💡Okay, so I need to compute the sum of sin² terms from 6 degrees up to 174 degrees, increasing by 6 degrees each time. That is, sin²6° + sin²12° + sin²18° + ... + sin²174°. Hmm, that seems like a lot of terms. Let me see how many terms there are first.Starting at 6°, each term increases by 6°, so the sequence is 6°, 12°, 18°, ..., 174°. To find the number of terms, I can use the formula for the nth term of an arithmetic sequence: a_n = a_1 + (n-1)d. Here, a_1 is 6°, d is 6°, and a_n is 174°. So, 174 = 6 + (n-1)*6. Let me solve for n:174 = 6 + 6(n-1)174 - 6 = 6(n-1)168 = 6(n-1)168 / 6 = n - 128 = n - 1n = 29So, there are 29 terms in this sum. That's good to know because it might help later on.Now, I remember that sin²x can be expressed using a double-angle identity. Let me recall it: sin²x = (1 - cos2x)/2. Yes, that's right. So, maybe I can rewrite each term in the sum using this identity. Let me try that.So, each term sin²(6k)° where k ranges from 1 to 29 can be written as [1 - cos(12k)°]/2. Therefore, the entire sum becomes:Sum = Σ [1 - cos(12k)°]/2 for k from 1 to 29.I can split this sum into two separate sums:Sum = (1/2) Σ 1 - (1/2) Σ cos(12k)° for k from 1 to 29.Calculating the first sum, Σ1 from k=1 to 29 is just 29. So, the first part is (1/2)*29 = 29/2.Now, the second sum is (1/2) times the sum of cos(12k)° from k=1 to 29. So, the entire expression becomes:Sum = 29/2 - (1/2) Σ cos(12k)° from k=1 to 29.So, now I need to compute the sum of cos(12k)° for k=1 to 29. Hmm, that seems tricky. Maybe I can use some trigonometric identities or properties of cosine to simplify this sum.I remember that the sum of cosines of angles in arithmetic progression can be expressed using a specific formula. Let me recall that formula. The sum Σ cos(a + (k-1)d) from k=1 to n is equal to [sin(n*d/2) / sin(d/2)] * cos(a + (n-1)d/2). Similarly, for sine terms, there's a similar formula.In our case, the angles are 12°, 24°, 36°, ..., 348°, which is an arithmetic sequence with a common difference of 12°, starting at 12° and ending at 348°, which is 12°*29. So, n = 29, a = 12°, d = 12°, and the last term is 348°.Wait, let me make sure. The first term is 12°, which is when k=1, so a = 12°, and each subsequent term increases by 12°, so d = 12°, and the number of terms is 29. So, applying the formula:Sum = [sin(n*d/2) / sin(d/2)] * cos(a + (n-1)d/2)Plugging in the values:Sum = [sin(29*12°/2) / sin(12°/2)] * cos(12° + (29 - 1)*12°/2)Simplify step by step.First, compute n*d/2: 29*12°/2 = 29*6° = 174°Then, sin(174°). Hmm, sin(174°) is equal to sin(180° - 6°) = sin6°, because sine is positive in the second quadrant and sin(180° - x) = sinx.Next, sin(d/2) = sin(6°), so the denominator is sin6°.Now, the first part of the formula is [sin174° / sin6°] = [sin6° / sin6°] = 1. That's interesting, it simplifies to 1.Now, the second part is cos(a + (n - 1)d/2). Let's compute that:a = 12°, (n - 1)d/2 = (28)*12°/2 = 28*6° = 168°So, a + (n - 1)d/2 = 12° + 168° = 180°Therefore, cos(180°) = -1.Putting it all together:Sum = [1] * (-1) = -1So, the sum of cos(12k)° from k=1 to 29 is -1.Wait, that seems surprising, but let me verify. If I consider the sum of cos(12°) + cos(24°) + ... + cos(348°), which is 29 terms, each 12° apart. Since 360° is a full circle, and 348° is just 12° less than 360°, so the angles are symmetrically distributed around the circle.But wait, actually, 12°, 24°, ..., 348° are all multiples of 12°, and 360° is 30*12°, but we have 29 terms, so it's from 12° to 348°, which is 29 terms. Hmm, but 360° is not included, so it's not a full circle.But regardless, using the formula, we found that the sum is -1. Let me see if that makes sense.Alternatively, I can think of this as the real part of a complex exponential sum. Let me try that approach to verify.Consider the sum S = Σ cis(12k°) from k=1 to 29, where cisθ = cosθ + i sinθ. Then, the real part of S is the sum of cos(12k°), which is what we're interested in.So, S = cis12° + cis24° + ... + cis348°. This is a geometric series with first term cis12°, common ratio cis12°, and 29 terms.The sum of a geometric series is S = a*(1 - r^n)/(1 - r), where a is the first term, r is the common ratio, and n is the number of terms.So, S = cis12°*(1 - (cis12°)^29)/(1 - cis12°)Compute (cis12°)^29 = cis(12°*29) = cis348°, since 12°*29 = 348°.So, S = cis12°*(1 - cis348°)/(1 - cis12°)But cis348° = cis(360° - 12°) = cis(-12°) = cos(-12°) + i sin(-12°) = cos12° - i sin12°So, 1 - cis348° = 1 - cos12° + i sin12°Similarly, 1 - cis12° = 1 - cos12° - i sin12°So, S = cis12°*(1 - cos12° + i sin12°)/(1 - cos12° - i sin12°)Let me compute the denominator and numerator:Let me denote A = 1 - cos12°, B = sin12°, so numerator is A + iB, denominator is A - iB.So, S = cis12°*(A + iB)/(A - iB)Multiply numerator and denominator by (A + iB):(A + iB)/(A - iB) = [(A + iB)^2]/(A² + B²)Compute numerator: (A + iB)^2 = A² - B² + 2iABDenominator: A² + B²So, S = cis12°*(A² - B² + 2iAB)/(A² + B²)But A = 1 - cos12°, B = sin12°, so A² + B² = (1 - 2cos12° + cos²12°) + sin²12° = 1 - 2cos12° + (cos²12° + sin²12°) = 1 - 2cos12° + 1 = 2(1 - cos12°)Similarly, A² - B² = (1 - 2cos12° + cos²12°) - sin²12° = 1 - 2cos12° + (cos²12° - sin²12°) = 1 - 2cos12° + cos24°, since cos²x - sin²x = cos2x.And 2AB = 2*(1 - cos12°)*sin12°So, putting it all together:S = cis12°*[ (1 - 2cos12° + cos24°) + i*(2*(1 - cos12°)*sin12°) ] / [2(1 - cos12°)]Let me factor out 1 - cos12° from the numerator:= cis12°*[ (1 - cos12°)(1 - cos12° + cos24°/(1 - cos12°)) + i*(2sin12°(1 - cos12°)) ] / [2(1 - cos12°)]Wait, this seems complicated. Maybe there's a better way.Alternatively, since S = cis12°*(1 - cis348°)/(1 - cis12°), and we can compute this as:Let me compute 1 - cis348° = 1 - cis(360° - 12°) = 1 - cis(-12°) = 1 - cos12° + i sin12°, as before.And 1 - cis12° = 1 - cos12° - i sin12°So, S = cis12°*(1 - cos12° + i sin12°)/(1 - cos12° - i sin12°)Let me denote the denominator as D = 1 - cos12° - i sin12°, and the numerator as N = 1 - cos12° + i sin12°So, N/D = (1 - cos12° + i sin12°)/(1 - cos12° - i sin12°)Multiply numerator and denominator by the conjugate of the denominator:= [(1 - cos12° + i sin12°)(1 - cos12° + i sin12°)] / [(1 - cos12°)^2 + (sin12°)^2]Compute the denominator:(1 - cos12°)^2 + (sin12°)^2 = 1 - 2cos12° + cos²12° + sin²12° = 2(1 - cos12°), as before.Compute the numerator:(1 - cos12° + i sin12°)^2 = (1 - cos12°)^2 - (sin12°)^2 + 2i(1 - cos12°)(sin12°)= (1 - 2cos12° + cos²12°) - sin²12° + 2i(1 - cos12°)(sin12°)= 1 - 2cos12° + (cos²12° - sin²12°) + 2i(1 - cos12°)(sin12°)= 1 - 2cos12° + cos24° + 2i(1 - cos12°)(sin12°)So, S = cis12° * [ (1 - 2cos12° + cos24°) + 2i(1 - cos12°)(sin12°) ] / [2(1 - cos12°)]Now, let's compute this:First, note that cis12° = cos12° + i sin12°So, multiplying cis12° with the numerator:= [cos12° + i sin12°] * [ (1 - 2cos12° + cos24°) + 2i(1 - cos12°)(sin12°) ] / [2(1 - cos12°)]This will result in a complex number. Let me compute the real and imaginary parts separately.Let me denote:A = 1 - 2cos12° + cos24°B = 2(1 - cos12°)(sin12°)So, the numerator becomes [cos12° + i sin12°]*(A + iB) = cos12°*A - sin12°*B + i(sin12°*A + cos12°*B)Therefore, S = [cos12°*A - sin12°*B + i(sin12°*A + cos12°*B)] / [2(1 - cos12°)]Now, let's compute A and B:A = 1 - 2cos12° + cos24°B = 2(1 - cos12°)(sin12°)Let me compute A:Using the identity cos24° = 2cos²12° - 1So, A = 1 - 2cos12° + 2cos²12° - 1 = -2cos12° + 2cos²12° = 2cos²12° - 2cos12° = 2cos12°(cos12° - 1)Similarly, B = 2(1 - cos12°)(sin12°) = 2sin12°(1 - cos12°)So, A = 2cos12°(cos12° - 1) = -2cos12°(1 - cos12°)And B = 2sin12°(1 - cos12°)So, plugging back into the expression for S:Real part: [cos12°*A - sin12°*B] / [2(1 - cos12°)]= [cos12°*(-2cos12°(1 - cos12°)) - sin12°*(2sin12°(1 - cos12°))] / [2(1 - cos12°)]Factor out -2(1 - cos12°):= [ -2(1 - cos12°)(cos²12° + sin²12°) ] / [2(1 - cos12°)]Since cos²x + sin²x = 1, this simplifies to:= [ -2(1 - cos12°)*1 ] / [2(1 - cos12°)] = -1Similarly, the imaginary part:[sin12°*A + cos12°*B] / [2(1 - cos12°)]= [sin12°*(-2cos12°(1 - cos12°)) + cos12°*(2sin12°(1 - cos12°))] / [2(1 - cos12°)]Factor out 2(1 - cos12°):= [ -2cos12°sin12°(1 - cos12°) + 2cos12°sin12°(1 - cos12°) ] / [2(1 - cos12°)]= [0] / [2(1 - cos12°)] = 0So, the imaginary part is 0, and the real part is -1. Therefore, S = -1.Therefore, the sum of cos(12k)° from k=1 to 29 is -1.So, going back to our original expression:Sum = 29/2 - (1/2)*(-1) = 29/2 + 1/2 = 30/2 = 15.Wait, that contradicts my earlier conclusion where I thought it was 31/2. Hmm, let me check.Wait, in the initial problem, the sum is from 6° to 174°, which is 29 terms, each 6° apart. When I converted sin²x to (1 - cos2x)/2, the angles became 12°, 24°, ..., 348°, which is 29 terms as well.But in the formula, when I used the sum of cosines, I got -1, so the sum becomes 29/2 - (1/2)*(-1) = 29/2 + 1/2 = 30/2 = 15.But earlier, I thought the answer was 31/2. Hmm, where did I go wrong?Wait, let me recount the number of terms. Starting at 6°, each term is 6°, so 6°, 12°, ..., 174°. Let's see: 174 - 6 = 168, divided by 6 is 28, so 28 intervals, meaning 29 terms. That's correct.When I converted sin²x, I had 29 terms of (1 - cos2x)/2, so the sum becomes 29*(1/2) - (1/2) sum cos2x. So, 29/2 - (1/2) sum cos2x.But when I calculated the sum of cos2x, which is cos12° + cos24° + ... + cos348°, I got -1. Therefore, the total sum is 29/2 - (1/2)*(-1) = 29/2 + 1/2 = 30/2 = 15.Wait, but in the initial problem, the user wrote the sum as sin²6° + sin²12° + ... + sin²174°, which is 29 terms. So, according to my calculation, the sum is 15.But in the initial problem, the user had a different approach, where they considered the sum y = cos0° + cos12° + ... + cos348°, which is 30 terms, and found that y = 0, leading to the sum of cos12° + ... + cos348° = -1. Then, they plugged that into the sum, getting 15 - (1/2)*(-1) = 15 + 1/2 = 31/2.Wait, so in the initial problem, the user considered y as the sum from k=0 to 29 of cos12k°, which is 30 terms, including cos0°. Then, they found that y = 0, so the sum from k=1 to 29 of cos12k° = y - cos0° = -1.Therefore, in their calculation, the sum of cos12° + ... + cos348° is -1, so the sum becomes 15 - (1/2)*(-1) = 15 + 1/2 = 31/2.But in my calculation, I considered the sum from k=1 to 29 of cos12k°, which is 29 terms, and found that it's -1. Therefore, the sum is 29/2 - (1/2)*(-1) = 15.So, which one is correct?Wait, the confusion arises from the number of terms. Let me clarify.In the initial problem, the user considered the sum y = cos0° + cos12° + ... + cos348°, which is 30 terms (k=0 to 29). They found that y = 0, so the sum from k=1 to 29 is y - cos0° = -1.Therefore, in their calculation, the sum of cos12° + ... + cos348° is -1, leading to the total sum being 15 + 1/2 = 31/2.But in my calculation, I considered the sum from k=1 to 29 of cos12k°, which is 29 terms, and using the formula, I found that it's -1. Therefore, the sum is 29/2 - (1/2)*(-1) = 15.Wait, but in the initial problem, the user had 30 terms in y, but in our problem, the sum of cos2x is from k=1 to 29, which is 29 terms. So, which one is correct?Wait, let's think about it. When we converted sin²6° + sin²12° + ... + sin²174°, each term becomes (1 - cos12°)/2, (1 - cos24°)/2, ..., (1 - cos348°)/2. So, the sum is 29*(1/2) - (1/2)(cos12° + cos24° + ... + cos348°). That is, 29/2 - (1/2) sum_{k=1}^{29} cos(12k°).So, the sum of cos(12k°) from k=1 to 29 is indeed 29 terms, and according to my calculation, it's -1. Therefore, the total sum is 29/2 - (1/2)*(-1) = 29/2 + 1/2 = 30/2 = 15.But in the initial problem, the user considered y = sum_{k=0}^{29} cos(12k°) = 0, so sum_{k=1}^{29} cos(12k°) = -1, leading to the total sum being 15 + 1/2 = 31/2.Wait, so which is correct? Let me check the number of terms again.In the initial problem, the user wrote:"Compute sin²6° + sin²12° + sin²18° + ... + sin²174°."So, starting at 6°, each term increases by 6°, up to 174°. So, the number of terms is (174 - 6)/6 + 1 = (168)/6 + 1 = 28 + 1 = 29 terms.Therefore, when converting each sin²x to (1 - cos2x)/2, we have 29 terms, each contributing (1 - cos2x)/2. So, the sum is 29*(1/2) - (1/2) sum_{k=1}^{29} cos(12k°).Therefore, the sum of cos(12k°) from k=1 to 29 is indeed 29 terms, and according to my calculation, it's -1. Therefore, the total sum is 29/2 - (1/2)*(-1) = 15.But in the initial problem, the user considered y = sum_{k=0}^{29} cos(12k°) = 0, so sum_{k=1}^{29} cos(12k°) = -1, leading to the total sum being 15 + 1/2 = 31/2.Wait, but in the initial problem, the user had 30 terms in y, including cos0°, which is 1. So, y = 0 implies that sum_{k=1}^{29} cos(12k°) = -1, which is correct because y = cos0° + sum_{k=1}^{29} cos(12k°) = 1 + (-1) = 0.Therefore, in their calculation, the sum of cos(12k°) from k=1 to 29 is -1, so the total sum is 15 - (1/2)*(-1) = 15 + 1/2 = 31/2.But in my calculation, I considered the sum from k=1 to 29 of cos(12k°) as -1, leading to the total sum being 15.Wait, but according to the initial problem, the user had 15 - (1/2)*(-1) = 15 + 1/2 = 31/2.But according to my calculation, it's 15.So, which one is correct?Wait, let me think again.The initial problem had:Sum = sin²6° + sin²12° + ... + sin²174°= Σ_{k=1}^{29} sin²(6k)°= Σ_{k=1}^{29} [1 - cos(12k)°]/2= 29/2 - (1/2) Σ_{k=1}^{29} cos(12k)°Now, the user considered y = Σ_{k=0}^{29} cos(12k)° = 0, so Σ_{k=1}^{29} cos(12k)° = y - cos0° = -1.Therefore, the sum becomes 29/2 - (1/2)*(-1) = 29/2 + 1/2 = 30/2 = 15.Wait, but in the initial problem, the user wrote:"Compute sin²6° + sin²12° + ... + sin²174°."Then, they converted each term to (1 - cos2x)/2, leading to:Sum = 15 - (1/2)(cos12° + ... + cos348°)Then, they considered y = cos0° + cos12° + ... + cos348° = 0, so cos12° + ... + cos348° = -1.Therefore, Sum = 15 - (1/2)*(-1) = 15 + 1/2 = 31/2.Wait, but in my calculation, I considered the sum from k=1 to 29 of cos(12k°) as -1, leading to the sum being 15.But according to the initial problem, the user had 30 terms in y, including cos0°, so the sum from k=1 to 29 is -1, leading to the total sum being 15 + 1/2 = 31/2.Wait, so which is correct?Wait, let's clarify the number of terms.In the initial problem, the user wrote:"Compute sin²6° + sin²12° + ... + sin²174°."Which is 29 terms.When converting to cos terms, each sin²x becomes (1 - cos2x)/2, so the sum becomes 29*(1/2) - (1/2) sum_{k=1}^{29} cos(12k°).Therefore, the sum is 29/2 - (1/2) sum_{k=1}^{29} cos(12k°).Now, the user considered y = sum_{k=0}^{29} cos(12k°) = 0, so sum_{k=1}^{29} cos(12k°) = -1.Therefore, the sum becomes 29/2 - (1/2)*(-1) = 29/2 + 1/2 = 30/2 = 15.But in the initial problem, the user wrote:"Compute sin²6° + sin²12° + ... + sin²174°."Then, they converted each term to (1 - cos2x)/2, leading to:Sum = 15 - (1/2)(cos12° + ... + cos348°)Wait, but 15 is 30/2, which would imply that they considered 30 terms in the sum. But the original sum is 29 terms.Wait, perhaps the user made a mistake in their initial calculation.Wait, let me check:In the initial problem, the user wrote:"Compute sin²6° + sin²12° + ... + sin²174°."Then, they converted each term to (1 - cos2x)/2, leading to:Sum = (1 - cos12°)/2 + (1 - cos24°)/2 + ... + (1 - cos348°)/2.Then, they wrote:= 15 - (1/2)(cos12° + cos24° + ... + cos348°)Wait, but 15 is 30/2, which would imply that they have 30 terms, each contributing 1/2. But the original sum is 29 terms, so 29*(1/2) = 29/2.Therefore, the user made a mistake in their initial calculation by writing 15 instead of 29/2.Therefore, the correct sum should be 29/2 - (1/2) sum cos(12k°).Then, since sum cos(12k°) from k=1 to 29 is -1, the total sum is 29/2 - (1/2)*(-1) = 29/2 + 1/2 = 30/2 = 15.Therefore, the correct answer is 15.But in the initial problem, the user concluded 31/2, which is 15.5, which is incorrect because they mistakenly wrote 15 instead of 29/2.Therefore, the correct answer is 15.Wait, but let me verify this with another approach.Another way to compute the sum of sin²x for equally spaced angles is to use the identity that the sum of sin²x over a full period is equal to n/2, where n is the number of terms, if the angles are symmetrically distributed.But in our case, the angles are from 6° to 174°, which is 29 terms, each 6° apart. The full circle is 360°, so 29 terms would not complete a full circle.But wait, 6°*29 = 174°, so it's not a full circle. Therefore, the sum might not be as straightforward.Alternatively, we can use the identity that the sum of sin²x over an arithmetic sequence can be expressed as (n/2) - (1/2) sum cos2x.Which is exactly what we did.So, sum sin²x = n/2 - (1/2) sum cos2x.In our case, n = 29, so sum sin²x = 29/2 - (1/2) sum cos(12k°).We found that sum cos(12k°) from k=1 to 29 is -1, so the total sum is 29/2 - (1/2)*(-1) = 29/2 + 1/2 = 30/2 = 15.Therefore, the correct answer is 15.But in the initial problem, the user concluded 31/2, which is incorrect because they mistakenly wrote 15 instead of 29/2.Therefore, the correct answer is 15.

💡Okay, so I have this problem about a test with m questions and n students. The scoring rule is a bit unique: for each question, if exactly x students get it wrong, then each of the students who got it right gets x points, and those who got it wrong get zero. The total score for each student is just the sum of their scores across all m questions. I need to find the maximum possible value of the sum of the highest score and the lowest score, which are p₁ and pₙ respectively, after sorting all the students' scores in descending order.Alright, let me break this down. First, for each question, the number of students who got it wrong is x, so the number who got it right is n - x. Each correct answer gives x points to the student. So, for each question, the total points awarded are x*(n - x). That makes sense because each of the n - x students gets x points.So, the total points across all questions would be the sum from k=1 to m of x_k*(n - x_k), where x_k is the number of students who got the k-th question wrong. Let me write that down:Total score S = Σ (from k=1 to m) [x_k*(n - x_k)].Expanding that, it's S = Σ [n*x_k - x_k²] = n*Σ x_k - Σ x_k².Hmm, okay. So, the total score is a function of the sum of x_k and the sum of x_k squared.Now, I need to find the maximum possible value of p₁ + pₙ. Since p₁ is the highest score and pₙ is the lowest, I need to think about how to maximize their sum.Let me think about the extremes. If one student gets all questions right, then for each question, x_k = n - 1, because only one student got it right. So, each question would give that student (n - 1) points, and the total score for that student would be m*(n - 1). The other students would get zero points for each question, so their total scores would be zero. So, in this case, p₁ = m*(n - 1) and pₙ = 0, so p₁ + pₙ = m*(n - 1).But wait, is this the maximum? Maybe there's a way to have pₙ not be zero but still have p₁ + pₙ be larger than m*(n - 1)? Let me think.Suppose instead that one student gets all questions right, and another student gets all questions wrong. Then, p₁ = m*(n - 1) and pₙ = 0, same as before. But if I have some other distribution where pₙ is not zero, maybe p₁ can be higher? But no, because if someone else gets some questions right, that would take away points from p₁, right? Because for each question, the points are distributed based on how many got it wrong.Wait, actually, if another student gets some questions right, it doesn't necessarily take away points from p₁. Because for each question, the points are given to all who got it right. So, if Student A gets all questions right, and Student B gets some questions right, then for those questions where both got it right, Student A still gets the x points, and Student B also gets the x points. So, actually, p₁ could potentially be higher if other students also get some questions right, but that would require that x is higher for some questions.Wait, no. Because x is the number of students who got it wrong. So, if more students get it right, x is smaller, which means each correct answer gives fewer points. So, if Student A is getting all questions right, and Student B is also getting some questions right, then for those questions where both got it right, x is smaller, so the points per question are smaller for Student A. So, actually, p₁ would be smaller in that case.So, maybe the maximum p₁ is achieved when only one student gets all questions right, and everyone else gets all questions wrong. Then, p₁ = m*(n - 1), and pₙ = 0, so p₁ + pₙ = m*(n - 1).But let me check if there's another scenario where p₁ + pₙ could be higher. Suppose that for some questions, x is larger, giving more points to the correct answers, but for other questions, x is smaller, giving fewer points. Maybe balancing it in a way that p₁ is high, and pₙ is also not too low.Wait, but pₙ is the minimum score. So, if I have some students scoring higher, but the minimum score is still low, maybe p₁ + pₙ can be higher? Let's think.Suppose that for each question, exactly one student gets it wrong. So, x_k = 1 for each question. Then, each correct answer gives 1 point. So, for each question, n - 1 students get 1 point each. So, the total points per question is 1*(n - 1) = n - 1. Across m questions, the total score S = m*(n - 1).Now, how are these points distributed among the students? If for each question, a different student gets it wrong, then each student gets m - 1 points, because they got all questions right except one. So, each student's score would be m - 1, so p₁ = p₂ = ... = pₙ = m - 1. Then, p₁ + pₙ = (m - 1) + (m - 1) = 2m - 2.Compare that to the previous scenario where p₁ + pₙ = m*(n - 1). Which one is larger?Well, 2m - 2 versus m*(n - 1). Since n ≥ 2, m*(n - 1) is at least m, which is larger than 2m - 2 when m ≥ 2. Because 2m - 2 = 2(m - 1), and m*(n - 1) ≥ m*(2 - 1) = m. So, if m ≥ 2, m*(n - 1) ≥ m ≥ 2(m - 1) when m ≤ 2. Wait, no, actually, for m=2, m*(n -1) = 2(n -1), and 2m - 2 = 2. So, 2(n -1) is larger than 2 when n ≥ 2. For m=3, m*(n -1)=3(n -1), and 2m -2=4. So, 3(n -1) is larger than 4 when n ≥ 3, but for n=2, it's 3(1)=3 vs 4, so 4 is larger.Wait, so depending on m and n, which one is larger? Hmm, maybe this approach isn't the best.Alternatively, maybe the maximum p₁ + pₙ is m*(n -1). Because in the first scenario, p₁ is as high as possible, and pₙ is as low as possible, but their sum is m*(n -1). In the second scenario, the sum is 2m -2, which is less than m*(n -1) when n ≥ 3.But wait, maybe there's a way to have p₁ + pₙ be higher than m*(n -1). Let me think.Suppose that for some questions, x is large, giving high points to the correct answers, and for other questions, x is small, giving low points. Maybe arrange it so that p₁ gets the high points from some questions, and pₙ gets the low points from others, but overall, their sum is higher.Wait, but pₙ is the minimum score, so even if pₙ gets some points from some questions, it's still constrained by the minimum. So, maybe pₙ can't be too high if p₁ is already high.Alternatively, maybe if p₁ and pₙ are both high, but that would require that the other students have scores in between, but since pₙ is the minimum, it's constrained.Wait, perhaps the maximum p₁ + pₙ is indeed m*(n -1), achieved when one student gets all questions right, and everyone else gets all wrong. Because in that case, p₁ is maximized, and pₙ is minimized, but their sum is m*(n -1). If we try to make pₙ higher, p₁ would have to be lower, so their sum might not increase.Alternatively, maybe there's a way to have p₁ and pₙ both be high, but I don't see how because pₙ is the minimum. So, if pₙ is high, that means all students have high scores, which would require that for each question, x is high, meaning that many students got it wrong, but then the points per question would be high. But if many students got it wrong, then the points per question are high, but the number of students getting it right is low, so the total points are high, but distributed among fewer students.Wait, maybe if for each question, exactly one student gets it wrong, then each question gives 1 point to n -1 students. So, each student gets m -1 points, as in the earlier case. So, p₁ + pₙ = 2(m -1). But if n is large, say n=10, and m=2, then 2(m -1)=2, while m*(n -1)=2*9=18, which is much larger.So, in that case, m*(n -1) is larger. So, perhaps the maximum is indeed m*(n -1).Wait, but let me think again. Suppose that for some questions, x is n -1, so one student gets it right, and for other questions, x is 1, so n -1 students get it right. Then, the student who got all the x=n -1 questions right would have a high score, and the student who got all the x=1 questions right would have a moderate score.But wait, if a student got all x=n -1 questions right, that means they got all questions right where only one student got it right. And for the other questions where x=1, they also got it right, so they got all questions right, which would make their score m*(n -1). But then, the student who got all the x=1 questions right would have a score of m*1 = m. So, p₁ = m*(n -1), pₙ = m, so p₁ + pₙ = m*(n -1) + m = m*n.Wait, that's higher than m*(n -1). So, maybe that's a better scenario.Wait, but can that happen? Let me see.Suppose that for each question, either x=1 or x=n -1. For the questions where x=1, n -1 students got it right, and for the questions where x=n -1, 1 student got it right.If we have k questions with x=1 and (m -k) questions with x=n -1.Then, the total points S = k*(n -1)*1 + (m -k)*1*(n -1) = k*(n -1) + (m -k)*(n -1) = m*(n -1).Wait, so the total points are still m*(n -1). Now, how are these points distributed?Suppose that Student A got all the x=n -1 questions right, so they got (m -k)*(n -1) points. And Student B got all the x=1 questions right, so they got k*1 points. The other students would have varying scores.But wait, if Student A got all the x=n -1 questions right, that means for those (m -k) questions, only Student A got them right, and everyone else got them wrong. For the k questions where x=1, n -1 students got them right, so Student A also got them right, along with n -2 others.So, Student A's total score is (m -k)*(n -1) + k*1.Student B's total score is k*1, because for the (m -k) questions where x=n -1, Student B got them wrong, so got 0 points.Wait, but Student B could have gotten some of the x=1 questions right. Wait, no, if x=1 for k questions, then n -1 students got them right. So, if Student B is one of those n -1, then they get k*1 points. But if Student B is the one who got the x=n -1 questions wrong, then they get 0 for those.Wait, this is getting complicated. Maybe I need to structure it more carefully.Let me define:- Let there be k questions where x=1 (so n -1 students got them right).- Let there be (m -k) questions where x=n -1 (so 1 student got them right).Now, let's assign which students got which questions right.Suppose Student A is the one who got all the x=n -1 questions right. So, for the (m -k) questions, only Student A got them right, giving them (m -k)*(n -1) points.For the k questions where x=1, n -1 students got them right. Let's say Student A is one of them, so they get k*1 points from these.Therefore, Student A's total score is (m -k)*(n -1) + k*1.Now, let's look at Student B. For the (m -k) questions where x=n -1, Student B got them wrong, so 0 points. For the k questions where x=1, Student B got them right, so k*1 points.Therefore, Student B's total score is k*1.Similarly, the other students (from Student C to Student n) would have varying scores. For the (m -k) questions where x=n -1, they got them wrong, so 0 points. For the k questions where x=1, they got them right, so k*1 points each.Wait, but if n -1 students got the k questions right, and Student A and Student B are two of them, then the other n -2 students also got those k questions right. So, their total scores are also k*1.Therefore, the scores are:- Student A: (m -k)*(n -1) + k- Student B: k- Students C to n: k eachSo, the scores are ordered as:p₁ = (m -k)*(n -1) + kp₂ = kp₃ = k...pₙ = kTherefore, p₁ + pₙ = [(m -k)*(n -1) + k] + k = (m -k)*(n -1) + 2k.Simplify this:= m(n -1) -k(n -1) + 2k= m(n -1) -k(n -1 - 2)= m(n -1) -k(n -3)Hmm, interesting. So, p₁ + pₙ = m(n -1) -k(n -3).Now, to maximize p₁ + pₙ, we need to minimize k(n -3). Since k is the number of questions where x=1, it can range from 0 to m.If n > 3, then (n -3) is positive, so to minimize k(n -3), we need to minimize k. The minimum k is 0. So, if k=0, then p₁ + pₙ = m(n -1).If n = 3, then (n -3)=0, so p₁ + pₙ = m(n -1) regardless of k.If n < 3, but n ≥ 2, so n=2. Then, (n -3)= -1, so p₁ + pₙ = m(n -1) -k(-1) = m(n -1) +k.Since k can be up to m, the maximum p₁ + pₙ would be m(n -1) + m = m(n -1 +1)=m*n.Wait, that's interesting. So, for n=2, p₁ + pₙ can be m*n.But wait, let's check that.If n=2, then for each question, x can be 0,1, or 2. But x=0 would mean everyone got it right, so no points. x=2 would mean everyone got it wrong, so no points. So, only x=1 is meaningful, giving 1 point to n -1=1 student.So, for n=2, each question can give 1 point to one student or the other.If we set k= m, meaning all questions have x=1, then each question gives 1 point to one student. So, if we alternate which student gets the point, each student gets m/2 points, but since m is integer, maybe one gets ceil(m/2) and the other floor(m/2). So, p₁ + pₙ would be ceil(m/2) + floor(m/2) = m.But wait, earlier I thought p₁ + pₙ could be m*n=2m, but that doesn't seem right.Wait, maybe my earlier reasoning was flawed. Let me think again.If n=2, and we have k questions where x=1, so one student gets 1 point each, and (m -k) questions where x=1 as well, because n=2, x can't be n -1=1, which is same as x=1. Wait, no, for n=2, x can be 0,1, or 2. If x=1, then one student gets 1 point. If x=2, both get 0. If x=0, both get 0.Wait, so for n=2, the only way to give points is to have x=1, which gives 1 point to one student.So, if we have k questions where x=1, then one student gets k points, and the other gets 0. So, p₁ =k, p₂=0, so p₁ + pₙ =k.But if we set k=m, then p₁ + pₙ =m. But earlier, I thought p₁ + pₙ could be m*n=2m, but that's not possible because the maximum any student can get is m points, since each question can give at most 1 point to a student.Wait, so for n=2, the maximum p₁ + pₙ is m, achieved when one student gets all m questions right (x=1 for each question), and the other gets none.But earlier, I thought p₁ + pₙ could be 2m, but that's not possible because the total points S = m*1 = m, so p₁ + p₂ = m, so p₁ + pₙ = m.Wait, so my earlier reasoning was wrong. For n=2, p₁ + pₙ can't exceed m.But earlier, when I considered k=0, p₁ + pₙ = m(n -1)=m(1)=m, which is consistent.So, maybe for n=2, p₁ + pₙ =m.But earlier, when I thought about n=2 and k=m, I thought p₁ + pₙ could be 2m, but that's incorrect because the total points are only m.So, in that case, the maximum p₁ + pₙ is m.Wait, so going back, for n ≥3, the maximum p₁ + pₙ is m(n -1), and for n=2, it's m.But wait, when n=2, m(n -1)=m(1)=m, which matches. So, maybe the general formula is m(n -1).But wait, earlier when I considered n=2 and k=m, p₁ + pₙ =m, which is equal to m(n -1). So, maybe the formula holds for all n ≥2.But wait, when n=2, if we have some questions with x=1 and some with x=1, it's the same as all x=1. So, p₁ + pₙ =m.But earlier, I thought that for n=2, if we have some questions with x=1 and some with x=1, but that's redundant. So, maybe the maximum p₁ + pₙ is always m(n -1).Wait, but in the case where n=2 and m=2, if we have both questions with x=1, then one student gets 2 points, and the other gets 0, so p₁ + pₙ=2. But m(n -1)=2(1)=2, which matches.Alternatively, if we have one question with x=1 and one with x=1, same thing.Wait, so maybe the maximum p₁ + pₙ is indeed m(n -1) for all n ≥2.But earlier, I thought that for n=2, if we have some questions with x=1 and others with x=1, but that's the same as all x=1, so p₁ + pₙ=m.So, perhaps the maximum p₁ + pₙ is m(n -1).Wait, but let me think again. Suppose that for n=3, m=2.If we set k=0, meaning all questions have x=2, so one student gets 2 points per question, total 4 points. The other two students get 0 each. So, p₁ + pₙ=4 +0=4.Alternatively, if we set k=1, meaning one question with x=1 and one with x=2.For the x=1 question, two students get 1 point each. For the x=2 question, one student gets 2 points.So, Student A gets 2 points from the x=2 question, and 1 point from the x=1 question, total 3.Student B gets 1 point from the x=1 question, and 0 from the x=2 question, total 1.Student C gets 1 point from the x=1 question, and 0 from the x=2 question, total 1.So, p₁=3, p₂=1, p₃=1, so p₁ + pₙ=3 +1=4, same as before.Alternatively, if we set k=2, meaning both questions have x=1.Then, for each question, two students get 1 point each. So, each student gets 2 points, so p₁ + pₙ=2 +2=4.Wait, so in all cases, p₁ + pₙ=4, which is m(n -1)=2*2=4.So, regardless of k, p₁ + pₙ=4.Wait, so maybe for n=3, m=2, p₁ + pₙ=4, which is m(n -1).Similarly, for n=2, m=2, p₁ + pₙ=2, which is m(n -1)=2*1=2.So, it seems that regardless of how we set k, p₁ + pₙ=m(n -1).Wait, but earlier, when I considered n=2 and k=m, I thought p₁ + pₙ could be m, which is m(n -1)=m(1)=m.So, maybe the maximum p₁ + pₙ is indeed m(n -1).But wait, earlier I thought that if n=2, and we have some questions with x=1 and others with x=1, but that's the same as all x=1, so p₁ + pₙ=m.But in the case where n=3, m=2, regardless of k, p₁ + pₙ=4.Wait, so maybe the maximum p₁ + pₙ is always m(n -1), regardless of how we distribute the x's.But let me think of another example.Suppose n=4, m=3.If we set k=0, meaning all questions have x=3, so one student gets 3 points per question, total 9 points. The other three students get 0 each. So, p₁ + pₙ=9 +0=9.Alternatively, if we set k=1, meaning one question with x=1 and two with x=3.For the x=1 question, three students get 1 point each. For the x=3 questions, one student gets 3 points each.So, Student A gets 3*2=6 points from the x=3 questions, and 1 point from the x=1 question, total 7.Students B, C, D get 1 point each from the x=1 question, and 0 from the x=3 questions, total 1 each.So, p₁=7, p₂=1, p₃=1, p₄=1, so p₁ + pₙ=7 +1=8.Alternatively, if we set k=3, meaning all questions have x=1.Each question gives 1 point to three students. So, each student gets 3 points, so p₁ + pₙ=3 +3=6.Wait, so in this case, when k=0, p₁ + pₙ=9, which is m(n -1)=3*3=9.When k=1, p₁ + pₙ=8, which is less than 9.When k=3, p₁ + pₙ=6, which is less than 9.So, the maximum p₁ + pₙ is indeed 9, achieved when k=0.So, this seems to confirm that the maximum p₁ + pₙ is m(n -1).Therefore, the maximum possible value of p₁ + pₙ is m(n -1).

💡Okay, so I have this problem here: Suppose that m and n are positive integers such that 75m = n³. I need to find the minimum possible value of m + n. The answer choices are 15, 30, 50, 60, and 5700. Hmm, let me try to figure this out step by step.First, I know that 75 is a number that can be factored into primes. Let me do that. 75 divided by 5 is 15, and 15 divided by 5 is 3. So, 75 is 5² times 3¹. So, 75 = 5² × 3.Now, the equation given is 75m = n³. Since n is an integer, n³ must be a perfect cube. That means that the prime factors of n³ must all have exponents that are multiples of 3. So, if I factor 75m into primes, each prime's exponent must be a multiple of 3.Looking at the prime factors of 75, which are 5² and 3¹, I can see that 5 has an exponent of 2 and 3 has an exponent of 1. To make 75m a perfect cube, I need to multiply by m such that the exponents of 5 and 3 become multiples of 3.For the prime 5: The exponent is 2. To get to the next multiple of 3, which is 3, I need to add 1 more 5. So, m must have at least one 5.For the prime 3: The exponent is 1. To get to the next multiple of 3, which is 3, I need to add 2 more 3s. So, m must have at least two 3s.Therefore, m must be at least 5¹ × 3². Let me calculate that: 5 × 9 = 45. So, m = 45.Now, let's check if this works. If m is 45, then 75m = 75 × 45. Let me compute that: 75 × 45. Hmm, 75 times 40 is 3000, and 75 times 5 is 375, so 3000 + 375 = 3375. So, 75m = 3375.Now, is 3375 a perfect cube? Let me see. The cube of 15 is 3375 because 15 × 15 = 225, and 225 × 15 = 3375. So, n³ = 3375 implies n = 15.So, m is 45 and n is 15. Therefore, m + n = 45 + 15 = 60.Looking at the answer choices, 60 is option D. So, the minimum possible value of m + n is 60.Wait, let me double-check to make sure I didn't miss anything. Is there a smaller m that could work? If I choose m smaller than 45, say m = 15, then 75 × 15 = 1125. Is 1125 a perfect cube? Let me see. The cube of 10 is 1000, and the cube of 11 is 1331. So, 1125 is between 1000 and 1331, but it's not a perfect cube. So, m = 15 doesn't work.What about m = 30? 75 × 30 = 2250. Is 2250 a perfect cube? The cube of 13 is 2197, and the cube of 14 is 2744. So, 2250 is between those, but not a perfect cube. So, m = 30 doesn't work either.What about m = 50? 75 × 50 = 3750. Is 3750 a perfect cube? The cube of 15 is 3375, and the cube of 16 is 4096. So, 3750 is between them, but not a perfect cube. So, m = 50 doesn't work.So, it seems that m = 45 is indeed the smallest m that makes 75m a perfect cube, giving n = 15 and m + n = 60. Therefore, the answer should be 60.**Final Answer**boxed{D}

💡First, I need to determine how many student tickets and general admission tickets were sold. I'll start by defining variables for each type of ticket.Let ( x ) represent the number of student tickets sold, and ( y ) represent the number of general admission tickets sold.From the problem, I know two key pieces of information:1. The total number of tickets sold is 525.2. The total revenue from ticket sales is 2876.Based on this information, I can set up the following system of equations:1. ( x + y = 525 )2. ( 4x + 6y = 2876 )To solve this system, I'll use the substitution method. First, I'll solve the first equation for ( x ):[ x = 525 - y ]Next, I'll substitute this expression for ( x ) into the second equation:[ 4(525 - y) + 6y = 2876 ]Expanding the equation:[ 2100 - 4y + 6y = 2876 ]Combining like terms:[ 2y = 776 ]Finally, I'll solve for ( y ):[ y = frac{776}{2} = 388 ]Therefore, the number of general admission tickets sold is 388.

💡First, I need to determine the total number of students, which is given as 10.Next, I'll calculate the total marks of all students by multiplying the number of students by the average mark:10 students * 70 average = 700 total marks.When 5 students are excluded, there are 5 students remaining. The total marks of these remaining students can be found by multiplying the number of remaining students by their average mark:5 students * 90 average = 450 total marks.To find the total marks of the excluded students, I'll subtract the total marks of the remaining students from the total marks of all students:700 total marks - 450 total marks = 250 total marks for excluded students.Finally, I'll calculate the average mark of the excluded students by dividing the total marks of the excluded students by the number of excluded students:250 total marks / 5 students = 50 average mark.

💡Okay, so I have this problem about a polygon Q where each interior angle is 9 times its corresponding exterior angle. I need to find the sum of the interior angles, S, and determine if Q can be a regular polygon. The options are given, and I have to choose between them. Let me try to figure this out step by step.First, I remember that for any polygon, the sum of the exterior angles is always 360 degrees. That's a key fact. So, if I denote each exterior angle as 'e', then the sum of all exterior angles would be 360 degrees. Now, the problem says each interior angle is 9 times its corresponding exterior angle. So, if an exterior angle is 'e', the interior angle would be 9e. That makes sense because in a polygon, each interior and exterior angle at a vertex add up to 180 degrees. So, if the interior angle is 9e, then the exterior angle should be e, and 9e + e = 10e = 180 degrees. Wait, that seems important.Let me write that down:9e + e = 180 degreesSo, 10e = 180 degreesTherefore, e = 18 degrees.So, each exterior angle is 18 degrees. Since the sum of all exterior angles is 360 degrees, I can find the number of sides, n, of the polygon by dividing 360 by the measure of each exterior angle.n = 360 / en = 360 / 18n = 20.So, the polygon has 20 sides. Now, the sum of the interior angles of a polygon is given by the formula (n-2)*180 degrees. Let me calculate that.Sum of interior angles, S = (20 - 2)*180S = 18*180S = 3240 degrees.Okay, so the sum of the interior angles is 3240 degrees. That narrows down the options to A, B, or C because D says 3600 degrees, which doesn't match.Now, the second part is whether the polygon can be regular. A regular polygon has all sides equal and all angles equal. In this case, each interior angle is 9 times its corresponding exterior angle, which we found to be 18 degrees. So, each interior angle is 9*18 = 162 degrees.If all interior angles are equal, then the polygon is regular. But wait, does the problem say that each interior angle is 9 times its corresponding exterior angle, or does it say that each interior angle is 9 times some exterior angle? Hmm, it says "each interior angle is 9 times its corresponding exterior angle." So, for each vertex, the interior angle is 9 times the exterior angle at that vertex.In a regular polygon, all exterior angles are equal, so if each interior angle is 9 times its corresponding exterior angle, and all exterior angles are equal, then all interior angles would also be equal. Therefore, the polygon can be regular.But wait, hold on. The problem doesn't specify that the polygon is regular, just that each interior angle is 9 times its corresponding exterior angle. So, could the polygon be irregular? Let me think.If the polygon is regular, then yes, each interior angle is 9 times the exterior angle, and all are equal. But if the polygon is irregular, could each interior angle still be 9 times its corresponding exterior angle? Let's see.In an irregular polygon, the exterior angles can vary, but the sum must still be 360 degrees. If each interior angle is 9 times its corresponding exterior angle, then each exterior angle would still have to satisfy 9e + e = 180, so e = 18 degrees for each. Wait, that would mean all exterior angles are 18 degrees, which would make the polygon regular because all exterior angles are equal.Hmm, so if each interior angle is 9 times its corresponding exterior angle, then each exterior angle must be 18 degrees, which would mean all exterior angles are equal, making the polygon regular.But wait, is that necessarily true? Let me think again. If each interior angle is 9 times its corresponding exterior angle, then for each vertex, 9e_i + e_i = 180, so e_i = 18 degrees for each i. Therefore, all exterior angles are 18 degrees, so the polygon must be regular because all exterior angles are equal.Therefore, the polygon must be regular. So, the sum S is 3240 degrees, and Q is regular. So, the answer should be A.But wait, looking back at the options, A says S=3240 and Q is regular, B says S=3240 and Q is not regular, C says S=3240 and Q may or may not be regular, D says S=3600 and Q may or may not be regular.Wait, but earlier, I thought that if each interior angle is 9 times its corresponding exterior angle, then all exterior angles must be equal, making the polygon regular. So, Q must be regular. Therefore, A is the answer.But wait, let me double-check. If the polygon is regular, then yes, each interior angle is 9 times the exterior angle. But can a polygon be irregular and still have each interior angle 9 times its corresponding exterior angle?Suppose we have a polygon where some exterior angles are larger and some are smaller, but each interior angle is still 9 times its corresponding exterior angle. But wait, if each interior angle is 9 times its corresponding exterior angle, then for each vertex, 9e_i + e_i = 180, so e_i = 18 degrees for each i. Therefore, all exterior angles must be 18 degrees, which means the polygon must be regular.Therefore, Q must be regular. So, the answer is A.Wait, but in the initial problem, it's not specified whether the polygon is convex or not. If it's a convex polygon, then all exterior angles are positive and less than 180 degrees, which is the case here. But if it's a concave polygon, some exterior angles can be negative. Hmm, does that affect anything?Wait, in the case of a concave polygon, the exterior angle is considered as negative if the interior angle is greater than 180 degrees. But in our case, each interior angle is 9 times its corresponding exterior angle. If the exterior angle is negative, then the interior angle would be negative as well, which doesn't make sense because interior angles are positive. Therefore, all exterior angles must be positive, meaning the polygon must be convex.Therefore, the polygon is convex and regular. So, the answer is A.Wait, but in the options, A is S=3240 and Q is regular, which matches our conclusion. So, I think A is the correct answer.But wait, let me check the calculation again. Number of sides n = 360 / 18 = 20. Sum of interior angles = (20-2)*180 = 18*180 = 3240. Yes, that's correct.And since each exterior angle is 18 degrees, all equal, so the polygon is regular. Therefore, A is correct.But wait, the initial problem didn't specify that the polygon is convex, but in reality, if a polygon has all exterior angles positive and equal, it must be convex and regular. So, yes, A is correct.Wait, but in the initial problem, it's not specified whether the polygon is convex or not. But in our case, since each interior angle is 9 times the exterior angle, and interior angles are positive, exterior angles must be positive as well, so the polygon must be convex. Therefore, it's regular.So, final answer is A.