = 5 ). Then ( 5^{4} = 625 ). ( 625 mod 5 = 0 ), which is not 1, as expected. So, the condition that ( a ) is not divisible by ( p ) is indeed necessary.I think I've got a good grasp on why Fermat's Little Theorem holds. Both proofs make sense, and the examples confirm the theorem's validity. It's fascinating how different approaches can lead to the same result in number theory.One last thing—Fermat's Little Theorem is widely used in cryptography, particularly in algorithms like RSA. Understanding its proof helps in appreciating why these cryptographic systems work. It's also a beautiful example of how properties of prime numbers can lead to such elegant and powerful theorems.In summary, by considering the permutation of residues modulo ( p ) and using mathematical induction with the binomial theorem, we've successfully proven Fermat's Little Theorem. It's a fundamental result in number theory with far-reaching applications.