🧠:[<think>Okay, let's try to solve this equation: 2^{1 - 2|x|} + 2 * 4^{1 + |x|} = 17. Hmm, exponential equations can sometimes be tricky, but maybe I can simplify it by using substitution or properties of exponents. Let me think.First, I notice that both terms have bases that are powers of 2. The first term is 2^{1 - 2|x|}, and the second term is 4^{1 + |x|}. Since 4 is 2 squared, maybe I can rewrite everything with base 2. Let me try that.So, 4^{1 + |x|} is the same as (2^2)^{1 + |x|}, which simplifies to 2^{2*(1 + |x|)} = 2^{2 + 2|x|}. Therefore, the original equation becomes:2^{1 - 2|x|} + 2 * 2^{2 + 2|x|} = 17.Now, let's simplify each term. The first term is already 2^{1 - 2|x|}, and the second term is 2 multiplied by 2^{2 + 2|x|}. Multiplying 2 by 2^{2 + 2|x|} is like adding exponents: 2^1 * 2^{2 + 2|x|} = 2^{1 + 2 + 2|x|} = 2^{3 + 2|x|}. Wait, is that correct?Wait, hold on. 2 times 2^{2 + 2|x|} is 2^{1} * 2^{2 + 2|x|}, so when you multiply exponents with the same base, you add the exponents. So that would be 2^{1 + 2 + 2|x|} = 2^{3 + 2|x|}. Yes, that's right. So the equation becomes:2^{1 - 2|x|} + 2^{3 + 2|x|} = 17.Hmm, now both terms are powers of 2. Let me set a substitution to simplify this. Let me let y = 2^{|x|}. But wait, actually, looking at the exponents, 1 - 2|x| and 3 + 2|x|. If I let t = |x|, then the exponents become 1 - 2t and 3 + 2t. Then, the equation becomes:2^{1 - 2t} + 2^{3 + 2t} = 17.Alternatively, maybe substitute u = 2^{2t}. Let's try that. Let me see. Let u = 2^{2t}. Then, 2^{1 - 2t} would be 2^1 * 2^{-2t} = 2 * (2^{2t})^{-1} = 2/u. Similarly, 2^{3 + 2t} is 2^3 * 2^{2t} = 8 * u. So substituting, the equation becomes:2/u + 8u = 17.That seems manageable. Let's write that out:(2)/u + 8u = 17.Multiply both sides by u to eliminate the denominator:2 + 8u^2 = 17u.Bring all terms to one side:8u^2 - 17u + 2 = 0.Now, we can try to solve this quadratic equation for u. Let's compute the discriminant:D = (-17)^2 - 4*8*2 = 289 - 64 = 225.Square root of 225 is 15. So solutions are:u = [17 ± 15]/(2*8) = [17 + 15]/16 or [17 - 15]/16.So, u = (32)/16 = 2, or u = (2)/16 = 1/8.Therefore, u = 2 or u = 1/8.But u was defined as 2^{2t}, so:Case 1: u = 2 = 2^{2t} => 2^{2t} = 2^1 => 2t = 1 => t = 1/2.Case 2: u = 1/8 = 2^{-3} => 2^{2t} = 2^{-3} => 2t = -3 => t = -3/2.But t was defined as |x|, which is always non-negative. So t = -3/2 is impossible. Therefore, only t = 1/2 is valid.Thus, t = |x| = 1/2. Therefore, x = ±1/2.Wait, so the solutions are x = 1/2 and x = -1/2. Let me check if these solutions actually satisfy the original equation.Let's check x = 1/2 first.Compute 2^{1 - 2|x|} + 2 * 4^{1 + |x|}.First term: 2^{1 - 2*(1/2)} = 2^{1 - 1} = 2^0 = 1.Second term: 2 * 4^{1 + 1/2} = 2 * 4^{3/2} = 2 * (2^2)^{3/2} = 2 * 2^{3} = 2 * 8 = 16.So total: 1 + 16 = 17. Which matches the right-hand side. Good.Now check x = -1/2.Same as above because of the absolute value: |x| is still 1/2. So same calculation: 1 + 16 = 17. Correct.Therefore, both solutions are valid. So the solutions are x = ±1/2.Wait, but let me double-check the substitution steps to make sure I didn't make a mistake.Original equation: 2^{1 - 2|x|} + 2 * 4^{1 + |x|} = 17.Expressed everything in base 2:2^{1 - 2|x|} + 2 * (2^2)^{1 + |x|} = 2^{1 - 2|x|} + 2 * 2^{2 + 2|x|} = 2^{1 - 2|x|} + 2^{3 + 2|x|}.Then set t = |x|, so equation becomes 2^{1 - 2t} + 2^{3 + 2t} = 17.Then substituted u = 2^{2t}, leading to 2/u + 8u = 17. Then multiplied by u: 2 + 8u^2 = 17u, leading to quadratic equation 8u^2 -17u +2=0.Solutions u=(17±15)/16, so u=2 or u=1/8. Then since u=2^{2t}, for u=2, 2t=1 => t=1/2. For u=1/8, 2t=-3, t=-3/2, which is invalid because t=|x|. So only t=1/2, leading to x=±1/2.All steps check out, and verifying the solutions in the original equation works. So the solutions are indeed x=±1/2.I don't think I made any mistakes here. The key was to express everything in terms of the same base, make a substitution to turn it into a quadratic equation, solve for u, then backtrack to find t and then x. The absolute value leads to two solutions, but since we set t=|x|, we have to consider both positive and negative x, but since |x| is positive, we get x=±1/2. All steps seem valid. So, confident that the solutions are x=1/2 and x=-1/2.Final AnswerThe solutions are boxed{frac{1}{2}} and boxed{-frac{1}{2}}.</think>To solve the equation (2^{1-2|x|} + 2 cdot 4^{1+|x|} = 17), we start by expressing all terms with base 2. First, rewrite (4^{1+|x|}) as ((2^2)^{1+|x|}):[4^{1+|x|} = 2^{2(1+|x|)} = 2^{2 + 2|x|}]Substituting this back into the original equation, we get:[2^{1 - 2|x|} + 2 cdot 2^{2 + 2|x|} = 17]Simplify the second term:[2 cdot 2^{2 + 2|x|} = 2^{3 + 2|x|}]Thus, the equation becomes:[2^{1 - 2|x|} + 2^{3 + 2|x|} = 17]Let (t = |x|). Then the equation transforms to:[2^{1 - 2t} + 2^{3 + 2t} = 17]Next, substitute (u = 2^{2t}). This gives:[2^{1 - 2t} = frac{2}{u} quad text{and} quad 2^{3 + 2t} = 8u]Substituting these into the equation, we get:[frac{2}{u} + 8u = 17]Multiply both sides by (u) to clear the fraction:[2 + 8u^2 = 17u]Rearranging into a standard quadratic form:[8u^2 - 17u + 2 = 0]Solving this quadratic equation using the discriminant (D = (-17)^2 - 4 cdot 8 cdot 2 = 225), we find:[u = frac{17 pm 15}{16}]This gives the solutions:[u = frac{32}{16} = 2 quad text{or} quad u = frac{2}{16} = frac{1}{8}]Since (u = 2^{2t}), we solve for (t):- For (u = 2), (2^{2t} = 2) implies (2t = 1) so (t = frac{1}{2}).- For (u = frac{1}{8}), (2^{2t} = 2^{-3}) implies (2t = -3), which is invalid as (t = |x| geq 0).Thus, (t = frac{1}{2}) gives (|x| = frac{1}{2}), leading to (x = pm frac{1}{2}).Verifying both solutions in the original equation confirms they are correct. Therefore, the solutions are:[boxed{frac{1}{2}} quad text{and} quad boxed{-frac{1}{2}}],<think>Okay, so I need to solve the equation (2^{1 - 2|x|} + 2 cdot 4^{1 + |x|} = 17). Hmm, let me start by understanding the components here. Both terms on the left side are exponential functions with base 2 and 4, which is a power of 2. Maybe I can rewrite everything in terms of base 2 to make it easier. Let me recall that (4 = 2^2), so (4^{1 + |x|}) can be written as ((2^2)^{1 + |x|}), which is (2^{2(1 + |x|)}). That simplifies to (2^{2 + 2|x|}). So substituting that back into the original equation, the second term becomes (2 cdot 2^{2 + 2|x|}). Then combining the exponents, since when you multiply same bases you add exponents, right? So (2 cdot 2^{2 + 2|x|} = 2^{1 + 2 + 2|x|} = 2^{3 + 2|x|}). Wait, no, hold on. Let me check that again. If it's 2 times (2^{2 + 2|x|}), that should be (2^{1} cdot 2^{2 + 2|x|}) which is (2^{1 + 2 + 2|x|} = 2^{3 + 2|x|}). Yeah, that seems right.So now the original equation becomes:(2^{1 - 2|x|} + 2^{3 + 2|x|} = 17).Hmm, okay. So both terms are powers of 2. Maybe I can set (y = 2^{ -2|x|}). Let me see. Let me think. Let me denote (y = 2^{2|x|}). Then, note that (2^{1 - 2|x|} = 2 cdot 2^{-2|x|} = 2 cdot frac{1}{2^{2|x|}} = frac{2}{y}). Similarly, (2^{3 + 2|x|} = 2^3 cdot 2^{2|x|} = 8 cdot y). So substituting these into the equation, we get:(frac{2}{y} + 8y = 17).That's a rational equation. Let's multiply both sides by y to eliminate the denominator:(2 + 8y^2 = 17y).Then rearrange the terms:(8y^2 - 17y + 2 = 0).Okay, so now we have a quadratic equation in terms of y. Let's try to solve this quadratic. The quadratic formula is (y = frac{17 pm sqrt{(-17)^2 - 4 cdot 8 cdot 2}}{2 cdot 8}). Let's compute the discriminant first:Discriminant D = (17^2 - 4 cdot 8 cdot 2 = 289 - 64 = 225).So square root of D is 15. Then,(y = frac{17 pm 15}{16}).So two solutions:1. (y = frac{17 + 15}{16} = frac{32}{16} = 2).2. (y = frac{17 - 15}{16} = frac{2}{16} = frac{1}{8}).So y can be 2 or 1/8. But remember that (y = 2^{2|x|}), and since (2^{2|x|}) is always positive, both solutions are valid. Now, let's solve for |x|.First case: (y = 2 = 2^{2|x|}). Taking log base 2 on both sides:(log_2 2 = 2|x|) => (1 = 2|x|) => (|x| = frac{1}{2}). So x is ±1/2.Second case: (y = frac{1}{8} = 2^{2|x|}). Again, take log base 2:(log_2 frac{1}{8} = 2|x|). Since (1/8 = 2^{-3}), so:(-3 = 2|x|) => (|x| = -frac{3}{2}). Wait, absolute value can't be negative. That's impossible. So this case has no solution.Therefore, the only solutions come from y = 2, which gives |x| = 1/2, so x = 1/2 or x = -1/2.But let me check this solution in the original equation to make sure.Let's test x = 1/2:First term: (2^{1 - 2|1/2|} = 2^{1 - 2*(1/2)} = 2^{1 - 1} = 2^0 = 1).Second term: (2 cdot 4^{1 + |1/2|} = 2 cdot 4^{1 + 1/2} = 2 cdot 4^{3/2}). 4^{3/2} is (4^{1/2})^3 = 2^3 = 8. So 2*8=16.So total: 1 + 16 = 17. Which matches the right-hand side. Good.Test x = -1/2:Same as above, because absolute value is same. So same result. 17. Correct.So both solutions work.But wait, let me check if I made a mistake in the substitution step. Let me go through the substitution again.Original equation: (2^{1 - 2|x|} + 2 cdot 4^{1 + |x|} = 17).Express 4 as 2^2: 2* (2^2)^{1 + |x|} = 2 * 2^{2(1 + |x|)} = 2^{1 + 2(1 + |x|)}.Wait, hold on. Wait, 2 * 4^{1 + |x|} = 2*(2^2)^{1 + |x|} = 2*(2^{2(1 + |x|)}) = 2^{1} * 2^{2(1 + |x|)} = 2^{1 + 2(1 + |x|)} = 2^{1 + 2 + 2|x|} = 2^{3 + 2|x|}. So that part is correct.Then the first term is 2^{1 - 2|x|}, correct.So the substitution steps are correct.Then setting y = 2^{2|x|}, so 2^{1 - 2|x|} = 2 / y, and 2^{3 + 2|x|} = 8y. Then equation becomes 2/y + 8y =17. Multiply by y: 2 +8y^2=17y, leading to 8y^2 -17y +2=0. Solving quadratic, solutions y=2 and y=1/8. Then check y=2 gives |x|=1/2, valid. y=1/8 gives |x| negative, invalid. So only solutions are x=1/2 and x=-1/2. So that's correct.Alternatively, maybe I can check by substituting x=0. Let's see, x=0. Then:First term: 2^{1 -0} =2^1=2.Second term: 2*4^{1+0}=2*4=8.Total: 2+8=10, which is less than 17. So x=0 is not a solution.How about x=1? Let's check:First term:2^{1 -2*1}=2^{-1}=0.5Second term:2*4^{1+1}=2*16=32Total:0.5+32=32.5 >17. So not a solution.Similarly x= -1 same. So seems like x=1/2 and -1/2 are the only solutions.Alternatively, maybe there's another approach. Let me try to think again.Another way: Let me set t = |x|, so the equation becomes in terms of t, where t ≥0.So equation:2^{1 - 2t} + 2*4^{1 + t} =17.Again, 4^{1 + t} =2^{2(1 + t)}.So equation becomes:2^{1 -2t} + 2*2^{2 + 2t} =17.Simplify exponents:2^{1 -2t} + 2^{3 + 2t} =17.Same as before, then substitution y=2^{2t}, so 2^{1 -2t} =2 *2^{-2t}=2 / y, and 2^{3 +2t}=8* y. Then equation is 2/y +8y=17. Which is same as before. So solving quadratic, same solutions.Therefore, same result. So conclusion remains t= |x| =1/2. So x=±1/2. So yeah, seems correct.Alternatively, maybe I can let u =2^{2|x|}, so u=2^{2|x|}, then 2^{ -2|x|}=1/u.But same substitution as before.So, the solutions are x=1/2 and x=-1/2.Therefore, the answer is x=±1/2, boxed as boxed{pm dfrac{1}{2}}.Final AnswerThe solutions to the equation are boxed{pm dfrac{1}{2}}.</think>To solve the equation (2^{1 - 2|x|} + 2 cdot 4^{1 + |x|} = 17), we start by expressing all terms in base 2. First, rewrite (4^{1 + |x|}) as (2^{2(1 + |x|)}). Substituting this into the equation, we get:[2^{1 - 2|x|} + 2 cdot 2^{2(1 + |x|)} = 17]Simplifying the second term:[2 cdot 2^{2(1 + |x|)} = 2^{1 + 2(1 + |x|)} = 2^{3 + 2|x|}]Thus, the equation becomes:[2^{1 - 2|x|} + 2^{3 + 2|x|} = 17]Let (y = 2^{2|x|}). Then, we can rewrite the terms as:[2^{1 - 2|x|} = frac{2}{y} quad text{and} quad 2^{3 + 2|x|} = 8y]Substituting these into the equation, we get:[frac{2}{y} + 8y = 17]Multiplying through by (y) to clear the denominator:[2 + 8y^2 = 17y]Rearranging into a quadratic equation:[8y^2 - 17y + 2 = 0]Solving this quadratic equation using the quadratic formula (y = frac{-b pm sqrt{b^2 - 4ac}}{2a}) with (a = 8), (b = -17), and (c = 2):[y = frac{17 pm sqrt{289 - 64}}{16} = frac{17 pm 15}{16}]This gives two solutions:[y = frac{32}{16} = 2 quad text{and} quad y = frac{2}{16} = frac{1}{8}]For (y = 2):[2^{2|x|} = 2 implies 2|x| = 1 implies |x| = frac{1}{2} implies x = pm frac{1}{2}]For (y = frac{1}{8}):[2^{2|x|} = frac{1}{8} implies 2|x| = -3 quad text{(not possible since absolute value cannot be negative)}]Thus, the only valid solutions are (x = pm frac{1}{2}). Verifying these solutions in the original equation confirms they are correct.The solutions to the equation are (boxed{pm dfrac{1}{2}}).]

🔑:Solution:2^{2|x|}=t geq 1, quad 8 t^{2}-17 t+2=0, quad t=2, quad 2^{2|x|}=2, quad 2|x|=1, quad x= pm frac{1}{2} quad text{Answer}: x= pm frac{1}{2}

🧠:[<think>Okay, let me try to solve this problem. Let's see. The problem is about complex numbers z₁ and z₂ on a circle centered at i with radius 1. The real part of the conjugate of z₁ multiplied by z₂ is zero. The principal argument of z₁ is π/6, and we need to find z₂. Hmm.First, let me recall some complex number concepts. The circle centered at i with radius 1 means all points z such that |z - i| = 1. So both z₁ and z₂ lie on this circle. The principal argument of z₁ is π/6. The principal argument is the angle in (-π, π], so π/6 is in the first quadrant. That means z₁ is in the first quadrant. Since it's on the circle centered at i, which is at (0,1) in the complex plane, the circle has radius 1. So z₁ is a point on this circle with argument π/6. Wait, but the argument is measured from the positive real axis. Hmm, but since the circle is centered at i, maybe we need to adjust the coordinates?Wait, maybe I should parametrize z₁ and z₂. Let me think. Any point on the circle |z - i| = 1 can be written as i + e^{iθ}, where θ is the angle parameter. Because the center is at i, and the radius is 1, so adding e^{iθ} to i gives all points on the circle. Alternatively, it can be written as i + cosθ + i sinθ, which is cosθ + i(1 + sinθ). So in terms of real and imaginary parts, z = x + iy where x = cosθ and y = 1 + sinθ.Given that the principal argument of z₁ is π/6. The argument of a complex number z = x + iy is arctan(y/x) if x > 0. Since the argument is π/6, that means arctan(y/x) = π/6, so y/x = tan(π/6) = 1/√3. Therefore, y = x/√3. But z₁ is on the circle |z - i| = 1, so x² + (y - 1)^2 = 1. Substituting y = x/√3 into this equation:x² + (x/√3 - 1)^2 = 1.Let me compute this:x² + (x²/3 - 2x/√3 + 1) = 1.Combine terms:x² + x²/3 - 2x/√3 + 1 = 1.Simplify:(4x²)/3 - 2x/√3 = 0.Multiply both sides by 3 to eliminate denominators:4x² - 6x/√3 = 0.Simplify 6/√3 as 2√3:4x² - 2√3 x = 0.Factor out x:x(4x - 2√3) = 0.So x = 0 or 4x - 2√3 = 0 => x = (2√3)/4 = √3/2.But if x = 0, then y = 0/√3 = 0. Then z₁ would be 0 + i0, but the problem states non-zero complex numbers, so z₁ can't be zero. Therefore, x = √3/2. Then y = (√3/2)/√3 = 1/2. Therefore, z₁ is (√3/2) + i(1/2). Wait, but z₁ is on the circle |z - i| = 1, so let's check:(√3/2)^2 + (1/2 - 1)^2 = (3/4) + (-1/2)^2 = 3/4 + 1/4 = 1. Yes, that works. So z₁ is √3/2 + i/2. Alternatively, in exponential form, since its argument is π/6, but wait, its coordinates are (√3/2, 1/2), which is indeed on the unit circle centered at the origin with argument π/6. But here, the circle is centered at i. So actually, z₁ is not on the unit circle centered at the origin, but on the circle centered at i. However, we found that z₁ is √3/2 + i/2. Let me check the distance from i:|z₁ - i| = |√3/2 + i/2 - i| = |√3/2 - i/2| = sqrt( (√3/2)^2 + (-1/2)^2 ) = sqrt(3/4 + 1/4) = sqrt(1) = 1. Correct, so it is on the circle. So z₁ is correctly determined.Now, we have to find z₂ such that the real part of conjugate(z₁) multiplied by z₂ is zero. So Re( (overline{z_1} cdot z_2) ) = 0. Let me write down conjugate(z₁). Since z₁ = √3/2 + i/2, then conjugate(z₁) is √3/2 - i/2. Let's denote (overline{z_1} = sqrt{3}/2 - i/2). Then, multiply this by z₂. Let z₂ = x + iy (which is also on the circle |z - i| = 1). So:(overline{z_1} cdot z_2 = (sqrt{3}/2 - i/2)(x + iy)).Multiply this out:= (√3/2)x + (√3/2)(iy) - (i/2)x - (i/2)(iy)= (√3/2)x + i(√3/2)y - i(x/2) - (i² y)/2Since i² = -1, this becomes:= (√3/2)x + i(√3 y/2 - x/2) + y/2So the real part of this is (√3/2)x + y/2, and the imaginary part is (√3 y/2 - x/2).Given that the real part is zero, so:(√3/2)x + (y)/2 = 0.Multiply both sides by 2:√3 x + y = 0.So, √3 x + y = 0. That's one equation.But z₂ is also on the circle |z - i| = 1, so x² + (y - 1)^2 = 1.So we have two equations:1. √3 x + y = 02. x² + (y - 1)^2 = 1We can substitute y = -√3 x into the second equation:x² + (-√3 x - 1)^2 = 1Compute (-√3 x - 1)^2:= (√3 x + 1)^2 = 3x² + 2√3 x + 1So the equation becomes:x² + 3x² + 2√3 x + 1 = 1Combine like terms:4x² + 2√3 x + 1 = 1Subtract 1 from both sides:4x² + 2√3 x = 0Factor out x:x(4x + 2√3) = 0So x = 0 or 4x + 2√3 = 0 => x = - (2√3)/4 = -√3/2If x = 0, then y = -√3 * 0 = 0. Then z₂ = 0 + i0, but the problem states non-zero complex numbers, so z₂ can't be zero. Therefore, x = -√3/2, then y = -√3*(-√3/2) = (√3)^2 / 2 = 3/2. Wait, hold on:Wait, equation 1: y = -√3 x. So if x = -√3/2, then y = -√3*(-√3/2) = (√3 * √3)/2 = 3/2. Therefore, z₂ is (-√3/2) + i*(3/2). Let me check if this is on the circle |z - i| = 1:Compute |z₂ - i| = |-√3/2 + i*(3/2 - 1)| = |-√3/2 + i*(1/2)|Which is sqrt( (√3/2)^2 + (1/2)^2 ) = sqrt( 3/4 + 1/4 ) = sqrt(1) = 1. Correct. So z₂ is (-√3/2) + (3/2)i.But let me check if there's another possibility. Because when solving the equation √3 x + y = 0 and the circle equation, we found x = -√3/2 and y = 3/2. So that's the only non-zero solution. Therefore, z₂ is (-√3/2) + (3/2)i.But let me also verify the condition Re( (overline{z_1} cdot z_2) ) = 0 with these values. Let's compute (overline{z_1}) * z₂:(overline{z_1}) = √3/2 - i/2z₂ = -√3/2 + i*(3/2)Multiplying these:(√3/2)(-√3/2) + (√3/2)(i*3/2) + (-i/2)(-√3/2) + (-i/2)(i*3/2)First term: (√3 * -√3)/4 = -3/4Second term: (i * 3√3)/4Third term: (i * √3)/4Fourth term: (-i² * 3)/4 = (3)/4 since i² = -1.So combine the terms:Real parts: -3/4 + 3/4 = 0Imaginary parts: (3√3/4 + √3/4)i = (4√3/4)i = √3 iSo the product is 0 + √3 i, whose real part is indeed zero. Perfect, that checks out.Therefore, z₂ is (-√3/2) + (3/2)i. To write this in a different form, maybe in terms of exponential or something? Let me see.Alternatively, since z₁ is √3/2 + i/2, which can be written as cos(π/6) + i sin(π/6), but shifted to the circle centered at i. Wait, but actually, z₁ is not on the unit circle, but on the circle centered at i. However, when we calculated, z₁ had coordinates (√3/2, 1/2), which is indeed shifted from the origin. But maybe there's another way to represent z₂.Alternatively, since z₁ and z₂ are on the circle centered at i with radius 1, maybe there's a geometric interpretation. The condition Re( (overline{z_1} cdot z_2) ) = 0 means that the inner product of z₁ and z₂ is purely imaginary. Wait, the inner product in complex numbers is (overline{z_1} cdot z_2), so if the real part is zero, that means they are orthogonal in the complex plane? Hmm, is that right?Orthogonal in the complex plane usually would mean that the Hermitian inner product is zero, which is when (overline{z_1} cdot z_2 = 0). But here, only the real part is zero. So their inner product is purely imaginary. So maybe they are orthogonal in some rotated sense? Not sure if that helps.But in any case, we found algebraically that z₂ is (-√3/2, 3/2), so expressed as a complex number, that's -√3/2 + (3/2)i. Let me check if that's one of the standard angles.The point z₂ is (-√3/2, 3/2). Wait, but let me write z₂ - i, since the circle is centered at i. So z₂ - i = (-√3/2, 3/2 - 1) = (-√3/2, 1/2). So in terms of the circle, the point is (-√3/2, 1/2) relative to center i. So in terms of angle, the angle θ such that cosθ = -√3/2 and sinθ = 1/2. That corresponds to θ = 150 degrees or 5π/6. Because cos(5π/6) = -√3/2 and sin(5π/6) = 1/2. So z₂ - i = e^{i5π/6}, hence z₂ = i + e^{i5π/6}.Alternatively, since z₁ - i = e^{iπ/6}, as we saw z₁ is i + e^{iπ/6}, and z₂ is i + e^{i5π/6}. So there might be a symmetry here. The arguments are π/6 and 5π/6, which are symmetric about the imaginary axis. Maybe that's a result of the orthogonality condition?Alternatively, since Re( (overline{z_1} cdot z_2) ) = 0, which is equivalent to (overline{z_1} cdot z_2) being purely imaginary. So z₂ is a scalar multiple of i or -i times z₁. But since they are on the same circle, maybe rotated by 90 degrees? But in this case, since z₂ is not just a rotation of z₁, because the circle is centered at i.But in any case, the algebraic solution gives z₂ = -√3/2 + (3/2)i, which is the answer.But let me confirm once more. Let me compute (overline{z_1} cdot z_2):z₁ = √3/2 + i/2(overline{z_1}) = √3/2 - i/2z₂ = -√3/2 + i3/2Multiplying:(√3/2)(-√3/2) + (√3/2)(i3/2) + (-i/2)(-√3/2) + (-i/2)(i3/2)Compute each term:First term: (√3 * -√3)/(2*2) = -3/4Second term: (√3 * i3)/(2*2) = (3√3 i)/4Third term: (-i * -√3)/(2*2) = (√3 i)/4Fourth term: (-i * i3)/(2*2) = (-i²3)/4 = (3)/4 because i² = -1.So total real parts: -3/4 + 3/4 = 0Total imaginary parts: (3√3/4 + √3/4)i = (4√3/4)i = √3 iHence, (overline{z_1} cdot z_2 = 0 + √3 i), so real part is zero. Perfect. So this satisfies the condition.Therefore, the answer is z₂ = -√3/2 + (3/2)i. To write this in the box as required, using LaTeX:boxed{-dfrac{sqrt{3}}{2} + dfrac{3}{2}i}Final Answerboxed{-dfrac{sqrt{3}}{2} + dfrac{3}{2}i}</think>Given non-zero complex numbers ( z_1 ) and ( z_2 ) on a circle centered at ( i ) with radius 1, where the real part of ( overline{z_1} cdot z_2 ) is zero and the principal value of the argument of ( z_1 ) is ( frac{pi}{6} ), we need to find ( z_2 ).1. Determine ( z_1 ): - The argument of ( z_1 ) is ( frac{pi}{6} ), so ( z_1 ) can be written as ( sqrt{3}/2 + i/2 ). - Verify ( z_1 ) lies on the circle ( |z - i| = 1 ): [ left| left( frac{sqrt{3}}{2} + frac{i}{2} right) - i right| = left| frac{sqrt{3}}{2} - frac{i}{2} right| = sqrt{left( frac{sqrt{3}}{2} right)^2 + left( -frac{1}{2} right)^2} = 1 ]2. Condition on ( overline{z_1} cdot z_2 ): - The real part of ( overline{z_1} cdot z_2 ) is zero. - Calculate ( overline{z_1} ): [ overline{z_1} = frac{sqrt{3}}{2} - frac{i}{2} ] - Let ( z_2 = x + iy ). The product ( overline{z_1} cdot z_2 ) must have a real part of zero: [ text{Re}left( left( frac{sqrt{3}}{2} - frac{i}{2} right)(x + iy) right) = 0 ] Simplifying, we get: [ frac{sqrt{3}}{2}x + frac{y}{2} = 0 implies sqrt{3}x + y = 0 ]3. Solve for ( z_2 ): - ( z_2 ) must also lie on the circle ( |z - i| = 1 ): [ x^2 + (y - 1)^2 = 1 ] - Substitute ( y = -sqrt{3}x ) into the circle equation: [ x^2 + (-sqrt{3}x - 1)^2 = 1 implies 4x^2 + 2sqrt{3}x = 0 implies x = -frac{sqrt{3}}{2} ] - Corresponding ( y ): [ y = -sqrt{3} left( -frac{sqrt{3}}{2} right) = frac{3}{2} ] - Thus, ( z_2 = -frac{sqrt{3}}{2} + frac{3}{2}i ).4. Verification: - Check ( |z_2 - i| = 1 ): [ left| -frac{sqrt{3}}{2} + frac{3}{2}i - i right| = left| -frac{sqrt{3}}{2} + frac{1}{2}i right| = sqrt{left( frac{sqrt{3}}{2} right)^2 + left( frac{1}{2} right)^2} = 1 ]The final answer is:[boxed{-dfrac{sqrt{3}}{2} + dfrac{3}{2}i}],<think>Okay, so I need to solve this complex numbers problem. Let me first try to understand what's given and what's being asked.We have two non-zero complex numbers, z₁ and z₂, lying on a circle centered at i with radius 1. So, the circle equation in the complex plane should be |z - i| = 1. That means both z₁ and z₂ are points on this circle.Next, it says the real part of the product of the conjugate of z₁ and z is zero. Wait, the original problem says "the real part of (bar{z}_{1} cdot z) is zero". Hmm, but the way it's phrased is a bit confusing. Let me check again. The problem states: "The real part of (bar{z}_{1} cdot z) is zero, and the principal value of the argument of (z_{1}) is (frac{pi}{6}), then (z_{2} = ) ___."Wait, so maybe there's a typo here? Because the original question mentions z₂ in the conclusion but mentions z in the condition. Maybe that z is supposed to be z₂? Let me see. It's possible that in the problem statement, there's a typo, and it should be (bar{z}_1 cdot z_2) instead of (bar{z}_1 cdot z). Otherwise, we have an undefined z. Let me check the original problem again. Hmm, the user wrote: "the real part of (bar{z}_{1} cdot z) is zero". Maybe it's supposed to be z₂? That would make sense because z₂ is the one we're supposed to find. Alternatively, maybe it's a different z. But since z is not defined in the problem, perhaps it's a typo. Let me assume that it's supposed to be z₂. So, Re((bar{z}_1 cdot z_2)) = 0. Otherwise, the problem doesn't mention z otherwise. So, I think that's a safe assumption. So, we have Re((bar{z}_1 z_2)) = 0.Additionally, the principal value of the argument of z₁ is π/6. So, arg(z₁) = π/6. And z₂ is another point on the same circle, and we need to find z₂ such that the real part of (bar{z}_1 z_2) is zero.Okay, let's start breaking it down step by step.First, since z₁ is on the circle centered at i with radius 1, its equation is |z - i| = 1. Also, the argument of z₁ is π/6. The principal value of the argument is the angle in (-π, π], so π/6 is already in that interval. So, z₁ can be represented in polar form. However, since the circle is centered at i, not the origin, maybe it's better to express z₁ in terms of coordinates or parametrize the circle.Alternatively, maybe it's easier to write z₁ as a point on the circle |z - i| = 1. Let me recall that any point on this circle can be written as i + e^{iθ}, where θ is the angle parameter. Wait, no. If the circle is centered at i (which is 0 + i*1) with radius 1, then parametrization would be z = i + e^{iθ}, where θ is from 0 to 2π. Because the center is at i, so adding e^{iθ} (which is a unit circle centered at the origin) to i gives the unit circle centered at i. So, yes, z = i + e^{iθ}. So, both z₁ and z₂ can be written as i + e^{iθ₁} and i + e^{iθ₂}, respectively.But wait, alternatively, another parametrization could be z = i + cosθ + i sinθ. So, separating real and imaginary parts: z = cosθ + i(1 + sinθ). So, the real part is cosθ and the imaginary part is 1 + sinθ.Alternatively, since the center is at (0,1) in the complex plane (since it's i), and radius 1, the circle is all points (x, y) such that x² + (y - 1)² = 1.Given that, and the argument of z₁ is π/6. The argument is the angle made with the positive real axis. So, if the argument is π/6, then the point z₁ is in the first quadrant, making an angle of π/6 with the x-axis. However, z₁ is on the circle centered at (0,1), so even though the argument is π/6, the actual coordinates might be different.Wait, let's clarify. The principal argument of z₁ is π/6, which is the angle between the positive real axis and the line from the origin to z₁. But z₁ is on the circle |z - i| = 1, so its position is relative to the center at i. Hmm, so maybe we need to find the coordinates of z₁ such that its argument is π/6, and it lies on the circle |z - i| = 1.So, let's first find z₁. Let me represent z₁ in rectangular form: z₁ = x + yi, where x and y are real numbers. The argument of z₁ is π/6, so tan(arg(z₁)) = y/x = tan(π/6) = 1/√3. So, y = x/√3. Also, since z₁ lies on the circle |z - i| = 1, so |x + yi - i| = 1 => |x + (y - 1)i| = 1. So, sqrt(x² + (y - 1)^2) = 1. Squaring both sides: x² + (y - 1)^2 = 1.But we also have y = x/√3. So, substitute y into the circle equation: x² + (x/√3 - 1)^2 = 1. Let's expand this.First, x² + (x/√3 - 1)^2 = x² + (x² / 3 - 2x/√3 + 1) = x² + x²/3 - 2x/√3 + 1 = (4x²)/3 - (2x)/√3 + 1 = 1.So, (4x²)/3 - (2x)/√3 + 1 = 1. Subtract 1 from both sides: (4x²)/3 - (2x)/√3 = 0. Factor out 2x/√3: (2x/√3)( (2x)/√3 - 1 ) = 0.So, either 2x/√3 = 0 => x = 0, or (2x)/√3 - 1 = 0 => (2x)/√3 = 1 => x = √3/2.If x = 0, then y = 0/√3 = 0. So, z₁ would be 0 + 0i = 0, but the problem states that z₁ is non-zero. So, x = 0 is invalid. Therefore, x = √3/2. Then, y = (√3/2)/√3 = 1/2. Therefore, z₁ = √3/2 + (1/2)i.Let me verify if this point is on the circle |z - i| = 1. So, |√3/2 + (1/2)i - i| = |√3/2 - (1/2)i|. The modulus is sqrt( (√3/2)^2 + (-1/2)^2 ) = sqrt( 3/4 + 1/4 ) = sqrt(1) = 1. Yes, that works. Also, the argument of z₁ is arctan( (1/2)/(√3/2) ) = arctan(1/√3) = π/6, which matches the given information. Perfect.So, z₁ is determined: z₁ = √3/2 + i/2.Now, z₂ is another point on the same circle |z - i| = 1. We need to find z₂ such that the real part of (bar{z}_1 z₂) is zero. So, Re( (bar{z}_1 z₂) ) = 0.Let me compute (bar{z}_1). Since z₁ = √3/2 + i/2, the conjugate is (bar{z}_1) = √3/2 - i/2.So, (bar{z}_1 z₂) = (√3/2 - i/2) z₂. We need the real part of this product to be zero.Let me denote z₂ as a point on the circle |z - i| = 1. Let's parameterize z₂ similarly. Let me write z₂ as i + e^{iθ}, but I need to check if that's the best approach. Alternatively, since we already have z₁ in rectangular form, perhaps it's easier to write z₂ as x + yi and use the circle condition.Alternatively, since z₂ is on the circle |z - i| = 1, we can write z₂ = i + e^{iθ} = cosθ + i(1 + sinθ). So, real part is cosθ, imaginary part is 1 + sinθ.Alternatively, express z₂ as (x, y) such that x² + (y - 1)^2 = 1. Then, we can write z₂ = x + yi with x and y satisfying this equation.But perhaps using the parametrization with angle θ. Let me try that.So, z₂ = i + e^{iθ} = cosθ + i(1 + sinθ). Therefore, Re(z₂) = cosθ, Im(z₂) = 1 + sinθ.But we need to compute Re( (bar{z}_1 z₂) ). Let's compute (bar{z}_1 z₂):(bar{z}_1) = √3/2 - i/2.z₂ = cosθ + i(1 + sinθ).Multiplying these:(√3/2)(cosθ) + (√3/2)(i)(1 + sinθ) - (i/2)(cosθ) - (i^2)/2 (1 + sinθ)Simplify:√3/2 cosθ + i √3/2 (1 + sinθ) - i/2 cosθ + 1/2 (1 + sinθ) [since i^2 = -1]Combine the real parts and the imaginary parts:Real part: √3/2 cosθ + 1/2 (1 + sinθ)Imaginary part: [ √3/2 (1 + sinθ) - 1/2 cosθ ] iSo, the real part of (bar{z}_1 z₂) is √3/2 cosθ + 1/2 (1 + sinθ). We are told that this real part is zero. Therefore:√3/2 cosθ + 1/2 (1 + sinθ) = 0Multiply both sides by 2 to eliminate denominators:√3 cosθ + 1 + sinθ = 0So, we have the equation:√3 cosθ + sinθ = -1This is a linear combination of sine and cosine. Let me recall that a cosθ + b sinθ can be written as R cos(θ - φ), where R = sqrt(a² + b²) and φ = arctan(b/a) or something like that.Let me compute R and φ for √3 cosθ + sinθ.Here, a = √3, b = 1.So, R = sqrt( (√3)^2 + 1^2 ) = sqrt(3 + 1) = 2.And φ = arctan(b/a) = arctan(1/√3) = π/6.Therefore, √3 cosθ + sinθ = 2 cos(θ - π/6)So, our equation becomes:2 cos(θ - π/6) = -1Therefore:cos(θ - π/6) = -1/2The solutions to this equation are:θ - π/6 = 2π/3 + 2πk or θ - π/6 = 4π/3 + 2πk for integer k.Therefore, θ = π/6 + 2π/3 + 2πk = π/6 + 4π/6 + 2πk = 5π/6 + 2πkor θ = π/6 + 4π/3 + 2πk = π/6 + 8π/6 + 2πk = 9π/6 + 2πk = 3π/2 + 2πkSo, θ = 5π/6 + 2πk or 3π/2 + 2πk.But since θ is an angle parameter, we can take the principal solutions in [0, 2π): θ = 5π/6 and θ = 3π/2.Therefore, there are two possible solutions for θ: 5π/6 and 3π/2.Therefore, z₂ can be expressed as:For θ = 5π/6:z₂ = cos(5π/6) + i(1 + sin(5π/6))Compute cos(5π/6) = -√3/2, sin(5π/6) = 1/2.So, z₂ = -√3/2 + i(1 + 1/2) = -√3/2 + i(3/2)For θ = 3π/2:z₂ = cos(3π/2) + i(1 + sin(3π/2)) = 0 + i(1 + (-1)) = 0 + i(0) = 0But the problem states that z₂ is non-zero, so z₂ = 0 is invalid. Therefore, the only valid solution is θ = 5π/6, giving z₂ = -√3/2 + (3/2)i.Wait, but let me confirm this. If θ = 3π/2, then z₂ = cos(3π/2) + i(1 + sin(3π/2)) = 0 + i(1 - 1) = 0, which is indeed zero, but the problem says non-zero complex numbers z₁ and z₂. Therefore, z₂ cannot be zero, so the only valid solution is θ = 5π/6, leading to z₂ = -√3/2 + (3/2)i.Therefore, z₂ = -√3/2 + (3/2)i. Let me check if this is correct.First, check that z₂ is on the circle |z - i| = 1:Compute |-√3/2 + (3/2)i - i| = |-√3/2 + (1/2)i|. The modulus is sqrt( ( (√3)/2 )^2 + (1/2)^2 ) = sqrt( 3/4 + 1/4 ) = sqrt(1) = 1. Correct.Also, check the real part of (bar{z}_1 z₂):(bar{z}_1) = √3/2 - i/2z₂ = -√3/2 + 3i/2Multiply them:(√3/2)(-√3/2) + (√3/2)(3i/2) + (-i/2)(-√3/2) + (-i/2)(3i/2)Compute term by term:First term: (√3 * -√3)/(2*2) = (-3)/4Second term: (√3 * 3i)/4 = (3√3 i)/4Third term: (i * √3)/4 = (√3 i)/4Fourth term: (-i * 3i)/4 = (-3i²)/4 = (-3*(-1))/4 = 3/4So, adding all terms:Real parts: (-3/4) + (3/4) = 0Imaginary parts: (3√3 i)/4 + (√3 i)/4 = (4√3 i)/4 = √3 iSo, the product is 0 + √3 i, which has a real part of 0. Perfect, so that satisfies the condition.Therefore, the answer is z₂ = -√3/2 + (3/2)i.Alternatively, to write this in a more standard form, we can write it as (-√3 + 3i)/2. But both are equivalent.Alternatively, since the problem is likely expecting an answer in terms of radicals and fractions, I think that's the correct form.But let me check once again if there could be another possible solution. We had θ = 5π/6 and θ = 3π/2. But θ = 3π/2 gives z₂ = 0, which is invalid. So, the only valid solution is θ = 5π/6, giving z₂ = -√3/2 + (3/2)i.Therefore, the answer should be z₂ = (-√3 + 3i)/2, which can be written as ( boxed{-dfrac{sqrt{3}}{2} + dfrac{3}{2}i} ).But let me check if there's another approach that might lead to the same answer. Maybe using geometric interpretations.Since z₁ and z₂ are on the circle |z - i| = 1, which is centered at i with radius 1. The condition Re((bar{z}_1 z_2)) = 0 can be interpreted as the dot product of z₂ and z₁ being zero, considering complex numbers as vectors. Wait, actually, Re((bar{z}_1 z_2)) is equivalent to the dot product of the vectors corresponding to z₁ and z₂ in the plane. Because if z₁ = a + bi and z₂ = c + di, then (bar{z}_1 z_2 = (a - bi)(c + di) = (ac + bd) + (ad - bc)i), so Re((bar{z}_1 z_2)) = ac + bd, which is the dot product of (a, b) and (c, d). Therefore, Re((bar{z}_1 z_2)) = 0 implies that the vectors z₁ and z₂ are perpendicular (orthogonal) in the plane.So, geometrically, z₂ lies on the circle centered at i with radius 1, and the vector from the origin to z₂ is perpendicular to the vector from the origin to z₁. But in this problem, z₁ is given with argument π/6, which is a specific direction from the origin, and z₂ is orthogonal to that.But since both z₁ and z₂ are on the circle |z - i| = 1, their positions are not arbitrary. So, perhaps using this geometric interpretation, we can find the intersection points of the circle |z - i| = 1 and the line (or lines) perpendicular to z₁.But z₁ is in the direction of π/6, so the line perpendicular to z₁ would be in the direction of π/6 + π/2 = 2π/3. But since we are dealing with vectors from the origin, the line through the origin in the direction of 2π/3. However, the circle |z - i| = 1 is not centered at the origin, so the intersection points might not lie on that line.Alternatively, since the dot product is zero, the vectors z₁ and z₂ are perpendicular. But z₁ is known (√3/2 + i/2), so z₂ must lie on the line perpendicular to z₁ at the origin. However, since z₂ is also on the circle |z - i| = 1, we need to find the intersection of these two loci: the circle |z - i| = 1 and the line through the origin perpendicular to z₁.But maybe that's overcomplicating. Let me check.The line perpendicular to z₁ would have the equation Re((bar{z}_1 z)) = 0, which is the same as the condition given for z₂. Wait, the problem states that Re((bar{z}_1 z_2)) = 0, which is equivalent to z₂ lying on the line through the origin perpendicular to z₁. However, z₂ is also on the circle |z - i| = 1, so the intersection points of that line with the circle are the solutions. But since the circle is not centered at the origin, there can be two intersection points, but in our case, one of them is zero (which is invalid), and the other is the solution we found.Therefore, this confirms that z₂ must be the other intersection point, which is non-zero, and that's exactly the point we found: -√3/2 + (3/2)i.Alternatively, perhaps using complex numbers transformations. Let's see. Since z₁ is known, and we need Re((bar{z}_1 z_2)) = 0. Let me denote w = z₂. Then, the condition is Re((bar{z}_1 w)) = 0. This is equivalent to (bar{z}_1 w) being purely imaginary, so (bar{z}_1 w = i k) for some real k. Therefore, w = (i k)/(bar{z}_1).But z₁ is given as √3/2 + i/2, so |z₁| = sqrt( (√3/2)^2 + (1/2)^2 ) = sqrt( 3/4 + 1/4 ) = sqrt(1) = 1. Therefore, 1/(bar{z}_1) = z₁. Wait, because if |z₁| = 1, then (bar{z}_1 = 1/z₁). So, 1/(bar{z}_1) = z₁. Therefore, w = i k z₁.But since w must lie on the circle |w - i| = 1, so |i k z₁ - i| = 1. Let's compute:|i k z₁ - i| = |i(k z₁ - 1)| = |k z₁ - 1| = 1.So, |k z₁ - 1| = 1. Let's write this as |k z₁ - 1| = 1.But z₁ is a complex number with |z₁| = 1 (since |z₁ - i| = 1 and |z₁| = 1? Wait, no. Wait, z₁ is on the circle |z - i| = 1, but |z₁| isn't necessarily 1. Wait, let's check: z₁ is √3/2 + i/2. Then, |z₁| = sqrt( (√3/2)^2 + (1/2)^2 ) = sqrt( 3/4 + 1/4 ) = sqrt(1) = 1. Oh, interesting! So, even though z₁ is on the circle |z - i| = 1, it's also on the unit circle centered at the origin. That's a coincidence? Wait, maybe not. Let me check.Wait, the circle |z - i| = 1 is centered at (0,1) with radius 1. The unit circle centered at the origin is |z| = 1. The intersection points of these two circles are the solutions to both equations. Let's compute them.The intersection points satisfy |z - i| = 1 and |z| = 1.Let z = x + yi. Then:x² + (y - 1)^2 = 1andx² + y² = 1Subtract the second equation from the first:(x² + y² - 2y + 1) - (x² + y²) = 1 - 1 => -2y + 1 = 0 => y = 1/2.Substitute y = 1/2 into the second equation:x² + (1/2)^2 = 1 => x² = 1 - 1/4 = 3/4 => x = ±√3/2.Therefore, the intersection points are √3/2 + i/2 and -√3/2 + i/2. So, z₁ is given as having argument π/6, which is the point √3/2 + i/2. The other intersection point is -√3/2 + i/2, which has argument -π/6, but since we are dealing with the principal value, it's 11π/6.But in our problem, z₂ is another point on |z - i| = 1 such that Re((bar{z}_1 z_2)) = 0. But from this approach, if z₂ is on both |z - i| = 1 and |z| = 1, then z₂ would be -√3/2 + i/2. But in our previous calculation, we found z₂ = -√3/2 + 3i/2, which is different.Wait, so why the discrepancy? Because in this approach, assuming that z₂ is on the unit circle, but in reality, z₂ is only required to be on the circle |z - i| = 1. The condition Re((bar{z}_1 z_2)) = 0 does not necessarily mean z₂ is on the unit circle unless |z₁| = 1. Wait, but in our case, |z₁| = 1, as we saw earlier, so maybe z₂ must lie on the unit circle as well? But no, because Re((bar{z}_1 z_2)) = 0 just means that z₂ is on the line perpendicular to z₁ through the origin. So, the intersection points of this line with the circle |z - i| = 1 are the solutions. But since z₁ is on both circles, the other intersection point is different.Wait, in our previous calculation, we found z₂ = -√3/2 + 3i/2. Let's compute |z₂|. The modulus is sqrt( (-√3/2)^2 + (3/2)^2 ) = sqrt( 3/4 + 9/4 ) = sqrt(12/4) = sqrt(3) ≈ 1.732, which is not 1. Therefore, z₂ is not on the unit circle. So, the earlier approach of intersecting the two circles only gives points that are on both circles, but our solution is not on the unit circle. Therefore, this method doesn't capture it. Therefore, the previous parametrization approach is the correct way.Therefore, going back, z₂ is at -√3/2 + 3i/2, which is on the circle |z - i| = 1, and satisfies Re((bar{z}_1 z_2)) = 0.Therefore, the answer is z₂ = -√3/2 + (3/2)i.But let me check once again if there's an alternative representation.Alternatively, since z₂ is on the circle |z - i| = 1, we can represent it as i + e^{iθ}. Let's see:z₂ = i + e^{iθ} = e^{iθ} + i.Then, compute Re((bar{z}_1 z₂)):(bar{z}_1) = √3/2 - i/2.Multiply by z₂ = e^{iθ} + i:= (√3/2 - i/2)(e^{iθ} + i)= √3/2 e^{iθ} + √3/2 i - i/2 e^{iθ} - i^2 / 2= √3/2 e^{iθ} + (√3/2)i - (i/2)e^{iθ} + 1/2Separate real and imaginary parts:Real part: √3/2 Re(e^{iθ}) + 1/2 + (√3/2) Im(i) - (1/2) Im(e^{iθ} i )Wait, this might be more complicated. Alternatively, expand the product:= √3/2 e^{iθ} + √3/2 i - i/2 e^{iθ} + 1/2Group terms with e^{iθ}:= ( √3/2 - i/2 ) e^{iθ} + √3/2 i + 1/2Then, take the real part:Real part = Re[ ( √3/2 - i/2 ) e^{iθ} ] + Re[ √3/2 i + 1/2 ]The second term: Re[ √3/2 i + 1/2 ] = 1/2.The first term: Let me compute ( √3/2 - i/2 ) e^{iθ}.Let’s denote ( √3/2 - i/2 ) = e^{-iπ/6}, since it has modulus 1 and angle -π/6.Therefore, ( √3/2 - i/2 ) e^{iθ} = e^{-iπ/6} e^{iθ} = e^{i(θ - π/6)}.The real part of this is cos(θ - π/6).Therefore, the real part of the entire expression is cos(θ - π/6) + 1/2.Set this equal to zero:cos(θ - π/6) + 1/2 = 0 => cos(θ - π/6) = -1/2Which is the same equation as before. So, solving θ - π/6 = 2π/3 or 4π/3, leading to θ = 5π/6 or 3π/2, as before. Therefore, this confirms the previous solution.Thus, z₂ = i + e^{iθ} where θ = 5π/6 or 3π/2. For θ = 5π/6:e^{iθ} = cos(5π/6) + i sin(5π/6) = -√3/2 + i/2Therefore, z₂ = i + (-√3/2 + i/2) = -√3/2 + i + i/2 = -√3/2 + (3/2)iFor θ = 3π/2:e^{iθ} = cos(3π/2) + i sin(3π/2) = 0 - iTherefore, z₂ = i + (0 - i) = 0, which is invalid.Therefore, this approach gives the same result.Thus, the answer is z₂ = -√3/2 + (3/2)i.Final AnswerThe complex number ( z_{2} ) is (boxed{-dfrac{sqrt{3}}{2} + dfrac{3}{2}i}).</think>Given non-zero complex numbers ( z_1 ) and ( z_2 ) on a circle centered at ( i ) with radius 1, we need to find ( z_2 ) such that the real part of ( bar{z}_1 cdot z_2 ) is zero, and the principal value of the argument of ( z_1 ) is ( frac{pi}{6} ).1. Determine ( z_1 ): - The argument of ( z_1 ) is ( frac{pi}{6} ), so ( z_1 ) can be written as ( sqrt{3}/2 + i/2 ). - Verify ( z_1 ) lies on the circle ( |z - i| = 1 ): [ left| sqrt{3}/2 + i/2 - i right| = left| sqrt{3}/2 - i/2 right| = 1 ]2. Condition for ( z_2 ): - The real part of ( bar{z}_1 cdot z_2 ) must be zero. - Calculate ( bar{z}_1 ): [ bar{z}_1 = sqrt{3}/2 - i/2 ] - Parameterize ( z_2 ) on the circle ( |z - i| = 1 ) as ( z_2 = i + e^{itheta} ).3. Solve for ( theta ): - The real part of ( bar{z}_1 cdot z_2 ) leads to the equation: [ sqrt{3} cos theta + sin theta = -1 ] - This simplifies to: [ 2 cos(theta - pi/6) = -1 implies cos(theta - pi/6) = -1/2 ] - Solutions are ( theta = 5pi/6 ) or ( 3pi/2 ). The solution ( theta = 3pi/2 ) gives ( z_2 = 0 ), which is invalid.4. Determine ( z_2 ): - For ( theta = 5pi/6 ): [ z_2 = i + e^{i cdot 5pi/6} = i + left( -sqrt{3}/2 + i/2 right) = -sqrt{3}/2 + 3i/2 ]Thus, the complex number ( z_2 ) is (boxed{-dfrac{sqrt{3}}{2} + dfrac{3}{2}i}).]

🔑:Solve: Fill in -frac{sqrt{3}}{2}+frac{3}{2} mathrm{i}. Reason: Since arg z_{1}=frac{pi}{6}, the vector overrightarrow{O Z} corresponding to z_{1} forms an angle of frac{pi}{3} with the y-axis, so left|z_{1}right|=1, z_{1}=frac{sqrt{3}}{2}+frac{1}{2} mathrm{i}.Also, the real part of bar{z}_{1} cdot z_{2} is zero, so arg z_{2}-frac{pi}{6}=frac{pi}{2} or frac{3}{2} pi. Since z_{2} lies on the circle |z-mathrm{i}|=1, frac{3}{2} pi should be discarded, i.e., arg z_{2}=frac{2}{3} pi. z_{2} lies on the diameter passing through z_{1} and the center of the circle, left|z_{2}right|=sqrt{3},thus z_{2}=sqrt{3}left(cos frac{2 pi}{3}+mathrm{i} sin frac{2 pi}{3}right)=-frac{sqrt{3}}{2}+frac{3}{2} mathrm{i}.

🧠:[<think>Alright, so I need to find the smallest natural number n such that if a prime number p divides n, then n is divisible by p. Wait, no, wait. Let me read that again. The problem says: "Find the smallest natural number n, for which the following condition is satisfied: if the number p is prime and n is divisible by p-1, then n is divisible by p." Hmm, okay. So the condition is that for every prime p, if p-1 divides n, then p must divide n. So n must be such that whenever p-1 is a divisor of n, then p itself must divide n. And I need the smallest such n.Let me rephrase it to make sure I understand. For all primes p, if p-1 divides n, then p divides n. So n has to include all primes p where p-1 divides n. So for example, if n is divisible by 1 (which every number is), then since 1 corresponds to p-1=1, so p=2, then n must be divisible by 2. Similarly, if n is divisible by 2, which would correspond to p-1=2, so p=3, then n must be divisible by 3. If n is divisible by 4, then p-1=4, so p=5, so n must be divisible by 5. Similarly, if n is divisible by 6, then p-1=6, so p=7, so n must be divisible by 7. Etc.Wait, so in other words, n must be divisible by all primes p such that p-1 divides n. So for each divisor d of n, if d+1 is prime, then d+1 must divide n. Therefore, n must contain all primes of the form d+1 where d divides n. So the problem reduces to constructing a number n such that for all divisors d of n, d+1 is prime implies that d+1 divides n. And we need the smallest such n.So, the strategy is to find the smallest n such that if d divides n and d+1 is prime, then d+1 divides n.So first, let's start with n=1. Then the divisors of 1 are just 1. Then d=1, d+1=2, which is prime. Therefore, n=1 must be divisible by 2. But 1 is not divisible by 2, so n=1 doesn't satisfy the condition.Next, n=2. Divisors of 2 are 1 and 2. For d=1, d+1=2 is prime, so n must be divisible by 2, which it is. For d=2, d+1=3, which is prime. So n=2 must be divisible by 3. But 2 is not divisible by 3, so n=2 fails.Next, n=3. Divisors are 1 and 3. For d=1, d+1=2 must divide 3? 3 is not divisible by 2. So n=3 fails.n=4. Divisors are 1,2,4. For d=1, 2 must divide 4. It does. For d=2, 3 must divide 4. 4 is not divisible by 3. So n=4 fails.n=5. Divisors 1,5. d=1: 2 divides 5? No. So fails.n=6. Divisors 1,2,3,6. For d=1: 2 divides 6, yes. d=2: 3 divides 6, yes. d=3: 4 is not prime. d=6: 7 must divide 6. No. So 6 fails because 7 doesn't divide 6.So n=6: primes p=2,3,7. p=7-1=6 divides 6, so 7 must divide 6. Which it doesn't. So n=6 fails.Next, n=7. Divisors 1,7. For d=1, 2 must divide 7. 7 isn't divisible by 2. So n=7 fails.n=8. Divisors 1,2,4,8. For d=1: 2 divides 8, yes. d=2: 3 divides 8? No. So n=8 fails.n=9. Divisors 1,3,9. d=1: 2 divides 9? No. Fails.n=10. Divisors 1,2,5,10. d=1: 2 divides 10. Yes. d=2: 3 divides 10? No. Fails.n=12. Divisors 1,2,3,4,6,12. For each d:d=1: 2 divides 12, yes.d=2: 3 divides 12, yes.d=3: 4 is not prime.d=4: 5 divides 12? 12/5=2.4. No. So n=12 must be divisible by 5 because d=4 (since 4 divides 12 and 4+1=5 is prime). But 12 isn't divisible by 5, so n=12 fails.Wait, so n=12 needs to be divisible by 5 because 4 divides 12 and 5 is prime, so 5 must divide 12. But 12 isn't divisible by 5. Therefore, n=12 fails.So moving on. Let's check n=15. Divisors 1,3,5,15.d=1: 2 divides 15? No. So fails.n=16. Divisors 1,2,4,8,16. d=1: 2 divides 16, yes. d=2: 3 divides 16? No. Fails.n=18. Divisors 1,2,3,6,9,18.d=1: 2 divides 18, yes.d=2: 3 divides 18, yes.d=3: 4 is not prime.d=6: 7 divides 18? No.d=9: 10 is not prime.d=18: 19 divides 18? No.So n=18 fails because 6 divides 18, so 7 must divide 18, but it doesn't. Similarly, 18 would require 19 as well.Hmm. So maybe n needs to be a multiple of 2,3,5,7, etc., but how to find the minimal one.Wait, perhaps we need to consider the concept of a "Carmichael number" or something similar, but not exactly. Wait, in this problem, the condition is that if p-1 divides n, then p divides n, for every prime p.This is similar to the concept in group theory where the order of an element divides the group order, but not sure.Alternatively, perhaps n must be a multiple of the least common multiple of all primes p where p-1 divides n. But n is dependent on itself here. So it's a bit of a recursive condition.Alternatively, think of n as needing to contain all primes p for which p-1 divides n. Therefore, n must be divisible by all primes p such that p-1 is a divisor of n.So n is the product of all primes p where p-1 divides n. But this seems circular because p depends on n. So how do we resolve this?Perhaps we can approach this by building n step by step, adding primes as necessary.Let's start with n=1. As before, fails because 1 is divisible by 1 (p=2) but 1 isn't divisible by 2.So n must at least include 2. Let n=2. But n=2 is divisible by 1 (so p=2 is included), but n=2 is also divisible by 2-1=1? Wait, maybe my initial step is wrong.Wait, the condition is if p is prime and n is divisible by p-1, then n is divisible by p.So for n=2, check primes p where p-1 divides 2. The divisors of 2 are 1 and 2. So p-1 can be 1 or 2. Therefore, p=2 (1+1) and p=3 (2+1). So for p=2, since 2-1=1 divides 2, n=2 must be divisible by 2. It is. For p=3, since 3-1=2 divides 2, n=2 must be divisible by 3. It's not. So n=2 is invalid. Therefore, to satisfy p=3, n must be a multiple of 3. So let's set n=6.Now check n=6. Divisors of 6 are 1,2,3,6. Therefore, primes p where p-1 is 1,2,3,6. So p=2,3,4,7. But 4 isn't prime, so p=2,3,7. For p=2: p-1=1 divides 6, so n must be divisible by 2. Yes. For p=3: p-1=2 divides 6, so n must be divisible by 3. Yes. For p=7: p-1=6 divides 6, so n must be divisible by 7. 6 isn't divisible by 7, so n=6 is invalid. Therefore, we need to include 7 in n. So set n=6*7=42.Check n=42. Divisors of 42 are 1,2,3,6,7,14,21,42. So for each divisor d, check if d+1 is prime. For each d in divisors:d=1: 2 is prime, so 2 must divide 42. Yes.d=2: 3 is prime, must divide 42. Yes.d=3: 4 is not prime.d=6: 7 is prime, must divide 42. Yes.d=7: 8 is not prime.d=14: 15 is not prime.d=21: 22 is not prime.d=42: 43 is prime. So 43 must divide 42. No. So n=42 fails because d=42 (divisor) gives p=43, which is prime, so 43 must divide 42, which it doesn't. Therefore, we need to include 43 in n. So set n=42*43=1806.Now check n=1806. Divisors of 1806: Let's factorize 1806. 1806 divided by 2 is 903. 903 divided by 3 is 301. 301 divided by 7 is 43. So 1806=2*3*7*43. Therefore, the divisors are all combinations of these primes. So divisors include 1,2,3,6,7,14,21,42,43,86,129,258,301,602,903,1806.For each divisor d, check if d+1 is prime. Then n must be divisible by d+1.Let's go through each divisor:d=1: 2 is prime. 2 divides 1806. Yes.d=2: 3 is prime. 3 divides 1806. Yes.d=3: 4 is not prime.d=6: 7 is prime. 7 divides 1806. Yes.d=7: 8 is not prime.d=14: 15 is not prime.d=21: 22 is not prime.d=42: 43 is prime. 43 divides 1806. Yes.d=43: 44 is not prime.d=86: 87 is not prime (87=3*29).d=129: 130 is not prime.d=258: 259 is prime? Let's check 259. 259 divided by 7 is 37, since 7*37=259? 7*30=210, 7*37=259. Yes, 259=7*37. So 259 is composite.d=301: 302 is not prime (even).d=602: 603 is divisible by 3 (6+0+3=9), so composite.d=903: 904 is even, composite.d=1806: 1807. Let's check if 1807 is prime. 1807 divided by 13 is 139 (13*139=1807). Therefore, 1807 is composite.So all the d+1 primes are 2,3,7,43. All of these primes divide 1806. 2,3,7,43 are all factors. So n=1806 satisfies the condition. But is there a smaller n?Wait, when we had n=42, the problem was that d=42, leading to p=43. So if n includes 43, then when n=42*43=1806, then d=42*43=1806, but p=1807. Wait, but in n=1806, the divisors are up to 1806. So when d=1806, d+1=1807 must be prime. But 1807 is not prime, so we don't need n to be divisible by 1807. So n=1806 is okay because even though d=1806 is a divisor, d+1=1807 is not prime, so there's no requirement for n to be divisible by 1807. So n=1806 actually satisfies the condition.But wait, let's check all the divisors again:For d=1: 2, prime, divides 1806.d=2: 3, prime, divides 1806.d=3: 4, not prime.d=6:7, prime, divides 1806.d=7:8, not prime.d=14:15, not prime.d=21:22, not prime.d=42:43, prime, divides 1806.d=43:44, not prime.d=86:87, not prime.d=129:130, not prime.d=258:259, composite.d=301:302, even.d=602:603, composite.d=903:904, even.d=1806:1807, composite.Therefore, all primes p where p-1 divides n=1806 are 2,3,7,43, and all of these divide n=1806. Therefore, n=1806 satisfies the condition. But was there a smaller number?Before 1806, we tried n=42, which required 43. Then n=42*43=1806. Let's check if there's a number between 42 and 1806 that might satisfy the condition. For example, let's see n=42*43=1806 is the product of 2,3,7,43. If we tried to make a number that includes 2,3,7,43, but maybe a smaller multiple?Wait, since 2,3,7,43 are all primes, their product is the LCM. So LCM(2,3,7,43)=2*3*7*43=1806. So that's the minimal number containing all those primes. So 1806 is the LCM. Therefore, there's no smaller number. Because if you take any subset, you might miss one.Wait, let's see. Suppose we tried n=2*3*7*43=1806. Suppose we remove 43. Then n=2*3*7=42. But then, as before, when n=42, d=42 is a divisor, so p=43 must divide n=42, which it doesn't. Therefore, we need 43. Similarly, if we have n=2*3*7*43, but remove 7, then n=2*3*43=258. Check if n=258 works.Divisors of 258: 1,2,3,6,43,86,129,258.Check d+1 for each:d=1:2 prime, divides 258. Yes.d=2:3 prime, divides 258. Yes.d=3:4, not prime.d=6:7 prime, divides 258? 258 divided by 7 is 36.85... So no. Therefore, since d=6 divides 258, p=7 must divide 258. It doesn't, so n=258 fails. Therefore, 7 must be included.Similarly, if we try n=2*3*7=42, but we saw that fails due to 43. So n must include 2,3,7,43. Therefore, LCM(2,3,7,43)=1806. Hence, 1806 is the minimal n.But to confirm, let's check if maybe a number smaller than 1806 could include these primes. For example, if there is a number less than 1806 that includes 2,3,7,43 and maybe other primes. But since 2,3,7,43 are all primes, their LCM is the product. So any number that includes all these primes must be at least 2*3*7*43=1806. So 1806 is the minimal.But just to be thorough, let's check another candidate. Let's suppose n=840. 840 is 2^3*3*5*7. Divisors of 840 include numbers up to 840. Let's check if 840 satisfies the condition.Take a divisor d=4 (since 4 divides 840). Then d+1=5, which is prime. So 5 must divide 840. 5 does divide 840. Similarly, d=6:7 divides 840. Yes. d=10:11 must divide 840? 10 divides 840, so 11 must divide 840. But 840/11=76.36... which is not integer. Therefore, 11 does not divide 840, so n=840 fails because d=10 divides 840 (10 is a divisor) and d+1=11 is prime, but 11 doesn't divide 840.Therefore, n=840 fails. Therefore, to satisfy the condition, if d=10 divides n, which it does in n=840, then 11 must divide n. So n must be multiple of 11. So n=840*11=9240. But then check for d=12:13 must divide 9240. 9240 divided by 13 is 710.769... Not integer. So n=9240 would need to include 13 as well, etc. So this approach of multiplying by primes as needed could lead to a larger number. Whereas 1806 is constructed by the primes required from its divisors.Therefore, 1806 seems to be the minimal n. Let me check another angle. If n=1806 is the minimal, then all previous numbers must fail. Let's check n=5460, which is 2*2*3*5*7*13. Wait, but maybe not. Let me see. Wait, n=1806 is 2*3*7*43. Are there any primes missing? For example, when we checked 1806, all the p primes required are 2,3,7,43, which are included. Are there any other primes p where p-1 divides 1806?Let's check for primes p where p-1 divides 1806. So p-1 must be a divisor of 1806. The divisors of 1806 are 1,2,3,6,7,14,21,42,43,86,129,258,301,602,903,1806. So p-1 can be any of these, so p would be 2,3,4,7,8,22,43,44,87,130,259,302,602, 903+1=904, 1806+1=1807. Out of these, the primes are 2,3,7,43, 259 is composite (as 259=7*37), 302 is even, 44 is composite, 87 is composite, 130 is even, 903+1=904 is even, 1807=13*139, composite. So the primes p are 2,3,7,43. All of these are included in n=1806, so n=1806 satisfies the condition.Therefore, n=1806 is the minimal such number.Final AnswerThe smallest natural number ( n ) is boxed{1806}.</think>To find the smallest natural number ( n ) such that if a prime number ( p ) satisfies ( n ) being divisible by ( p-1 ), then ( n ) is also divisible by ( p ), we need to ensure that ( n ) includes all primes ( p ) for which ( p-1 ) is a divisor of ( n ).1. Understanding the condition: For every prime ( p ), if ( p-1 ) divides ( n ), then ( p ) must divide ( n ). This means ( n ) must be divisible by all primes ( p ) such that ( p-1 ) is a divisor of ( n ).2. Checking smaller numbers: - Numbers like 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 12, 15, 16, 18, etc., fail because they do not satisfy the condition for all primes ( p ).3. Constructing the solution: - Start with ( n = 2 ), which fails because it does not include 3. - Next, ( n = 6 ), which fails because it does not include 7. - Continue this process, adding necessary primes until all conditions are met.4. Key steps: - ( n = 42 ) fails because it does not include 43. - ( n = 1806 ) (the product of 2, 3, 7, and 43) is checked. The divisors of 1806 are 1, 2, 3, 6, 7, 14, 21, 42, 43, 86, 129, 258, 301, 602, 903, 1806. For each divisor ( d ), ( d+1 ) is checked for primality: - ( d = 1 rightarrow 2 ) (prime, divides 1806) - ( d = 2 rightarrow 3 ) (prime, divides 1806) - ( d = 6 rightarrow 7 ) (prime, divides 1806) - ( d = 42 rightarrow 43 ) (prime, divides 1806) - Other divisors result in non-prime values for ( d+1 ).5. Verification: - All primes ( p ) such that ( p-1 ) divides 1806 are 2, 3, 7, and 43, all of which divide 1806. Thus, 1806 satisfies the condition.Thus, the smallest natural number ( n ) is (boxed{1806}).,<think>Alright, so I need to find the smallest natural number n such that if a prime p divides n, then p-1 divides n. Wait, no, the problem says: if p is prime and n is divisible by p-1, then n is divisible by p. Hmm. Let me parse that again.The condition is: For every prime p, if n is divisible by p-1, then n must also be divisible by p. So, in other words, whenever p is a prime and p-1 divides n, then p must divide n. So n has to include all primes p for which p-1 divides n. So n has to be a multiple of all such primes. The question is to find the smallest such n.This seems related to concepts like Carmichael numbers or Korselt's criterion, where certain numbers must be divisible by primes under specific conditions. Let me recall. Korselt's criterion states that a Carmichael number n is composite and square-free and for all prime divisors p of n, it is true that p-1 divides n - 1. But this is a different condition. Here, if p-1 divides n, then p must divide n. So n must contain all primes p where p-1 divides n.So the task is to construct the smallest n where for every prime p, p-1 | n implies p | n. So n must be a multiple of all primes p such that p-1 divides n.This seems recursive. Because the primes p that must divide n depend on n itself. For example, if n is even, then p-1 divides n. For p=2, p-1=1, which divides any n. So p=2 must divide n. Therefore, n must be even. So 2 divides n.Similarly, if n is even, then 1 divides n, so 2 divides n. If n is divisible by 2, then for p=3, p-1=2. So if 2 divides n, then 3 must divide n. Therefore, n must be divisible by 3. Then since n is divisible by 3, we need to check primes p where p-1 divides n. For example, p=4 would be 3, but 4 isn't prime. Wait, primes p where p-1 divides n. So p-1 divides n. So if n is divisible by 3, then p-1=3 implies p=4, which is not prime. So next prime would be p=7, since 7-1=6. If 6 divides n, then p=7 must divide n. So if n is divisible by 6, then 7 must divide n. Hmm.Wait, let's try to build n step by step. Start with n=1. Then check the condition. For n=1, the primes p such that p-1 divides 1. Since p-1 divides 1, p-1 must be 1, so p=2. Then the condition requires that if 1 divides 1 (which it does), then 2 divides 1. But 2 does not divide 1. Therefore, n=1 does not satisfy the condition. So n=1 is invalid.Next, n=2. Check primes p where p-1 divides 2. p-1 divides 2. So possible p-1 are 1 and 2. Therefore, p=2 and p=3. Then, for p=2: if 1 divides 2, then 2 divides 2. Which it does. For p=3: if 2 divides 2, then 3 divides 2. But 3 does not divide 2. So n=2 does not satisfy the condition. Therefore, n=2 is invalid.Next, n=3. Primes p where p-1 divides 3. p-1 can be 1 or 3. So p=2 and p=4 (which isn't prime). So p=2. Check if 1 divides 3, which it does, then 2 must divide 3. But 2 does not divide 3. So n=3 is invalid.n=4. Primes p where p-1 divides 4. p-1 can be 1,2,4. So p=2,3,5. Check each:- p=2: 1 divides 4, so 2 must divide 4. Yes.- p=3: 2 divides 4? 2 divides 4, so 3 must divide 4. No. 3 does not divide 4. Therefore, n=4 invalid.n=5. Primes p where p-1 divides 5. p-1=1,5. p=2,6 (non-prime). So p=2. 1 divides 5, so 2 must divide 5. No. Invalid.n=6. Primes p where p-1 divides 6. p-1=1,2,3,6. So p=2,3,4 (non-prime),7. So p=2,3,7. Check:- p=2: 1 divides 6, must have 2 divides 6. Yes.- p=3: 2 divides 6? Yes, so 3 must divide 6. Yes.- p=7: 6 divides 6? Yes, so 7 must divide 6. No. Therefore, n=6 invalid.n=7. Primes p where p-1 divides 7. p-1=1,7. So p=2,8 (non-prime). Check p=2: 1 divides 7, so 2 divides 7? No. Invalid.n=8. Primes p where p-1 divides 8. p-1=1,2,4,8. So p=2,3,5,9 (non-prime). Check:- p=2: 1 divides 8, 2 divides 8. Yes.- p=3: 2 divides 8? Yes, so 3 divides 8? No. Invalid.n=9. p-1 divides 9. p-1=1,3,9. So p=2,4 (non-prime),10 (non-prime). So p=2. Check: 1 divides 9, so 2 divides 9? No. Invalid.n=10. p-1 divides 10. p-1=1,2,5,10. So p=2,3,6 (non-prime),11. Check:- p=2: 1 divides 10, 2 divides 10. Yes.- p=3: 2 divides 10? Yes, so 3 divides 10? No. Invalid.- p=11: 10 divides 10? Yes, so 11 divides 10? No. Invalid.n=12. Let's check n=12. p-1 divides 12. p-1=1,2,3,4,6,12. So p=2,3,4 (non),5,7,13. Check:- p=2: 1 divides 12, so 2 divides 12. Yes.- p=3: 2 divides 12? Yes, 3 divides 12. Yes.- p=5: 4 divides 12? 4 divides 12, so 5 must divide 12. 5 does not divide 12. Invalid. So n=12 is invalid because p=5: 4 divides 12, so 5 must divide 12, which it doesn't.Therefore, need to include 5 in n. So maybe n=12*5=60? Let's check n=60. But let's check step by step.Wait, perhaps the minimal n is 12, but since p=5, where p-1=4 divides 12, but 5 does not divide 12, then n must be multiple of 5 as well. So n must be LCM(12,5)=60. Let's check n=60. Then, primes p where p-1 divides 60. So p-1 can be any divisor of 60. The divisors of 60 are 1,2,3,4,5,6,10,12,15,20,30,60. Therefore p=2,3,4(non),5,6(non),7,11,13,16(non),21(non),31,61. So primes p=2,3,5,7,11,13,31,61. Check for each p:- p=2: 1 divides 60, 2 divides 60. Yes.- p=3: 2 divides 60, 3 divides 60. Yes.- p=5: 4 divides 60, 5 divides 60. Yes.- p=7: 6 divides 60, 7 divides 60? 60 divided by 7 is not integer. 7*8=56, 7*9=63. So 7 does not divide 60. So n=60 is invalid because when p=7, p-1=6 divides 60, but 7 does not divide 60.Thus, n=60 is invalid. Therefore, we need to include 7 in n. So n= LCM(60,7)=420. Let's check n=420. Now, primes p where p-1 divides 420.Divisors of 420 are 1,2,3,4,5,6,7,10,12,14,15,20,21,28,30,35,42,60,70,84,105,140,210,420. Then p-1 is one of these, so p is 2,3,4(non),5,6(non),7,8(non),11,13,15(non),16(non),21(non),22(non),29,31,36(non),43,61,71,85(non),106(non),211,421. So primes p=2,3,5,7,11,13,29,31,43,61,71,211,421.Check each:- p=2: 1 divides 420, 2 divides 420. Yes.- p=3: 2 divides 420, 3 divides 420. Yes.- p=5: 4 divides 420, 5 divides 420. Yes.- p=7: 6 divides 420, 7 divides 420. Yes.- p=11: 10 divides 420, 11 divides 420? 420 /11=38.18... Not integer. So 11 does not divide 420. Therefore, n=420 invalid.So need to include 11 in n. So n= LCM(420,11)=4620. Check n=4620. Now, primes p where p-1 divides 4620.Divisors of 4620: Let's factorize 4620. 4620 = 420*11=2^2*3*5*7*11. The divisors are all combinations of these primes. The divisors would be numbers of the form 2^a*3^b*5^c*7^d*11^e, where a=0,1,2; b=0,1; c=0,1; d=0,1; e=0,1. The total number of divisors is (2+1)(1+1)(1+1)(1+1)(1+1)=3*2*2*2*2=48. That's a lot. But the key point is that for each divisor d of 4620, p = d +1 must be prime, and then p must divide 4620.So let's consider all divisors d of 4620, compute p = d +1, check if p is prime, and then check if p divides 4620.But this is going to be tedious. Let's see. Let's start with the small divisors:d=1: p=2, which is prime. 2 divides 4620. Yes.d=2: p=3, prime. 3 divides 4620. Yes.d=3: p=4, not prime. Ignore.d=4: p=5, prime. 5 divides 4620. Yes.d=5: p=6, not prime.d=6: p=7, prime. 7 divides 4620. Yes.d=7: p=8, not prime.d=10: p=11, prime. 11 divides 4620. Yes.d=11: p=12, not prime.d=12: p=13, prime. 13 divides 4620? 4620 /13=355.38... No. 13*355=4615, 4620-4615=5. So 13 does not divide 4620. Therefore, n=4620 is invalid because p=13, p-1=12 divides 4620, but 13 does not divide 4620.Therefore, we need to include 13 in n. So n= LCM(4620,13)=60060. Now check n=60060. Now, divisors of 60060 are 2^2*3*5*7*11*13. So divisors are combinations of these primes. For each divisor d, p=d+1, check if prime and if p divides 60060.Again, tedious, but let's check key points. The next prime to check would be where d=12 (p=13, included), d=16 (p=17), d=22 (p=23), etc.Wait, let's take some key divisors:d=16: p=17, prime. 17 divides 60060? 60060 /17= 60060 ÷17. 17*3500=59500, 60060-59500=560. 560/17≈32.94. So no, 17 does not divide 60060. Therefore, if d=16 divides 60060, then p=17 must divide 60060. But 16 divides 60060? 60060 divided by16=3753.75. Not integer. So 16 is not a divisor of 60060. So p=17 would only need to divide n if 16 divides n. But since 16 does not divide 60060, we don't need p=17.Similarly, check d=22: divisor 22, p=23. 22 divides 60060? 60060 /22=2730, which is integer. So 22 divides 60060. Then p=23 must divide 60060. 60060 /23=2611.3... Not integer. So 23 does not divide 60060. Therefore, n=60060 is invalid because p=23, p-1=22 divides n, but p=23 does not divide n.Therefore, we need to include 23 in n. So n= LCM(60060,23)=60060*23=1381380. Now check n=1381380. Divisors now include up to 1381380. But again, we need to check if for every divisor d of n, p=d+1 is prime, then p divides n.This is getting into a loop where each time we find a new prime that must be included, which requires multiplying n by that prime, which introduces new divisors, which might require new primes.This seems similar to the concept of a "Giuga number", where for each prime divisor p, p divides (n/p - 1). But I'm not sure. Alternatively, this problem resembles the requirement that n must be a multiple of all primes p such that p-1 divides n. So n must be divisible by the set P = { p | p is prime and p-1 divides n }. Therefore, n must be the LCM of all such primes. But since the set P depends on n, it's a self-referential problem.Therefore, to find the minimal n, we need to find a number n such that for every prime p where p-1 divides n, p divides n. So n must be the LCM of all primes p where p-1 divides n. This seems like a circular definition, but perhaps we can construct n step by step.Let's try building n incrementally:Start with n=1. But p=2 (since p-1=1 divides n=1) requires 2 divides n=1. Not valid. So n must include 2. Set n=2.Check if n=2 satisfies the condition. Primes p where p-1 divides 2: p=2 (p-1=1) and p=3 (p-1=2). So p=2: already divides n=2. p=3: p-1=2 divides n=2, so p=3 must divide n=2. But 3 does not divide 2. Therefore, n=2 invalid. So n must include 3. Now n=6.Check n=6. Primes p where p-1 divides 6: p-1=1,2,3,6. So p=2,3,4 (non-prime),7. So p=2,3,7. Check if they divide 6. p=2,3 do, p=7 doesn't. So n=6 invalid. Need to include 7. Now n=42.Check n=42. Primes p where p-1 divides 42. Divisors of 42:1,2,3,6,7,14,21,42. So p=2,3,4(non),7,8(non),15(non),22(non),43. Check p=2,3,7,43. 2,3,7 divide 42. p=43: p-1=42 divides n=42, so 43 must divide n=42. No. So n=42 invalid. Need to include 43. Now n=42*43=1806.Check n=1806. Divisors of 1806: Let's factorize 1806. 1806 ÷2=903. 903 ÷3=301. 301 ÷7=43. So prime factors are 2,3,7,43. Therefore, divisors are products of these primes. For each divisor d of 1806, p=d+1 must be prime and divide n.Check primes p:- d=1: p=2. Divides 1806. Yes.- d=2: p=3. Divides 1806. Yes.- d=3: p=4. Not prime.- d=6: p=7. Divides 1806. Yes.- d=7: p=8. Not prime.- d=14: p=15. Not prime.- d=21: p=22. Not prime.- d=42: p=43. Divides 1806. Yes.- d=43: p=44. Not prime.- d=86: p=87. Not prime.- d=129: p=130. Not prime.- d=258: p=259. Let's check 259: 259 ÷7=37. 7*37=259. So 259 is composite.- d=301: p=302. Not prime.- d=602: p=603. Not prime.- d=903: p=904. Not prime.- d=1806: p=1807. Check if prime. 1807 ÷13=139. 13*139=1807. So composite.Therefore, all p where p-1 divides 1806 and p is prime are 2,3,7,43. All of these divide 1806. Therefore, n=1806 satisfies the condition. But wait, is there a smaller n?Wait, how did we get here? Starting from n=1, then 2, 6, 42, 1806. But perhaps there's a smaller number. Let's check.Wait, n=1806 is 2*3*7*43. Let's check if any smaller multiple of these primes would work. For example, 2*3*7*43=1806. Is there a number less than 1806 that includes all primes p where p-1 divides it?Suppose we consider n=42. We saw that p=43 is required because p-1=42 divides n=42, but 43 does not divide 42. So n=42 invalid. Then n=43*42=1806 is the next.Alternatively, check n=LCM(2,3,7,43)=1806. So seems minimal.But let's check n=546. Wait, 546=2*3*7*13. Let's check if this works. The divisors would be 1,2,3,6,7,13,14,21,26,39,42,78,91,182,273,546. Then primes p where p-1 divides 546:- d=1: p=2. Divides 546. Yes.- d=2: p=3. Divides. Yes.- d=3: p=4. Not prime.- d=6: p=7. Divides. Yes.- d=7: p=8. Not prime.- d=13: p=14. Not prime.- d=14: p=15. Not prime.- d=21: p=22. Not prime.- d=26: p=27. Not prime.- d=39: p=40. Not prime.- d=42: p=43. 43 does not divide 546. So p=43 needs to divide 546, but doesn't. Hence, n=546 invalid.Similarly, n=1806 is the minimal n that includes 2,3,7,43. Let's check if any smaller number could include all required primes. For example, starting from 1806, if we remove any prime factor, say 43, then n=2*3*7=42. But as before, 42 requires 43, which isn't present. Therefore, 43 is necessary. Similarly, 7 is necessary because 6 divides n, so 7 must divide n. 3 is necessary because 2 divides n, so 3 must divide n. 2 is necessary because 1 divides n, so 2 must divide n. Therefore, all four primes are necessary, so LCM(2,3,7,43)=1806 is indeed minimal.Wait, but let me check another candidate. Let's say n= 2*3*7*43=1806. Suppose there is a prime p where p-1 divides 1806, but p does not divide 1806. Then 1806 would not satisfy the condition. But according to the earlier check, the primes p where p-1 divides 1806 are 2,3,7,43, which all divide 1806. Let's check if there are more primes.Divisors of 1806 are 1,2,3,6,7,14,21,42,43,86,129,258,301,602,903,1806. Then p=d+1 for each divisor d:- d=1: p=2 (prime, divides)- d=2: p=3 (prime, divides)- d=3: p=4 (non-prime)- d=6: p=7 (prime, divides)- d=7: p=8 (non-prime)- d=14: p=15 (non-prime)- d=21: p=22 (non-prime)- d=42: p=43 (prime, divides)- d=43: p=44 (non-prime)- d=86: p=87 (non-prime)- d=129: p=130 (non-prime)- d=258: p=259 (non-prime, as 259=7*37)- d=301: p=302 (non-prime)- d=602: p=603 (non-prime)- d=903: p=904 (non-prime)- d=1806: p=1807 (non-prime, as 1807=13*139)Therefore, all primes p where p-1 divides 1806 are 2,3,7,43, which all divide 1806. Hence, n=1806 satisfies the condition.Is there a smaller n? Let's see. Suppose n=2*3*7*43=1806. Suppose there is a smaller number. Let's consider that maybe some primes could be combined differently. For example, if there is a prime q where q-1 divides a smaller n, but q is smaller than 43. But wait, when n was 42, which is 2*3*7, the next prime needed was 43, because 42 divides n=42, so p=43 must divide n. Since 43 is prime and greater than 42, and n=42*43=1806. Therefore, it seems unavoidable.Alternatively, perhaps there is a number between 43 and 1806 that works. But considering that once you include 43, you have to check if adding 43 introduces any new primes. For example, n=1806. Divisors include 1,2,3,6,7,14,21,42,43,... So p-1=42 gives p=43, which is included. p-1=43-1=42. 42 divides 1806, so 43 is included. No other primes p where p-1 divides 1806 except those already included. Hence, n=1806 is closed under the condition.Thus, I think n=1806 is the minimal such number. Therefore, the answer is boxed{1806}.Final Answerboxed{1806}</think>To find the smallest natural number ( n ) such that if a prime ( p ) satisfies ( p - 1 ) divides ( n ), then ( p ) must divide ( n ), we need to ensure ( n ) includes all primes ( p ) for which ( p - 1 ) is a divisor of ( n ).1. Initial Checks: - For ( n = 1 ) to ( n = 12 ), we find that each fails the condition due to missing primes required by their divisors. - For example, ( n = 6 ) fails because ( p = 7 ) (where ( 7 - 1 = 6 )) does not divide 6.2. Step-by-Step Construction: - Start with ( n = 2 ), which fails because ( p = 3 ) must divide ( n ). - Next, ( n = 6 ), which fails because ( p = 7 ) must divide ( n ). - Continue this process, each time adding the required prime to ( n ).3. Critical Check at ( n = 1806 ): - ( n = 1806 ) factors as ( 2 times 3 times 7 times 43 ). - Divisors of 1806 are checked, and for each divisor ( d ), ( p = d + 1 ) is verified as a prime and a divisor of 1806. - Primes ( p = 2, 3, 7, 43 ) are confirmed to divide 1806, and all other potential primes from divisors do not introduce new primes.4. Verification: - All primes ( p ) such that ( p - 1 ) divides 1806 are confirmed to be included in the factors of 1806. - No smaller number than 1806 meets the condition as shown through incremental checks and factor analysis.Thus, the smallest natural number ( n ) that satisfies the given condition is (boxed{1806}).]

🔑:Let n satisfy this condition. Since n is divisible by 1=2-1, it must be divisible by 2, but then it is divisible by 3=2+1, by7=2 cdot 3+1 and by 43=2 cdot 3 cdot 7+1. Therefore, n must be divisible by 1806=2 cdot 3 cdot 7 cdot 43. Hence, the minimum n (if it exists) is no less than 1806.On the other hand, for 1806 the condition of the problem is satisfied. Here are all the divisors of the number 1806: 1, 2, 3, 6, 7, 14, 21, 42, 43, 86, 129, 301, 1806. Now increase each of them by one, to get: 2, 3, 4, 7, 8, 15, 22, 43, 44, 87, 130, 302, 1807.It remains to select all the prime numbers from the second set and verify that they are included in the first set (1807 is not prime, it is divisible by 13).## Answern=1806.

🧠:[<think>Okay, so I need to figure out how many of these three statements are necessarily true when choosing five numbers from 1 to 9. Let's take them one by one. First, statement (1): "there must be two numbers that are coprime." Hmm, coprime means their greatest common divisor is 1, right? So, if I can find a set of five numbers where no two are coprime, then this statement is false. Otherwise, it's true. Let me think. If I pick all even numbers. The numbers from 1 to 9 are 1,2,3,4,5,6,7,8,9. The even numbers are 2,4,6,8. But there are only four even numbers. So I can't pick five even numbers. But maybe including 1? Wait, 1 is coprime with every number. So if I include 1, then 1 and any other number are coprime. Therefore, if I include 1 in the five numbers, then statement (1) is automatically true. So maybe the only way to avoid having coprime numbers is to not include 1 and pick numbers with a common divisor. But since there are only four even numbers, if I pick five numbers without 1, I have to include at least one odd number. But the odd numbers from 3 to 9 are 3,5,7,9. Each of these is either prime or 9, which is 3 squared. Wait, if I try to pick five numbers without 1, maybe all even numbers plus one odd number. Let's say 2,4,6,8, and then 3. Then check if there's a pair coprime. For example, 3 and 2 are coprime, since gcd(3,2)=1. Similarly, 3 and 4: gcd(3,4)=1. So even adding 3 would create coprimality with the evens. What if instead of 3, I pick 9? Then 9 and 2: gcd(9,2)=1. Still coprime. Hmm. What if I pick 5? Then 5 and 2: coprime. 7? 7 and 2: coprime. So any odd number added to the four evens will result in at least one coprime pair. Therefore, it's impossible to have five numbers without 1 and not have a coprime pair. Alternatively, if you include 1, then 1 is coprime with everyone. So in any case, if you pick five numbers, you have to have at least two that are coprime. Therefore, statement (1) must be true. Okay, so (1) is correct.Now statement (2): "there must be one number that is a multiple of another." So we need to check if in any five numbers selected from 1 to 9, there is at least one pair where one divides the other. If we can find a set of five numbers where no one divides another, then statement (2) is false. Let me try to construct such a set. The idea is to pick numbers that are all prime or powers of different primes, or numbers that are not multiples of each other. Let's see. For example, primes: 2,3,5,7. Then 4 is a multiple of 2, so can't include 4. 6 is multiple of 2 and 3. 8 is multiple of 2. 9 is multiple of 3. So if I take 2,3,5,7, and then maybe 9? Wait, 9 is 3 squared, so 3 divides 9. So that's bad. Alternatively, take 4,5,6,7, but 4 is a multiple of 2, but 4 isn't prime. Wait, maybe pick numbers that are all in the same range but not multiples. Let me think. The strategy here is similar to the concept of an "antichain" in posets, where no two elements are comparable under the partial order of divisibility.According to Dilworth's theorem, the size of the largest antichain in the divisibility poset of {1,...,9} would determine the maximum set without such a pair. Let's see. The numbers from 1 to 9 can be grouped by their prime factors. To avoid multiples, we can pick numbers that are not divisors of each other. For example, choosing numbers from 5 to 9: 5,6,7,8,9. Wait, but 6 and 3 are divisors, but 3 isn't in the set. Let's check this set: 5,6,7,8,9. Is there a multiple? 6 is 2*3, 8 is 2^3. So 6 and 8 have a common divisor 2, but neither divides the other. 6 and 9: 6=2*3, 9=3^2. 3 divides 9, but 3 isn't in the set. Wait, but 6 and 9: gcd(6,9)=3, but neither divides the other. 8 and 4: 4 isn't in the set. So 5,6,7,8,9: does any number divide another? Let's check. 5 divides none. 6: factors are 2,3. 6 divides 6*2=12, which isn't in the set. 7: prime. 8: factors 2. 8 divides 16, not in the set. 9: factors 3. 9 divides 18, not in the set. So in this set, no number divides another. Therefore, this is a set of five numbers with no multiples. Therefore, statement (2) is false. So (2) is incorrect.But wait, 6 and 9: 6 is 2*3 and 9 is 3*3. So 3 divides both, but 3 isn't in the set. However, neither 6 divides 9 nor 9 divides 6. So that's okay. So yes, 5,6,7,8,9 is a valid set with five numbers where no number is a multiple of another. Therefore, conclusion (2) is not necessarily true. So (2) is incorrect.Now statement (3): "there must be one number whose double is a multiple of another." Hmm. So this means that in the five numbers, there exists two numbers a and b such that 2a is a multiple of b, or 2b is a multiple of a. Alternatively, either a divides 2b or b divides 2a. So, for example, if one number is double another, or half of another (if even). Wait, but since we are dealing with integers, if a is half of b, then b must be even. So, in other words, in the set, either there is a pair where one is double the other, or a pair where one is even and the other is half of it.Alternatively, if there's a number that is even, and another number is half of it. Or if there's an odd number that is half of an even number (but that's not possible because half of an even number is integer, but if the original number is odd, then 2*odd would be even, which would have to be in the set. Wait, maybe the idea is that either a pair (a, b) exists such that 2a = b or 2b = a. But since all numbers are from 1 to 9, 2a must be <=9, so a can be up to 4 (since 2*4=8). Similarly, 2b = a would require that a is even and b is a/2. So possible pairs are (1,2), (2,4), (3,6), (4,8), (5,10) but 10 is not in the set. So the possible pairs where one is double the other are (1,2), (2,4), (3,6), (4,8). So if we can avoid all these pairs, then statement (3) is false. Let's try to construct a set of five numbers that doesn't include any of these pairs.For example, take numbers that are not in these pairs. Let's see. Let's take 1: but 1 and 2 can't be both included. If we exclude 2, then we can have 1. Similarly, 3 and 6 can't both be included. Let's try to construct a set. Let's take 1,3,5,7,9. All odd numbers. Let's check: 1 and 3: 2*1=2, which is not in the set. 2*3=6, not in the set. 2*5=10, not in the set. Similarly, 7 and 9: 14 and 18, not in the set. So in this set, no number's double is present. But also, since all numbers are odd, even if you double them, they become even, which are not in the set. Therefore, in this set of 1,3,5,7,9, there is no pair where one's double is a multiple of another. Wait, but wait. The statement says "there must be one number whose double is a multiple of another." So not necessarily that the double is in the set, but that the double is a multiple of another number. Wait, let me check again.Wait, maybe I misinterpreted the statement. It says "there must be one number whose double is a multiple of another." So that is, there exists two numbers a and b in the set such that 2a is a multiple of b. So 2a = k*b for some integer k. Similarly, 2b = k*a. So, 2a divisible by b, or 2b divisible by a. So for example, if a=3 and b=6: 2*3=6, which is a multiple of 6? Wait, 6 is 6, so 6 divides 6. So 2a = 6, which is a multiple of b=6. So 6 divides 6. So in that case, (3,6) would satisfy this because 2*3=6 is a multiple of 6. Similarly, if a=2 and b=4: 2*2=4 is a multiple of 4. So even if the numbers are in a ratio of 1:2, then this condition is satisfied.But maybe even if they are not in ratio 1:2. For example, if a=3 and b=2: 2*3=6, which is a multiple of 2. So even if a is 3 and b is 2, 6 is a multiple of 2, so that works. So actually, the statement is more general than just having a pair where one is double the other. It could be that doubling one number gives a multiple of another. So even if you have numbers like 3 and 6: 2*3=6, which is a multiple of 6. Or 3 and 2: 2*3=6, which is a multiple of 2.So, if I take the set of all odd numbers: 1,3,5,7,9. Let's check if in this set, for any a, 2a is a multiple of some b in the set. Let's see:For a=1: 2*1=2. Is 2 a multiple of any number in the set? The set is {1,3,5,7,9}. 2 is a multiple of 1 and 2. But 2 is not in the set. So 2 is a multiple of 1. Since 1 is in the set, then 2*1=2 is a multiple of 1. So, does that count? Because the problem says "there must be one number whose double is a multiple of another." So here, number 1's double is 2, which is a multiple of 1. But 2 is not in the set. Wait, but 1's double is 2, which is a multiple of 1. Wait, 2 is a multiple of 1, yes. But the original numbers are from 1 to 9. The problem says "there must be one number whose double is a multiple of another." So, the double doesn't have to be in the set. Wait, 2a is a multiple of b. If b divides 2a, then 2a is a multiple of b. So, for a=1, 2a=2. Then any b that divides 2 is either 1 or 2. Since 1 is in the set, then b=1 divides 2. So 2 is a multiple of 1. Therefore, this satisfies the condition. Therefore, even in the set {1,3,5,7,9}, since 1 is present, the double of 1 is 2, which is a multiple of 1. So according to this, statement (3) is always true? But that can't be right.Wait, the problem says "there must be one number whose double is a multiple of another." So, if there's a number a in the set such that 2a is a multiple of another number b in the set. So, 2a is divisible by b. So for example, in the set {1,3,5,7,9}, take a=1: 2*1=2, which is divisible by b=1 (since 2/1=2). So yes, a=1 and b=1. But 1 is in the set. So does that count? The problem says "another" number. So, does "another" mean different? If so, then b has to be different from a. Then in the case of a=1, b=1 is same number, which may not count. So maybe the problem requires two distinct numbers.Looking back at the original problem: "there must be one number whose double is a multiple of another." The wording is a bit ambiguous. If "another" implies a different number, then 1's double being a multiple of 1 would not count because it's the same number. If "another" just means any other number (including itself), then it would count. But usually, in combinatorial problems, when they say "another," they mean a different one. So perhaps we need two distinct numbers a and b where 2a is a multiple of b.Therefore, in the set {1,3,5,7,9}, if we require a and b to be distinct, then 2a must be a multiple of b. For a=1, 2*1=2. Is 2 a multiple of any other number in the set (3,5,7,9)? No. For a=3: 2*3=6. 6 is a multiple of 3 (but 3 is in the set). But again, same number. If "another" requires different, then no. For a=5: 2*5=10, which is a multiple of 5 (same number). 10 is not in the set. For a=7: 2*7=14, multiple of 7 (same number). For a=9: 2*9=18, multiple of 9 (same number). So if "another" requires different numbers, then in the set {1,3,5,7,9}, there is no pair of distinct numbers a and b where 2a is a multiple of b. Hence, this would be a counterexample, making statement (3) false.But earlier, if we allow the same number, then trivially, every number's double is a multiple of itself. So in that case, the statement is always true. But the problem is in Chinese, and the translation says "another." The Chinese term for "another" is likely to imply a different one. Therefore, in the problem statement, it's probably two distinct numbers. So {1,3,5,7,9} would be a set where no two distinct numbers satisfy that 2a is a multiple of b. Let's check:For a=1, 2*1=2: not a multiple of 3,5,7,9.For a=3, 2*3=6: not a multiple of 1,5,7,9. Wait, 6 is a multiple of 1, but 1 is in the set. But 6 is not in the set. Wait, but we are checking if 6 is a multiple of any number in the set. 6 is a multiple of 1 and 3. But 3 is in the set. But 6 is not in the set, so does that matter? Wait, the problem states "there must be one number whose double is a multiple of another." So, the double (which is 6) does not need to be in the set. The question is whether 6 is a multiple of another number in the set. So 6 is a multiple of 3 (since 3 is in the set). Therefore, a=3, double is 6, which is a multiple of 3. But 3 is in the set. So even though 6 is not in the set, 6 is a multiple of 3 which is in the set. Therefore, does this count? If so, then in the set {1,3,5,7,9}, the number 3's double is 6, which is a multiple of 3. Since 3 is in the set, then this would satisfy the condition. But since the problem says "another" number, if "another" means different, then 3's double being a multiple of 3 (same number) wouldn't count. But if "another" allows the same number, then it does. This is a crucial point. Let's re-examine the problem statement:"(3) there must be one number whose double is a multiple of another."The wording is a bit ambiguous. If "another" means a different number, then the double must be a multiple of a different number in the set. If "another" can be the same number, then it's trivial because any number's double is a multiple of itself. However, in standard mathematical problems, such statements usually refer to two distinct numbers unless specified otherwise. So, assuming "another" means a different number, then in the set {1,3,5,7,9}:- For a=1: 2*1=2. 2 is a multiple of 1 (same number) but not of any other number in the set. So if "another" requires different, then no.- For a=3: 2*3=6. 6 is a multiple of 3 (same number) and 1 (but 1 is in the set). But again, same number. If different, 6 is not a multiple of 1? Wait, 6 is a multiple of 1, but 1 is in the set. So if 1 is considered, then 2*3=6 is a multiple of 1, which is another number. Therefore, in this case, a=3 and b=1. 6 is a multiple of 1. Since 1 is in the set, then this satisfies the condition. Wait, even if "another" means different, 3 and 1 are different. So 6 is a multiple of 1. Therefore, even in this set, there exists a number (3) whose double (6) is a multiple of another number (1). Therefore, statement (3) would be true. Wait, but 1 is in the set. So 2*3=6, which is a multiple of 1, which is another number in the set. Therefore, even in this set, statement (3) holds. Therefore, the previous attempt to construct a counterexample fails. Similarly, if we take any number in the set:- For a=5: 2*5=10. 10 is a multiple of 1 and 5. 1 is in the set, so 10 is a multiple of 1. Therefore, a=5 and b=1.- For a=7: 2*7=14, multiple of 1 and 7. 1 is in the set.- For a=9: 2*9=18, multiple of 1 and 9. 1 is in the set.Therefore, in any case, if 1 is in the set, then every number's double is a multiple of 1. Therefore, if 1 is included, statement (3) is automatically true. So the question is, can we construct a set of five numbers without 1, such that for every number a in the set, 2a is not a multiple of any other number in the set?Let's try to construct such a set. Suppose we exclude 1. Then, pick numbers from 2 to 9. Let's try to select numbers such that for any a in the set, 2a is not a multiple of any other number in the set. Let's consider numbers where 2a is not in the set and also not a multiple of any other number in the set.For example, take the set {3,4,5,7,9}. Let's check:- 2*3=6. 6 is not in the set. Is 6 a multiple of any other number in the set? 6 is a multiple of 3. But 3 is in the set. So 2*3=6 is a multiple of 3, which is in the set. Therefore, 3's double is a multiple of 3. But since 3 is in the set, and we need "another" number, which would be 3 itself. If "another" allows the same number, then yes, but if it requires a different one, then no. But 3 is the same as a. So if "another" requires different, then this does not count.But wait, 6 is a multiple of 3, which is in the set. So if "another" means different, then since 3 is in the set, and 6 is a multiple of 3, then a=3 and b=3. If "another" is allowed to be the same, then it's okay. But if it must be different, then no.Alternatively, maybe other numbers:Take the set {4,5,6,7,9}:- 2*4=8. 8 is not in the set. 8 is a multiple of 4, which is in the set. So 4's double is a multiple of 4. Again, same number.- 2*5=10: multiple of 5 and 1. 5 is in the set.- 2*6=12: multiple of 6,3,4, etc. 6 is in the set.- 2*7=14: multiple of 7.- 2*9=18: multiple of 9.So, in this set, all the doubles are multiples of the same number. If "another" allows the same number, then yes. If not, then no.Alternatively, try another set. Let's take {2,5,7,8,9}:- 2*2=4: not in the set. 4 is a multiple of 2. 2 is in the set.- 2*5=10: multiple of 5 and 2. 5 and 2 are in the set.- 2*7=14: multiple of 7 and 2.- 2*8=16: multiple of 8 and 2.- 2*9=18: multiple of 9 and 2.So in this case, for each a, 2a is a multiple of a and possibly 2. So if 2 is in the set, then 2a is a multiple of 2. For example, 2*5=10 is a multiple of 2. So here, even if we exclude 1, if 2 is in the set, then for any a in the set, 2a is a multiple of 2, which is in the set. Therefore, if 2 is in the set, then for every a, 2a is a multiple of 2, hence satisfying the condition. So in this case, the set {2,5,7,8,9} would have for each a, 2a is a multiple of 2 (which is in the set). So even if "another" requires a different number, since 2 is in the set, any a's double is a multiple of 2, so a and 2 are different numbers. Therefore, this would satisfy the condition.Wait, for example, take a=5: 2*5=10, which is a multiple of 2. Since 2 is in the set, then 5's double is a multiple of 2. So here, a=5 and b=2. Different numbers. Therefore, statement (3) is satisfied.Similarly, in the set {3,4,5,7,9}, if 2 is not in the set, but 4 is. Let's check:- 2*3=6, multiple of 3. If 3 is in the set, but 6 is not. So 6 is a multiple of 3. But 3 is in the set. So 3's double is a multiple of 3. If "another" allows same number, yes. If not, no. But if "another" is required to be different, then no. However, let's check other numbers:- 2*4=8, which is a multiple of 4. 4 is in the set. Same number.- 2*5=10, multiple of 5. 5 is in the set.- 2*7=14, multiple of 7.- 2*9=18, multiple of 9.So in this case, if we have a set without 1 and 2, but includes 3,4,5,7,9, then for each a, 2a is a multiple of a. So if "another" must be different, then this doesn't count. But if "another" can be the same, it does. However, in the original problem statement, it's likely that "another" implies a different number. Therefore, to construct a valid counterexample, we need a set where for every a, 2a is not a multiple of any other number (different from a) in the set.But in the set {3,4,5,7,9}:- a=3: 2*3=6. 6 is a multiple of 3 (same number) and 1 (not in the set). So no different number.- a=4: 2*4=8. 8 is a multiple of 4 (same number) and 2 (not in the set). So no different number.- a=5: 2*5=10. Multiple of 5 and 2 (not in the set). No different.- a=7: 2*7=14. Multiple of 7 and 2. 2 not in the set.- a=9: 2*9=18. Multiple of 9 and 2. 2 not in the set.Therefore, in this set, for each a, 2a is a multiple of a (same number) or multiples of numbers not in the set. Since "another" implies a different number, this set does not satisfy the condition. Therefore, statement (3) is false. Therefore, this set would be a counterexample to statement (3). Wait, but earlier, if a=3 and b=3, is that considered? If the problem allows the same number, then yes. But if not, then no. But since the problem says "another," it's more likely that they are different. So in that case, this set {3,4,5,7,9} is a valid counterexample where no two distinct numbers satisfy that one's double is a multiple of the other. Hence, statement (3) is false. Therefore, statement (3) is not necessarily true.But wait, let's check again. Let me take each pair:In {3,4,5,7,9}:- 3 and 4: 2*3=6, which is not a multiple of 4. 2*4=8, which is not a multiple of 3.- 3 and 5: 2*3=6, not multiple of 5. 2*5=10, not multiple of 3.- 3 and 7: Similarly, 2*3=6, 2*7=14, neither multiple of each other.- 3 and 9: 2*3=6, not multiple of 9. 2*9=18, not multiple of 3 (wait, 18 is a multiple of 3. But 18 is not in the set. But does that matter? The problem says "there must be one number whose double is a multiple of another." So, for a=9, 2*9=18. Is 18 a multiple of any other number in the set {3,4,5,7,9}? 18 is a multiple of 3 and 9. 3 and 9 are in the set. So 2*9=18 is a multiple of 3 and 9. Therefore, even though 18 isn't in the set, the fact that 18 is a multiple of 3 and 9 (which are in the set) means that 9's double is a multiple of 3 and 9. Therefore, in this case, a=9 and b=3 or b=9. Since 3 is different from 9, then a=9 and b=3 satisfies the condition because 18 is a multiple of 3. Therefore, even in this set, there exists a pair: 9 and 3. 2*9=18, which is a multiple of 3. Therefore, statement (3) holds.Wait, this complicates things. So even if the double is not in the set, as long as it's a multiple of another number in the set, it counts. So in the set {3,4,5,7,9}, take a=9. 2*9=18, which is a multiple of 3 (since 3 is in the set). Therefore, 9's double is a multiple of 3. Therefore, even in this set, such a pair exists: 9 and 3. Hence, statement (3) is true. Therefore, this set is not a counterexample.Wait, this changes everything. So even if the double is not in the set, if it's a multiple of another number in the set, then it counts. Therefore, even in {3,4,5,7,9}, since 2*9=18 is a multiple of 3, which is in the set, then this satisfies the condition. Therefore, statement (3) is necessarily true.But how can that be? Let me think again. The problem states: "there must be one number whose double is a multiple of another." So, for number a, 2a is a multiple of b, where b is in the set. It does not require that 2a is in the set, just that 2a is a multiple of b. So even if 2a is not in the set, as long as b divides 2a, it's okay. So in the set {3,4,5,7,9}, for a=9, 2a=18. 18 is divisible by 3 and 9. 3 is in the set, so b=3. Therefore, 9's double is a multiple of 3. Therefore, such a pair exists. Similarly, in the set {3,5,7,9}, a=9 and b=3. Therefore, it's impossible to avoid such pairs.Wait, so maybe no matter what five numbers you choose, there will always be such a pair. Let's try to see.Suppose we try to construct a set where for every a in the set, 2a is not a multiple of any other b in the set. Let's see. Let's pick numbers such that for any a, 2a is not divisible by any other b in the set. Let's try with primes. Suppose we take 3,5,7,11, but 11 is not in 1-9. Let's think of 3,5,7, and then others. Wait, if we pick 3,5,7,9, and maybe 4. Let's check:Set {3,4,5,7,9}:- For 3: 2*3=6. 6 is divisible by 3. So if 3 is in the set, then 6 is divisible by 3. So 3's double is a multiple of 3. But since 3 is in the set, this counts if "another" allows the same number. If not, we need a different number. Let's assume "another" is different. Then, 6 is divisible by 3, which is in the set. So a=3 and b=3. If different, then no. But if we need a different number, then 6 is not divisible by 4,5,7,9. So for a=3, 2a=6, not divisible by 4,5,7,9. Therefore, no different b.For a=4: 2*4=8. 8 is divisible by 4. Same number. If different, 8 is divisible by 2, which is not in the set. So no.For a=5: 2*5=10. Divisible by 5 and 2. 5 is in the set. So same number. If different, 10 divisible by 2, which is not in the set. So no.For a=7: 2*7=14. Divisible by 7 and 2. 7 is in the set. Same number. Different number 2 not in the set.For a=9: 2*9=18. Divisible by 9,3,2, etc. 9 and 3 are in the set. So 18 is divisible by 3 (different number). So a=9 and b=3. Therefore, 2*9=18 is divisible by 3, which is in the set. Therefore, even in this set, there exists a pair (9,3) where 9's double is a multiple of 3. Therefore, statement (3) holds.Therefore, no matter how we choose five numbers from 1-9, there will always be at least one pair where one number's double is a multiple of another number in the set. Therefore, statement (3) is necessarily true.But wait, how? Let's see another example. Suppose we take the set {2,4,5,7,9}:- For a=2: 2*2=4, which is in the set. So 4 is a multiple of 4 (same number). If different, 4 is a multiple of 2 (which is in the set). So 2's double is 4, which is a multiple of 2. Therefore, a=2 and b=2 (same) or b=2 (same). But if different, then 4 is a multiple of 2, which is different. Therefore, 2 and 4: 2's double is 4, which is a multiple of 2. Wait, 4 is a multiple of 2. So 2*2=4 is a multiple of 2. So yes, in this case, even if different numbers are required, 4 is a multiple of 2. So a=2 and b=4. Because 4 is in the set. 2*2=4, which is a multiple of 4. Wait, 4 is a multiple of 2. So 2*2=4 is a multiple of both 2 and 4. Therefore, even if different numbers, there's a pair (2,4) where 2's double is a multiple of 4. Therefore, statement (3) holds.Alternatively, take a=4: 2*4=8. If 8 is not in the set, 8 is a multiple of 4 and 2. Both 4 and 2 are in the set. So 8 is a multiple of 4 (same number) or 2 (different). So a=4 and b=2. Therefore, 2*4=8 is a multiple of 2. So yes, statement (3) holds.Another example: set {5,6,7,8,9}:- a=5: 2*5=10, multiple of 5 and 2. 5 is in the set.- a=6: 2*6=12, multiple of 6,3,4,2. 6,3,4,2 are in the set or not? In this set, 6,7,8,9 are present. So 2*6=12 is a multiple of 6 (same) and 3 (not in the set), 4 (not in the set), 2 (not in the set). So only 6. So same number. If different numbers, 12 is not a multiple of any others. - a=7: 2*7=14, multiple of 7 and 2. 7 is in the set.- a=8: 2*8=16, multiple of 8 and 2. 8 is in the set.- a=9: 2*9=18, multiple of 9 and 3,2. 9 is in the set, 3 is not.But for a=6, 2*6=12 is a multiple of 6 (same) and 3,4,2. 3,4,2 are not in the set. Therefore, the only multiple is 6. But in the set, there's a=6 and b=6. If "another" means different, then 6's double is not a multiple of any other number in the set. However, there are other numbers in the set. For example, a=5: 2*5=10, which is a multiple of 5 (same) and 2 (not in the set). a=7: 2*7=14, multiple of 7. a=8: multiple of 8. a=9: multiple of 9. But also, a=5: 10 is a multiple of 5. a=6: 12 is a multiple of 6. But in this set, does there exist at least one pair where a's double is a multiple of b where a and b are different?Wait, let's check for a=8: 2*8=16. 16 is a multiple of 8, which is same. But 16 is also a multiple of 2, which is not in the set. However, 16 is not a multiple of any other number in the set (5,6,7,9). Similarly, a=6: 12 is multiple of 6 and 3,4,2. None of those except 6 are in the set. a=5: 10 is multiple of 5 and 2. 5 is in the set. a=7: 14 is multiple of 7. a=9: 18 is multiple of 9, 3, etc. 9 is in the set. So in this set, for each a, the double is a multiple of a itself or numbers not in the set. But does any a's double be a multiple of a different number in the set?Wait, take a=9: 2*9=18. 18 is a multiple of 9 and 3, 6. 3 and 6 are in the set? 3 is not in the set {5,6,7,8,9}. 6 is in the set. So 18 is a multiple of 6. Therefore, a=9 and b=6. 2*9=18 is a multiple of 6. Therefore, even in this set, there exists a pair: 9 and 6. Therefore, statement (3) holds.Wow, so no matter how I choose the five numbers, there will always be at least one pair where one's double is a multiple of another. Therefore, statement (3) must be true. Therefore, going back:Statement (1): Correct.Statement (2): Incorrect (as shown by the example {5,6,7,8,9} where no number is a multiple of another).Statement (3): Correct.But wait, let's verify again with another example. Suppose I pick the set {3,5,7,8,9}:- For a=3: 2*3=6. Not multiple of any other number except 3.- For a=5: 2*5=10. Multiple of 5 and 2. 5 is in the set.- For a=7: 2*7=14. Multiple of 7.- For a=8: 2*8=16. Multiple of 8 and 2. 8 is in the set.- For a=9: 2*9=18. Multiple of 9, 3, 6. 9 is in the set.But also, 18 is a multiple of 3 and 6. 3 is not in the set. 6 is not in the set. Wait, in this set, {3,5,7,8,9}, 18 is a multiple of 9 and 3. 3 is in the set. Therefore, a=9 and b=3. 2*9=18 is a multiple of 3. So even here, statement (3) holds.Another example: {2,3,5,7,9}:- a=2: 2*2=4. Multiple of 2.- a=3: 6. Multiple of 3 and 2.- a=5: 10. Multiple of 5 and 2.- a=7: 14. Multiple of 7 and 2.- a=9: 18. Multiple of 9,3,2.In all cases, the double is a multiple of 2 or the number itself. Since 2 is in the set, for any a, 2a is a multiple of 2. Therefore, for a=3, 2*3=6 is a multiple of 2. So here, a=3 and b=2. Therefore, statement (3) holds.Therefore, it seems impossible to construct a set of five numbers from 1 to 9 where for every a in the set, 2a is not a multiple of any other b in the set. Therefore, statement (3) is necessarily true.But wait, let me think again. Suppose I pick numbers that are all odd and greater than 1, like {3,5,7,9, something}. Wait, but from 1-9, the odd numbers are 1,3,5,7,9. If I exclude 1 and pick {3,5,7,9}, but I need five numbers, so I have to include 1 or an even number. If I include an even number, say 2, then 2's double is 4, which is a multiple of 2. But also, for other numbers like 3: 2*3=6 is a multiple of 2. Therefore, in this case, a=3 and b=2. So statement (3) holds.If I include 1, then 1's double is 2, which is a multiple of 1. So even with 1, it's satisfied.Alternatively, if I include only odd numbers greater than 1: {3,5,7,9, something}. But there are only four such numbers. To make five, I have to include 1 or an even number. If I include 1, then 1's double is 2, which is a multiple of 1. If I include an even number like 4, then 4's double is 8, which is a multiple of 4 (if included) or 2 (if not). But 4 is in the set. So 8 is a multiple of 4.Wait, no, in the set {3,5,7,9,4}, 4's double is 8. 8 is not in the set, but 8 is a multiple of 4, which is in the set. Therefore, a=4 and b=4. If "another" requires different, then it's not. However, for a=3: 2*3=6 is a multiple of 3 (same number). For a=5: 10 multiple of 5. For a=7: 14 multiple of 7. For a=9: 18 multiple of 9. For a=4: 8 multiple of 4. So if "another" requires different numbers, this set doesn't satisfy. But wait, 18 is also a multiple of 3. So a=9's double is 18, which is a multiple of 3. 3 is in the set. Therefore, a=9 and b=3. Different numbers. Therefore, statement (3) holds.So in all cases, there is always a pair where one number's double is a multiple of another number. Therefore, statement (3) is necessarily true.Therefore, among the three statements:(1) Correct.(2) Incorrect.(3) Correct.So the number of correct conclusions is 2. Therefore, the answer is C.But wait, let me check again for statement (3). Earlier, I thought of the set {3,4,5,7,9} and considered a=9, b=3. So even in that set, statement (3) is true. Therefore, no matter how you choose five numbers, you can't avoid having at least one pair where one's double is a multiple of another. Therefore, statement (3) is always true.Thus, statements (1) and (3) are correct, while statement (2) is not. Hence, the number of correct conclusions is 2. Answer C.Final Answerboxed{C}</think>To determine the number of correct conclusions among the given statements, we analyze each statement individually:1. Statement (1): There must be two numbers that are coprime. - Coprime numbers have a greatest common divisor (gcd) of 1. - If we try to avoid coprime pairs, we cannot pick five numbers without including 1 or having at least one coprime pair. Including 1 automatically makes pairs with 1 and any other number coprime. Without 1, picking five numbers from the even numbers (2, 4, 6, 8) and one odd number (3, 5, 7, 9) will still result in coprime pairs. Hence, statement (1) is true.2. Statement (2): There must be one number that is a multiple of another. - We can construct a set of five numbers where no number is a multiple of another. For example, the set {5, 6, 7, 8, 9} contains numbers where none is a multiple of another. Hence, statement (2) is false.3. Statement (3): There must be one number whose double is a multiple of another. - For any set of five numbers, we need to check if there exists a pair where one number's double is a multiple of another. For example, in the set {3, 4, 5, 7, 9}, the number 9's double (18) is a multiple of 3. This pattern holds for any set of five numbers, ensuring at least one such pair exists. Hence, statement (3) is true.Conclusion:- Statements (1) and (3) are correct, while statement (2) is incorrect. Therefore, the number of correct conclusions is 2.[boxed{C}],<think>Okay, let's try to tackle this problem step by step. The question is about choosing five numbers from 1 to 9 and determining which of the three given conclusions must hold. The options are that either 0, 1, 2, or all 3 of the conclusions are correct. First, let me restate the problem to make sure I understand it. We need to pick any five numbers from the set {1,2,3,4,5,6,7,8,9}. Then, among the three conclusions presented, we need to figure out how many are necessarily true for any such selection. The conclusions are:1. There must be at least two numbers that are coprime.2. There must be at least one number that is a multiple of another.3. There must be at least one number whose double is a multiple of another.We need to check each conclusion and see if they must hold regardless of which five numbers we pick. The answer choices are (A) 0, (B) 1, (C) 2, or (D) 3. Let me start with conclusion (1): "There must be two numbers that are coprime." Coprime numbers are numbers whose greatest common divisor (GCD) is 1. So, the question is whether in any five-number selection from 1 to 9, there will always be at least two numbers that share no common divisors other than 1.To approach this, maybe I should try to see if it's possible to choose five numbers where all pairs have a common divisor greater than 1. If such a selection is impossible, then conclusion (1) must hold. Let me think about the numbers from 1 to 9. The primes in this range are 2, 3, 5, 7. 1 is a special case because it's coprime with every number. However, if we include 1 in our selection, then 1 will be coprime with all other numbers. So, if 1 is among the five numbers, conclusion (1) is automatically true. Therefore, the only way conclusion (1) could be false is if we can choose five numbers excluding 1, where every pair has a common divisor greater than 1.Wait, but 1 is in the original set. So, if someone picks 1, then it's coprime with all others. But the problem states "choose five numbers from 1,2,...,9". So, 1 is an option, but maybe we can avoid picking it. Let's see.So, if we try to pick five numbers without 1, such that every pair has a GCD greater than 1. Let's see if that's possible.Looking at the numbers 2 through 9. Let's list them:2, 3, 4, 5, 6, 7, 8, 9.Now, primes here are 2, 3, 5, 7. The composite numbers are 4 (2x2), 6 (2x3), 8 (2x2x2), 9 (3x3).If we try to pick numbers such that every pair shares a common divisor >1, we need all numbers to share a common prime factor, or at least, every pair shares at least one prime factor.But in the set from 2 to 9, the primes are 2,3,5,7. Let's see if we can pick five numbers where all are even. The even numbers in 2-9 are 2,4,6,8. There are four even numbers. If we pick all four even numbers, we need one more number. But the remaining numbers are 3,5,7,9. If we pick 9, which is 3x3, then 9 is not even, so GCD(9,2)=1, GCD(9,4)=1, GCD(9,6)=3, GCD(9,8)=1. So, 9 and 6 share a common divisor 3, but 9 and 2,4,8 are coprime. Therefore, if we take 2,4,6,8,9, then 9 and 2,4,8 are coprime. Thus, this selection has coprime pairs. So, conclusion (1) holds here.Alternatively, if we try to pick five numbers all divisible by 3. The numbers divisible by 3 in 2-9 are 3,6,9. Only three numbers. So we can't pick five numbers all divisible by 3. Similarly for 5 and 7; they have even fewer multiples.Alternatively, maybe not all numbers share the same prime, but every pair shares at least one prime. For example, maybe using multiple primes. Let's see. For example, 2,4,6,8 (all even) and 9. But as before, 9 and 2,4,8 are coprime. So that doesn't work. Alternatively, 3,6,9 (multiples of 3) and then 2,4, but 2 and 3 are coprime. So that would introduce coprime pairs. So maybe it's impossible to have five numbers without 1 where all pairs have GCD >1.Alternatively, let's try to use numbers with different primes but ensuring that every pair shares at least one prime. For example, 2,4,6,8, and 9. As before, 9 is coprime with 2,4,8. So that introduces coprimality. Alternatively, 6,3,9, and even numbers. Wait, if I take 2,4,6,8, and 3. Then 3 and 2,4,6,8. 3 is coprime with 2,4,8. So again, coprime pairs. Similarly, if I take 5, but 5 is prime and coprime with most numbers.Wait, perhaps if we take numbers that are multiples of 2 or 3. Let me try that. For example, numbers divisible by 2 or 3. Let's see:Numbers divisible by 2: 2,4,6,8.Numbers divisible by 3: 3,6,9.So numbers divisible by 2 or 3: 2,3,4,6,8,9. So six numbers. If we pick five numbers from these six, can we avoid coprime pairs? Let's see.Take 2,4,6,8,3. Then 3 and 2,4,6,8. 3 and 2: GCD=1. 3 and 4: GCD=1. So 3 is coprime with 2 and 4. Therefore, there are coprime pairs.Alternatively, take 2,4,6,8,9. As before, 9 and 2,4,8 have GCD=1. So that's coprime.Alternatively, take 3,6,9, and 2,4. Then 3 and 2 are coprime. So that's a problem.Alternatively, take 6, which is divisible by both 2 and 3. So if we take 6, then perhaps numbers that are multiples of 6. But multiples of 6 in 1-9 are 6. Only one number. So that doesn't help.Alternatively, maybe pick numbers with factors 2 and 3. But even so, unless the numbers share a common factor, they can still be coprime.It seems challenging to pick five numbers from 2-9 without having at least two coprime numbers. Since even if we pick numbers from the set divisible by 2 or 3, there are still coprime pairs like 2 and 3, 2 and 9, 3 and 4, etc.Alternatively, if we try to use numbers that are multiples of 2, but 2,4,6,8. Then adding another number which is 9, but as we saw, 9 and 2,4,8 are coprime. So that introduces coprimes. If we add 5 instead, 5 is coprime with 2,4,6,8. If we add 7, same thing. If we add 3, then 3 and 2,4,6,8: 3 and 2 are coprime, 3 and 4 are coprime, 3 and 6 share 3, 3 and 8 coprime. So 3 is coprime with 2,4,8.Therefore, no matter which fifth number we add to the four even numbers, that fifth number will be coprime with some of the even numbers. Hence, in this case, conclusion (1) must hold.Alternatively, if we try to pick numbers that are all odd. Let's see. The odd numbers from 1-9 are 1,3,5,7,9. There are five odd numbers. If we pick all five, then 1 is included. Since 1 is coprime with every number, so 1 and 3 are coprime, 1 and 5, etc. Therefore, in this case, conclusion (1) holds.But wait, what if we exclude 1? The problem says "choose five numbers from 1,2,...,9". So, if we pick five numbers, we can choose whether to include 1 or not. If we exclude 1 and pick five odd numbers from 3,5,7,9, but there are only four such numbers (3,5,7,9). So we can't pick five odd numbers without including 1. Therefore, any selection of five numbers must include at least one even number or 1.Wait, hold on. If we have to pick five numbers, and we don't pick 1, then we can only pick from 2-9, which are eight numbers. But among these eight numbers, four are even (2,4,6,8) and four are odd (3,5,7,9). So, if we don't pick 1, we can pick five numbers from these eight. The maximum number of odd numbers we can pick is four (3,5,7,9). Therefore, if we don't pick 1, then to get five numbers, we need to pick at least one even number. Similarly, if we do pick 1, then we can pick four more numbers from 2-9.So, if we pick 1, then we have four more numbers. If we don't pick 1, we have five numbers from 2-9. Let's analyze both cases.Case 1: Selection includes 1. Then 1 is coprime with all other numbers. Therefore, conclusion (1) is true.Case 2: Selection does not include 1. Then, we have five numbers from 2-9. As we saw earlier, the even numbers are 2,4,6,8 (four numbers). If we pick all four even numbers and one odd number, that odd number must be 3,5,7, or9. But each of these odd numbers is prime (except 9, which is 3^2). Let's check:If the odd number is 3: Then 3 and 2,4,6,8. GCD(3,2)=1, so 3 and 2 are coprime. Similarly, GCD(3,4)=1, GCD(3,6)=3, GCD(3,8)=1. So, 3 is coprime with 2,4,8. Therefore, conclusion (1) holds.If the odd number is 5: 5 is coprime with 2,4,6,8 (since 5 is prime and doesn't divide any of them). So 5 and 2 have GCD 1, 5 and 4 GCD 1, etc. So conclusion (1) holds.If the odd number is 7: Similarly, 7 is coprime with all even numbers. So 7 and 2,4,6,8: GCD 1 with each. Therefore, conclusion (1) holds.If the odd number is 9: 9 is 3^2. GCD(9,2)=1, GCD(9,4)=1, GCD(9,6)=3, GCD(9,8)=1. So 9 and 6 have GCD 3, but 9 is coprime with 2,4,8. So, 9 and 2 are coprime. Therefore, conclusion (1) holds.Alternatively, if in case 2, instead of picking four even numbers and one odd, we pick three even numbers and two odd numbers. Then, among the two odd numbers, they could be 3 and 9, which have GCD 3, but others like 3 and 5 are coprime, 5 and 7 are coprime, etc. So, if we have two odd numbers, they might or might not be coprime. For example, 3 and 9 are not coprime, but 3 and 5 are coprime. However, even if the two odd numbers are coprime, the question is whether there must exist at least one coprime pair. If even among the five numbers, there is at least one coprime pair, then conclusion (1) holds.Wait, in this case, suppose we have three even numbers and two odd numbers. The three even numbers: since they are all even, they all share a common divisor of 2. So any two even numbers have GCD at least 2. The two odd numbers: if they are coprime or not. For example, if we pick 3 and 5, they are coprime. If we pick 3 and 9, they are not. But even if the two odd numbers are not coprime, say 9 and 3, then we still have coprime pairs among the even and odd numbers. For example, 3 and 2 (if 2 is included in the even numbers). But wait, if the even numbers are 4,6,8, and the odd numbers are 3 and 9, then 3 and 4: GCD 1, so coprime. Similarly, 9 and 4: GCD 1. Therefore, even if the two odd numbers share a common factor, they can still form coprime pairs with the even numbers. Therefore, in such a selection, there will be coprime pairs.Therefore, regardless of how we pick five numbers from 2-9 (without 1), we must have at least one coprime pair. Therefore, conclusion (1) is always true.So conclusion (1) is correct. Therefore, conclusion (1) must hold.Now moving on to conclusion (2): "There must be one number that is a multiple of another." So, in any five-number selection from 1-9, there must be at least one pair where one number is a multiple of the other.To check this, perhaps we can try to find a selection of five numbers where no number is a multiple of another. If such a selection exists, then conclusion (2) is false. Otherwise, conclusion (2) is true.Let me try to construct such a set. The idea is to pick numbers such that none divides another. For example, prime numbers, but even primes can be divided by their multiples. Let's think.First, note that 1 is a divisor of every number, so if 1 is included, then 1 divides all other numbers. Therefore, any set containing 1 will automatically have numbers (all others) that are multiples of 1. Therefore, in such a case, conclusion (2) holds. Therefore, if we include 1 in our five-number selection, conclusion (2) is true. Hence, to check if conclusion (2) is not always true, we need to find a five-number selection without 1 where no number divides another.Let me try to find such a selection.Looking at numbers 2-9. Let's list the numbers and their multiples within 2-9:- 2: multiples in 2-9 are 4,6,8- 3: multiples in 2-9 are 6,9- 4: multiples in 2-9 are 8- 5: multiples in 2-9 are none (only 10 is next multiple, which is outside)- 6: multiples in 2-9 are none (12 is next)- 7: multiples in 2-9 are none (14 is next)- 8: multiples in 2-9 are none (16 is next)- 9: multiples in 2-9 are none (18 is next)So, to avoid multiples, we can try to pick numbers that are not multiples of each other. Let's try to pick numbers from 2,3,5,7. These are primes. However, 2 and 3 have multiples (4,6,8,9). If we pick 2,3,5,7, let's see. 2 and 3: their multiples are 4,6,8,9. If we add another number, say 5 and 7. Wait, 2,3,5,7, and another. If we pick 8, which is a multiple of 2. So, 8 is a multiple of 2. Therefore, to avoid multiples, we can't pick 8. Similarly, 9 is a multiple of 3. So, perhaps pick 2,3,5,7, and then a number that is not a multiple of any of these. But in 2-9, the remaining numbers are 4,6,8,9. 4 is multiple of 2, 6 is multiple of 2 and 3, 8 is multiple of 2, 9 is multiple of 3. So, if we need to pick a fifth number, we have to pick one of 4,6,8,9, but each is a multiple of 2 or 3. Therefore, if we have already picked 2 and 3, then 4,6,8,9 are multiples of 2 or 3, so they would form multiples with 2 or 3. Therefore, such a fifth number would be a multiple of 2 or 3. Hence, in this case, the fifth number would be a multiple of 2 or 3. Therefore, the set would include a multiple.Alternatively, what if we don't pick both 2 and 3? For instance, if we avoid 2 and 3, then their multiples (4,6,8,9) can be included without being multiples of the other numbers. Let's try.If we pick numbers from 5,7,4,6,8,9. Let's see:Suppose we pick 4,5,6,7,8. Let's check if any number is a multiple of another:- 4 and 8: 8 is a multiple of 4. So, this pair invalidates the set.Alternatively, pick 5,7,4,6,9. Check:- 4 and 8: Not here. 5,7,4,6,9. 4 and 6: 6 is not a multiple of 4 nor vice versa. 4 and 9: No. 5 and the rest: 5 is prime, no multiples. 7 and the rest: same. 6 and 9: 6 is not a multiple of 9, nor 9 a multiple of 6. However, 6 and 3: but 3 isn't in the set. Wait, 6 is in the set. If we have 6, does it have a multiple? 6's multiple would be 12, which is outside. So 6 is only divided by 2 and 3, which are not in the set (if we avoid 2 and 3). So, in this case, 4,5,6,7,9. Let's check each pair:- 4 and 5: neither is a multiple.- 4 and 6: GCD is 2, but neither is a multiple (4 and 6: 6/4=1.5, not integer).- 4 and 7: No.- 4 and 9: No.- 5 and 6: No.- 5 and 7: No.- 5 and 9: No.- 6 and 7: No.- 6 and 9: GCD 3, but neither is a multiple (9/6=1.5).- 7 and 9: No.So, in this set {4,5,6,7,9}, there are no numbers where one is a multiple of the other. Therefore, conclusion (2) is false because we have found a selection where no number is a multiple of another. Therefore, conclusion (2) is not necessarily true.Wait, but hold on. Let's confirm again. 4,5,6,7,9. None of them divides another. 4 and 6: GCD 2, but neither divides the other. 6 and 9: GCD 3, but neither divides the other. 4 and 5,7,9: no. 5 and others: no. 7 and others: no. So, yes, this set seems to satisfy the condition. Therefore, conclusion (2) is not necessarily true. So conclusion (2) is incorrect.But wait, the problem is in Chinese Jiangsu Province Junior High School Competition, 1988. Maybe there's a mistake in my reasoning.Wait, in the set {4,5,6,7,9}, 6 is 2x3. But 2 and 3 are not in the set. So, 6 is not a multiple of any other number in the set. Similarly, 4 is 2x2, but 2 is not in the set. 9 is 3x3, but 3 is not in the set. So, yes, none of the numbers in this set divide another. So this is a valid counterexample. Therefore, conclusion (2) is false. Hence, conclusion (2) is not necessarily true.Therefore, conclusion (2) is incorrect.Now, moving on to conclusion (3): "There must be one number whose double is a multiple of another." So, in any five-number selection from 1-9, there must be at least one pair where one number's double is a multiple of the other. In other words, there exist two numbers a and b in the set such that 2a is a multiple of b, or 2b is a multiple of a.Alternatively, this can be phrased as there exists a pair where one number is at least half of another number. Wait, not exactly. Let me parse the statement again."One number whose double is a multiple of another." So, there exists two numbers x and y in the set such that 2x is a multiple of y, i.e., y divides 2x. So, y | 2x. So, either y divides x, or y divides 2x but not x. Which would mean that y is a divisor of 2x, but not of x, so y must be 2 times a divisor of x. For example, if x is even, y could be x/ something. Wait, maybe examples would help.For example, if x=3 and y=6, then 2x=6, which is a multiple of y=6. So 6 divides 6. Alternatively, x=2, y=4: 2x=4, which is y. So y divides 2x. Another example: x=4, y=8: 2*4=8, so y=8 divides 8. Or x=5, y=10, but 10 is not in the set. Wait, maybe x=5, y=5: 2*5=10, which is a multiple of 5, but since y must be in the set. So, if y=5, then 5 divides 10, which is 2*5. So, if a number is in the set, then 2 times it is a multiple of itself. But this is trivial. However, the problem states "one number whose double is a multiple of another". So, the same number can be considered? For example, 5's double is 10, which is a multiple of 5. But 10 is not in the set. Wait, but if y is another number, not x. Wait, the wording is ambiguous. Let me check the original problem statement:"(3) there must be one number whose double is a multiple of another."So, does it mean that there exists two distinct numbers a and b such that 2a is a multiple of b? Or can a and b be the same number?If a and b can be the same number, then for any number a, 2a is a multiple of a. So, trivially, for every number, 2a is a multiple of a, but since the numbers are from 1 to 9, 2a may not be in the set. However, if the problem is interpreted as "there exists a number in the set such that its double is a multiple of another number in the set", then perhaps a and b need to be distinct. Otherwise, the statement would be trivially true for all numbers (since 2a is a multiple of a, but 2a might not be in the set). For example, take a=5: 2*5=10, which is not in the set, so there is no number in the set that is 10. Hence, if the problem requires that 2a is a multiple of some other number b in the set, then it's possible that such a pair exists.Wait, perhaps the problem is saying that there is a number x in the set such that 2x is a multiple of some other number y in the set. That is, there exists x and y in the set where y divides 2x.Alternatively, maybe it's phrased as "one number whose double is a multiple of another", meaning x's double is a multiple of y. So 2x is divisible by y, i.e., y | 2x. So, y divides 2x. So, this can happen if y divides x, then y divides 2x. Or if y is 2 times a divisor of x.For example, if x=3 and y=6, then 6 divides 2*3=6. So, yes. If x=2 and y=4, 4 divides 2*2=4. If x=4 and y=8, 8 divides 2*4=8. If x=5 and y=10 (but 10 not in the set). So, the valid pairs are where y is a divisor of 2x, and y is in the set.So, to disprove conclusion (3), we need to find a set of five numbers from 1-9 where for every x in the set, 2x is not a multiple of any other number y in the set (y ≠x if necessary). If such a set exists, then conclusion (3) is false.Let me try to construct such a set.First, let's consider numbers for which 2x is not in the set. For numbers x ≥5, 2x ≥10, which is outside the set. So, for x=5,6,7,8,9, 2x=10,12,14,16,18, which are not in 1-9. So, for these x, 2x is outside the set. Therefore, if y is in the set, then y divides 2x only if y divides 2x. Since 2x is outside the set, but y is in the set, so y can still divide 2x. For example, x=5, y=2: 2 divides 10. But 2 is in the set. So, if 2 is in the set, then y=2 divides 2x=10. Therefore, even if x=5, if 2 is in the set, then 2 divides 10, so 2x is a multiple of y=2. Therefore, if 2 is in the set, and 5 is in the set, then conclusion (3) holds.Therefore, to prevent conclusion (3), we need to avoid such pairs. Let me attempt to construct a set where for all x in the set, 2x is not a multiple of any y in the set (distinct y).Let's try to pick numbers such that for any x in the set, 2x is not a multiple of any other y in the set. Let's see.Start by avoiding small numbers. Let's take numbers from 5-9: 5,6,7,8,9. Let's check:For x=5, 2x=10. 10 is not in the set, but does any y in the set divide 10? The divisors of 10 in 1-9 are 1,2,5,10. Since 5 is in the set, y=5 divides 10. Therefore, 5 is in the set, so 5 divides 2*5=10. Therefore, there exists a number (5) whose double (10) is a multiple of another number (5) in the set. But since 10 is not in the set, but 5 divides 10. Wait, does the problem require that the multiple is also in the set? The problem says "one number whose double is a multiple of another". The double (2x) doesn't need to be in the set, but the multiple is in the sense that 2x is a multiple of y. Since y is in the set, even if 2x is not, as long as y divides 2x.But in this case, 2x=10 is a multiple of y=5, but 10 is not in the set. So, does this count? The problem says "one number whose double is a multiple of another". If another number is in the set, then even if the double is outside, but as long as y divides 2x, then yes. So, in this case, 5 is a number whose double is 10, which is a multiple of 5 (another number in the set). So, this would satisfy the condition. Therefore, even in the set {5,6,7,8,9}, conclusion (3) holds because 2*5=10 is a multiple of 5. Wait, but 5 is the same number. If the problem allows the same number, then every number x satisfies that 2x is a multiple of x. But since the double is 2x, which is a multiple of x. However, the problem says "another", which might imply a different number. The original Chinese wording might clarify, but since we don't have that, we have to assume based on the translation.If "another" means a different number, then in the set {5,6,7,8,9}, 2*5=10 is a multiple of 5, but since 10 is not in the set, there's no other number y in the set such that y divides 10 except 5 itself. So, if "another" means a different number, then this does not satisfy the condition. But if "another" can be the same number, then it does. Given that 2x is a multiple of x, but the problem says "another", which usually implies a different one. So, perhaps in this case, the answer would be that even in {5,6,7,8,9}, there is no number whose double is a multiple of another distinct number in the set. Let's check:Take each number:- 5: 2*5=10. Numbers in the set that divide 10 are 5. So, only itself.- 6: 2*6=12. Numbers in the set that divide 12 are 6 (12/6=2). So, only itself.- 7: 2*7=14. Divisors in the set: 7. Only itself.- 8: 2*8=16. Divisors in the set: 8. Only itself.- 9: 2*9=18. Divisors in the set: 9. Only itself.Therefore, if "another" means a different number, then in this set {5,6,7,8,9}, there is no number whose double is a multiple of another distinct number. Therefore, this would be a counterexample, making conclusion (3) false.But is {5,6,7,8,9} a valid set? Let's confirm. These are five numbers from 1-9, none is 1, but they are all from 5-9. So yes, it's a valid selection. Therefore, if we can have a set where for each number x, 2x is only divisible by x itself (not present in the set as a different number), then conclusion (3) would be false.However, let's verify again. For example, take x=6: 2*6=12. Are there any other numbers in {5,6,7,8,9} that divide 12? 12's divisors are 1,2,3,4,6,12. In the set, only 6 divides 12. So, only 6 itself. Similarly, x=8: 2*8=16. Divisors are 1,2,4,8,16. In the set, only 8. So, same. Therefore, in this set, there's no pair where one number's double is a multiple of another distinct number. Therefore, conclusion (3) is not necessarily true.But wait, what if the set includes smaller numbers? For example, if we include 3 and 6. Then 2*3=6, which is in the set. So, 6 is a multiple of 6, but 6 is the same number. If we need another number, then 3's double is 6, which is in the set, so 6 is a multiple of 3. Therefore, if we have 3 and 6 in the set, then 3's double is 6, which is a multiple of 3. Wait, no: 6 is a multiple of 3, but the double of 3 is 6, which is a multiple of 3. Wait, 6 is in the set, so 3's double is 6, which is a multiple of 3. Therefore, this satisfies the condition. Hence, in this case, conclusion (3) holds.But if we can create a set where no such pairs exist, then conclusion (3) is false. The previous example {5,6,7,8,9} seems to be such a set. Therefore, conclusion (3) is not necessarily true.Wait, but perhaps there is something I missed. Let's check again. Take the set {5,6,7,8,9}:- For 5: 2*5=10. 10 is not in the set, and none of the other numbers divide 10 (5 is in the set, but that's the same number).- For 6: 2*6=12. None of the numbers except 6 divide 12 (but 6 is the same number).- For 7: 2*7=14. None.- For 8: 2*8=16. None.- For 9: 2*9=18. None.Therefore, in this set, there is no number whose double is a multiple of another distinct number in the set. Therefore, conclusion (3) is not necessarily true.However, wait a minute. If we include 1 in the set, then 1's double is 2. If 2 is in the set, then 2 is a multiple of 1. But 1's double is 2. So, if 1 and 2 are in the set, then 1's double is 2, which is a multiple of 1. But since 1 is in the set, 2 is a multiple of 1. However, if 1 is not in the set, then this doesn't apply.But in the set {5,6,7,8,9}, 1 is not included. So, it's a valid counterexample.Therefore, conclusion (3) is not necessarily true.But let's try another example. Suppose we pick {3,4,5,7,8}. Let's check:- 3: 2*3=6. Not in the set. Divisors of 6 in the set: 3 (only if 6 were present). So, no.- 4: 2*4=8. 8 is in the set. So, 8 is a multiple of 4. Therefore, 4's double is 8, which is in the set. So, here, 8 is a multiple of 4. Therefore, in this set, conclusion (3) holds.Therefore, this set does have such a pair. Therefore, conclusion (3) can be true, but the question is whether it must be true. Since we found a set where it's not true, {5,6,7,8,9}, then conclusion (3) is not necessarily true.But wait, in the set {5,6,7,8,9}, let's check again:- 5: double is 10, not in the set; 10's divisors in the set are 5 (same number).- 6: double is 12; divisors in the set: 6.- 7: 14; 7.- 8: 16; 8.- 9: 18; 9.Therefore, no two distinct numbers where one's double is a multiple of the other. Therefore, conclusion (3) is false here. So, conclusion (3) is not necessarily true.But wait, is there another interpretation? The problem says "one number whose double is a multiple of another". It could mean that the double is a multiple of another number, regardless of whether the double is in the set. For example, if a number x is in the set, and another number y is in the set such that y divides 2x. In the set {5,6,7,8,9}:- For x=5, 2x=10. Check if any y in the set divides 10. The divisors of 10 are 1,2,5,10. In the set, only 5. So y=5. So, 5 divides 10, which is 2*5. But y is the same as x. If the problem allows y=x, then this would satisfy. But if "another" is meant to be a different number, then it doesn't.If the problem allows y=x, then every number trivially satisfies this because 2x is a multiple of x. However, in this case, the statement would be trivially true for all sets, which contradicts our previous analysis. But since the problem is from a competition, it's more likely that "another" refers to a different number. Otherwise, conclusion (3) would always be true, which conflicts with the existence of option (A)-(D) where it's possible that only some conclusions are correct.Therefore, assuming that "another" means a different number, then conclusion (3) is not necessarily true, as shown by the counterexample {5,6,7,8,9}. Therefore, conclusion (3) is false.But wait, let me confirm this with another example. Suppose we pick the set {3,5,7,8,9}. Let's check:- 3: 2*3=6. Divisors in the set: none.- 5: 2*5=10. Divisors: 5.- 7: 14. Divisors:7.- 8: 16. Divisors:8.- 9:18. Divisors:9.Again, no two distinct numbers where one's double is a multiple of the other. So this also serves as a counterexample.Alternatively, another example: {5,7,8,9,4}. 4's double is 8, which is in the set. So, 8 is a multiple of 4. Therefore, here, conclusion (3) holds. But this is a different set. Therefore, depending on the set, conclusion (3) can be true or false. Therefore, it's not necessarily always true.Therefore, conclusion (3) is not necessarily correct.But wait, another angle: perhaps in any five-number selection, you cannot avoid having at least one pair where one number's double is a multiple of another. Maybe my previous counterexample is invalid.Wait, let's re-examine {5,6,7,8,9}. For each number:- 5: 2*5=10. Is there a y in the set (other than 5) that divides 10? 10's divisors are 1,2,5. None of these except 5 are in the set. So, no.- 6: 2*6=12. Divisors are 1,2,3,4,6. In the set, only 6. So, no.- 7: 2*7=14. Divisors:1,2,7. In the set, only 7.- 8: 2*8=16. Divisors:1,2,4,8. In the set, only 8.- 9: 2*9=18. Divisors:1,2,3,6,9. In the set, 6 and 9. Wait, 6 is in the set. So, 18 is divisible by 6 and 9. So, 2*9=18. 18 is divisible by 6 and 9. But 6 and 9 are in the set. Therefore, for x=9, 2x=18. 18 is divisible by y=6 and y=9. Therefore, y=6 is in the set, and 6 divides 18. Therefore, in the set {5,6,7,8,9}, for x=9, 2x=18 is divisible by y=6. Therefore, 6 divides 18, which is 2*9. Therefore, in this set, there exists a number (9) whose double (18) is a multiple of another number (6) in the set. Therefore, conclusion (3) actually holds here.Wait, this contradicts my previous analysis. So, I must have made a mistake earlier.Let me check this again. In the set {5,6,7,8,9}:- Take x=9: 2*9=18. Now, 18 is divisible by 6 (since 18/6=3). 6 is in the set. Therefore, y=6 divides 18=2*9. Therefore, in this set, there exists a pair (9 and 6) where 2*9 is a multiple of 6. Therefore, conclusion (3) holds.Ah, so my previous assertion that this set is a counterexample is incorrect. I missed that 2*9=18 is a multiple of 6, which is in the set. Therefore, even in this set, conclusion (3) holds.Therefore, perhaps conclusion (3) is actually always true. Let's verify this.So, in any five-number selection, is there necessarily a pair where one number's double is a multiple of another number in the set?Let me try to see if it's possible to construct a set where this is not the case. Let's attempt to pick five numbers where for every x in the set, 2x is not a multiple of any other number in the set.Suppose we try to pick numbers such that no number is half of another number or such that no number is double another number. But given that the numbers are from 1-9, let's see.First, list all pairs where one number is double another:- 1 and 2- 2 and 4- 3 and 6- 4 and 8These are the pairs where one is double the other. Additionally, for the condition of 2x being a multiple of y, even if y is not exactly 2x, but a divisor of 2x.For example, if x=5 and y=10 (not in set), or y=2,5. So, if 2 is in the set, then y=2 divides 2x=10. Similarly, x=3, y=6: 6 divides 6=2*3. So, avoiding such pairs requires not having both x and y where y divides 2x.Let me try to construct a set avoiding such pairs.Let's try to pick numbers from the higher end: {5,7,8,9}. Let's add another number. If we add 6, then 6 and 3 (but 3 is not in the set). However, 2*9=18, which is divisible by 6. So, if 6 and 9 are in the set, conclusion (3) holds. Similarly, if we have 8, 2*4=8, but 4 might not be in the set. Let's try {5,7,8,9,6}. Then, as before, 2*9=18, divisible by 6. So, this set still has the pair 9 and 6.If we remove 6 and add another number, say 5,7,8,9,4. Then, 2*4=8, which is in the set. So, 8 is a multiple of 4. So, this also satisfies conclusion (3).If we try 5,7,8,9,3. Then, 2*3=6, which is not in the set, but 6's divisors in the set are 3 (since 6=2*3). But 2 is not in the set. Wait, 3 is in the set, so 3 divides 6=2*3. Therefore, 3's double is 6, which is a multiple of 3. But since 6 is not in the set, does this count? Wait, the wording is "one number whose double is a multiple of another". So, if the double is not in the set, but another number divides the double. For example, x=3, double is 6, which is not in the set, but y=3 divides 6. But y is the same as x. If "another" means a different number, then this doesn't count. However, if y is allowed to be the same as x, then this would count, but that's trivial.Assuming "another" means a different number, then in the set {3,5,7,8,9}:- x=3: 2*3=6. Divisors are 3 and 2. In the set, only 3. So, no other number.- x=5: 2*5=10. Divisors 5,2. In the set, only 5.- x=7: 2*7=14. Divisors 7,2. In the set, only 7.- x=8: 2*8=16. Divisors 8,4,2. In the set, only 8.- x=9: 2*9=18. Divisors 9,6,3,2. In the set, 9 and 3. So, 3 is in the set. 3 divides 18. Therefore, 2*9=18 is a multiple of 3. Therefore, here, 9 and 3 are in the set. Therefore, conclusion (3) holds.Therefore, even in this set, conclusion (3) holds because 2*9=18 is a multiple of 3.Wait, so even if we try to exclude numbers that form such pairs, if we have 3 and 9 in the set, then 2*9=18 is a multiple of 3. Therefore, conclusion (3) holds.Therefore, maybe it's impossible to create a set of five numbers without such a pair.Let me try another approach. Let's attempt to use the pigeonhole principle. The numbers from 1-9 can be grouped in a way that if we pick five numbers, two of them must fall into a certain relationship.Alternatively, think of numbers and their doubles. But since the doubles might be outside the set, this might not help.Alternatively, consider the prime factors. For each number x, if we can associate it with a unique prime or something. Not sure.Alternatively, consider the following: For each number x in the set, if there exists another number y in the set such that y divides 2x. To prevent this, for every x in the set, none of the other numbers y (excluding x) divide 2x.Let me try to construct such a set.Start with 5. 2*5=10. To prevent any y from dividing 10, we need to exclude numbers that divide 10. The divisors of 10 in 1-9 are 1,2,5,10. Excluding 1 and 10, we need to exclude 2 and 5. But 5 is already in the set. So, if we have 5, we need to exclude 2.Next, pick 7. 2*7=14. Divisors are 1,2,7,14. So in the set, we need to exclude 2 and 7. Since 7 is in the set, we need to exclude 2.Next, pick 9. 2*9=18. Divisors are 1,2,3,6,9,18. So, in the set, need to exclude 2,3,6. Since we have 9, we need to exclude 2,3,6.Next, pick 8. 2*8=16. Divisors are 1,2,4,8,16. So, exclude 2,4. But 4 is not in the set yet. So, need to exclude 2.Next, pick 5,7,9,8, and another number. The remaining numbers are 1,2,3,4,6. But we have to exclude 2,3,4,6. So, only 1 is left. But 1 is coprime with all, but if we include 1, then 1 and any other number, but we are focusing on conclusion (3). If we include 1, then 2*1=2. If 2 is excluded, but 1's double is 2, which is not in the set. However, y=1 divides 2*1=2, but since 2 is not in the set, does this count? If y=1 is in the set, then 1 divides 2*1=2, but 2 is not in the set. So, is there a y in the set that divides 2x? In this case, x=1, 2x=2. y=1 divides 2, but 1 is in the set, so 1 divides 2. However, the problem states "another number", which might mean a different number. So, 1 divides 2*1=2, but since y=1 is the same as x=1, if "another" is required, then this doesn't count.Therefore, including 1 in the set {1,5,7,8,9}:- x=1: 2*1=2. No other number y divides 2 (since 2 is not in the set).- x=5: 10. No y in the set (except 5).- x=7:14. No y except 7.- x=8:16. No y except 8.- x=9:18. Divisors in the set: 9 and 1. 1 is in the set, but 1 divides everything. However, if "another" means a different number, then 1 and 9: 1 divides 18. So, 9's double is 18, which is a multiple of 1. But 1 is a different number. Therefore, in this case, there is a pair where 9's double is a multiple of 1. Since 1 is in the set, this satisfies the condition. Therefore, conclusion (3) holds here.Therefore, even if we include 1, we can't avoid having a pair where one's double is a multiple of another (different) number. Because 1 divides any number, so if 1 is in the set, then for any x in the set, 1 divides 2x. Therefore, if 1 is in the set, conclusion (3) holds. If we don't include 1, then in the set {5,7,8,9, another number}, we have to include a number that might create the necessary pair. For example, if the fifth number is 3, then 2*9=18 is divisible by 3. If it's 6, then 2*9=18 divisible by 6. If it's 4, then 2*8=16 divisible by 4. If it's 2, then 2 divides many doubles. Wait, let's try to construct a set without 1 and without such pairs.Try {5,7,9,8,4}:- 4: 2*4=8, which is in the set. So, 8 is a multiple of 4. Therefore, conclusion (3) holds.Another try: {5,7,9,8,3}:- 3: 2*3=6. Not in the set. But 2*9=18, which is divisible by 3 (which is in the set). So, conclusion (3) holds.Another try: {5,7,9,8,6}:- 6: 2*6=12. Not in the set. But 2*9=18 divisible by 6. So, conclusion (3) holds.Another try: {5,7,9,8,2}:- 2: 2*2=4. Not in the set. But 2*5=10 divisible by 2. So, conclusion (3) holds.Another try: {5,7,9,8,1}:- 1: 2*1=2. Not in the set. But 1 divides all doubles. Therefore, conclusion (3) holds.So, it seems that no matter how we pick five numbers, we can't avoid having at least one pair where one number's double is a multiple of another number. This is because:- If 1 is in the set, then 1 divides every double. Therefore, for any x, 2x is a multiple of 1, which is in the set. So, conclusion (3) holds.- If 1 is not in the set, then we have five numbers from 2-9. Let's consider the numbers 2-9. If the set contains an even number, say 2, then 2 divides many doubles. For example, 2 divides 2x for any x. But if x=3, 2x=6, which is divisible by 2. But if 2 is not in the set, but another even number is present. For example, 4: 2*4=8, which is divisible by 4. If 4 is in the set, then 8 (if included) would be divisible by 4. Alternatively, if the set contains a number x and another number y where y divides 2x. For example, in a set with 5 and 2, 2 divides 2*5=10. If 2 is not in the set, but say 6 is in the set, then 6 divides 2*3=6, but 3 may not be in the set.Wait, this is getting convoluted. Let me think differently.Perhaps using the pigeonhole principle. Let's categorize numbers based on their prime factors and see if selecting five numbers forces two numbers into a certain relationship.Alternatively, consider that in the set of numbers from 2-9, excluding 1, we can pair numbers in a way that one's double is a multiple of the other. For example, 2 and 4, 2 and 6, 2 and 8, 3 and 6, 3 and 9, 4 and 8. So, there are several such pairs.If we pick five numbers, perhaps we can't avoid including at least one such pair. For example, if we try to pick numbers not in these pairs. But the numbers not involved in such pairs are 5 and 7. So, 5 and 7 are primes, and their doubles are 10 and 14, which are outside the set. So, if we pick 5,7 and three other numbers, the three other numbers would have to come from {2,3,4,6,8,9}, which includes the pairs mentioned earlier.But {2,3,4,6,8,9} has several pairs where one's double is a multiple of the other. If we pick three numbers from this set, along with 5 and 7, maybe we can avoid the pairs. But wait, we need to pick five numbers total. If we include 5 and 7, we need three more numbers. If we pick three numbers from {2,3,4,6,8,9} such that none of them are in a pair where one is double the other or such that none's double is a multiple of another. For example, pick 2,3,4. Then 2 and 4 are in a pair. So, conclusion (3) holds. Pick 2,3,6. 2 and 6: 6 is a multiple of 2's double (2*3=6). So, conclusion (3) holds. Pick 3,6,9: 3 and 6, 3 and 9. Conclusion (3) holds. Pick 4,8,9: 4 and 8. Conclusion (3) holds. Pick 2,8,9: 2 and 8: 8 is a multiple of 2's quadruple. Wait, 2*4=8. If 4 is not in the set, but 8 is. So, 8 is a multiple of 2*4, but 4 is not in the set. So, 2's double is 4, which is not in the set, but 8 is a multiple of 4. But since 4 is not in the set, does this count? 8 is a multiple of 2*4, but since 4 is not in the set, does 8 being a multiple of 2*4 matter? No, because y needs to be in the set. So, if y=8 and x=2: 2*2=4, which is not a multiple of 8. If x=8: 2*8=16, which is not in the set. So, in this case, maybe 2 and 8 don't form a pair. However, 2*8=16, which is not in the set. So, does 8 divide 2* something? For example, if x=8, 2*8=16, which is not in the set. If x=2, 2*2=4, which is not in the set. So, maybe 2 and 8 don't form a problematic pair.Wait, but if we have 2 and 8 in the set, does 8 divide 2* something in the set? 2*8=16, which is not in the set. Or, 8 divides 2*x, where x is in the set. So, 8 divides 2*x => x must be 4, 8, etc. If 4 is not in the set, then x=8. So, 8 divides 2*8=16, but 16 is not in the set. So, maybe 2 and 8 do not form a pair.Similarly, 8 and 9: 2*9=18, which is divisible by 9 and 6, but 6 is not in the set. 9 is in the set, but 18 is not.Therefore, if we pick 2,8,9 along with 5,7, then:- 2: 2*2=4 (not in the set)- 8: 2*8=16 (not in the set)- 9: 2*9=18 (not in the set)- 5: 2*5=10 (not in the set)- 7: 2*7=14 (not in the set)But, does any y in the set divide 2x for any x in the set?For x=2: 2*2=4. Divisors in the set: 2 divides 4? But 4 is not in the set. So, no.For x=8: 2*8=16. Divisors in the set: 2 and 8. 2 divides 16? Yes. 2 is in the set. So, 2 divides 16=2*8. Therefore, there exists a number (8) whose double (16) is a multiple of y=2. Therefore, in this set {2,5,7,8,9}, conclusion (3) holds because 2*8=16 is divisible by 2. Hence, y=2 divides 16=2*8. Therefore, even in this set, conclusion (3) holds.Therefore, no matter how we pick five numbers from 1-9, there will always be at least one pair where one number's double is a multiple of another number in the set. This is because:- If 1 is in the set, then 1 divides any double.- If 1 is not in the set, but 2 is in the set, then 2 divides the double of any even number.- If neither 1 nor 2 is in the set, then consider higher numbers. For example, if 3 is in the set, then 3 divides the double of 3 (which is 6) or other numbers. If 4 is in the set, then 4 divides 8, etc.But when I tried the set {5,6,7,8,9}, it seemed like there was a pair (9 and 6). Therefore, conclusion (3) is actually always true. My previous assumption that {5,6,7,8,9} was a counterexample was incorrect because I overlooked the pair (9,6).Therefore, conclusion (3) must hold. Hence, conclusion (3) is correct.Wait, but then why did the set {5,6,7,8,9} have the pair 9 and 6? Because 6 divides 2*9=18. Yes. Therefore, in any five-number selection, if we have a number x and another number y where y divides 2x, then conclusion (3) holds. To verify this, even if we pick numbers that are all odd and greater than 1, say {3,5,7,9, another odd}. Let's say {3,5,7,9,11} but 11 is not in the original set. Within the original set 1-9, adding another odd number would require picking from 3,5,7,9. But since we can only pick five numbers, the fifth number would have to be another odd number or an even number. If we pick four odd numbers and one even number:For example, {3,5,7,9,2}. Then, 2*9=18, which is divisible by 2 and 3. Since 2 is in the set, conclusion (3) holds. If we pick {3,5,7,9,4}, then 2*4=8, which is divisible by 4. Therefore, conclusion (3) holds.If we try to pick five odd numbers, we have to include 1, since there are only four odd numbers from 3-9. So {1,3,5,7,9}. Here, 1 divides any double, so conclusion (3) holds.Therefore, no matter how we pick five numbers, we can't avoid having at least one pair where one number's double is a multiple of another number. Therefore, conclusion (3) is always true.But earlier, I thought that {5,6,7,8,9} had such a pair, which it does (6 divides 2*9=18). Therefore, conclusion (3) is correct.This contradicts my previous analysis where I thought conclusion (3) was not necessarily true, but upon re-examination, it turns out that conclusion (3) must hold.Therefore, summarizing:- Conclusion (1): Must hold.- Conclusion (2): Does not necessarily hold (counterexample {4,5,6,7,9}).- Conclusion (3): Must hold.Therefore, the number of correct conclusions is 2 (conclusions 1 and 3). So the answer is (C).Final Answerboxed{C}</think>To determine the number of correct conclusions among the three given, we analyze each conclusion step by step:1. Conclusion (1): There must be two numbers that are coprime. - We need to check if any five numbers chosen from 1 to 9 must include at least one coprime pair. - If 1 is included, it is coprime with all other numbers. If 1 is excluded, we need to check if five numbers from 2 to 9 can be chosen without any coprime pairs. However, any selection of five numbers from 2 to 9 will always include coprime pairs. For example, picking four even numbers and one odd number will result in the odd number being coprime with some even numbers. Therefore, conclusion (1) is correct.2. Conclusion (2): There must be one number that is a multiple of another. - We need to check if any five numbers chosen can avoid having a multiple pair. - A counterexample is the set {4, 5, 6, 7, 9}, where no number is a multiple of another. Therefore, conclusion (2) is incorrect.3. Conclusion (3): There must be one number whose double is a multiple of another. - We need to check if any five numbers chosen must include a pair where one number's double is a multiple of another. - For any five numbers, there will always be such a pair. For example, in the set {5, 6, 7, 8, 9}, 2*9 = 18 is a multiple of 6. This holds for any selection of five numbers, making conclusion (3) correct.Thus, conclusions (1) and (3) are correct, making the number of correct conclusions 2.[boxed{C}]]

🔑:[Solution] Since five numbers are chosen from the nine consecutive natural numbers 1 sim 9, there must be two positive integers that are adjacent, otherwise it would include 1. Adjacent positive integers are always coprime, and 1 is coprime with any natural number, so (1) holds.Choosing 5,6,7,8,9 can negate (2).Divide these nine natural numbers into three groups: (1,2,4,8), (3,6,9), and (5,7).In the first two groups, any two numbers chosen from each group will always have one number that is a multiple of the other's double.Thus, if all five numbers are taken from the first two groups, conclusion (3) clearly holds;and if four or three of the five numbers are taken from the first two groups, at least two numbers will be taken from one of the groups, so conclusion (3) holds.Therefore, the answer is (C).

🧠:[<think>Okay, so I need to find the derivative of two functions. Let's start with the first one: y equals (x + 3) divided by (x squared plus 3) at the point where x is 3. Alright, derivatives. Hmm, since this is a rational function, I think I should use the quotient rule here. The quotient rule is for when you have a function divided by another function, right? The formula is (low d high minus high d low) over (low squared). Let me write that down to remember: if y = u/v, then y' = (v*u' - u*v') / v². So in this case, u is the numerator, which is x + 3, and v is the denominator, x² + 3. First, I need to find the derivatives of u and v. Let's compute u'. The derivative of x + 3 with respect to x is 1 + 0, which is 1. Then v', the derivative of x² + 3 is 2x + 0, so 2x. Now plug these into the quotient rule formula. So y' would be [v*u' - u*v'] divided by v squared. Substituting, that's [(x² + 3)(1) - (x + 3)(2x)] all over (x² + 3)². Let me simplify the numerator step by step. First, expand (x² + 3)(1): that's just x² + 3. Then subtract (x + 3)(2x). Let's compute that part. Multiply x by 2x to get 2x², and 3 times 2x is 6x. So (x + 3)(2x) is 2x² + 6x. Therefore, the numerator becomes (x² + 3) - (2x² + 6x). Let's subtract those terms: x² + 3 - 2x² - 6x. Combining like terms, x² - 2x² is -x², and then -6x + 3. So the numerator simplifies to -x² - 6x + 3. Therefore, the derivative y' is (-x² - 6x + 3) divided by (x² + 3) squared. Now, the question asks for the derivative at x = 3. So I need to plug in x = 3 into this derivative. Let me compute the numerator and denominator separately. First, numerator at x=3: -(3)² -6*(3) +3. That's -9 -18 +3. Let's compute that: -9 -18 is -27, plus 3 is -24. Denominator at x=3: (3² +3)². Compute inside first: 9 +3 is 12, so 12 squared is 144. Therefore, the derivative at x=3 is -24 / 144. Simplify that fraction. Both numerator and denominator are divisible by 24. -24 divided by 24 is -1, and 144 divided by 24 is 6. So it simplifies to -1/6. Wait, is that right? Let me double-check my steps. First, the quotient rule: correct. Then derivatives of u and v: u' is 1, v' is 2x, that's correct. Then substituting into the formula: numerator (x² +3)*1 - (x +3)*2x. Yes. Then expanding: x² +3 -2x² -6x, which simplifies to -x² -6x +3. Correct. Then substituting x=3: numerator is -9 -18 +3 = -24. Denominator is (9 +3)^2 = 144. Then -24/144 = -1/6. Yup, that seems right. Okay, so part (1) answer is -1/6. Now moving on to part (2): Find the derivative of the function y = (x + 1/x)^5. Alright, this is a composite function, so I need to use the chain rule here. The chain rule states that the derivative of a composite function is the derivative of the outer function evaluated at the inner function multiplied by the derivative of the inner function. So, the outer function is something raised to the 5th power, and the inner function is x + 1/x. Let me denote u = x + 1/x, so y = u^5. Then dy/dx = 5u^4 * du/dx. First, compute du/dx. The derivative of x is 1, and the derivative of 1/x is -1/x². Therefore, du/dx = 1 - 1/x². So putting it together, dy/dx = 5*(x + 1/x)^4*(1 - 1/x²). Let me check if that's correct. Yes, chain rule applied correctly. The outer function derivative is 5u^4, then times the derivative of the inner function u, which is 1 - 1/x². So that should be the derivative. Alternatively, maybe I can expand the function, but that would be more complicated because (x + 1/x)^5 would involve binomial expansion, which is tedious. So using the chain rule is definitely the way to go here. Therefore, the derivative is 5*(x + 1/x)^4*(1 - 1/x²). I can leave it like that, but maybe simplify the last term. Let's see: 1 - 1/x² can be written as (x² -1)/x². So then the derivative would be 5*(x + 1/x)^4*(x² -1)/x². But unless the problem asks for a simplified form, both are correct. The problem just says "find the derivative," so either form is acceptable. However, perhaps writing it as 5*(x +1/x)^4*(1 -1/x²) is simpler. Let me check if I made any mistakes. Let's do a quick verification. Let’s pick a simple value for x and compute both the original function and the derivative. Let's take x =1. Then y = (1 +1/1)^5 =2^5=32. The derivative at x=1: 5*(2)^4*(1 -1/1)=5*16*0=0. Alternatively, compute numerically. If x=1, derivative is 0. Let's check via another method. Maybe compute the function at x=1.01 and x=0.99 to approximate the derivative. Wait, x=1. Let's compute y at x=1.01: (1.01 + 1/1.01)^5. 1.01 + ~0.9901 is approx 2.0001. Then raised to 5th power is approx 32.008. Then at x=0.99: (0.99 +1/0.99)^5. 1/0.99 is approx 1.0101, so 0.99 +1.0101=2.0001, same as before. So y(1.01) ≈32.008, y(0.99)≈32.008. The difference is about 0.016 over 0.02, so slope approx 0.016/0.02=0.8, but according to our derivative, it's 0. Wait, that's conflicting. Hmm. Wait, but that can't be. Wait, the derivative at x=1 is 0? Let me check again. Original derivative: 5*(x +1/x)^4*(1 -1/x²). At x=1, that's 5*(2)^4*(1 -1)=5*16*0=0. So derivative is 0. But when I computed y at x=1.01 and x=0.99, both give approximately 32.008. So the change is symmetric, but if the function has a local maximum or minimum at x=1, the derivative would be zero. Let's see. Let's check x=2. x=2, y=(2 +0.5)^5=2.5^5=97.65625. The derivative there would be 5*(2.5)^4*(1 -1/4)=5*(39.0625)*(0.75)=5*29.296875=146.484375. So that's a positive derivative. At x=0.5, y=(0.5 +2)^5=2.5^5 same as above. The derivative would be 5*(2.5)^4*(1 -1/(0.25))=5*(39.0625)*(1 -4)=5*39.0625*(-3)= -585.9375. So that's a negative derivative. So at x=1, the derivative is zero, which makes sense if the function has a critical point there. Let's plot the function mentally: when x approaches zero from the right, 1/x becomes large, so y becomes large. When x approaches infinity, x dominates, so y behaves like x^5. At x=1, the function is 32, and the derivative is zero, so maybe a local minimum or maximum. Wait, but when I checked x=1.01 and x=0.99, both give slightly higher than 32? Wait, 1.01 +1/1.01 is 1.01 + ~0.990099= ~2.000099. So raised to the 5th power is (2 +0.000099)^5≈32 +5*2^4*0.000099=32 +5*16*0.000099=32 +0.00792≈32.00792. Similarly, x=0.99 gives similar value. So actually, the function is symmetric around x=1? Wait, x +1/x. If we replace x with 1/x, then 1/x +x, so the function is symmetric in that sense. So x=1 is the minimum point? Because when x is 1, x +1/x is 2, which is the minimum value of x +1/x for x>0. The expression x +1/x has its minimum at x=1, where it's 2. So the function (x +1/x)^5 would have its minimum at x=1, hence the derivative is zero there, which matches our calculation. So the derivative at x=1 is indeed zero, and the small increment around x=1 gives almost the same y, hence the approximate derivative is zero, which is consistent. So our derivative formula seems correct. Therefore, I think the derivative is correctly found as 5*(x +1/x)^4*(1 -1/x²). So that should be the answer for part (2).Final Answer(1) boxed{-dfrac{1}{6}}(2) boxed{5left(x + dfrac{1}{x}right)^4 left(1 - dfrac{1}{x^2}right)}</think>(1) To find the derivative of ( y = frac{x+3}{x^2+3} ) at the point where ( x = 3 ), we use the quotient rule. The quotient rule states that if ( y = frac{u}{v} ), then ( y' = frac{v cdot u' - u cdot v'}{v^2} ). Here, ( u = x + 3 ) and ( v = x^2 + 3 ). First, compute the derivatives of ( u ) and ( v ):- ( u' = 1 )- ( v' = 2x )Applying the quotient rule:[y' = frac{(x^2 + 3)(1) - (x + 3)(2x)}{(x^2 + 3)^2}]Simplifying the numerator:[(x^2 + 3) - (2x^2 + 6x) = x^2 + 3 - 2x^2 - 6x = -x^2 - 6x + 3]Thus, the derivative is:[y' = frac{-x^2 - 6x + 3}{(x^2 + 3)^2}]Evaluating at ( x = 3 ):- Numerator: ( -(3)^2 - 6(3) + 3 = -9 - 18 + 3 = -24 )- Denominator: ( (3^2 + 3)^2 = (9 + 3)^2 = 12^2 = 144 )Therefore, the derivative at ( x = 3 ) is:[frac{-24}{144} = -frac{1}{6}](2) To find the derivative of ( y = left(x + frac{1}{x}right)^5 ), we use the chain rule. Let ( u = x + frac{1}{x} ), then ( y = u^5 ).First, compute the derivative of ( u ):[frac{du}{dx} = 1 - frac{1}{x^2}]Applying the chain rule:[frac{dy}{dx} = 5u^4 cdot frac{du}{dx} = 5left(x + frac{1}{x}right)^4 left(1 - frac{1}{x^2}right)] Final Answer(1) boxed{-dfrac{1}{6}}(2) boxed{5left(x + dfrac{1}{x}right)^4 left(1 - dfrac{1}{x^2}right)},<think>Okay, so I have two calculus problems to solve here. Let me start with the first one: finding the derivative of y = (x + 3)/(x² + 3) at the point where x = 3. Hmm, derivatives. Right, this is a quotient, so I should use the quotient rule. The quotient rule is if you have a function f(x)/g(x), the derivative is (f’(x)g(x) - f(x)g’(x))/[g(x)]². Let me recall that properly. Yeah, numerator is derivative of the top times the bottom minus the top times derivative of the bottom, all over the bottom squared. So let's identify f(x) and g(x) here. The numerator is x + 3, so f(x) = x + 3. The denominator is x² + 3, so g(x) = x² + 3. Then f’(x) would be the derivative of x + 3, which is 1 + 0 = 1. And g’(x) is the derivative of x² + 3, which is 2x + 0 = 2x. So applying the quotient rule: y’ = [f’(x)g(x) - f(x)g’(x)] / [g(x)]² = [1*(x² + 3) - (x + 3)*(2x)] / (x² + 3)². Let me simplify that numerator.Expanding the terms: 1*(x² + 3) is x² + 3. Then (x + 3)*(2x) is 2x² + 6x. So subtracting those: (x² + 3) - (2x² + 6x) = x² + 3 - 2x² -6x = -x² -6x +3. So the derivative simplifies to (-x² -6x +3)/(x² +3)^2. Now, I need to evaluate this at x = 3. Let's plug in 3 for x. First, compute the numerator: - (3)^2 -6*(3) +3. That's -9 -18 +3 = -24. Then the denominator is (3² +3)^2. 3² is 9, so 9 +3 =12. Then 12 squared is 144. So putting it together, the derivative at x =3 is -24/144. Simplifying that, divide numerator and denominator by 24: -1/6. So the answer for part (1) is -1/6.Wait, let me check my steps again. When expanding the numerator: f’(x)g(x) is (1)(x² +3) = x² +3. Then f(x)g’(x) is (x +3)(2x) = 2x² +6x. So subtracting: x² +3 -2x² -6x = -x² -6x +3. That seems right. Then plugging in x=3: numerator is -9 -18 +3 = -24. Denominator is (9 +3)^2 = 12² =144. -24/144 = -1/6. Yep, that looks correct.Now moving on to the second problem: find the derivative of y = (x + 1/x)^5. Okay, so this is a composite function. The outer function is something raised to the fifth power, and the inner function is x + 1/x. So I should use the chain rule here. The chain rule states that if y = f(g(x)), then y’ = f’(g(x)) * g’(x). So here, f(u) = u^5 and u = x + 1/x. Therefore, the derivative would be 5*(x +1/x)^4 multiplied by the derivative of (x +1/x).So first, compute the derivative of the inner function u = x +1/x. The derivative of x is 1, and the derivative of 1/x is -1/x². So u’ = 1 -1/x². Therefore, combining these, the derivative of y is 5*(x +1/x)^4*(1 -1/x²). Let me check that. Yes, that seems straightforward. Let me verify if I can simplify that further. Maybe expand the terms or something, but the problem just asks for the derivative, so probably that is sufficient. Let me write it again: y’ = 5*(x + 1/x)^4*(1 - 1/x²). Alternatively, you can factor the terms in the derivative. Let's see:1 -1/x² can be written as (x² -1)/x². So then, y’ = 5*(x +1/x)^4*(x² -1)/x². But unless the problem requires a specific form, either expression is correct. Since the question just asks for the derivative, both forms are acceptable, but the first one is perhaps simpler. Let me check if my calculation is correct.Start again: y = (x +1/x)^5. Let u =x +1/x, so dy/dx = 5u^4 * du/dx. Then du/dx is derivative of x is 1, derivative of 1/x is -1/x². So yes, 1 -1/x². So dy/dx =5*(x +1/x)^4*(1 -1/x²). That seems correct. So unless there's a simplification required, that's the answer. Alternatively, if I expand (x +1/x)^4, but that would be tedious. Since the problem doesn't specify, leaving it in factored form is better. So the answer is 5*(x +1/x)^4*(1 -1/x²). Let me check if that's the same as 5*(x +1/x)^4*( (x² -1)/x² ). Yes, because 1 -1/x² = (x² -1)/x². So both are equivalent. But again, unless specified, either form is fine. Since the problem says "find the derivative," and doesn't specify the form, either is acceptable. Maybe the first form is preferable. Hmm, so final answer for part (2) is 5*(x +1/x)^4*(1 -1/x²). Alternatively written as 5(x + 1/x)^4 (1 - 1/x²). I think that's all. Let me recap to make sure I didn't miss anything.First problem: used the quotient rule, calculated derivatives, substituted x=3, simplified the fraction, got -1/6. Second problem: used the chain rule, took derivative of outer function, multiplied by derivative of inner function, simplified to 5*(x +1/x)^4*(1 -1/x²). Yep, that seems right. I don't think I made any mistakes here.Final Answer(1) boxed{-dfrac{1}{6}}(2) boxed{5left(x + dfrac{1}{x}right)^4 left(1 - dfrac{1}{x^2}right)}</think>(1) To find the derivative of ( y = frac{x + 3}{x^2 + 3} ) at the point where ( x = 3 ), we use the quotient rule. The quotient rule states that if ( y = frac{f(x)}{g(x)} ), then the derivative ( y' ) is given by:[y' = frac{f'(x)g(x) - f(x)g'(x)}{[g(x)]^2}]Here, ( f(x) = x + 3 ) and ( g(x) = x^2 + 3 ). Calculating the derivatives, we get ( f'(x) = 1 ) and ( g'(x) = 2x ). Applying the quotient rule:[y' = frac{(1)(x^2 + 3) - (x + 3)(2x)}{(x^2 + 3)^2}]Simplifying the numerator:[(x^2 + 3) - (2x^2 + 6x) = -x^2 - 6x + 3]Thus, the derivative is:[y' = frac{-x^2 - 6x + 3}{(x^2 + 3)^2}]Evaluating this at ( x = 3 ):[text{Numerator: } -(3)^2 - 6(3) + 3 = -9 - 18 + 3 = -24][text{Denominator: } (3^2 + 3)^2 = (9 + 3)^2 = 12^2 = 144]Therefore, the derivative at ( x = 3 ) is:[frac{-24}{144} = -frac{1}{6}](2) To find the derivative of the function ( y = left(x + frac{1}{x}right)^5 ), we use the chain rule. The chain rule states that if ( y = f(g(x)) ), then ( y' = f'(g(x)) cdot g'(x) ). Here, the outer function ( f(u) = u^5 ) and the inner function ( u = x + frac{1}{x} ). First, compute the derivative of the outer function:[frac{d}{du} u^5 = 5u^4]Next, compute the derivative of the inner function ( u = x + frac{1}{x} ):[frac{d}{dx} left(x + frac{1}{x}right) = 1 - frac{1}{x^2}]Applying the chain rule:[y' = 5left(x + frac{1}{x}right)^4 left(1 - frac{1}{x^2}right)] Final Answer(1) boxed{-dfrac{1}{6}}(2) boxed{5left(x + dfrac{1}{x}right)^4 left(1 - dfrac{1}{x^2}right)}]

🔑:(1) To find the derivative of y=frac{x+3}{x^2+3}, we will use the quotient rule, which states that (frac{u}{v})'=frac{u'v-uv'}{v^2}. Here, u=x+3 and v=x^2+3.First, find the derivatives of u and v:u'=(x+3)'=1v'=(x^2+3)'=2xNow, apply the quotient rule:y'=(frac{x+3}{x^2+3})'=frac{(x+3)'(x^2+3)-(x+3)(x^2+3)'}{(x^2+3)^2}y'=frac{(1)(x^2+3)-(2x)(x+3)}{(x^2+3)^2}y'=frac{x^2+3-2x^2-6x}{(x^2+3)^2}y'=frac{-x^2-6x+3}{(x^2+3)^2}Now, we need to find the derivative at the point where x=3:y'|_{x=3}=frac{-(3)^2-6(3)+3}{((3)^2+3)^2}=-frac{1}{6}boxed{y'=-frac{1}{6}}(2) To find the derivative of y=(x+frac{1}{x})^5, we will use the chain rule, which states that (u^n)'=nu^{n-1}u'. Here, u=x+frac{1}{x} and n=5.First, find the derivative of u:u'=(x+frac{1}{x})'=1-frac{1}{x^2}Now, apply the chain rule:y'=5(x+frac{1}{x})^4(1-frac{1}{x^2})boxed{y'=5(x+frac{1}{x})^4(1-frac{1}{x^2})}

🧠:[<think>Okay, so I need to figure out how many integers between 123 and 789 inclusive have at least two identical digits. The options are given, but I have to work through it step by step. Hmm, let's start by understanding the problem.First, the numbers are between 123 and 789. So that's from 123 up to and including 789. But wait, 789 is a three-digit number, and 123 is the starting point. So we're dealing with three-digit numbers here. Wait, 789 is a three-digit number, and 123 is the first three-digit number that's above 122. So all numbers considered here are three-digit numbers. But I need to confirm if 123 and 789 are included. The problem says "between 123 and 789," but sometimes "between" can be ambiguous. However, in math problems like this, usually, they specify if it's inclusive or not. Since the options are given, maybe we can check at the end if the answer is there. Let me proceed assuming inclusive.So, the task is to find the number of three-digit integers from 123 to 789 inclusive that have at least two identical digits. The digits are in base 10, so each digit is from 0 to 9, but the first digit (hundreds place) can't be 0.But wait, starting from 123, the first digit can be 1 up to 7 because 789 is the upper limit. Wait, 789 is the upper limit, so the hundreds digit can be 1, 2, 3, 4, 5, 6, or 7. However, numbers starting with 7 can only go up to 789. Let me verify:If hundreds digit is 1: numbers from 100 to 199, but since we start at 123, the numbers here would be 123 to 199.If hundreds digit is 2: 200-299Similarly up to hundreds digit 7: 700-789So the range is from 123 to 789. Therefore, the hundreds digit can be 1-7, but with different restrictions. For hundreds digit 1, the tens and units digits start at 2 and 3, respectively? Wait, no. Wait, 123 is the starting number. So 123 is the first number, which is 1 (hundreds), 2 (tens), 3 (units). So for hundreds digit 1, the tens digit starts at 2 if the hundreds digit is 1? Wait, no. Let's think.If we take numbers from 123 to 199 inclusive, that's for hundreds digit 1. The first number is 123, so tens digit starts at 2, but once the hundreds digit is 1, the tens digit can be from 2 up to 9, but actually, numbers like 130, 131, etc., are included as long as they are between 123 and 199. Wait, no. Wait, the numbers from 123 to 199 include all numbers where hundreds digit is 1, tens digit from 2 up to 9, but also numbers where the hundreds digit is 1, tens digit is 1 and units digit is higher than 3? Wait, no. Wait, 123 is the starting point. So 123 is 1-2-3, and the next number is 124, then 125, up to 129, then 130. Wait, but 130 is less than 199, so it's included. Wait, but 110 is not included because it's less than 123. So in the hundreds digit 1, the numbers start at 123 and go up to 199. So actually, the tens digit can be from 2 to 9 when the hundreds digit is 1, but also for tens digit 2, units digits go from 3 to 9, and for tens digits higher than 2, units digits can be from 0 to 9. Wait, no. Wait, 123 is the starting point, so when hundreds digit is 1, the number has to be at least 123. So:If hundreds digit is 1:- If tens digit is 2, units digit can be 3-9 (7 numbers)- If tens digit is 3-9, units digit can be 0-9 (each tens digit gives 10 numbers)So total numbers when hundreds digit is 1:For tens digit 2: 7 numbers (123-129)For tens digits 3-9: 7 tens digits (3-9), each with 10 units digits: 7*10=70Total: 7 + 70 = 77 numbersSimilarly, for hundreds digits 2-6, the numbers are all numbers from 200-299, 300-399, ..., 600-699. Each of these is 100 numbers each. Then for hundreds digit 7, the numbers are from 700 to 789, which is 90 numbers (700-789 inclusive). Wait, 789-700+1=90.But wait, 789 is included, so from 700 to 789: that's 90 numbers. Let me confirm:789 - 700 = 89, so 89 + 1 = 90 numbers.So total numbers in the range:For hundreds digit 1: 77For hundreds digits 2-6: 5*100=500For hundreds digit 7: 90Total numbers: 77 + 500 + 90 = 667 numbers.Wait, but let's verify:123 to 199: 199 - 123 + 1 = 77. Correct.200 to 699: 699 - 200 + 1 = 500. But here, hundreds digits 2-6 correspond to 200-699. Wait, 200-299 is 100 numbers (hundreds digit 2), 300-399 is hundreds digit 3, etc., up to 600-699 (hundreds digit 6). Then hundreds digit 7 is 700-789: 789 - 700 +1=90. So total numbers from 123 to 789 inclusive: 77 + 500 + 90=667.So total numbers in the range: 667. Now, the question is how many of these have at least two identical digits. It might be easier to calculate the total number of numbers without any identical digits and subtract that from the total to get the numbers with at least two identical digits.Yes, complementary counting. Total numbers - numbers with all distinct digits = numbers with at least two identical digits.So first, total numbers: 667.Now, compute numbers with all distinct digits.We need to compute for each hundreds digit (1-7), the number of numbers with all distinct digits.So let's break it down by hundreds digit:Case 1: Hundreds digit is 1.We need numbers from 123 to 199 with all distinct digits.But in this case, hundreds digit is 1. So the tens and units digits must be different from 1 and from each other.But since the numbers start at 123, for hundreds digit 1, tens digit starts at 2 (as in 123). Wait, no. Wait, 123 is 1-2-3, but numbers like 102 would be excluded because they are below 123. So actually, when hundreds digit is 1, the tens digit can be from 0 to 9, but the number must be at least 123. Therefore, the tens digit must be at least 2, or if tens digit is 1, the units digit must be at least 3? Wait, no. Wait, if hundreds digit is 1 and tens digit is 0, the number would be 100-109, which are all below 123. So they are excluded. Similarly, hundreds digit 1, tens digit 1: numbers 110-119, which are below 123. Tens digit 2: 120-129. But 120-122 are below 123, so only 123-129 are included. So, in the hundreds digit 1 case, the allowed tens digits are:- Tens digit 2: units digits 3-9 (7 numbers, as before)- Tens digits 3-9: units digits 0-9, but with all digits distinct.But since hundreds digit is 1, tens digits can be 3-9 (7 possibilities), and units digits can be 0-9 except 1 and the tens digit. So for each tens digit (3-9):Tens digit choices: 3,4,5,6,7,8,9 (7 digits)For each such tens digit, units digits can be 0-9 except 1 and the chosen tens digit. So units digits: 10 - 2 = 8 choices.Therefore, for tens digits 3-9: 7 * 8 = 56 numbers.But wait, the units digits can include numbers like 130, which is 1-3-0, all distinct. But 130 is included in our range (123-199). So yes, these are valid.But hold on, when hundreds digit is 1 and tens digit is 3, units digits can be 0,2,4-9 (excluding 1 and 3). So 8 choices.Therefore, total numbers with all distinct digits in hundreds digit 1:Tens digit 2: units digits 3-9 (7 numbers, each with digits 1-2-x, where x is 3-9, all distinct)Tens digits 3-9: 7 digits * 8 units digits = 56Total: 7 + 56 = 63 numbers with all distinct digits in hundreds digit 1.Case 2: Hundreds digits 2-6.Each of these hundreds digits (2,3,4,5,6). For each, the tens and units digits can be from 0-9, but they must be different from each other and from the hundreds digit.So for hundreds digit d (where d is 2-6):- Hundreds digit: d- Tens digit: can be 0-9 except d (9 choices)- Units digit: can be 0-9 except d and the tens digit (8 choices)Therefore, for each hundreds digit 2-6: 9 * 8 = 72 numbers.Since there are 5 hundreds digits (2-6), total: 5 * 72 = 360.Case 3: Hundreds digit 7.Numbers from 700 to 789. So hundreds digit is 7. The tens digit can be 0-8 (since 789 is the upper limit), so tens digits 0-8. However, digits must be distinct.So hundreds digit is 7.Tens digit: 0-8 (9 choices). But must not be 7.Wait, tens digit can't be 7 because hundreds digit is already 7, so tens digit can be 0-6,8-8? Wait, 789 is the maximum number, so tens digit can be 0-8, but since hundreds digit is 7, tens digit must be different from 7.Wait, 700-789. So tens digit can be 0-8 (since 789 is the highest, which is 7-8-9). But to have distinct digits, tens digit can be 0-8 excluding 7. So tens digits: 0,1,2,3,4,5,6,8 (8 choices). Wait, 0-8 is 9 digits, but excluding 7, it's 8 digits.Wait, let's check:If hundreds digit is 7, tens digit can be from 0 to 8 (because 789 is allowed, which has tens digit 8). However, to have distinct digits, tens digit cannot be 7, so 0-6 and 8 (total 8 choices). Then for each tens digit, units digit can be any digit except 7 and the tens digit. So units digits: 0-9 except 7 and tens digit. So units digits: 8 choices.But also, note that when hundreds digit is 7 and tens digit is 8 (as in 789), units digit can be 0-9 except 7 and 8, so digits 0-6,9. But 789 is allowed, but wait, 789 has units digit 9, which is allowed because 7,8,9 are all distinct. Wait, but in the calculation, when hundreds digit is 7, tens digit is 8, units digit can be 0-9 except 7 and 8, which would be 8 choices (digits 0-6,9). So 8 choices.Therefore, for hundreds digit 7:Tens digits: 8 choices (0-6,8)For each tens digit, units digits: 8 choices (excluding 7 and the tens digit)Therefore, total numbers with all distinct digits in hundreds digit 7: 8 * 8 = 64.But wait, wait. Let's check this. For example, if hundreds digit is 7, tens digit is 8 (allowed, as in 789), then units digit can be 0-9 except 7 and 8. So units digits allowed: 0-6,9 (8 digits). So 8 choices. Similarly, if tens digit is 0, units digits can be 1-9 except 7, which is 8 choices (since 0 is already used in tens digit). Wait, no: hundreds digit is 7, tens digit is 0, units digit can be 1-9 except 7. So units digits: 1-6,8-9. That's 8 choices. Wait, 1-6 is 6 digits, 8-9 is 2 digits, total 8. Correct.Similarly, tens digit is 5: units digits can be 0-9 except 7 and 5, so 8 choices.Therefore, 8 * 8 = 64 numbers with all distinct digits in hundreds digit 7.But wait, but the upper limit is 789. So even though when hundreds digit is 7, tens digit 8 allows units digits 0-6,9, the maximum units digit when tens digit is 8 is 9 (as in 789). So in our case, even though units digits could go up to 9, in the hundreds digit 7 and tens digit 8 case, units digits can only go up to 9, which is allowed. So the calculation holds.Therefore, total numbers with all distinct digits:Case 1: 63Case 2: 360Case 3: 64Total: 63 + 360 + 64 = 487Therefore, total numbers with all distinct digits are 487. Then, numbers with at least two identical digits are total numbers (667) minus 487, which is 667 - 487 = 180.But wait, option a is 180. But let me check if the calculations are correct.Wait, total numbers: 667.Numbers with all distinct digits:Case 1: hundreds digit 1: 63Case 2: hundreds digits 2-6: 5 * 72 = 360Case 3: hundreds digit 7: 64Total: 63 + 360 + 64 = 487Therefore, 667 - 487 = 180. So answer is 180, which is option a.But wait, the options are given as a. 180, b. 184, c. 186, d. 189, e. 191. So 180 is an option. But let me verify once again because sometimes when I do these problems, it's easy to miscount.Let me recheck each case.Case 1: Hundreds digit 1.Numbers from 123-199 with all distinct digits.We split into tens digit 2 and tens digits 3-9.Tens digit 2: units digits 3-9, which is 7 numbers (123,124,...,129). All of these have hundreds digit 1, tens digit 2, units digits 3-9. All digits distinct. Correct. So 7 numbers.Tens digits 3-9: 7 tens digits (3,4,5,6,7,8,9). For each, units digits can be 0-9 except 1 and the tens digit. So units digits: 10 - 2 = 8. Therefore, 7 * 8 = 56. So total for hundreds digit 1: 7 + 56 = 63. Correct.Case 2: Hundreds digits 2-6.Each hundreds digit d (2-6). For each, tens digits: 0-9 except d, so 9 choices. Units digits: 0-9 except d and the tens digit, so 8 choices. So 9*8=72 per hundreds digit. 5 hundreds digits: 5*72=360. Correct.Case 3: Hundreds digit 7.Tens digits: 0-8 except 7, so 0-6 and 8, which is 8 choices. For each tens digit, units digits: 0-9 except 7 and the tens digit. So 8 choices. 8*8=64. Correct.Total distinct: 63+360+64=487.Total numbers: 77 (hundreds 1) + 500 (hundreds 2-6) + 90 (hundreds 7) = 667. 77+500=577, 577+90=667. Correct.667 - 487 = 180. So answer is 180, which is option a.But the options given are a. 180, b. 184, c. 186, d. 189, e. 191. So according to this, the answer is a. 180.But wait, hold on. Let me check if the total numbers is correct.From 123 to 789 inclusive.Numbers from 123-199: 199 - 123 + 1 = 77. Correct.Numbers from 200-699: Each hundred series has 100 numbers. 699 - 200 +1 = 500. Correct.Numbers from 700-789: 789 - 700 +1 = 90. Correct. 77+500+90=667. Correct.Numbers with all distinct digits:Case 1: 63Case 2: 360Case 3: 64Total: 63+360=423, 423+64=487. Correct.So 667 - 487=180. So answer is 180. But let me think again: could there be a mistake in the calculation?Wait, for hundreds digit 1, in the tens digits 3-9, units digits can't be 1 or the tens digit. So for example, if tens digit is 3, units digits can be 0,2,4-9 (excluding 1 and 3). That is 8 digits. So 7*8=56. Then plus 7 (tens digit 2). So 63. Correct.For hundreds digits 2-6: 5*72=360. Correct.For hundreds digit 7: 8*8=64. Correct.Total distinct: 487. Total numbers:667. So 667-487=180. Therefore, answer is a. 180.Wait, but why are the other answer choices higher? Maybe I made a mistake in counting the total numbers? Let me check again.123 to 789 inclusive.From 123 to 199:199-123+1=77. Correct.From 200 to 699: Each hundred is 100 numbers. 699-200+1=500. Correct.From 700 to 789:789-700+1=90. Correct.Total:77+500+90=667. Correct.So total numbers:667. Then numbers with all distinct digits:487. So 667-487=180. So answer is 180. So option a. But let me check with another approach.Alternative approach: Compute directly the numbers with at least two identical digits. Maybe I missed some cases?But complementary counting is usually safer. However, let me verify with hundreds digit 7.Wait, in hundreds digit 7, numbers from 700 to 789. So when we calculated numbers with all distinct digits as 8*8=64, is that correct?Tens digits available:0-8 except 7: 0,1,2,3,4,5,6,8 (8 digits). Then units digits:0-9 except 7 and the tens digit. So if tens digit is 0, units digits can be 1-9 except 7. That's 8 digits. Similarly, if tens digit is 8, units digits can be 0-6,9 (excluding 7 and 8). Also 8 digits. So yes, 8*8=64. Correct.Another check: for hundreds digit 1, numbers with all distinct digits.Total numbers with hundreds digit 1:77 (123-199). Numbers with all distinct digits:63. So numbers with at least two identical digits:77-63=14.Similarly, for hundreds digits 2-6: each has 100 numbers. Numbers with all distinct digits:72 each. So numbers with duplicates:100-72=28 per hundreds digit. 5 hundreds digits:28*5=140.For hundreds digit 7:90 numbers. Numbers with all distinct digits:64. So numbers with duplicates:90-64=26.Total numbers with duplicates:14+140+26=180. Correct.Therefore, answer is 180. So option a. 180.Final Answerboxed{A}</think>To determine how many integers between 123 and 789 have at least two identical digits, we use complementary counting. We first calculate the total number of integers in the range and then subtract the number of integers with all distinct digits.Total Numbers:- From 123 to 199: (199 - 123 + 1 = 77)- From 200 to 699: (699 - 200 + 1 = 500)- From 700 to 789: (789 - 700 + 1 = 90)- Total: (77 + 500 + 90 = 667)Numbers with All Distinct Digits:1. Hundreds digit 1 (123-199): - Tens digit 2: Units digits 3-9 (7 numbers) - Tens digits 3-9: Each with 8 possible units digits (7*8=56) - Total: (7 + 56 = 63)2. Hundreds digits 2-6: - Each hundreds digit (2-6) has 9 choices for tens digit and 8 choices for units digit: (9*8=72) - Total for 5 hundreds digits: (5*72 = 360)3. Hundreds digit 7 (700-789): - Tens digits 0-8 (excluding 7): 8 choices - Units digits: 8 choices for each tens digit - Total: (8*8 = 64)Total Numbers with All Distinct Digits:- (63 + 360 + 64 = 487)Numbers with At Least Two Identical Digits:- Total numbers - Numbers with all distinct digits: (667 - 487 = 180)Thus, the number of integers between 123 and 789 with at least two identical digits is (boxed{A}).]

🔑:To determine how many integers between 123 and 789 have at least two identical digits, we can follow these steps:1. Calculate the total number of integers between 123 and 789: [ 789 - 123 + 1 = 667 ]2. Calculate the number of integers between 123 and 789 with all distinct digits: - We need to count the number of three-digit numbers with distinct digits in the range 123 to 789. - A three-digit number overline{abc} has distinct digits if a, b, c are all different.3. Count the number of three-digit numbers with distinct digits: - The first digit a can be any digit from 1 to 7 (since 789 is the upper limit). - The second digit b can be any digit from 0 to 9 except a. - The third digit c can be any digit from 0 to 9 except a and b. Therefore, the number of distinct three-digit numbers overline{abc} is: [ 7 times 9 times 8 = 504 ]4. Calculate the number of integers with at least two identical digits: - Subtract the number of integers with all distinct digits from the total number of integers. [ 667 - 504 = 163 ]5. Adjust for the range 123 to 789: - The solution provided in the community has a different approach, which involves calculating the number of integers with at least two identical digits directly. - According to the solution, the number of integers in [100, 789] with at least two repeated digits is: [ 28 times 7 + 2 times 8 - 18 = 194 ] - The number of integers in [100, 122] with at least two repeated digits is: [ 28 times 1 + 2 times 2 - 18 = 14 ] - Therefore, the number of integers in [123, 789] with at least two repeated digits is: [ 194 - 14 = 180 ]The final answer is boxed{180}