💡First, I need to determine how many placards each person took. The problem states that every person picked a couple of placards, which typically means 2 placards per person.Next, I know that the basket can hold 823 placards. Since the organizers had to replace the basket after 39 minutes, it implies that all 823 placards were taken during that time.To find out how many people entered the stadium, I can divide the total number of placards by the number of placards each person took:823 placards ÷ 2 placards per person = 411.5 people.However, since it's not possible to have half a person, I need to round down to the nearest whole number. Therefore, 411 people entered the stadium and took placards from the basket before it had to be replaced.

💡Okay, so I need to find the volume of Q₃ given that Q₀ is a regular octahedron with a volume of 8. The process to get from Qᵢ to Qᵢ₊₁ is to replace each face with an inward-pointing regular tetrahedron. The base of each tetrahedron is the midpoint triangle of the face, and the side length of this triangle is 1/3 of the original triangle's side length.First, let me recall what a regular octahedron is. It has 8 triangular faces, 6 vertices, and 12 edges. Each face is an equilateral triangle. Since the volume of Q₀ is 8, I can use that as my starting point.Now, to get Q₁, each face of Q₀ is replaced by an inward-pointing tetrahedron. The base of each tetrahedron is the midpoint triangle of the original face. The midpoint triangle divides each face into four smaller triangles, each with 1/3 the side length of the original. So, each original face is split into four smaller faces, each of which is a triangle with side length 1/3 of the original.Wait, hold on. The problem says the base of each tetrahedron is the midpoint triangle, and the side length of this triangle is 1/3 of the original. So, does that mean each face is being replaced by a tetrahedron whose base is a triangle with side length 1/3 of the original face's side length?Yes, that seems to be the case. So, each original face is being replaced by a tetrahedron whose base is a smaller triangle with side length 1/3 of the original. The tetrahedron is inward-pointing, so it's like carving out a small tetrahedron from each face.I need to figure out how this affects the volume. Each time we add a tetrahedron, we're removing some volume from the original polyhedron. So, the volume of Q₁ will be the volume of Q₀ minus the volumes of all these small tetrahedrons.First, let's find the volume of one such small tetrahedron. Since the side length is 1/3 of the original face, the volume scales with the cube of the side length. So, if the original tetrahedron (if the face were a tetrahedron) had volume V, the new one would have volume V*(1/3)³ = V/27.But wait, the original face is a triangle, not a tetrahedron. So, perhaps I need to think differently. Maybe I should consider the volume of the tetrahedron being added. Since the base is a triangle with side length 1/3 of the original, and it's a regular tetrahedron, so all edges are equal.So, the volume of a regular tetrahedron is (edge length)³ / (6√2). So, if the edge length is 1/3 of the original, then the volume is (1/3)³ / (6√2) = 1/(27*6√2) = 1/(162√2). But wait, the original tetrahedron's volume isn't given. Hmm.Alternatively, maybe I can relate the volume of the small tetrahedron to the original octahedron. Since each face is being replaced, and the original octahedron has volume 8, perhaps I can find the volume of one face's tetrahedron in terms of the original.Wait, perhaps it's better to think in terms of proportions. Since each face is being replaced by a tetrahedron whose base is a triangle with side length 1/3 of the original, the area of the base is (1/3)² = 1/9 of the original face's area. But since it's a tetrahedron, the volume is proportional to the cube of the side length, so 1/27.But wait, the original face is a triangle, not a tetrahedron. So, maybe the volume of the tetrahedron is 1/27 of the volume of a tetrahedron with the same face as the original.But I don't know the volume of a tetrahedron with the same face as the original octahedron's face. Maybe I need another approach.Alternatively, perhaps I can consider that each face is being replaced by a tetrahedron whose base is a smaller triangle, and the height of the tetrahedron is such that it's inward-pointing. Since the base is a midpoint triangle, the height of the tetrahedron would be related to the original octahedron's geometry.Wait, maybe I can compute the volume of the small tetrahedron relative to the original octahedron.Let me think about the original octahedron. A regular octahedron can be thought of as two square pyramids glued together at their bases. Each pyramid has a square base and four triangular faces. The volume of the octahedron is 8, so each pyramid has volume 4.The height of each pyramid can be found using the volume formula for a pyramid: (base area * height)/3 = volume. The base area is the area of the square. Let me denote the edge length of the octahedron as 'a'. Then, the base of each pyramid is a square with side length 'a', so area a².The volume of each pyramid is (a² * h)/3 = 4. So, h = (4 * 3)/a² = 12/a².But I also know that in a regular octahedron, the height of each pyramid is related to the edge length. The height h of the pyramid is equal to the distance from the center of the octahedron to one of its vertices. For a regular octahedron with edge length 'a', the height h is given by h = √(2)/2 * a.Wait, let me verify that. The height of a regular octahedron can be found using the formula for the distance from the center to a vertex. For a regular octahedron with edge length 'a', the distance from the center to a vertex is indeed √(2)/2 * a. So, h = √(2)/2 * a.But from the volume, we have h = 12/a². So, equating the two expressions:√(2)/2 * a = 12/a²Multiply both sides by a²:√(2)/2 * a³ = 12So, a³ = (12 * 2)/√(2) = 24/√(2) = 12√2Therefore, a³ = 12√2, so a = (12√2)^(1/3)But maybe I don't need the exact value of 'a'. Instead, perhaps I can find the volume of the small tetrahedron in terms of the original octahedron's volume.Each face of the octahedron is an equilateral triangle with side length 'a'. The area of each face is (√3/4) * a².The midpoint triangle has side length a/3, so its area is (√3/4) * (a/3)² = (√3/4) * a² / 9 = (√3/36) * a².So, the area of the midpoint triangle is 1/9 of the original face's area.But the tetrahedron added has this midpoint triangle as its base. The height of the tetrahedron would be the distance from the midpoint triangle to the opposite vertex of the tetrahedron. Since it's a regular tetrahedron, all edges are equal, so the height can be found using the formula for the height of a regular tetrahedron: h_tetra = √(2/3) * edge length.Wait, the edge length of the tetrahedron is a/3, so the height is √(2/3) * (a/3) = √(2)/3√3 * a.But wait, is that correct? The height of a regular tetrahedron is indeed h = √(6)/3 * edge length. So, h = √(6)/3 * (a/3) = √(6)/9 * a.So, the height of the small tetrahedron is √(6)/9 * a.Therefore, the volume of the small tetrahedron is (base area * height)/3 = [(√3/36) * a²] * [√(6)/9 * a] / 3.Let's compute that:First, multiply the base area and height:(√3/36) * a² * (√6/9) * a = (√3 * √6) / (36 * 9) * a³ = (√18) / 324 * a³ = (3√2)/324 * a³ = √2 / 108 * a³.Then, divide by 3:(√2 / 108 * a³) / 3 = √2 / 324 * a³.So, the volume of each small tetrahedron is √2 / 324 * a³.But from earlier, we have a³ = 12√2. So, substituting:Volume = √2 / 324 * 12√2 = (√2 * 12√2) / 324 = (12 * 2) / 324 = 24 / 324 = 2 / 27.So, each small tetrahedron has a volume of 2/27.Wait, that seems a bit high. Let me double-check my calculations.Starting from the volume of the small tetrahedron:Volume = (base area * height)/3.Base area = (√3/4) * (a/3)² = (√3/4) * a² / 9 = √3 a² / 36.Height = √(6)/3 * (a/3) = √6 a / 9.So, Volume = (√3 a² / 36) * (√6 a / 9) / 3.First, multiply the terms:√3 * √6 = √18 = 3√2.a² * a = a³.Denominator: 36 * 9 * 3 = 972.So, Volume = (3√2 * a³) / 972 = (√2 * a³) / 324.Since a³ = 12√2, then:Volume = (√2 * 12√2) / 324 = (12 * 2) / 324 = 24 / 324 = 2 / 27.Yes, that seems correct. So, each small tetrahedron has a volume of 2/27.But wait, the original octahedron has volume 8. If each face is replaced by a tetrahedron of volume 2/27, and there are 8 faces, then the total volume subtracted would be 8 * (2/27) = 16/27. So, the volume of Q₁ would be 8 - 16/27 = (216/27 - 16/27) = 200/27 ≈ 7.407.But that seems like a lot of volume being subtracted. Maybe I made a mistake in calculating the volume of the small tetrahedron.Wait, another approach: since the side length is scaled by 1/3, the volume scales by (1/3)^3 = 1/27. So, if the original tetrahedron (if the face were a tetrahedron) had volume V, the new one would have V/27.But the original face is a triangle, not a tetrahedron. So, perhaps I need to consider the volume of the tetrahedron relative to the original octahedron.Alternatively, maybe I can think of the volume removed as a fraction of the original volume.Wait, perhaps it's better to consider that each face is being replaced by a tetrahedron whose base is a triangle with side length 1/3 of the original, so the area is 1/9, and the height is scaled accordingly.But I think my earlier calculation is correct: each small tetrahedron has volume 2/27.But let's see: the original octahedron has volume 8. If each face is replaced by a tetrahedron of volume 2/27, then 8 faces would subtract 16/27, leaving 200/27 ≈ 7.407. That seems plausible.But let me think about the process again. When we replace each face with a tetrahedron, we're not just subtracting the volume of the tetrahedron, but also changing the shape. However, in this case, since the tetrahedron is inward-pointing, it's like carving out a small piece from each face, so the volume subtracted is indeed the volume of the tetrahedron.But wait, actually, the tetrahedron is added inward, so it's not subtracting volume but adding. Wait, no, because it's replacing the face. So, the original volume is 8, and then we're adding these inward-pointing tetrahedrons, which effectively subtract volume from the original octahedron.Wait, no, actually, when you replace a face with a tetrahedron, you're adding volume. But since it's inward-pointing, it's like subtracting volume. Hmm, I'm confused.Wait, let me clarify. If you have a polyhedron and you replace a face with a tetrahedron, depending on the direction, it can either add or subtract volume. In this case, it's inward-pointing, so it's like digging a hole into the original polyhedron, thus subtracting volume.So, yes, the volume of Q₁ would be 8 minus the volume of all the small tetrahedrons added inward.So, if each small tetrahedron has volume 2/27, and there are 8 faces, then total volume subtracted is 16/27, so Q₁ has volume 8 - 16/27 = 200/27.But wait, let me think again. The volume of the small tetrahedron is 2/27. But the original octahedron's volume is 8. So, 2/27 is about 0.074, and 8 * 0.074 is about 0.592. So, subtracting that from 8 gives about 7.408, which seems reasonable.But I'm not sure if the volume of the small tetrahedron is 2/27. Let me try another approach.The volume of a regular tetrahedron is (edge length)^3 / (6√2). So, if the edge length is 1/3 of the original, then the volume is (1/3)^3 / (6√2) = 1/(27*6√2) = 1/(162√2). But this is relative to a unit edge length. Since the original octahedron has edge length 'a', the volume of the small tetrahedron would be (a/3)^3 / (6√2) = a³/(27*6√2) = a³/(162√2).But the volume of the original octahedron is 8, which is (sqrt(2)/3) * a³. So, 8 = (sqrt(2)/3) * a³ => a³ = 24 / sqrt(2) = 12√2.So, substituting back, the volume of the small tetrahedron is (12√2) / (162√2) = 12 / 162 = 2 / 27. So, yes, that confirms it. Each small tetrahedron has volume 2/27.Therefore, Q₁ has volume 8 - 8*(2/27) = 8 - 16/27 = (216 - 16)/27 = 200/27.Now, moving on to Q₂. Each face of Q₁ is replaced by an inward-pointing tetrahedron. But now, Q₁ has more faces than Q₀. Each original face of Q₀ was replaced by a tetrahedron, which has 3 new faces (since a tetrahedron has 4 faces, but one is the base which is glued to the original face, so 3 new faces per tetrahedron).Wait, no. When you replace a face with a tetrahedron, you're removing one face and adding three new faces. So, each original face becomes three new faces. Therefore, the number of faces increases by a factor of 3 each time.Wait, let me think. Original Q₀ has 8 faces. When you replace each face with a tetrahedron, each tetrahedron has 3 new faces (since one face is the base, which is glued to the original face). So, for each original face, you now have 3 new faces. Therefore, Q₁ has 8 * 3 = 24 faces.Similarly, Q₂ would have 24 * 3 = 72 faces, and Q₃ would have 72 * 3 = 216 faces.But wait, that might not be entirely accurate because each new tetrahedron adds 3 new faces, but the original face is removed. So, the total number of faces after replacement is original faces - 1 + 3 = original faces + 2 per face. Wait, no, that's not correct.Actually, each face is replaced by a tetrahedron, which has 4 faces, but one is glued to the original face, so the number of new faces per original face is 3. Therefore, the total number of faces after replacement is original number of faces * 3.So, Q₀: 8 faces.Q₁: 8 * 3 = 24 faces.Q₂: 24 * 3 = 72 faces.Q₃: 72 * 3 = 216 faces.Yes, that seems correct.Now, for each face in Q₁, we replace it with a tetrahedron. Each such tetrahedron has a base which is the midpoint triangle of the face, with side length 1/3 of the original face's side length.But wait, the original face in Q₁ is a triangle with side length 1/3 of the original octahedron's face. So, the midpoint triangle of Q₁'s face would have side length 1/3 of that, which is 1/9 of the original octahedron's face.But wait, no. The side length of the base of the tetrahedron added to Q₁ is 1/3 of the face it's replacing. So, since Q₁'s faces are triangles with side length 1/3 of Q₀'s faces, the base of the tetrahedron added to Q₁ would have side length 1/3 of that, which is 1/9 of Q₀'s face.Therefore, the volume of each tetrahedron added to Q₁ is (1/9)^3 = 1/729 of the original tetrahedron's volume. But wait, the original tetrahedron's volume was 2/27, so the new one would be (2/27) * (1/27) = 2/729.Wait, no. The volume scales with the cube of the side length. So, if the side length is 1/3, the volume is (1/3)^3 = 1/27 of the original tetrahedron's volume.But the original tetrahedron added to Q₀ had volume 2/27. So, the new tetrahedron added to Q₁ would have volume (2/27) * (1/27) = 2/729.But wait, that might not be correct because the original tetrahedron's volume was relative to the original octahedron. Maybe I need to think differently.Alternatively, since each face in Q₁ has side length 1/3 of Q₀'s face, the volume of the tetrahedron added to Q₁ would be (1/3)^3 = 1/27 of the volume of the tetrahedron added to Q₀.Since the tetrahedron added to Q₀ had volume 2/27, the one added to Q₁ would have volume (2/27) * (1/27) = 2/729.Therefore, each tetrahedron added to Q₁ has volume 2/729.Since Q₁ has 24 faces, the total volume subtracted to get Q₂ is 24 * (2/729) = 48/729 = 16/243.Therefore, the volume of Q₂ is Q₁'s volume minus 16/243.Q₁'s volume was 200/27, so Q₂'s volume is 200/27 - 16/243.To subtract these, convert 200/27 to 243 denominator: 200/27 = (200 * 9)/243 = 1800/243.So, 1800/243 - 16/243 = 1784/243.Simplify 1784/243: Let's see, 243 * 7 = 1701, 1784 - 1701 = 83. So, 1784/243 = 7 + 83/243.But we can leave it as 1784/243 for now.Now, moving on to Q₃. Each face of Q₂ is replaced by an inward-pointing tetrahedron. Q₂ has 72 faces.Each face in Q₂ has side length 1/9 of Q₀'s face, because each face in Q₁ was 1/3 of Q₀'s, and each face in Q₂ is 1/3 of Q₁'s, so 1/3 * 1/3 = 1/9.Therefore, the volume of each tetrahedron added to Q₂ is (1/9)^3 = 1/729 of the original tetrahedron's volume. But the original tetrahedron added to Q₀ was 2/27, so the new one would be (2/27) * (1/27) = 2/729, but wait, that's the same as the ones added to Q₁.Wait, no, because the side length is 1/9 of the original, so the volume is (1/9)^3 = 1/729 of the original tetrahedron's volume, which was 2/27. So, the new volume is (2/27) * (1/729) = 2/(27*729) = 2/19683.Wait, that seems too small. Alternatively, since each face in Q₂ has side length 1/9 of Q₀'s, the volume of the tetrahedron added to Q₂ is (1/9)^3 = 1/729 of the volume of the tetrahedron added to Q₀, which was 2/27. So, 2/27 * 1/729 = 2/(27*729) = 2/19683.But wait, that would mean each tetrahedron added to Q₂ has volume 2/19683.Since Q₂ has 72 faces, the total volume subtracted to get Q₃ is 72 * (2/19683) = 144/19683 = 16/2187.Wait, 144 divided by 19683: 19683 ÷ 144 = 136.6875, so 144/19683 = 16/2187 (since 144 ÷ 9 = 16, 19683 ÷ 9 = 2187).So, the volume subtracted is 16/2187.Therefore, the volume of Q₃ is Q₂'s volume minus 16/2187.Q₂'s volume was 1784/243. So, let's compute 1784/243 - 16/2187.First, find a common denominator. 243 is 3^5, 2187 is 3^7. So, the common denominator is 2187.Convert 1784/243 to 2187 denominator: 243 * 9 = 2187, so multiply numerator and denominator by 9: 1784 * 9 = 16056, so 16056/2187.Now, subtract 16/2187: 16056 - 16 = 16040.So, Q₃'s volume is 16040/2187.Simplify this fraction: Let's see if 16040 and 2187 have any common factors.2187 is 3^7, which is 2187 = 3^7.16040: Let's factor it.16040 ÷ 10 = 1604.1604 ÷ 4 = 401.401 is a prime number.So, 16040 = 10 * 4 * 401 = 2^3 * 5 * 401.2187 is 3^7, so no common factors. Therefore, 16040/2187 is already in simplest terms.But let me double-check my calculations because this seems a bit involved.Starting from Q₀: volume 8.Q₁: 8 - 8*(2/27) = 8 - 16/27 = 200/27 ≈ 7.407.Q₂: 200/27 - 24*(2/729) = 200/27 - 48/729 = 200/27 - 16/243.Convert 200/27 to 243 denominator: 200 * 9 = 1800, so 1800/243.1800/243 - 16/243 = 1784/243 ≈ 7.337.Q₃: 1784/243 - 72*(2/19683) = 1784/243 - 144/19683.Convert 1784/243 to 2187 denominator: 1784 * 9 = 16056, so 16056/2187.16056/2187 - 16/2187 = 16040/2187 ≈ 7.333.Wait, 16040/2187 is approximately 7.333, which is 22/3, but 22/3 is about 7.333, but 16040/2187 is exactly 16040 ÷ 2187 ≈ 7.333.But let me check if 16040/2187 can be simplified. As I factored earlier, 16040 = 2^3 * 5 * 401, and 2187 = 3^7, so no common factors. Therefore, 16040/2187 is the simplest form.But wait, the problem asks for the volume of Q₃. So, according to my calculations, it's 16040/2187.But let me check if I made any mistakes in the scaling factors.Each time we go from Qᵢ to Qᵢ₊₁, the number of faces triples, and the volume subtracted per face is (1/27) of the previous volume subtracted.Wait, no. The volume subtracted per face is scaled by (1/3)^3 = 1/27 each time.So, from Q₀ to Q₁: 8 faces, each subtracting 2/27, total subtracted 16/27.From Q₁ to Q₂: 24 faces, each subtracting (2/27)*(1/27) = 2/729, total subtracted 24*(2/729) = 48/729 = 16/243.From Q₂ to Q₃: 72 faces, each subtracting (2/729)*(1/27) = 2/19683, total subtracted 72*(2/19683) = 144/19683 = 16/2187.So, the total subtracted volumes are:Q₁: 16/27Q₂: 16/243Q₃: 16/2187Therefore, the total volume subtracted up to Q₃ is 16/27 + 16/243 + 16/2187.Let me compute that:Convert all to denominator 2187:16/27 = (16 * 81)/2187 = 1296/218716/243 = (16 * 9)/2187 = 144/218716/2187 = 16/2187Total subtracted: 1296 + 144 + 16 = 1456/2187.Therefore, the volume of Q₃ is 8 - 1456/2187.Convert 8 to 2187 denominator: 8 = 17496/2187.So, 17496/2187 - 1456/2187 = (17496 - 1456)/2187 = 16040/2187.Yes, that matches my earlier result.So, the volume of Q₃ is 16040/2187.But let me check if this fraction can be simplified. As before, 16040 factors into 2^3 * 5 * 401, and 2187 is 3^7, so no common factors. Therefore, 16040/2187 is the simplest form.Alternatively, we can write it as a mixed number, but since the problem asks for the volume, a fraction is fine.So, the final answer is 16040/2187.But wait, let me check if 16040 and 2187 have any common factors. 2187 ÷ 3 = 729, 16040 ÷ 3 = 5346.666, so not divisible by 3. 16040 ÷ 2 = 8020, 2187 ÷ 2 = 1093.5, so not divisible by 2. 16040 ÷ 5 = 3208, 2187 ÷ 5 = 437.4, so not divisible by 5. 16040 ÷ 7 = 2291.428, not integer. 16040 ÷ 11 = 1458.181, not integer. So, no common factors.Therefore, the volume of Q₃ is 16040/2187.But let me see if I can write this as a mixed number or decimal for better understanding.16040 ÷ 2187 ≈ 7.333...Because 2187 * 7 = 15309, 16040 - 15309 = 731.731/2187 ≈ 0.333...So, approximately 7.333..., which is 22/3 ≈ 7.333.But 16040/2187 is exactly equal to 7 + 731/2187.But 731/2187 can be simplified? Let's check.731 ÷ 17 = 43, because 17*43 = 731.2187 ÷ 17 = 128.647, not integer.So, 731 = 17*43, and 2187 is 3^7, so no common factors.Therefore, 731/2187 is in simplest terms.So, 16040/2187 = 7 + 731/2187.But since the problem asks for the volume, either form is acceptable, but likely as an improper fraction.Therefore, the volume of Q₃ is 16040/2187.But wait, let me check my earlier steps again because I might have made a mistake in the scaling factor.When moving from Q₀ to Q₁, each face is replaced by a tetrahedron with volume 2/27. So, 8 faces, total subtracted 16/27.From Q₁ to Q₂, each of the 24 faces is replaced by a tetrahedron with volume (2/27)*(1/27) = 2/729. So, 24*(2/729) = 48/729 = 16/243.From Q₂ to Q₃, each of the 72 faces is replaced by a tetrahedron with volume (2/729)*(1/27) = 2/19683. So, 72*(2/19683) = 144/19683 = 16/2187.Total subtracted: 16/27 + 16/243 + 16/2187.Convert to common denominator 2187:16/27 = 16*81/2187 = 1296/218716/243 = 16*9/2187 = 144/218716/2187 = 16/2187Total: 1296 + 144 + 16 = 1456/2187Therefore, volume of Q₃ = 8 - 1456/2187 = (8*2187 - 1456)/2187 = (17496 - 1456)/2187 = 16040/2187.Yes, that's correct.So, the final answer is 16040/2187.But let me check if this can be simplified further. As before, 16040 and 2187 have no common factors, so it's already in simplest form.Therefore, the volume of Q₃ is 16040/2187.

💡Alright, so I've got this problem here where I need to prove that among 55 numbers chosen from 1 to 100, there will always be two numbers with specific differences: 9, 10, 12, and 13. Additionally, I need to show that it's possible to choose 55 numbers without any two differing by 11. Hmm, okay, let's break this down step by step.Starting with part (a), which asks about a difference of 9. I remember something about the pigeonhole principle being useful in these types of problems. Maybe I can divide the numbers into groups where each group contains numbers that are 9 apart. For example, the first group could be 1, 10, 19, 28, and so on. If I have enough groups, and I pick more numbers than there are groups, then by the pigeonhole principle, at least two numbers from the same group will be picked, meaning their difference will be 9.Let me try to formalize this. If I create groups like {1,10,19,...}, {2,11,20,...}, up to {9,18,27,...}, that's 9 groups. Wait, but 9 times 11 is 99, and we have 100 numbers. Maybe I need to adjust the grouping. Alternatively, perhaps I can divide the numbers into blocks where each block has numbers that are spaced 9 apart. For example, the first block could be 1-9, the second 10-18, and so on. But I'm not sure if that's the right approach.Wait, maybe I should think about it differently. If I want to avoid having any two numbers with a difference of 9, I can pick at most one number from each of these pairs: (1,10), (2,11), (3,12), ..., (91,100). That gives me 90 pairs, but since we're picking 55 numbers, which is more than half of 90, that might not directly apply. Hmm, perhaps I need to consider overlapping pairs or something else.Wait, another idea: if I divide the numbers into 91 groups where each group consists of numbers that are 9 apart, like {1,10,19,...}, {2,11,20,...}, up to {9,18,27,...}. Each of these groups has either 11 or 12 numbers. If I pick 55 numbers, and there are 91 groups, then by the pigeonhole principle, at least one group must have at least two numbers picked. But wait, 55 divided by 91 is less than 1, so that doesn't help. Maybe I'm not grouping correctly.Perhaps I need to think about it as avoiding differences of 9. If I pick numbers such that no two are 9 apart, I can pick at most one number from each of these pairs: (1,10), (2,11), ..., (91,100). That gives me 90 pairs, and I can pick at most one from each pair, so maximum 90 numbers. But we're picking 55, which is less than 90, so that doesn't necessarily guarantee a pair with difference 9. Hmm, maybe this approach isn't working.Wait, maybe I need to consider the numbers modulo 9. If two numbers are congruent modulo 9, their difference is a multiple of 9. But we want a difference of exactly 9, not just a multiple. So maybe that's not directly helpful.Alternatively, perhaps I can use the concept of arithmetic progressions. If I have an arithmetic progression with common difference 9, then any two numbers in it differ by a multiple of 9. But again, we need exactly 9.Wait, maybe I should think about the maximum number of numbers I can pick without having any two differ by 9. If I can show that this maximum is less than 55, then picking 55 numbers must include such a pair.To construct such a set without any two numbers differing by 9, I can pick numbers from one residue class modulo 9. For example, pick all numbers congruent to 1 modulo 9: 1,10,19,...,91. That's 11 numbers. Similarly, for each residue class, I can pick 11 or 12 numbers. But wait, if I pick numbers from multiple residue classes, I might end up with differences of 9.Wait, no, if I pick numbers from only one residue class modulo 9, then their differences will be multiples of 9, but not necessarily exactly 9. For example, 1 and 10 differ by 9, which is exactly what we want to avoid. So actually, picking numbers from a single residue class modulo 9 would include pairs differing by 9. So that's not helpful.Hmm, maybe I need to pick numbers such that no two are in the same residue class modulo 9. But that would limit me to picking only 9 numbers, which is way less than 55. So that's not useful either.Wait, perhaps I need to partition the numbers into blocks where within each block, numbers are spaced 9 apart, and then use the pigeonhole principle based on the number of blocks.Let me try dividing the numbers into blocks of 9 consecutive numbers: {1-9}, {10-18}, {19-27}, ..., {91-99}, and then {100}. There are 12 such blocks (since 9*11=99, plus 100). If I pick more than 11 numbers from any block, then by the pigeonhole principle, I'll have two numbers within the same block, but their difference could be less than 9. Hmm, not directly helpful.Wait, maybe I need to consider that within each block of 9 numbers, if I pick more than one number, their difference could be 9 or more. But I need exactly 9.Alternatively, maybe I can think of the numbers as being arranged in a circle, but that might complicate things.Wait, another approach: consider the numbers from 1 to 100 and pair them such that each pair has a difference of 9. There are 91 such pairs: (1,10), (2,11), ..., (91,100). If I pick more than 91/2 numbers, which is 45.5, so 46 numbers, then by the pigeonhole principle, I must have picked both numbers from at least one pair. But we're picking 55 numbers, which is more than 46, so that would imply that we have at least one pair with difference 9. Wait, that seems promising.Let me verify: there are 91 pairs with difference 9. If I pick 55 numbers, the maximum number of pairs I can avoid is 55, but since there are 91 pairs, and each number can be in up to two pairs (except the first and last few), this might not directly apply. Hmm, maybe I need to think differently.Wait, actually, each number from 1 to 91 is part of a pair (n, n+9). So there are 91 pairs. If I pick 55 numbers, the maximum number of pairs I can avoid is 55, but since there are 91 pairs, and each number can be in only one pair (since n+9 is unique), then picking 55 numbers would mean that at least 55 - 91 = -36, which doesn't make sense. Hmm, maybe this approach isn't correct.Wait, perhaps I should use the concept of matching. If I have 91 pairs, and I pick 55 numbers, then the number of pairs covered is at least 55 - (100 - 91) = 55 -9=46. Wait, that doesn't seem right either.Maybe I need to think about it as a graph where each pair with difference 9 is an edge, and then finding a matching. But that might be overcomplicating.Wait, let's try a different angle. Suppose we want to pick as many numbers as possible without having any two differ by 9. How many can we pick? For each pair (n, n+9), we can pick at most one. Since there are 91 such pairs, the maximum number of numbers we can pick without having any two differ by 9 is 91. But we're picking 55, which is less than 91, so that doesn't help.Wait, no, actually, if we have 91 pairs, and we can pick at most one from each pair, then the maximum number is 91, but since we're picking 55, which is less than 91, it's possible to pick 55 without having any two differ by 9. But that contradicts the problem statement, which says we need to prove that there will be two numbers with difference 9. So maybe my reasoning is flawed.Wait, perhaps I'm misunderstanding the pairing. Actually, each number from 1 to 91 is part of a pair (n, n+9), but numbers from 92 to 100 are only part of one pair each (92-101, but 101 is beyond 100, so only 92-100 are single). So actually, there are 91 pairs, but the numbers 92-100 can only be paired with 83-91, which are already part of other pairs.So, if I want to pick numbers without any two differing by 9, I can pick at most one from each of the 91 pairs. But since we're picking 55 numbers, which is more than half of 91 (which is 45.5), then by the pigeonhole principle, we must have picked both numbers from at least one pair. Wait, that makes sense.Because if there are 91 pairs, and we pick 55 numbers, then the number of pairs covered is 55. Since 55 > 91/2 = 45.5, by the pigeonhole principle, at least one pair must have both numbers picked. Therefore, there must be at least one pair with difference 9.Okay, that seems to work. So for part (a), the answer is that among 55 numbers, there must be two with a difference of 9.Moving on to part (b), which is similar but for a difference of 10. I think a similar approach can be used. Let's see.Again, we can consider pairs of numbers differing by 10: (1,11), (2,12), ..., (90,100). There are 90 such pairs. If we pick 55 numbers, how does that affect the pairs?If we can pick at most one from each pair, the maximum number of numbers we can pick without having any two differ by 10 is 90. But since we're picking 55, which is less than 90, that doesn't directly imply anything. Wait, but actually, similar to part (a), if we have 90 pairs, and we pick 55 numbers, then the number of pairs covered is 55. Since 55 > 90/2 = 45, by the pigeonhole principle, at least one pair must have both numbers picked. Therefore, there must be at least one pair with difference 10.Wait, but 90 pairs, picking 55 numbers, so 55 > 45, so yes, at least one pair must be fully picked. Therefore, part (b) is also proven.For part (c), the difference is 12. Let's see if the same approach works. Pairs would be (1,13), (2,14), ..., (88,100). That's 88 pairs. Picking 55 numbers, which is more than half of 88 (44), so 55 > 44, so by pigeonhole principle, at least one pair must be fully picked, meaning two numbers differ by 12.Wait, but 88 pairs, picking 55 numbers, so 55 > 44, yes, so at least one pair must have both numbers picked. Therefore, part (c) is also proven.For part (d), the difference is 13. Let's check. Pairs would be (1,14), (2,15), ..., (87,100). That's 87 pairs. Picking 55 numbers, which is more than half of 87 (43.5), so 55 > 43.5, so by pigeonhole principle, at least one pair must be fully picked, meaning two numbers differ by 13.Wait, but 87 pairs, picking 55 numbers, so 55 > 43.5, yes, so at least one pair must have both numbers picked. Therefore, part (d) is also proven.Finally, part (e) asks to show that it's possible to pick 55 numbers without any two differing by 11. So, unlike the previous parts, here we need to construct such a set or show that it's possible.To do this, we can try to pick numbers in such a way that no two are 11 apart. One way to do this is to pick numbers from certain residue classes modulo 11. For example, if we pick numbers congruent to 1 modulo 11, that would be 1,12,23,...,99. That's 10 numbers. Similarly, we can pick from other residue classes, but we need to ensure that within each residue class, the numbers are spaced more than 11 apart.Wait, but actually, if we pick numbers from multiple residue classes, we might end up with numbers that are 11 apart across different classes. Hmm, maybe a better approach is to partition the numbers into blocks where within each block, numbers are spaced 11 apart, and then pick numbers from only certain blocks.Alternatively, we can use the concept of arithmetic progressions with difference greater than 11. For example, if we pick every 12th number, we can ensure that no two are 11 apart. But let's see.Wait, another idea: if we divide the numbers into 11 groups based on their residue modulo 11. Each group will have numbers like {1,12,23,...}, {2,13,24,...}, up to {11,22,33,...}. If we pick numbers from only 5 of these groups, we can ensure that no two numbers are in the same group, thus avoiding differences of 11. But wait, actually, numbers in different groups can still differ by 11 if they are in adjacent groups.Wait, no, if we pick numbers from non-adjacent groups, maybe that helps. For example, if we pick from groups 1,3,5,7,9,11, then the differences between numbers in these groups would be multiples of 11 plus the difference between the residues. But I'm not sure if that avoids differences of exactly 11.Wait, actually, if two numbers are in different groups, their difference could still be 11 if their residues differ by 11. For example, a number congruent to 1 modulo 11 and another congruent to 12 modulo 11 (which is equivalent to 1 modulo 11) would differ by 11. Wait, no, because 12 modulo 11 is 1, so they are in the same group. So actually, if we pick numbers from only certain groups, we can avoid having two numbers in the same group, thus avoiding differences of 11.Wait, but if we pick numbers from only 5 groups, say, groups 1,2,3,4,5, then numbers within these groups could still differ by 11 if they are in adjacent groups. For example, 1 and 12 are in groups 1 and 2, respectively, and their difference is 11. So that doesn't help.Hmm, maybe a better approach is to pick numbers such that no two are in the same or adjacent residue classes modulo 11. That way, their differences won't be exactly 11.Wait, but that might be too restrictive and limit the number of numbers we can pick. Alternatively, perhaps we can pick numbers from every other residue class modulo 11. For example, pick from groups 1,3,5,7,9,11. That's 6 groups. Each group has about 9 or 10 numbers. So 6 groups times 9 numbers each would give us 54 numbers, which is close to 55. Maybe we can adjust to get 55.Wait, let's calculate. There are 11 residue classes modulo 11. If we pick from 5 residue classes, each with 9 or 10 numbers, that would give us 45 to 55 numbers. So if we pick 5 residue classes, each with 11 numbers, that's 55 numbers. But wait, each residue class modulo 11 has either 9 or 10 numbers in 1-100. Specifically, residues 1-9 have 10 numbers each, and residues 10 and 11 have 9 numbers each.So if we pick 5 residue classes, say, residues 1,3,5,7,9, each has 10 numbers, that's 50 numbers. Then, we can pick 5 more numbers from residues 10 or 11, but we have to ensure that these don't form a difference of 11 with any other number.Wait, but if we pick from residues 10 or 11, their numbers are 10,21,32,...,99 for residue 10, and 11,22,33,...,99 for residue 11. If we pick from residue 10, we have to make sure that none of these numbers differ by 11 from any number in residues 1,3,5,7,9.But since residues 1,3,5,7,9 are spaced by 2 modulo 11, the differences between residue 10 and these residues would be 9,7,5,3,1 modulo 11, which are not 0, so their differences won't be multiples of 11. Therefore, picking from residue 10 won't create a difference of 11 with any number in residues 1,3,5,7,9.Similarly, picking from residue 11, which is equivalent to 0 modulo 11, would have differences of 10,8,6,4,2 modulo 11 with residues 1,3,5,7,9, which are also not 0. Therefore, picking from residue 11 won't create a difference of 11 with any number in residues 1,3,5,7,9.So, if we pick 50 numbers from residues 1,3,5,7,9 (10 numbers each), and then pick 5 more numbers from residue 10 or 11, we can have a total of 55 numbers without any two differing by 11.Wait, but residue 10 has 10 numbers, and residue 11 has 9 numbers. So we can pick 5 from residue 10 and 0 from residue 11, or 4 from residue 10 and 1 from residue 11, etc., as long as we don't pick more than 5 from residue 10 to keep the total at 55.Therefore, it is possible to pick 55 numbers without any two differing by 11.So, summarizing:(a) By considering pairs differing by 9, and using the pigeonhole principle, we find that picking 55 numbers must include at least one such pair.(b) Similarly, for differences of 10, the same approach applies.(c) For differences of 12, the same logic holds.(d) For differences of 13, the same method works.(e) By carefully selecting numbers from specific residue classes modulo 11, we can avoid having any two numbers differ by 11, thus showing it's possible.I think that covers all parts of the problem.

💡Okay, so I have this equation here: 2a equals 5b. And I need to find the value of a over b. Hmm, let's see. I think I remember from algebra that if I have an equation like this, I can solve for one variable in terms of the other. So maybe I can solve for a in terms of b or vice versa.Let me try solving for a first. If 2a equals 5b, then to get a by itself, I should divide both sides of the equation by 2. That would give me a equals (5b)/2. Okay, so a is equal to five halves of b. Now, if I want to find a over b, that would be (5b/2) divided by b. Wait, dividing by b is the same as multiplying by 1/b, right? So that would be (5b/2) times (1/b). The b in the numerator and the b in the denominator should cancel each other out, leaving me with 5/2. So, a over b is 5 over 2.Let me double-check that to make sure I didn't make a mistake. If 2a equals 5b, then dividing both sides by 2b should give me a/b equals 5/2. Yeah, that seems right. So the ratio of a to b is 5 to 2.Looking at the answer choices, option B is 5/2, which matches what I got. So I think B is the correct answer. I don't see any mistakes in my steps, so I feel confident about this.

💡Okay, so I have this problem where I'm given three non-zero real numbers (a), (b), and (c) that aren't all equal. They satisfy the equation:[frac{a^{2}}{2 a^{2}+b c}+frac{b^{2}}{2 b^{2}+c a}+frac{c^{2}}{2 c^{2}+a b}=1.]And I need to find the value of (a + b + c).Hmm, let me start by trying to understand the structure of this equation. Each term in the sum has a similar form: the numerator is the square of one variable, and the denominator is twice that square plus the product of the other two variables. So, for the first term, it's (a^2) over (2a^2 + bc), and similarly for the others.I wonder if there's a way to manipulate this equation to make it more manageable. Maybe I can introduce some substitutions or find a common pattern. Let me try defining new variables to simplify things.Let me set:[x = frac{2a^2}{bc}, quad y = frac{2b^2}{ca}, quad z = frac{2c^2}{ab}.]So, (x), (y), and (z) are defined in terms of (a), (b), and (c). Let me see if I can express the original equation in terms of (x), (y), and (z).Starting with the first term:[frac{a^2}{2a^2 + bc} = frac{a^2}{2a^2 + bc} = frac{1}{2 + frac{bc}{a^2}}.]But since (x = frac{2a^2}{bc}), then (frac{bc}{a^2} = frac{2}{x}). So, substituting back:[frac{1}{2 + frac{2}{x}} = frac{1}{2 + frac{2}{x}} = frac{x}{2x + 2} = frac{x}{2(x + 1)}.]Wait, that doesn't seem right. Let me check my substitution again.If (x = frac{2a^2}{bc}), then ( frac{bc}{a^2} = frac{2}{x} ). So, substituting into the denominator:[2 + frac{bc}{a^2} = 2 + frac{2}{x} = frac{2x + 2}{x} = frac{2(x + 1)}{x}.]Therefore, the first term becomes:[frac{a^2}{2a^2 + bc} = frac{1}{2 + frac{bc}{a^2}} = frac{1}{frac{2(x + 1)}{x}} = frac{x}{2(x + 1)}.]Similarly, the second term:[frac{b^2}{2b^2 + ca} = frac{1}{2 + frac{ca}{b^2}}.]Since (y = frac{2b^2}{ca}), then (frac{ca}{b^2} = frac{2}{y}). So, substituting:[frac{1}{2 + frac{2}{y}} = frac{y}{2(y + 1)}.]And the third term:[frac{c^2}{2c^2 + ab} = frac{1}{2 + frac{ab}{c^2}}.]Since (z = frac{2c^2}{ab}), then (frac{ab}{c^2} = frac{2}{z}). So, substituting:[frac{1}{2 + frac{2}{z}} = frac{z}{2(z + 1)}.]Putting it all together, the original equation becomes:[frac{x}{2(x + 1)} + frac{y}{2(y + 1)} + frac{z}{2(z + 1)} = 1.]Multiplying both sides by 2 to eliminate the denominators:[frac{x}{x + 1} + frac{y}{y + 1} + frac{z}{z + 1} = 2.]Hmm, that's a bit simpler. Now, let me think about how to handle this equation. Each term is of the form (frac{variable}{variable + 1}). Maybe I can rewrite each term as (1 - frac{1}{variable + 1}).Let me try that:[frac{x}{x + 1} = 1 - frac{1}{x + 1},][frac{y}{y + 1} = 1 - frac{1}{y + 1},][frac{z}{z + 1} = 1 - frac{1}{z + 1}.]Substituting back into the equation:[left(1 - frac{1}{x + 1}right) + left(1 - frac{1}{y + 1}right) + left(1 - frac{1}{z + 1}right) = 2.]Simplifying the left side:[3 - left(frac{1}{x + 1} + frac{1}{y + 1} + frac{1}{z + 1}right) = 2.]Subtracting 3 from both sides:[-left(frac{1}{x + 1} + frac{1}{y + 1} + frac{1}{z + 1}right) = -1.]Multiplying both sides by -1:[frac{1}{x + 1} + frac{1}{y + 1} + frac{1}{z + 1} = 1.]Okay, so now I have:[frac{1}{x + 1} + frac{1}{y + 1} + frac{1}{z + 1} = 1.]I need to find a relationship between (x), (y), and (z). Let me recall that earlier, I defined (x), (y), and (z) in terms of (a), (b), and (c):[x = frac{2a^2}{bc}, quad y = frac{2b^2}{ca}, quad z = frac{2c^2}{ab}.]If I multiply all three together:[xyz = left(frac{2a^2}{bc}right)left(frac{2b^2}{ca}right)left(frac{2c^2}{ab}right).]Let me compute this:First, multiply the numerators: (2a^2 times 2b^2 times 2c^2 = 8a^2b^2c^2).Then, multiply the denominators: (bc times ca times ab = a^2b^2c^2).So,[xyz = frac{8a^2b^2c^2}{a^2b^2c^2} = 8.]Therefore, (xyz = 8).So, I know that (xyz = 8) and that (frac{1}{x + 1} + frac{1}{y + 1} + frac{1}{z + 1} = 1).I need to find (x + y + z) or something related to it. Maybe I can express the sum of reciprocals in terms of (x + y + z).Let me consider the equation:[frac{1}{x + 1} + frac{1}{y + 1} + frac{1}{z + 1} = 1.]To combine these fractions, I can find a common denominator, which would be ((x + 1)(y + 1)(z + 1)).So, let's write each term with this common denominator:[frac{(y + 1)(z + 1)}{(x + 1)(y + 1)(z + 1)} + frac{(x + 1)(z + 1)}{(x + 1)(y + 1)(z + 1)} + frac{(x + 1)(y + 1)}{(x + 1)(y + 1)(z + 1)} = 1.]Combining the numerators:[(y + 1)(z + 1) + (x + 1)(z + 1) + (x + 1)(y + 1) = (x + 1)(y + 1)(z + 1).]Let me expand each term:First, expand the left side:1. ((y + 1)(z + 1) = yz + y + z + 1),2. ((x + 1)(z + 1) = xz + x + z + 1),3. ((x + 1)(y + 1) = xy + x + y + 1).Adding them together:[(yz + y + z + 1) + (xz + x + z + 1) + (xy + x + y + 1) = yz + y + z + 1 + xz + x + z + 1 + xy + x + y + 1.]Combine like terms:- Terms with (xy), (yz), (xz): (xy + yz + xz).- Terms with (x): (x + x = 2x).- Terms with (y): (y + y = 2y).- Terms with (z): (z + z + z = 3z).- Constants: (1 + 1 + 1 = 3).So, the left side becomes:[xy + yz + xz + 2x + 2y + 3z + 3.]Wait, that doesn't seem right. Let me recount:Wait, actually, when adding the three expanded terms:- (yz + y + z + 1),- (xz + x + z + 1),- (xy + x + y + 1).Adding them:- (yz + xz + xy) (the product terms),- (y + z + x + z + x + y) (the linear terms),- (1 + 1 + 1) (the constants).So, grouping:- Product terms: (xy + yz + xz),- Linear terms: (y + z + x + z + x + y = 2x + 2y + 2z),- Constants: (3).So, the left side is:[xy + yz + xz + 2x + 2y + 2z + 3.]Now, the right side is:[(x + 1)(y + 1)(z + 1).]Let me expand this:First, multiply ((x + 1)(y + 1)):[(x + 1)(y + 1) = xy + x + y + 1.]Then, multiply by ((z + 1)):[(xy + x + y + 1)(z + 1) = xy z + xy + x z + x + y z + y + z + 1.]So, the right side is:[xyz + xy + xz + x + yz + y + z + 1.]Now, set the left side equal to the right side:[xy + yz + xz + 2x + 2y + 2z + 3 = xyz + xy + xz + x + yz + y + z + 1.]Let me subtract the right side from both sides to bring everything to the left:[xy + yz + xz + 2x + 2y + 2z + 3 - (xyz + xy + xz + x + yz + y + z + 1) = 0.]Simplify term by term:- (xy - xy = 0),- (yz - yz = 0),- (xz - xz = 0),- (2x - x = x),- (2y - y = y),- (2z - z = z),- (3 - 1 = 2),- (-xyz).So, the equation becomes:[x + y + z + 2 - xyz = 0.]But we know from earlier that (xyz = 8). So, substituting that in:[x + y + z + 2 - 8 = 0,][x + y + z - 6 = 0,][x + y + z = 6.]Alright, so (x + y + z = 6).Now, let's recall the definitions of (x), (y), and (z):[x = frac{2a^2}{bc}, quad y = frac{2b^2}{ca}, quad z = frac{2c^2}{ab}.]So, (x + y + z = frac{2a^2}{bc} + frac{2b^2}{ca} + frac{2c^2}{ab} = 6).Let me write that out:[frac{2a^2}{bc} + frac{2b^2}{ca} + frac{2c^2}{ab} = 6.]I can factor out the 2:[2left(frac{a^2}{bc} + frac{b^2}{ca} + frac{c^2}{ab}right) = 6.]Divide both sides by 2:[frac{a^2}{bc} + frac{b^2}{ca} + frac{c^2}{ab} = 3.]Hmm, maybe I can write this as:[frac{a^3}{abc} + frac{b^3}{abc} + frac{c^3}{abc} = 3.]Because (frac{a^2}{bc} = frac{a^3}{abc}), and similarly for the others.So, combining the terms:[frac{a^3 + b^3 + c^3}{abc} = 3.]Multiply both sides by (abc):[a^3 + b^3 + c^3 = 3abc.]Okay, so now I have the equation (a^3 + b^3 + c^3 = 3abc). I remember there's an identity related to this. Let me recall it.The identity is:[a^3 + b^3 + c^3 - 3abc = (a + b + c)(a^2 + b^2 + c^2 - ab - bc - ca).]So, substituting our equation into this identity:[a^3 + b^3 + c^3 - 3abc = 0,][(a + b + c)(a^2 + b^2 + c^2 - ab - bc - ca) = 0.]This implies that either (a + b + c = 0) or (a^2 + b^2 + c^2 - ab - bc - ca = 0).Now, let's analyze both possibilities.First, if (a + b + c = 0), that would satisfy the equation. But what about the other factor?If (a^2 + b^2 + c^2 - ab - bc - ca = 0), then:[a^2 + b^2 + c^2 = ab + bc + ca.]I know that (a^2 + b^2 + c^2 geq ab + bc + ca) for real numbers, with equality if and only if (a = b = c). But the problem states that (a), (b), and (c) are not all equal. Therefore, (a^2 + b^2 + c^2 - ab - bc - ca neq 0).Hence, the only possibility is that (a + b + c = 0).Therefore, the value of (a + b + c) is 0.**Final Answer**boxed{0}

💡First, I need to determine the two parts of the total amount. The total amount is Rs. 2500, and the amount lent at 5% is Rs. 1500.0000000000007, which I can round to Rs. 1500 for simplicity.Next, I'll calculate the second part by subtracting the first part from the total amount:P2 = 2500 - 1500 = 1000.Now, I'll calculate the interest from each part. For the first part at 5% simple interest:I1 = 1500 * 5% * 1 = 75.For the second part at 6% simple interest:I2 = 1000 * 6% * 1 = 60.Finally, I'll add the two interests to find the total yearly annual income:Total Income = 75 + 60 = 135.