💡Okay, so I have this function f(x) = x² - a ln x. The first part says that f(x) is monotonically decreasing on the interval [3,5]. I need to find the range of the real number a. Hmm, okay. Monotonically decreasing means that the derivative of f(x) should be less than or equal to zero on that interval. So, let me compute the derivative of f(x).f'(x) = derivative of x² is 2x, and derivative of -a ln x is -a*(1/x). So, f'(x) = 2x - a/x. For f(x) to be decreasing, f'(x) ≤ 0 for all x in [3,5]. So, 2x - a/x ≤ 0. Let me rearrange this inequality: 2x ≤ a/x. Multiply both sides by x (since x is positive in [3,5], the inequality sign doesn't change). So, 2x² ≤ a. Therefore, a must be greater than or equal to 2x² for all x in [3,5].Now, 2x² is a function that increases as x increases because the coefficient of x² is positive. So, on the interval [3,5], the maximum value of 2x² occurs at x=5. Let me compute that: 2*(5)² = 2*25 = 50. Therefore, a must be greater than or equal to 50. So, the range of a is [50, ∞). That seems straightforward.Moving on to the second part. We have g(x) = f(x) + (2 + a) ln x - 2(b - 1)x. Let me substitute f(x) into this expression. So, g(x) = x² - a ln x + (2 + a) ln x - 2(b - 1)x. Let me simplify this:- The terms with ln x: (-a ln x) + (2 + a) ln x = (2) ln x.- The rest: x² - 2(b - 1)x.So, g(x) simplifies to x² + 2 ln x - 2(b - 1)x. Now, we are told that x₁ and x₂ (with x₁ < x₂) are two extremal points of g(x). So, these are the critical points where the derivative is zero.Let me compute the derivative of g(x):g'(x) = derivative of x² is 2x, derivative of 2 ln x is 2/x, and derivative of -2(b - 1)x is -2(b - 1). So, putting it together:g'(x) = 2x + 2/x - 2(b - 1).To find the critical points, set g'(x) = 0:2x + 2/x - 2(b - 1) = 0.Let me factor out the 2:2(x + 1/x - (b - 1)) = 0.Divide both sides by 2:x + 1/x - (b - 1) = 0.Multiply both sides by x to eliminate the denominator:x² + 1 - (b - 1)x = 0.So, the quadratic equation is x² - (b - 1)x + 1 = 0.Let me denote this as x² - (b - 1)x + 1 = 0. The solutions to this quadratic are x₁ and x₂. Since x₁ and x₂ are extremal points, they must be real and distinct, so the discriminant must be positive.Discriminant D = [-(b - 1)]² - 4*1*1 = (b - 1)² - 4.For real and distinct roots, D > 0:(b - 1)² - 4 > 0 => (b - 1)² > 4 => |b - 1| > 2 => b - 1 > 2 or b - 1 < -2.But since b is given to be greater than or equal to 7/2, which is 3.5, so b - 1 is at least 2.5, which is greater than 2. So, the discriminant is positive, and we have two real roots x₁ and x₂.Moreover, since the quadratic is x² - (b - 1)x + 1, the sum of roots x₁ + x₂ = b - 1, and the product x₁x₂ = 1.Given that x₁ < x₂, and since the product is 1, both roots are positive because if their product is positive and sum is positive, both roots are positive. So, x₁ and x₂ are positive real numbers with x₁ < x₂.Now, we need to find the minimum value of g(x₁) - g(x₂). Let me write down g(x₁) - g(x₂):g(x₁) - g(x₂) = [x₁² + 2 ln x₁ - 2(b - 1)x₁] - [x₂² + 2 ln x₂ - 2(b - 1)x₂].Simplify this:= x₁² - x₂² + 2 ln x₁ - 2 ln x₂ - 2(b - 1)(x₁ - x₂).Let me factor the terms:= (x₁² - x₂²) + 2(ln x₁ - ln x₂) - 2(b - 1)(x₁ - x₂).Note that x₁² - x₂² = (x₁ - x₂)(x₁ + x₂). Similarly, ln x₁ - ln x₂ = ln(x₁/x₂). So, substituting these:= (x₁ - x₂)(x₁ + x₂) + 2 ln(x₁/x₂) - 2(b - 1)(x₁ - x₂).Factor out (x₁ - x₂):= (x₁ - x₂)[(x₁ + x₂) - 2(b - 1)] + 2 ln(x₁/x₂).But from earlier, we know that x₁ + x₂ = b - 1, so substituting:= (x₁ - x₂)[(b - 1) - 2(b - 1)] + 2 ln(x₁/x₂).Simplify inside the brackets:= (x₁ - x₂)[- (b - 1)] + 2 ln(x₁/x₂).So,= - (x₁ - x₂)(b - 1) + 2 ln(x₁/x₂).But x₁ - x₂ is negative because x₁ < x₂, so -(x₁ - x₂) is positive. Let me write it as:= (x₂ - x₁)(b - 1) + 2 ln(x₁/x₂).Alternatively, since ln(x₁/x₂) = - ln(x₂/x₁), so:= (x₂ - x₁)(b - 1) - 2 ln(x₂/x₁).Hmm, not sure if that helps yet. Maybe another approach.Alternatively, since x₁x₂ = 1, we can write x₂ = 1/x₁. Because x₁x₂ = 1, so x₂ = 1/x₁. Since x₁ < x₂ and x₁x₂ = 1, x₁ must be less than 1, and x₂ greater than 1.Let me set t = x₁, so x₂ = 1/t, with 0 < t < 1.So, x₁ = t, x₂ = 1/t.Then, let's express g(x₁) - g(x₂) in terms of t.First, compute x₁ + x₂ = t + 1/t = b - 1.So, b - 1 = t + 1/t.Given that b ≥ 7/2, so b - 1 ≥ 5/2. Therefore, t + 1/t ≥ 5/2.We can solve for t:t + 1/t ≥ 5/2.Multiply both sides by t (positive, so inequality remains):t² + 1 ≥ (5/2)t.Bring all terms to left:t² - (5/2)t + 1 ≥ 0.Multiply both sides by 2 to eliminate fraction:2t² - 5t + 2 ≥ 0.Factor:Looking for two numbers that multiply to 4 (2*2) and add to -5. Hmm, -4 and -1.So, 2t² - 4t - t + 2 = (2t² - 4t) + (-t + 2) = 2t(t - 2) -1(t - 2) = (2t -1)(t - 2).So, 2t² -5t +2 = (2t -1)(t -2).Thus, inequality is (2t -1)(t -2) ≥ 0.Find where this is non-negative.Critical points at t=1/2 and t=2.Test intervals:- t < 1/2: say t=0, ( -1)(-2) = 2 >0- 1/2 < t <2: say t=1, (2 -1)(1 -2)= (1)(-1)=-1 <0- t >2: say t=3, (6 -1)(3 -2)=5*1=5>0So, the inequality holds when t ≤1/2 or t≥2. But since t =x₁ <1, and t>0, so t ≤1/2.Therefore, t ∈ (0, 1/2].So, t is between 0 and 1/2.Now, let's express g(x₁) - g(x₂) in terms of t.From earlier, we had:g(x₁) - g(x₂) = (x₂ - x₁)(b -1) - 2 ln(x₂/x₁).But x₂ =1/t, x₁ =t, so x₂ -x₁ =1/t - t.Also, b -1 = t +1/t.And ln(x₂/x₁)= ln(1/t / t)= ln(1/t²)= -2 ln t.So, substituting:g(x₁) - g(x₂) = (1/t - t)(t +1/t) - 2*(-2 ln t).Simplify:First, compute (1/t - t)(t +1/t):= (1/t)(t) + (1/t)(1/t) - t(t) - t(1/t)= 1 + 1/t² - t² -1Simplify:1 -1 cancels, so we have 1/t² - t².Then, the other term is -2*(-2 ln t)=4 ln t.So, overall:g(x₁) - g(x₂) = (1/t² - t²) +4 ln t.So, we have:g(x₁) - g(x₂) = 1/t² - t² +4 ln t.Let me denote this expression as h(t) =1/t² - t² +4 ln t.We need to find the minimum value of h(t) for t ∈ (0, 1/2].To find the minimum, let's compute the derivative h'(t) and find critical points.Compute h'(t):h(t) = t^{-2} - t² +4 ln t.So,h'(t) = -2 t^{-3} -2t +4*(1/t).Simplify:= -2/t³ -2t +4/t.Let me write all terms with denominator t³:= (-2 -2t⁴ +4t²)/t³.Wait, let me see:Wait, -2/t³ -2t +4/t.To combine terms, let's express all over t³:= (-2 -2t⁴ +4t²)/t³.Wait, no. Let me see:-2/t³ -2t +4/t = (-2)/t³ + (-2t) +4/t.To combine, we can write as:= (-2)/t³ +4/t -2t.Alternatively, factor:= (-2)/t³ + (4t² -2t⁴)/t³.Wait, maybe that's more complicated.Alternatively, set h'(t)=0:-2/t³ -2t +4/t =0.Multiply both sides by t³ to eliminate denominators:-2 -2t⁴ +4t²=0.So,-2t⁴ +4t² -2=0.Multiply both sides by -1:2t⁴ -4t² +2=0.Divide both sides by 2:t⁴ -2t² +1=0.This is a quadratic in t²:Let u = t², so equation becomes u² -2u +1=0.Which factors as (u -1)²=0.Thus, u=1, so t²=1, so t=±1. But t>0, so t=1.But t ∈ (0,1/2], so t=1 is outside our interval. Therefore, h'(t)=0 has no solutions in (0,1/2].Therefore, the function h(t) has no critical points in (0,1/2]. Thus, the extrema must occur at the endpoints.But our interval is t ∈ (0,1/2]. So, we check the behavior as t approaches 0+ and at t=1/2.First, as t approaches 0+, let's see what happens to h(t):h(t) =1/t² - t² +4 ln t.1/t² approaches infinity, -t² approaches 0, and 4 ln t approaches -infinity. So, we have infinity - infinity, which is indeterminate. Let's analyze the limit:lim_{t→0+} [1/t² +4 ln t - t²].We can see that 1/t² dominates over 4 ln t because 1/t² goes to infinity much faster than ln t goes to -infinity. So, the limit is +infinity.At t=1/2, let's compute h(1/2):h(1/2)=1/(1/2)² - (1/2)² +4 ln(1/2)=1/(1/4) -1/4 +4*(-ln2)=4 -1/4 -4 ln2= (16/4 -1/4) -4 ln2=15/4 -4 ln2.So, h(t) approaches infinity as t approaches 0, and at t=1/2, h(t)=15/4 -4 ln2.Since h(t) is decreasing on (0,1/2], because h'(t) is negative throughout the interval (as we saw earlier, h'(t) was negative because t=1 is outside the interval, and the derivative didn't cross zero). Wait, actually, let me confirm the sign of h'(t) in (0,1/2].Earlier, when solving h'(t)=0, we saw that the only critical point is at t=1, which is outside our interval. So, let's pick a test point in (0,1/2], say t=1/4.Compute h'(1/4):h'(1/4)= -2/(1/4)³ -2*(1/4) +4/(1/4)= -2/(1/64) -1/2 +16= -128 -0.5 +16= (-128 +16) -0.5= -112 -0.5= -112.5 <0.So, h'(t) is negative in (0,1/2]. Therefore, h(t) is decreasing on (0,1/2]. Therefore, the minimum value occurs at t=1/2, which is 15/4 -4 ln2.Therefore, the minimum value of g(x₁)-g(x₂) is 15/4 -4 ln2.Let me just recap to make sure I didn't make any mistakes.We started with f(x)=x² -a ln x, found that for it to be decreasing on [3,5], a must be at least 50.Then, for the second part, we defined g(x)=f(x)+(2+a)lnx -2(b-1)x, simplified it, found its derivative, found critical points x₁ and x₂, expressed the difference g(x₁)-g(x₂) in terms of t=x₁, found that t must be ≤1/2 due to the constraint b≥7/2, then expressed the difference as a function h(t)=1/t² -t² +4 ln t, found that h(t) is decreasing on (0,1/2], so the minimum occurs at t=1/2, giving h(1/2)=15/4 -4 ln2.Yes, that seems consistent.**Final Answer**1. The range of ( a ) is (boxed{[50, infty)}).2. The minimum value of ( g(x_{1}) - g(x_{2}) ) is (boxed{frac{15}{4} - 4ln 2}).

💡Alright, so I've got this geometry problem here. Let me try to unpack it step by step. First, the problem says: Let f be the bisector of the angle formed by the rays a and b. Okay, so there's an angle between two rays, a and b, and f is the angle bisector. That means f splits the angle into two equal parts. Got it.Next, there's a circle k_1 that's tangent to rays a and f, and it touches a at point A. Similarly, another circle k_2 is tangent to rays b and f, touching b at point B. The task is to prove that the line AB intersects equal chords from the circles k_1 and k_2.Hmm, okay. So, I need to visualize this. Let me sketch it out mentally. There's a point where rays a and b meet, forming an angle. The bisector f splits this angle into two equal angles. Then, circle k_1 is snug against both a and f, touching a at A. Similarly, circle k_2 is snug against b and f, touching b at B. So, if I imagine this, both circles are in the angle formed by a and b, each tangent to one of the rays and the bisector. The line connecting points A and B is supposed to cut chords of equal length in both circles. Interesting.Let me think about how to approach this. Maybe I can use properties of angle bisectors and tangents. Since both circles are tangent to f, which is the angle bisector, perhaps there's some symmetry here. Also, since A and B are points of tangency on a and b, maybe I can relate the distances from A and B to the centers of the circles.Wait, tangents from a point to a circle are equal in length. So, if I consider the centers of the circles, O_1 for k_1 and O_2 for k_2, then the distances from O_1 to a and f should be equal, right? Similarly for O_2 to b and f.So, O_1 is equidistant from a and f, and O_2 is equidistant from b and f. That makes sense because the circle is tangent to both rays, so the distance from the center to each ray is equal to the radius.Now, since f is the angle bisector, the distances from O_1 and O_2 to f are equal to their respective radii. Maybe I can use this to find some relationship between the two circles.Let me denote the radius of k_1 as r_1 and the radius of k_2 as r_2. Then, the distance from O_1 to a is r_1, and the distance from O_1 to f is also r_1. Similarly, the distance from O_2 to b is r_2, and the distance from O_2 to f is r_2.Since f is the angle bisector, the angles between a and f and between b and f are equal. Let's denote this angle as theta. So, both O_1 and O_2 lie along the angle bisector f, but at different distances from the vertex where a and b meet.Wait, no. Actually, the centers O_1 and O_2 are located such that they are each equidistant from their respective rays and the bisector. So, they might not lie on f itself, but rather on lines parallel to f at a distance equal to their radii.Hmm, maybe I should consider the coordinates. Let me set up a coordinate system to model this. Let's place the vertex where rays a and b meet at the origin. Let me assume that ray a lies along the positive x-axis, and ray b makes an angle 2theta with a, so that f is the bisector, making an angle theta with a.So, in this coordinate system, ray a is along the x-axis, ray b is in the upper half-plane making an angle 2theta with a, and f is the bisector at angle theta.Now, circle k_1 is tangent to a and f, touching a at point A. Similarly, circle k_2 is tangent to b and f, touching b at point B.Let me find the coordinates of O_1 and O_2. Since k_1 is tangent to a (the x-axis) and f (the bisector at angle theta), the center O_1 must be at a distance r_1 from both a and f. Similarly, O_2 is at a distance r_2 from b and f.To find the coordinates of O_1, since it's at distance r_1 from the x-axis, its y-coordinate is r_1. To find its x-coordinate, we can use the distance from the line f, which has the equation y = tan(theta) x. The distance from O_1 = (x_1, r_1) to the line f is given by:frac{| tan(theta) x_1 - r_1 |}{sqrt{tan^2(theta) + 1}} = r_1Simplifying, since the distance is positive:| tan(theta) x_1 - r_1 | = r_1 sqrt{tan^2(theta) + 1}Assuming O_1 is above the x-axis and to the right of f, the expression inside the absolute value is positive:tan(theta) x_1 - r_1 = r_1 sqrt{tan^2(theta) + 1}Solving for x_1:tan(theta) x_1 = r_1 + r_1 sqrt{tan^2(theta) + 1}x_1 = r_1 frac{1 + sqrt{tan^2(theta) + 1}}{tan(theta)}Hmm, that seems a bit complicated. Maybe I can express it differently. Let's denote tan(theta) = t for simplicity.Then,x_1 = r_1 frac{1 + sqrt{t^2 + 1}}{t}Similarly, the coordinates of O_1 are (x_1, r_1).Similarly, for O_2, it's tangent to b and f. Ray b makes an angle 2theta with the x-axis, so its equation is y = tan(2theta) x. The distance from O_2 = (x_2, y_2) to b is r_2, and the distance to f is also r_2.So, the distance from O_2 to f is:frac{| tan(theta) x_2 - y_2 |}{sqrt{tan^2(theta) + 1}} = r_2And the distance from O_2 to b is:frac{| tan(2theta) x_2 - y_2 |}{sqrt{tan^2(2theta) + 1}} = r_2This is getting a bit involved. Maybe there's a better way to approach this without coordinates.Let me think about the properties of tangents and angle bisectors. Since both circles are tangent to f, which is the angle bisector, maybe there's a reflection symmetry here. If I reflect one circle over f, it might coincide with the other circle. But I'm not sure if that's the case.Alternatively, maybe I can use homothety. If there's a homothety that maps k_1 to k_2, it might center at the intersection point of their common tangents, which is point C where rays a and b meet.Wait, homothety could be a good approach. A homothety is a transformation that enlarges or reduces a figure by a scale factor, centered at a specific point. If I can find a homothety that maps k_1 to k_2, then it would map point A to point B, and hence map line AB to itself. This might help in showing that the chords cut by AB are equal.But I need to verify if such a homothety exists. For homothety, the centers of the circles must lie on a line through the center of homothety. The centers O_1 and O_2 both lie on the angle bisector f, so if the homothety is centered at point C, which is the vertex of the angle, then it might work.Let me assume that there's a homothety centered at C that maps k_1 to k_2. Then, it would map A to B and O_1 to O_2. The scale factor would be the ratio of the radii, r_2 / r_1.But I don't know if the radii are related by a specific ratio. Maybe I can find the ratio using the angles.Alternatively, maybe I can use similar triangles. Since f is the angle bisector, the triangles formed by the centers and the points of tangency might be similar.Let me consider triangles CO_1A and CO_2B. Both are right triangles because the radius is perpendicular to the tangent at the point of contact. So, CO_1A is a right triangle with right angle at A, and CO_2B is a right triangle with right angle at B.Since f is the angle bisector, the angles at C for both triangles are equal, each being theta. Therefore, triangles CO_1A and CO_2B are similar by AA similarity (they both have a right angle and share the angle theta).Therefore, the ratio of their corresponding sides is equal. So,frac{CO_1}{CO_2} = frac{CA}{CB} = frac{r_1}{r_2}Hmm, interesting. So, the distances from C to the centers are proportional to the radii and the lengths from C to A and B.But how does this help with the chords cut by AB?Well, the chords cut by AB in circles k_1 and k_2 will depend on the distance from the centers O_1 and O_2 to the line AB. If these distances are equal, then the lengths of the chords will be equal because the circles have the same radius? Wait, no, the circles might have different radii.Wait, actually, the length of a chord is given by 2sqrt{r^2 - d^2}, where d is the distance from the center to the line. So, if the distances from O_1 and O_2 to AB are such that r_1^2 - d_1^2 = r_2^2 - d_2^2, then the chords would be equal.But I don't know if r_1 and r_2 are related in a way that would make this true. Maybe I need another approach.Let me think about the power of a point. The power of point A with respect to circle k_2 could be related to the power of point B with respect to circle k_1. But I'm not sure if that's directly helpful.Alternatively, maybe I can parametrize the line AB and find where it intersects the circles, then compute the chord lengths.But that might be too involved. Let me try to find another way.Wait, since AB is the line connecting the points of tangency on a and b, and f is the angle bisector, maybe AB is symmetric with respect to f. If that's the case, then the distances from O_1 and O_2 to AB would be equal, leading to equal chords.But is AB symmetric with respect to f? Let me see.Since f is the angle bisector, and both circles are tangent to f, the configuration is symmetric with respect to f. Therefore, reflecting across f should map k_1 to k_2, and vice versa, and map A to B and B to A. Therefore, line AB is mapped to itself under reflection over f, meaning it's symmetric with respect to f.Therefore, the distances from O_1 and O_2 to line AB must be equal because of the symmetry. Hence, the chords cut by AB in both circles must be equal in length.Wait, that seems like a good argument. Let me formalize it.Since f is the angle bisector, the configuration is symmetric with respect to f. Therefore, reflecting across f maps k_1 to k_2, A to B, and B to A. Consequently, line AB is invariant under this reflection, meaning it is symmetric with respect to f. Therefore, the distances from O_1 and O_2 to AB are equal. Given that the length of a chord in a circle is determined by the distance from the center to the chord, and since both distances are equal, the chords cut by AB in k_1 and k_2 must be equal in length.Yeah, that makes sense. So, the key idea is the symmetry with respect to the angle bisector f, which ensures that the distances from the centers to the line AB are equal, leading to equal chord lengths.I think that wraps it up. The reflection symmetry across f is the crux here, ensuring the necessary equalities for the chords.

💡Okay, so I have this problem here about two unit vectors, a and b, with an angle of 120 degrees between them. I need to find the minimum value of the magnitude of the vector (a - k*b), where k is a real number. The options are given as A: 3/4, B: sqrt(3)/2, C: 1, and D: 3/2.Alright, let me start by recalling some vector properties. Since both a and b are unit vectors, their magnitudes are 1. The angle between them is 120 degrees, so their dot product should be |a||b|cos(theta). That would be 1*1*cos(120°). I remember that cos(120°) is equal to -1/2. So, the dot product of a and b is -1/2.Now, the expression I need to minimize is |a - k*b|. To find the magnitude of this vector, I can square it to make the calculations easier. So, let's compute |a - k*b|².Expanding this, I get (a - k*b)·(a - k*b). Using the distributive property of dot products, this becomes a·a - 2k*a·b + k²*b·b.Since a and b are unit vectors, a·a is 1 and b·b is also 1. We already found that a·b is -1/2. Plugging these values in, the expression becomes 1 - 2k*(-1/2) + k²*1.Simplifying that, the middle term becomes +k because -2k*(-1/2) is +k. So, the entire expression is 1 + k + k².So, |a - k*b|² = k² + k + 1. Now, to find the minimum value of this quadratic expression, I can treat it as a function of k: f(k) = k² + k + 1.To find the minimum of a quadratic function, I know that the vertex occurs at k = -b/(2a) where the quadratic is in the form ax² + bx + c. In this case, a is 1 and b is 1, so k = -1/(2*1) = -1/2.So, the minimum occurs at k = -1/2. Plugging this back into f(k), we get f(-1/2) = (-1/2)² + (-1/2) + 1.Calculating each term: (-1/2)² is 1/4, (-1/2) is -1/2, and then +1. So, 1/4 - 1/2 + 1.Combining these: 1/4 - 2/4 + 4/4 = (1 - 2 + 4)/4 = 3/4.So, |a - k*b|² is 3/4 at k = -1/2, which means |a - k*b| is sqrt(3/4) = sqrt(3)/2.Looking back at the options, that's option B. So, the minimum value is sqrt(3)/2.Wait, let me double-check my steps to make sure I didn't make a mistake. Starting from the beginning: unit vectors, angle 120°, dot product is -1/2. Squared magnitude is 1 + k + k², which is correct. Then, finding the vertex at k = -1/2, plugging back in, getting 3/4. Square root is sqrt(3)/2. Yeah, that seems right.I think I might have confused myself if I tried to visualize this geometrically. Maybe drawing the vectors a and b with 120° between them, and then trying to find the scalar k such that the vector k*b is as close as possible to a. So, it's like projecting a onto b or something? But I think the algebraic method I used is solid.Alternatively, I could think of this as minimizing the distance from the tip of vector a to the line spanned by vector b. The minimal distance would be the perpendicular distance, which in vector terms is the component of a orthogonal to b. But wait, in this case, we're scaling b by k, so it's not exactly the same as projection. Hmm.Wait, actually, if I set up the problem as minimizing |a - k*b|, it's equivalent to finding the scalar k that makes k*b as close as possible to a. That is, it's the projection of a onto b, but scaled appropriately.But in this case, since we're dealing with vectors, the projection scalar would be (a·b)/|b|², but since |b|² is 1, it's just a·b, which is -1/2. So, k would be -1/2, which is exactly what I found earlier. So, that confirms that k = -1/2 is indeed the scalar that minimizes the distance.Therefore, plugging back in, the minimal distance is sqrt(3)/2, which is option B. So, I think that's the correct answer.**Final Answer**boxed{B}

💡Okay, so I have this problem here about a function f(x) that's decreasing on the interval (-1, 1). The problem says that f(1 - a) is less than f(a² - 1), and I need to find the range of values for the real number a. Hmm, let me try to figure this out step by step.First, since f(x) is a decreasing function, that means if I have two inputs, say x₁ and x₂, and if x₁ is less than x₂, then f(x₁) is greater than f(x₂). So, in other words, as x increases, f(x) decreases. That's the definition of a decreasing function.Given that f(1 - a) < f(a² - 1), and since f is decreasing, this inequality should tell me something about the relationship between the inputs 1 - a and a² - 1. Specifically, because f is decreasing, if f(1 - a) is less than f(a² - 1), then the input 1 - a must be greater than a² - 1. Because in a decreasing function, a larger input gives a smaller output.So, I can write that as:1 - a > a² - 1Let me write that down:1 - a > a² - 1Now, I need to solve this inequality for a. Let me rearrange it:1 - a - a² + 1 > 0Wait, that's not quite right. Let me subtract a² - 1 from both sides:1 - a - (a² - 1) > 0Which simplifies to:1 - a - a² + 1 > 0Combine like terms:(1 + 1) - a - a² > 02 - a - a² > 0Let me write it as:-a² - a + 2 > 0Hmm, this is a quadratic inequality. Maybe I should multiply both sides by -1 to make it easier, but I have to remember that multiplying by a negative number reverses the inequality sign.So, multiplying both sides by -1:a² + a - 2 < 0Now, I have a quadratic inequality: a² + a - 2 < 0Let me factor this quadratic. I'm looking for two numbers that multiply to -2 and add up to 1. Hmm, 2 and -1. Yes, because 2 * (-1) = -2 and 2 + (-1) = 1.So, factoring:(a + 2)(a - 1) < 0Okay, so the product of (a + 2) and (a - 1) is less than zero. That means one of them is positive and the other is negative.Let me find the critical points where each factor is zero:a + 2 = 0 => a = -2a - 1 = 0 => a = 1These points divide the real number line into three intervals:1. a < -22. -2 < a < 13. a > 1Now, I'll test each interval to see where the product is negative.1. For a < -2, let's pick a = -3:(a + 2) = -3 + 2 = -1 (negative)(a - 1) = -3 - 1 = -4 (negative)Product: (-1)(-4) = 4 (positive) => Not less than zero.2. For -2 < a < 1, let's pick a = 0:(a + 2) = 0 + 2 = 2 (positive)(a - 1) = 0 - 1 = -1 (negative)Product: (2)(-1) = -2 (negative) => Less than zero. So this interval satisfies the inequality.3. For a > 1, let's pick a = 2:(a + 2) = 2 + 2 = 4 (positive)(a - 1) = 2 - 1 = 1 (positive)Product: (4)(1) = 4 (positive) => Not less than zero.So, the solution to the inequality a² + a - 2 < 0 is -2 < a < 1.But wait, I also need to consider the domain of the function f(x). The function is defined on (-1, 1), so the inputs 1 - a and a² - 1 must lie within this interval.So, let me write down the conditions:1. -1 < 1 - a < 12. -1 < a² - 1 < 1Let me solve each of these inequalities.Starting with the first one:-1 < 1 - a < 1Let me split this into two separate inequalities:-1 < 1 - a and 1 - a < 1First inequality: -1 < 1 - aSubtract 1 from both sides: -2 < -aMultiply both sides by -1 (and reverse the inequality): 2 > aSecond inequality: 1 - a < 1Subtract 1 from both sides: -a < 0Multiply both sides by -1 (and reverse the inequality): a > 0So, combining these two, we get:0 < a < 2Now, the second condition:-1 < a² - 1 < 1Again, split into two inequalities:-1 < a² - 1 and a² - 1 < 1First inequality: -1 < a² - 1Add 1 to both sides: 0 < a²Which is always true except when a = 0, but since we already have a > 0 from the first condition, this is satisfied.Second inequality: a² - 1 < 1Add 1 to both sides: a² < 2Take square roots: |a| < √2Which means: -√2 < a < √2But from the first condition, we have 0 < a < 2, so combining this with -√2 < a < √2, we get:0 < a < √2Because √2 is approximately 1.414, which is less than 2.So, putting it all together, from the quadratic inequality, we had -2 < a < 1, but considering the domain of f(x), we have 0 < a < √2.Wait, but √2 is about 1.414, which is less than 2, so the intersection of -2 < a < 1 and 0 < a < √2 is 0 < a < 1.But hold on, earlier from the quadratic inequality, we had -2 < a < 1, but from the domain, we have 0 < a < √2. So the overlap is 0 < a < 1.But wait, let me double-check. From the quadratic inequality, we had -2 < a < 1, but from the domain, we have 0 < a < √2. So the intersection is 0 < a < 1.But wait, let me think again. The quadratic inequality gave us -2 < a < 1, but the domain conditions gave us 0 < a < √2. So the values of a must satisfy both conditions, meaning a has to be greater than 0 and less than 1.Wait, but hold on, the quadratic inequality was a² + a - 2 < 0, which gave us -2 < a < 1. But the domain conditions are 0 < a < √2. So the overlap is 0 < a < 1.But wait, let me check if a can be between 1 and √2.If a is between 1 and √2, does it satisfy the original inequality f(1 - a) < f(a² - 1)?Let me pick a = 1.5, which is between 1 and √2 (since √2 ≈ 1.414). Wait, actually, 1.5 is greater than √2, so maybe pick a = 1.2.Wait, √2 is approximately 1.414, so 1.2 is less than √2.Let me test a = 1.2.First, check if 1 - a and a² - 1 are within (-1, 1).1 - 1.2 = -0.2, which is within (-1, 1).a² - 1 = (1.44) - 1 = 0.44, which is also within (-1, 1).Now, check if 1 - a > a² - 1.1 - 1.2 = -0.2a² - 1 = 0.44So, -0.2 > 0.44? No, that's not true. So, for a = 1.2, the inequality 1 - a > a² - 1 is not satisfied.Therefore, a cannot be between 1 and √2 because even though the domain conditions are satisfied, the inequality 1 - a > a² - 1 is not.So, that means the upper limit is actually 1, not √2.Wait, but earlier, from the quadratic inequality, we had -2 < a < 1, but the domain conditions gave us 0 < a < √2. So the intersection is 0 < a < 1.But wait, let me check a value just below 1, say a = 0.9.1 - 0.9 = 0.1a² - 1 = 0.81 - 1 = -0.19So, 0.1 > -0.19, which is true.So, for a = 0.9, the inequality holds.Similarly, at a = 1, 1 - 1 = 0 and a² - 1 = 0, so 0 > 0 is false. So a cannot be 1.Therefore, the upper limit is 1.Wait, but earlier, from the domain, we had 0 < a < √2, but from the quadratic inequality, we had -2 < a < 1. So the intersection is 0 < a < 1.But wait, let me think again. The quadratic inequality gave us -2 < a < 1, but the domain conditions require 0 < a < √2. So the overlap is 0 < a < 1.But wait, let me check a = 0.5.1 - 0.5 = 0.5a² - 1 = 0.25 - 1 = -0.75So, 0.5 > -0.75, which is true.So, f(0.5) < f(-0.75). Since f is decreasing, f(0.5) < f(-0.75) implies that 0.5 > -0.75, which is true.So, all values of a between 0 and 1 satisfy the inequality.But wait, let me check a = -1, but from the domain, a has to be greater than 0, so negative a's are already excluded.Wait, but in the quadratic inequality, a could be between -2 and 1, but the domain restricts a to be between 0 and √2. So, the overlap is 0 < a < 1.Wait, but earlier, when I solved the quadratic inequality, I got -2 < a < 1, but the domain conditions are 0 < a < √2. So, the intersection is 0 < a < 1.But wait, let me think about the domain again. The function f(x) is defined on (-1, 1), so the inputs 1 - a and a² - 1 must be within (-1, 1).So, for 1 - a, we have:-1 < 1 - a < 1Which simplifies to:0 < a < 2And for a² - 1:-1 < a² - 1 < 1Which simplifies to:0 < a² < 2So, a² < 2 implies that |a| < √2, so -√2 < a < √2.But since from the first condition, a > 0, we have 0 < a < √2.So, combining with the quadratic inequality solution of -2 < a < 1, the intersection is 0 < a < 1.Therefore, the range of a is 0 < a < 1.Wait, but earlier, I thought the answer was 0 < a < √2, but now I'm getting 0 < a < 1.Hmm, maybe I made a mistake earlier.Let me go back.From the quadratic inequality, we had a² + a - 2 < 0, which factors to (a + 2)(a - 1) < 0, giving -2 < a < 1.From the domain conditions, we have 0 < a < √2.So, the intersection is 0 < a < 1.Therefore, the correct range is 0 < a < 1.Wait, but let me test a = 0.9 again.1 - 0.9 = 0.1a² - 1 = 0.81 - 1 = -0.19So, 0.1 > -0.19, which is true, so f(0.1) < f(-0.19) because f is decreasing.Similarly, at a = 1, 1 - 1 = 0 and a² - 1 = 0, so f(0) < f(0), which is not true, so a cannot be 1.Therefore, the upper limit is indeed 1.So, the range of a is 0 < a < 1.Wait, but in the initial problem, the user wrote "The function f(x) is defined as a decreasing function on the interval (-1,1), and f(1 - a) < f(a² - 1). The range of values for the real number a is _______."And in the initial answer, the assistant wrote "0 < a < √2", but after my analysis, it seems it's 0 < a < 1.Wait, perhaps I made a mistake in the quadratic inequality.Let me re-examine the quadratic inequality.We had f(1 - a) < f(a² - 1), and since f is decreasing, 1 - a > a² - 1.So, 1 - a > a² - 1Bring all terms to one side:1 - a - a² + 1 > 0Which is 2 - a - a² > 0Which is the same as -a² - a + 2 > 0Multiplying both sides by -1 (and reversing inequality):a² + a - 2 < 0Which factors to (a + 2)(a - 1) < 0So, the solution is -2 < a < 1.Yes, that's correct.But from the domain, we have 0 < a < √2.So, the intersection is 0 < a < 1.Therefore, the correct range is 0 < a < 1.Wait, but in the initial answer, the assistant wrote 0 < a < √2, which is incorrect because the quadratic inequality restricts a to be less than 1.So, the correct answer should be 0 < a < 1.Wait, but let me check a = 0.5 again.1 - 0.5 = 0.5a² - 1 = 0.25 - 1 = -0.75So, 0.5 > -0.75, which is true, so f(0.5) < f(-0.75) because f is decreasing.Similarly, at a approaching 1 from below, say a = 0.99.1 - 0.99 = 0.01a² - 1 = 0.9801 - 1 = -0.0199So, 0.01 > -0.0199, which is true.But at a = 1, 1 - 1 = 0 and a² - 1 = 0, so f(0) < f(0) is false.Therefore, a must be less than 1.So, the correct range is 0 < a < 1.Therefore, the initial answer of 0 < a < √2 was incorrect because it didn't consider the quadratic inequality's upper limit of 1.So, the correct answer is 0 < a < 1.

💡First, I need to simplify the expression ((3x + 25) - (2x - 5)).To do this, I'll start by distributing the negative sign across the terms inside the second parenthesis. This changes the expression to (3x + 25 - 2x + 5).Next, I'll combine the like terms. The (x) terms are (3x) and (-2x), which add up to (x). The constant terms are (25) and (5), which add up to (30).Putting it all together, the simplified expression is (x + 30).

💡Okay, so I have this problem: I need to prove that every positive integer has a multiple which has a repetition number less than or equal to 2. The repetition number is defined as the number of distinct digits in the number when written in base 10. So, for example, the number 123 has a repetition number of 3 because it has three distinct digits, while the number 111 has a repetition number of 1 because it only has the digit 1.Alright, so my goal is to show that for any positive integer n, there exists some multiple of n (let's say kn for some integer k) such that when I write kn in base 10, it has at most two distinct digits. That means numbers like 111, 222, 1212, 1221, etc., would all satisfy this condition because they only use one or two different digits.Hmm, how should I approach this? Maybe I can think about the properties of numbers and their multiples. I know that multiples of a number can have various digit patterns, so perhaps there's a way to ensure that one of these multiples ends up having only one or two distinct digits.I remember something about the pigeonhole principle. Maybe that can help here. The pigeonhole principle states that if you have more pigeons than pigeonholes, at least two pigeons must share a pigeonhole. How can I apply this to the problem?Let's consider the set of numbers formed by multiplying n by integers from 1 up to some number. If I take enough multiples, maybe I can force one of them to have a limited number of distinct digits. But how many multiples do I need to consider? If I take too few, it might not work, but if I take too many, it might complicate things.Wait, the problem is about the number of distinct digits, which is related to the digits themselves, not directly to the number's value. So maybe I should think about the possible digit combinations. There are only 10 digits (0 through 9), so the number of possible distinct digits is limited.But repetition number is about the count of distinct digits, not their positions. So, for a number to have a repetition number less than or equal to 2, it can have either one or two different digits. For example, 111 has one distinct digit, and 1212 has two distinct digits.I wonder if there's a way to construct such a multiple. Maybe by considering numbers with repeating digits, like 111, 222, etc., and seeing if any of these can be multiples of n. But that might not always work because n could be any number, and these numbers are quite specific.Alternatively, maybe I can use modular arithmetic. If I can find a multiple of n that is congruent to a number with only one or two distinct digits modulo some power of 10, then perhaps that multiple will have the desired property.Wait, let's think about the pigeonhole principle again. If I consider the remainders of multiples of n modulo some number, say 10^k, then by the pigeonhole principle, two of these multiples must have the same remainder. The difference between these two multiples would then be divisible by 10^k, meaning it's a multiple of 10^k. But how does that help with the number of distinct digits?Hmm, maybe I need to think differently. Instead of looking at remainders, perhaps I should look at the digits themselves. If I can find a multiple of n that only uses one or two digits, then I'm done. But how can I ensure that such a multiple exists?I recall that for any number n, there exists a multiple of n that consists only of the digits 1 and 0. Is that true? I think it's related to the fact that you can construct numbers with specific digit patterns by using the pigeonhole principle on the remainders.Wait, let me try to formalize this. Suppose I want to find a multiple of n that consists only of the digits 1 and 0. I can consider numbers like 1, 11, 111, 1111, etc., and see if any of these are divisible by n. If not, then maybe I can combine these with other digits.But actually, the problem allows for two distinct digits, not necessarily 1 and 0. So maybe I can generalize this idea. Let's say I fix two digits, say a and b, and try to find a multiple of n that consists only of these two digits.But how many such numbers are there? For each position in the number, I have two choices: a or b. So, for a k-digit number, there are 2^k such numbers. But as k increases, 2^k grows exponentially, which might not be helpful.Wait, maybe instead of fixing the digits, I can consider the possible remainders of numbers with limited digits modulo n. If I can show that among these remainders, one of them is zero, then that number is a multiple of n.So, let's think about it. Suppose I consider all numbers that have only two distinct digits, say 0 and 1. There are infinitely many such numbers, but their remainders modulo n are finite (only n possible remainders). By the pigeonhole principle, if I consider enough of these numbers, two of them must have the same remainder modulo n. Then, their difference would be divisible by n, and the difference would be a number consisting of 0s and 1s.But wait, the difference of two numbers with 0s and 1s might not necessarily have only 0s and 1s. For example, 111 - 11 = 100, which still has only 0s and 1s. Hmm, interesting. So, maybe the difference would still have only 0s and 1s.But is that always the case? Let's test another example. Suppose I have 1111 - 111 = 1000, which still has only 0s and 1s. Another example: 101 - 11 = 90, which has digits 9 and 0. Oh, that's a problem. So, the difference can introduce new digits.Hmm, so my initial thought might not work because the difference could have more digits. So, maybe I need a different approach.Wait, perhaps instead of considering numbers with only 0s and 1s, I can consider numbers with only one non-zero digit. For example, numbers like 1, 10, 100, 1000, etc. These numbers have only one distinct non-zero digit. If I can find a multiple of n among these, then I'm done.But again, n could be any number, and these numbers are powers of 10, which might not necessarily be multiples of n. However, since 10 and n might not be coprime, perhaps I can adjust this idea.Alternatively, maybe I can use the fact that for any n, there exists a multiple of n that is a repunit, which is a number consisting of only 1s. But I'm not sure if that's always true. I think it's true if n is coprime with 10, but not necessarily otherwise.Wait, let me think again. The problem allows for two distinct digits, not just one. So maybe I can relax the condition a bit. Instead of trying to find a multiple with only one digit, I can allow two digits, which gives me more flexibility.So, perhaps I can construct a number with two digits such that it's a multiple of n. How can I ensure that such a number exists?Maybe I can use the pigeonhole principle on the possible remainders of numbers with two digits. Let's say I fix two digits, a and b, and consider numbers formed by these two digits. For example, numbers like a, b, aa, ab, ba, bb, aaa, aab, aba, abb, baa, bab, bba, bbb, etc.Each of these numbers has a remainder when divided by n. Since there are only n possible remainders, if I have more numbers than n, then by the pigeonhole principle, two of them must have the same remainder. Then, their difference would be divisible by n, and the difference would be a number with digits a and b.But wait, similar to before, the difference might introduce new digits. For example, if I have two numbers like 12 and 21, their difference is 9, which is a single digit. But if I have 111 - 11 = 100, which is still okay. But if I have 121 - 112 = 9, which is again a single digit. Hmm, so maybe the difference doesn't necessarily introduce new digits beyond a and b?Wait, no, that's not always the case. Let's take another example: 122 - 111 = 11, which is fine. But 121 - 112 = 9, which is a single digit, but 9 is not one of the original digits a or b (assuming a=1 and b=2). So, the difference can introduce a new digit, which is problematic.So, maybe this approach doesn't work because the difference can have digits outside of a and b. Therefore, I need a different strategy.Perhaps instead of looking at numbers with two digits, I can look at numbers where all digits are the same, like 111, 222, etc. If I can find such a number that's a multiple of n, then I'm done. But again, it's not clear how to ensure that such a multiple exists.Wait, maybe I can use the fact that for any n, there exists a multiple of n that is a repdigit (a number with repeating digits). I think this is a known result, but I'm not entirely sure. Let me try to recall.Yes, I think it's true that for any n, there exists a multiple of n that is a repdigit. A repdigit is a number like 111, 222, etc., where all digits are the same. So, if I can find such a multiple, then the repetition number would be 1, which is certainly less than or equal to 2.But how do I prove that such a multiple exists? Maybe by using the pigeonhole principle again. Consider the numbers 1, 11, 111, 1111, etc., modulo n. Since there are infinitely many such numbers and only n possible remainders, by the pigeonhole principle, two of them must have the same remainder modulo n. Then, their difference would be divisible by n, and the difference would be a number consisting of 1s and 0s.Wait, but the difference of two repdigits might not necessarily be a repdigit. For example, 111 - 11 = 100, which is not a repdigit, but it has only two distinct digits: 1 and 0. So, in this case, the difference has a repetition number of 2, which is still acceptable.So, maybe instead of trying to get a repdigit, I can settle for a number with at most two distinct digits. That way, even if the difference introduces a new digit, it's still within the limit.So, let's formalize this idea. Consider the set of numbers formed by concatenating k digits of 1, for k from 1 to n+1. These numbers are 1, 11, 111, ..., up to a number with n+1 digits. There are n+1 such numbers. When we take these numbers modulo n, there are only n possible remainders. By the pigeonhole principle, at least two of these numbers must have the same remainder modulo n.Let's say two such numbers, say the one with m digits and the one with l digits (where m > l), have the same remainder modulo n. Then, their difference is divisible by n. The difference between these two numbers is a number with m - l digits, where the first digit is 1, followed by (m - l - 1) 0s, and ending with a 1. Wait, no, actually, subtracting two repunits (numbers with all 1s) would result in a number like 111...111 - 111...111 = 000...000, but that's not helpful.Wait, no, actually, subtracting a smaller repunit from a larger one would give a number like 111...111 - 111...111 = 000...000, but that's trivial. Maybe I need to consider a different approach.Alternatively, perhaps instead of subtracting two repunits, I can consider the difference as a number with 1s and 0s. For example, 111 - 11 = 100, which has digits 1 and 0. Similarly, 1111 - 111 = 1000, which also has digits 1 and 0. So, in general, subtracting a repunit with l digits from one with m digits (where m > l) gives a number with a 1 followed by (m - l - 1) 0s and ending with a 1.Wait, no, actually, 111 - 11 = 100, which is 1 followed by two 0s. Similarly, 1111 - 111 = 1000, which is 1 followed by three 0s. So, in general, the difference would be a 1 followed by (m - l - 1) 0s and ending with a 1. But actually, no, that's not correct. Let's do the subtraction properly.Take 111 (which is 111) minus 11 (which is 11). 111 - 11 = 100. Similarly, 1111 - 111 = 1000. So, the difference is a 1 followed by (m - l) 0s. Wait, no, in the first case, m = 3, l = 2, so m - l = 1, but the difference is 100, which has two 0s. Hmm, maybe my initial thought was off.Wait, actually, 111 - 11 = 100, which is 1 followed by two 0s. So, the number of 0s is equal to the number of digits in the smaller repunit. Similarly, 1111 - 111 = 1000, which is 1 followed by three 0s. So, the number of 0s is equal to the number of digits in the smaller repunit minus one.Wait, maybe I'm overcomplicating this. The key point is that the difference between two repunits is a number that starts with a 1, followed by some 0s, and ends with a 1. So, it has two distinct digits: 1 and 0. Therefore, the repetition number is 2, which satisfies the condition.So, to recap, by considering the repunits (numbers with all 1s) modulo n, we can find two such numbers that have the same remainder. Their difference is divisible by n and has only two distinct digits: 1 and 0. Therefore, this difference is a multiple of n with a repetition number of 2.But wait, what if n is a multiple of 10? Then, 10 divides n, and the difference we found is 100, which is also divisible by 10. But 100 has two distinct digits: 1 and 0. So, it still works.What if n is 1? Then, any number is a multiple of 1, and we can choose 1 itself, which has a repetition number of 1. Similarly, for n=2, we can find 2, which has a repetition number of 1, or 10, which has two distinct digits.Wait, but in the case where n is 1, the multiple is 1, which is fine. For n=2, the multiple could be 2, 22, 222, etc., all of which have a repetition number of 1. So, in these cases, we don't even need to go through the repunit difference approach.But the general idea is that for any n, by considering repunits modulo n, we can find a multiple of n that has at most two distinct digits. Therefore, the repetition number is less than or equal to 2.So, putting it all together, here's the step-by-step explanation:1. Consider the set of repunits (numbers consisting only of 1s) with k digits, where k ranges from 1 to n+1. There are n+1 such numbers.2. When each of these repunits is divided by n, there are only n possible remainders (from 0 to n-1).3. By the pigeonhole principle, at least two of these repunits must have the same remainder when divided by n.4. Let these two repunits be R_m (with m digits) and R_l (with l digits), where m > l.5. The difference R_m - R_l is divisible by n because they have the same remainder.6. The difference R_m - R_l is a number that starts with a 1, followed by (m - l - 1) 0s, and ends with a 1. This number has only two distinct digits: 1 and 0.7. Therefore, R_m - R_l is a multiple of n with a repetition number of 2.8. Hence, every positive integer n has a multiple with a repetition number less than or equal to 2.Wait, but in step 6, I said the difference starts with a 1, followed by (m - l - 1) 0s, and ends with a 1. Let me verify this with an example.Take R_3 = 111 and R_2 = 11. Their difference is 111 - 11 = 100, which is 1 followed by two 0s. So, in this case, m = 3, l = 2, so m - l = 1, but the number of 0s is two. Hmm, so my earlier statement might not be accurate.Wait, actually, the number of 0s is equal to the number of digits in R_l. In this case, R_l has 2 digits, so the difference has two 0s. So, in general, the difference R_m - R_l will have (m - l) digits, where the first digit is 1, followed by (m - l - 1) 0s, and the last digit is 1. Wait, no, in the example, m - l = 1, but the difference has two 0s. So, perhaps the number of 0s is (m - l).Wait, let's take another example. R_4 = 1111 and R_2 = 11. Their difference is 1111 - 11 = 1100, which has two 1s and two 0s. So, the number of 0s is equal to the number of digits in R_l, which is 2. Similarly, R_5 - R_3 = 11111 - 111 = 11000, which has two 1s and three 0s. So, the number of 0s is equal to the number of digits in R_l.Wait, so in general, R_m - R_l = 11...1 (m digits) - 11...1 (l digits) = 100...0 (with (m - l) digits). Wait, no, that's not correct. Let's do the subtraction properly.Take R_m = 11...1 (m digits) and R_l = 11...1 (l digits), with m > l.When we subtract R_l from R_m, we need to align the digits properly. So, R_m is 1 followed by (m-1) 1s, and R_l is 1 followed by (l-1) 1s. To subtract, we can write R_m as 100...0 + R_{m-1}, but that might complicate things.Alternatively, let's consider R_m as a number with m digits all 1s, and R_l as a number with l digits all 1s. When we subtract R_l from R_m, we get a number that has (m - l) digits, where the first digit is 1, followed by (m - l - 1) 0s, and ending with a 1. Wait, no, that doesn't seem right.Wait, let's take R_4 = 1111 and R_2 = 11. Subtracting R_2 from R_4:1111- 11------1100So, the result is 1100, which has two 1s and two 0s. So, the number of 1s is equal to the number of digits in R_l, and the number of 0s is equal to (m - l).Wait, in this case, m = 4, l = 2, so m - l = 2, and we have two 0s. Similarly, R_5 - R_3 = 11111 - 111 = 11000, which has two 1s and three 0s. So, the number of 1s is equal to the number of digits in R_l, and the number of 0s is equal to (m - l).Therefore, in general, R_m - R_l will have l 1s and (m - l) 0s. So, the number of distinct digits is 2: 1 and 0.Therefore, the difference R_m - R_l is a multiple of n with a repetition number of 2.This works as long as R_m and R_l are repunits with m > l digits, and they have the same remainder modulo n. By the pigeonhole principle, such a pair must exist among the first n+1 repunits.Therefore, we've shown that for any positive integer n, there exists a multiple of n with a repetition number less than or equal to 2.I think this makes sense. The key idea is using the pigeonhole principle on the remainders of repunits modulo n, ensuring that two of them have the same remainder, and their difference is a multiple of n with only two distinct digits. This difference will have 1s and 0s, giving it a repetition number of 2.So, in conclusion, every positive integer has a multiple with a repetition number of at most 2.