💡Alright, so I have this problem where I need to figure out which of the given graphs have inverses. There are five graphs labeled A to E, each described in a certain way. I remember that for a function to have an inverse, it needs to be one-to-one, meaning that each input has a unique output and vice versa. This is often checked using the horizontal line test: if any horizontal line intersects the graph more than once, the function doesn't have an inverse.Let me go through each graph one by one.**Graph A**: It's a downward facing parabola with the vertex at (0,3). Okay, a parabola is a U-shaped curve, and since it's facing downward, it opens downward. The vertex is the highest point on the graph. Now, parabolas are not one-to-one functions because they fail the horizontal line test. If I imagine drawing a horizontal line across the parabola, it would intersect the graph at two points for any y-value below the vertex. So, for example, if I draw a horizontal line at y=2, it would intersect the parabola at two different x-values. Therefore, Graph A does not have an inverse.**Graph B**: This is a piecewise linear function. It has a horizontal line from (-3,-2) to (3,-2) and then a vertical line segment connecting (3,-2) to (3,3). Hmm, okay, so the first part is a horizontal line, which means it's constant over that interval. That already tells me it's not one-to-one because all the x-values from -3 to 3 map to the same y-value, -2. If I apply the horizontal line test here, a horizontal line at y=-2 would intersect the graph infinitely many times along that horizontal segment. So, Graph B definitely doesn't have an inverse.**Graph C**: This is a linear function represented by the line y = x. Linear functions are straight lines, and y = x is a straight line passing through the origin with a slope of 1. Since it's a straight line, it's one-to-one because it passes both the vertical and horizontal line tests. For any x-value, there's only one y-value, and for any y-value, there's only one x-value. So, Graph C does have an inverse, which, interestingly, is the same function y = x because it's its own inverse.**Graph D**: It's a semicircle centered at the origin with radius 3, oriented upward from (-3,0) to (3,0). Wait, a semicircle is only half of a circle, so it's either the top half or the bottom half. Since it's oriented upward, it's the top half. Now, is a semicircle a function? Well, if it's the top half, then for each x-value between -3 and 3, there's only one y-value, so it does pass the vertical line test and is a function. But does it have an inverse? To have an inverse, it needs to pass the horizontal line test. If I draw a horizontal line across the semicircle, it will intersect the graph at two points for any y-value between 0 and 3. For example, y=1 would intersect at two x-values, one positive and one negative. Therefore, Graph D does not have an inverse.**Graph E**: This is a cubic function represented by the equation y = x^3 + x. Cubic functions are interesting because they can have one or more turning points, but in this case, let's analyze it. To check if it's one-to-one, I can look at its derivative. The derivative of y with respect to x is dy/dx = 3x^2 + 1. Since 3x^2 is always non-negative and adding 1 makes it always positive, the derivative is always positive. This means the function is strictly increasing; it never decreases. A strictly increasing function passes the horizontal line test because it never takes the same y-value more than once. Therefore, Graph E is one-to-one and has an inverse.So, summarizing my thoughts:- Graph A: Fails the horizontal line test, no inverse.- Graph B: Fails the horizontal line test, no inverse.- Graph C: Passes both tests, has an inverse.- Graph D: Fails the horizontal line test, no inverse.- Graph E: Passes both tests, has an inverse.Therefore, only Graphs C and E have inverses.

💡Okay, so I'm trying to understand why the Central Bank of the Russian Federation would be interested in using a loyalty program, especially one tied to the "Mir" payment system. I know that loyalty programs are common in retail and banking to encourage customer retention and spending. But why would a central bank, which is more about monetary policy and financial stability, care about this?First, I think about the role of the Central Bank. They're responsible for maintaining the stability of the financial system, controlling inflation, and managing the country's currency. So, how does a loyalty program fit into that? Maybe it's about promoting the use of domestic payment systems over international ones like Visa or MasterCard. If more people use "Mir" cards, it reduces reliance on foreign payment networks, which could be strategic, especially in times of geopolitical tensions or sanctions.Another angle could be economic stimulation. If the loyalty program offers cashback or discounts, it might encourage people to spend more, which could boost the economy. More spending means more transactions, which could help in tracking economic activity and might even increase tax revenues if more businesses are using formal payment methods.Now, thinking about the benefits for banks whose clients participate in the loyalty program. Banks might see increased customer loyalty because customers get rewards for using their cards. This could lead to more consistent use of the bank's services and potentially higher fees or interest income from increased card usage. Also, banks might gain better insights into customer spending habits, allowing them to offer more tailored financial products or services.But then, why does the payment system require registration of "Mir" cards for these bonuses? Why not just let anyone with a "Mir" card participate? Maybe it's about data collection. By requiring registration, the system can gather information on how customers use their cards, which could be valuable for marketing or improving the program. It might also help in preventing fraud by ensuring that only legitimate users receive the benefits.Also, registration could be a way to ensure that the benefits are distributed fairly and efficiently. Without registration, it might be harder to track who is eligible for what rewards, leading to potential misuse or administrative challenges. Plus, it might help in personalizing the rewards based on individual spending patterns, making the program more appealing to users.I'm also considering economic arguments here. If the payment system requires registration, it might be a way to control costs. By limiting the number of participants, they can manage the expenses associated with the loyalty program, like cashback rewards or discounts. It could also be a strategy to build a stronger relationship with registered users, encouraging them to use the "Mir" system more exclusively.Another thought is about data security. By requiring registration, the system can ensure that only authorized users have access to the benefits, reducing the risk of fraud or unauthorized transactions. This could enhance the overall security of the payment system, which is crucial for maintaining user trust and the integrity of the financial network.In summary, the Central Bank might be interested in the loyalty program to promote the use of the domestic payment system, stimulate economic activity, and enhance financial stability. For banks, the benefits could include increased customer loyalty, higher transaction volumes, and better customer insights. Requiring registration for "Mir" cards likely serves purposes related to data collection, fraud prevention, cost management, and personalization of the loyalty program.

💡Alright, so I have this problem about a triangle inscribed in a circle, and points marked on its sides. The triangle was erased, and I need to prove that it can be uniquely restored if and only if the cevians from the vertices through these marked points concur. Hmm, okay, let me try to unpack this.First, let me recall some concepts. A triangle inscribed in a circle means it's a cyclic triangle, right? So all its vertices lie on the circumference of the circle. Now, points A1, B1, C1 are marked on the sides of the triangle. After marking these points, the original triangle ABC was erased. So now, all we have are these three points on the sides, and the circle k.The problem is asking me to prove that the original triangle can be uniquely restored if and only if the cevians AA1, BB1, and CC1 concur. That means, if these three lines meet at a single point, then we can uniquely reconstruct the original triangle ABC. Conversely, if the triangle can be uniquely restored, then these cevians must concur.Okay, so I need to prove both directions: the "if" part and the "only if" part. Let me start by understanding what it means for the triangle to be uniquely restored. If the triangle can be uniquely restored, that means given the points A1, B1, C1 on the circle, there's only one possible triangle ABC that could have these points on its sides. If it's not unique, there might be multiple triangles that fit these points.Now, cevians are lines from a vertex to the opposite side. So AA1 is a line from A to A1 on BC, BB1 from B to B1 on AC, and CC1 from C to C1 on AB. If these three lines meet at a single point, that point is called the point of concurrency. Ceva's theorem comes to mind here, which gives a condition for three cevians to concur.Ceva's theorem states that for concurrent cevians AA1, BB1, CC1 in triangle ABC, the product of the ratios of the divided sides is equal to 1. That is, (BA1/A1C) * (CB1/B1A) * (AC1/C1B) = 1. So, if these cevians concur, this product equals 1. Maybe this will be useful in proving the uniqueness.But wait, the problem is about restoring the triangle from the points A1, B1, C1 on the circle. So, if I have three points on the circle, how can I reconstruct the triangle? I guess I need to find the vertices A, B, C such that A1 is on BC, B1 is on AC, and C1 is on AB.If the cevians concur, then perhaps there's a unique way to reconstruct the triangle. Conversely, if the triangle can be uniquely reconstructed, then the cevians must concur because otherwise, there might be multiple triangles satisfying the conditions.Let me think about the "if" part first: if the cevians concur, then the triangle can be uniquely restored. Suppose AA1, BB1, CC1 meet at a point T. Then, given points A1, B1, C1 on the circle, we can use this concurrency point to reconstruct the triangle.Wait, but how exactly? If I have the circle and points A1, B1, C1, and I know that the cevians concur at T, can I find A, B, C? Maybe by intersecting lines from T through A1, B1, C1 with the circle? Hmm, not exactly, because A, B, C are the vertices, not necessarily on the lines TA1, TB1, TC1.Alternatively, maybe using the properties of cyclic quadrilaterals or something related to power of a point. If T is the point of concurrency, then perhaps T has equal power with respect to the circle, or something like that.Wait, maybe I should consider the circumcircle k. Since ABC is inscribed in k, all points A, B, C lie on k. So, if I can find three points on k such that A1 is on BC, B1 is on AC, and C1 is on AB, and the cevians concur at T, then ABC is uniquely determined.But how do I ensure uniqueness? Maybe if T is inside the circle, and the cevians are constructed in a certain way, then the intersection points A, B, C are uniquely determined.Alternatively, perhaps using inversion. If I invert the figure with respect to the circle k, maybe the points A1, B1, C1 will map to points that can help reconstruct ABC.Wait, maybe that's complicating things. Let me think about the converse first: if the triangle can be uniquely restored, then the cevians must concur. Suppose that given A1, B1, C1 on the circle, there's only one triangle ABC such that A1 is on BC, B1 is on AC, and C1 is on AB. Then, in this case, the cevians AA1, BB1, CC1 must concur.Why? Because if they didn't concur, then perhaps we could have another triangle A'B'C' with the same points A1, B1, C1 on its sides, but with cevians not concurring, leading to a different triangle, contradicting the uniqueness.Wait, that seems a bit hand-wavy. Maybe I need a more concrete argument.Perhaps using the concept of Ceva's theorem again. If the triangle can be uniquely restored, then the ratios (BA1/A1C), (CB1/B1A), (AC1/C1B) must satisfy Ceva's condition, meaning their product is 1, which implies concurrency.But I'm not sure if that directly follows. Maybe I need to think about the dual problem. If the cevians don't concur, then there might be multiple triangles satisfying the conditions, hence the triangle isn't uniquely restored.Alternatively, maybe using projective geometry. If the cevians don't concur, then the points A1, B1, C1 might lie on a conic, but since they lie on the circle k, which is already a conic, perhaps there's a uniqueness condition.Wait, I'm getting a bit tangled here. Let me try to structure my thoughts.First, suppose that AA1, BB1, CC1 concur at a point T. Then, given points A1, B1, C1 on the circle k, can we reconstruct ABC uniquely?Yes, because T is the point of concurrency, and we can use the properties of Ceva's theorem to determine the ratios in which the sides are divided. Since the product of the ratios is 1, we can solve for the positions of A, B, C on the circle such that A1, B1, C1 lie on the respective sides.Moreover, since the triangle is inscribed in the circle, the positions of A, B, C are uniquely determined by the intersection points of the cevians with the circle.On the other hand, suppose that the triangle can be uniquely restored. Then, there must be a unique set of points A, B, C on the circle such that A1, B1, C1 lie on BC, AC, AB respectively. If the cevians AA1, BB1, CC1 did not concur, then by Ceva's theorem, the product of the ratios would not equal 1, implying that there could be another triangle A'B'C' with the same points A1, B1, C1 on its sides, but with cevians not concurring, leading to a different triangle, contradicting the uniqueness.Therefore, the triangle can be uniquely restored if and only if the cevians concur.Wait, but I'm not sure if this is rigorous enough. Maybe I need to use more precise theorems or properties.Perhaps using the concept of the circumcircle and the power of a point. If AA1, BB1, CC1 concur at T, then T has equal power with respect to the circle k, meaning that TA * TA1 = TB * TB1 = TC * TC1. This could help in uniquely determining the positions of A, B, C on the circle.Alternatively, maybe using the concept of poles and polars. If T is the point of concurrency, then the polar of T with respect to circle k would be the line BC, AC, or AB, depending on the cevian. This could help in reconstructing the sides of the triangle.But I'm not entirely sure how to tie this together. Maybe I need to look up some related theorems or properties that connect cevian concurrency with the uniqueness of triangle reconstruction from points on its sides.Wait, I recall something about the converse of Ceva's theorem. If the product of the ratios equals 1, then the cevians concur. So, if we can show that the product equals 1, then concurrency follows.But in this case, we're given that the triangle can be uniquely restored, which implies that the ratios must satisfy Ceva's condition, hence the cevians concur.Similarly, if the cevians concur, then Ceva's condition is satisfied, and the triangle can be uniquely determined by the intersection points on the circle.So, putting it all together, the uniqueness of the triangle restoration is equivalent to the concurrency of the cevians.I think I'm getting closer. Maybe I should try to formalize this argument.Suppose that AA1, BB1, CC1 concur at T. Then, by Ceva's theorem, (BA1/A1C) * (CB1/B1A) * (AC1/C1B) = 1. Given points A1, B1, C1 on the circle, we can use this ratio condition to uniquely determine the positions of A, B, C on the circle such that the cevians concur at T. Hence, the triangle ABC is uniquely restored.Conversely, suppose that the triangle ABC can be uniquely restored from points A1, B1, C1 on the circle. Then, the ratios (BA1/A1C), (CB1/B1A), (AC1/C1B) must satisfy Ceva's condition, implying that the cevians concur at a point T. Therefore, the cevians must concur.Thus, the triangle can be uniquely restored if and only if the cevians AA1, BB1, CC1 concur.I think this makes sense. The key idea is that the concurrency of the cevians ensures that the ratios satisfy Ceva's theorem, which in turn allows us to uniquely determine the vertices of the triangle on the circumcircle. Conversely, if the triangle can be uniquely restored, the ratios must satisfy Ceva's condition, implying concurrency.I should probably check if there are any edge cases or exceptions. For example, what if the cevians are parallel or something? But since the triangle is inscribed in a circle, the cevians can't be parallel because they all meet at the vertices, which are on the circle. So, as long as the cevians are not parallel, they will either concur or not.Wait, but in projective geometry, even parallel lines meet at a point at infinity. So, maybe even in that case, the cevians concur at a point at infinity, but I don't think that affects the uniqueness of the triangle restoration because the triangle would still be uniquely determined.Hmm, maybe I'm overcomplicating it. The main point is that the concurrency of the cevians is equivalent to the uniqueness of the triangle restoration, based on Ceva's theorem and the properties of cyclic triangles.Alright, I think I've got a reasonable understanding now. Time to put it all together in a clear, step-by-step explanation.

💡Okay, so I've got this equation here: √(36 - k²) - 6 = 0. I need to figure out for which value of k this equation holds true. The options are A) k=6, B) k=0, C) k=±6, and D) no values of k. Hmm, let's break this down step by step.First, I remember that when dealing with square roots, the expression inside the square root must be non-negative because you can't take the square root of a negative number in the set of real numbers. So, 36 - k² has to be greater than or equal to zero. That gives me 36 - k² ≥ 0, which simplifies to k² ≤ 36. Taking the square root of both sides, that means |k| ≤ 6, so k is between -6 and 6, inclusive. Okay, so k can't be more than 6 or less than -6.Now, looking back at the original equation: √(36 - k²) - 6 = 0. I want to solve for k. Let me isolate the square root term first. If I add 6 to both sides, I get √(36 - k²) = 6. That seems manageable.Next, to get rid of the square root, I can square both sides of the equation. Squaring the left side, (√(36 - k²))², simplifies to 36 - k². Squaring the right side, 6², is 36. So now I have 36 - k² = 36.Hmm, let's subtract 36 from both sides to see what happens. That gives me -k² = 0. If I multiply both sides by -1, I get k² = 0. Taking the square root of both sides, k = 0. So, k must be zero.Wait a second, let me make sure I didn't make any mistakes. I started with √(36 - k²) - 6 = 0, isolated the square root to get √(36 - k²) = 6, then squared both sides to get 36 - k² = 36. Subtracting 36 from both sides gives -k² = 0, so k² = 0, which means k = 0. That seems correct.But just to be thorough, let me plug k = 0 back into the original equation to verify. Plugging in, I get √(36 - 0²) - 6 = √36 - 6 = 6 - 6 = 0. Yep, that works out. So k = 0 is a valid solution.Looking at the answer choices, option B is k = 0, which matches my solution. Option A is k = 6, but if I plug that in, I get √(36 - 36) - 6 = √0 - 6 = 0 - 6 = -6, which isn't zero. Similarly, option C is k = ±6, but both 6 and -6 give me -6 when plugged into the equation, not zero. Option D says no values of k, but I found that k = 0 works, so D isn't correct either.Therefore, the correct answer must be B) k = 0.

💡To find the sum of the arithmetic series ( 10 + 11 + 12 + cdots + 100 ), we can use the formula for the sum of an arithmetic series:[S = frac{n(a_1 + a_n)}{2}]where:- ( n ) is the number of terms,- ( a_1 ) is the first term,- ( a_n ) is the last term.1. **Identify the first term (( a_1 )) and the last term (( a_n ))**: - The first term ( a_1 = 10 ). - The last term ( a_n = 100 ).2. **Determine the number of terms (( n ))**: - The common difference ( d = 1 ). - Using the formula for the ( n )-th term of an arithmetic sequence: [ a_n = a_1 + (n - 1)d ] Plugging in the known values: [ 100 = 10 + (n - 1) times 1 ] Simplifying: [ 100 - 10 = n - 1 implies 90 = n - 1 implies n = 91 ]3. **Calculate the sum (( S ))**: - Using the sum formula: [ S = frac{91(10 + 100)}{2} = frac{91 times 110}{2} = 91 times 55 = 5005 ]Therefore, the sum of the arithmetic series is ( 5005 ).

💡Okay, so I have this problem where I need to find the smallest and largest natural numbers ( n ) such that there are exactly 100 natural numbers between ( 0.5n ) and ( 0.6n ). Hmm, let me try to break this down step by step.First, let me understand what it means for there to be exactly 100 natural numbers between ( 0.5n ) and ( 0.6n ). If I think about it, the numbers between ( 0.5n ) and ( 0.6n ) would be the integers starting from the smallest integer greater than ( 0.5n ) up to the largest integer less than ( 0.6n ). So, if I denote the smallest integer greater than ( 0.5n ) as ( k ), then the numbers would be ( k, k+1, k+2, ldots, k+99 ), right? Because there are 100 numbers in total.So, mathematically, I can write this as:[ 0.5n < k leq k+99 < 0.6n ]But since ( k ) has to be an integer, I should consider the ceiling of ( 0.5n ) and the floor of ( 0.6n ). That is:[ k = lceil 0.5n rceil ]and[ k + 99 = lfloor 0.6n rfloor ]So, substituting ( k ) into the second equation:[ lceil 0.5n rceil + 99 = lfloor 0.6n rfloor ]Now, let me think about the difference between ( 0.6n ) and ( 0.5n ). That's ( 0.1n ). This difference represents the length of the interval between ( 0.5n ) and ( 0.6n ). Since we're looking for exactly 100 integers in this interval, the length of the interval must be just enough to contain 100 integers. But since ( 0.5n ) and ( 0.6n ) might not be integers themselves, the exact number of integers between them can vary depending on their fractional parts.So, to ensure there are exactly 100 integers between ( 0.5n ) and ( 0.6n ), the length of the interval ( 0.1n ) should be slightly more than 100 but not so much that it includes 101 integers. Wait, actually, if the interval is exactly 100, then it should contain 100 integers. But because of the fractional parts, it might sometimes include 100 or 101. Hmm, maybe I need to set up inequalities to capture this.Let me denote ( a = 0.5n ) and ( b = 0.6n ). The number of integers between ( a ) and ( b ) is given by:[ lfloor b rfloor - lceil a rceil + 1 = 100 ]So,[ lfloor 0.6n rfloor - lceil 0.5n rceil + 1 = 100 ]Which simplifies to:[ lfloor 0.6n rfloor - lceil 0.5n rceil = 99 ]Now, let me express ( lfloor 0.6n rfloor ) and ( lceil 0.5n rceil ) in terms of ( n ). Let's denote ( m = lfloor 0.5n rfloor ). Then ( lceil 0.5n rceil ) would be ( m ) if ( 0.5n ) is an integer, otherwise ( m + 1 ). Similarly, ( lfloor 0.6n rfloor ) would be some integer ( p ).So, substituting back, we have:[ p - (m + 1) = 99 ]Which gives:[ p - m = 100 ]But ( p = lfloor 0.6n rfloor ) and ( m = lfloor 0.5n rfloor ). So,[ lfloor 0.6n rfloor - lfloor 0.5n rfloor = 100 ]Now, let me think about the relationship between ( lfloor 0.6n rfloor ) and ( lfloor 0.5n rfloor ). Since ( 0.6n - 0.5n = 0.1n ), the difference between ( lfloor 0.6n rfloor ) and ( lfloor 0.5n rfloor ) is approximately ( 0.1n ). But since we're dealing with floor functions, the actual difference could be either ( lfloor 0.1n rfloor ) or ( lfloor 0.1n rfloor + 1 ) depending on the fractional parts.Given that ( lfloor 0.6n rfloor - lfloor 0.5n rfloor = 100 ), we can say that:[ 0.1n - 1 < 100 leq 0.1n ]Wait, that might not be precise. Let me think again.If ( lfloor 0.6n rfloor - lfloor 0.5n rfloor = 100 ), then:[ 0.6n - 1 < lfloor 0.6n rfloor leq 0.6n ]and[ 0.5n leq lfloor 0.5n rfloor < 0.5n + 1 ]So, subtracting these inequalities:[ 0.6n - 1 - (0.5n + 1) < lfloor 0.6n rfloor - lfloor 0.5n rfloor leq 0.6n - 0.5n ]Simplifying:[ 0.1n - 2 < 100 leq 0.1n ]So,[ 0.1n > 100 ]and[ 0.1n - 2 < 100 ]Which gives:[ n > 1000 ]and[ n < 1020 ]Wait, that doesn't seem right. Let me check my steps.Actually, when subtracting inequalities, I should be careful. Let me write it properly.From:[ lfloor 0.6n rfloor leq 0.6n ]and[ lfloor 0.5n rfloor geq 0.5n - 1 ]So,[ lfloor 0.6n rfloor - lfloor 0.5n rfloor leq 0.6n - (0.5n - 1) = 0.1n + 1 ]Similarly,[ lfloor 0.6n rfloor geq 0.6n - 1 ]and[ lfloor 0.5n rfloor leq 0.5n ]So,[ lfloor 0.6n rfloor - lfloor 0.5n rfloor geq (0.6n - 1) - 0.5n = 0.1n - 1 ]Therefore, combining these:[ 0.1n - 1 leq lfloor 0.6n rfloor - lfloor 0.5n rfloor leq 0.1n + 1 ]But we know that ( lfloor 0.6n rfloor - lfloor 0.5n rfloor = 100 ), so:[ 0.1n - 1 leq 100 leq 0.1n + 1 ]Which gives two inequalities:1. ( 0.1n - 1 leq 100 ) → ( 0.1n leq 101 ) → ( n leq 1010 )2. ( 100 leq 0.1n + 1 ) → ( 99 leq 0.1n ) → ( n geq 990 )So, combining these, ( 990 leq n leq 1010 ). But wait, is this correct? Let me test with specific values.Let's take ( n = 1000 ). Then ( 0.5n = 500 ) and ( 0.6n = 600 ). The integers between 500 and 600 are 501 to 599, which is 99 numbers. Hmm, that's only 99, but we need 100. So, ( n = 1000 ) is too small.Wait, so maybe my inequalities are off. Let me think again.If ( n = 1000 ), ( 0.5n = 500 ), ( 0.6n = 600 ). The integers strictly between 500 and 600 are 501 to 599, which is 99 numbers. So, to get 100 numbers, we need the interval to include 100 integers. So, perhaps the lower bound should be such that ( 0.5n ) is just below an integer, and ( 0.6n ) is just above another integer, so that the count is 100.Wait, maybe I should consider the number of integers in the interval ( (0.5n, 0.6n) ). The number of integers in this interval is ( lfloor 0.6n rfloor - lceil 0.5n rceil ). We need this to be 100.So,[ lfloor 0.6n rfloor - lceil 0.5n rceil = 100 ]Let me denote ( lceil 0.5n rceil = m ), so ( m - 1 < 0.5n leq m ). Similarly, ( lfloor 0.6n rfloor = p ), so ( p leq 0.6n < p + 1 ).Given ( p - m = 100 ), so ( p = m + 100 ).From ( m - 1 < 0.5n leq m ), we have:[ m - 1 < 0.5n leq m ]Multiplying by 2:[ 2m - 2 < n leq 2m ]From ( p leq 0.6n < p + 1 ), and ( p = m + 100 ):[ m + 100 leq 0.6n < m + 101 ]Multiplying by ( frac{10}{6} ):[ frac{10}{6}(m + 100) leq n < frac{10}{6}(m + 101) ]Simplifying:[ frac{5}{3}(m + 100) leq n < frac{5}{3}(m + 101) ]Now, combining this with the earlier inequality ( 2m - 2 < n leq 2m ), we have:[ frac{5}{3}(m + 100) leq 2m ]and[ 2m - 2 < frac{5}{3}(m + 101) ]Let me solve the first inequality:[ frac{5}{3}(m + 100) leq 2m ]Multiply both sides by 3:[ 5(m + 100) leq 6m ][ 5m + 500 leq 6m ][ 500 leq m ]Now, the second inequality:[ 2m - 2 < frac{5}{3}(m + 101) ]Multiply both sides by 3:[ 6m - 6 < 5(m + 101) ][ 6m - 6 < 5m + 505 ][ m < 511 ]So, combining both, ( 500 leq m < 511 ). Since ( m ) is an integer, ( m ) ranges from 500 to 510.Now, for each ( m ) in 500 to 510, we can find the corresponding ( n ).From ( frac{5}{3}(m + 100) leq n leq 2m ), let's compute the range for ( n ).For ( m = 500 ):[ frac{5}{3}(600) = 1000 leq n leq 1000 ]So, ( n = 1000 ). But earlier, we saw that ( n = 1000 ) gives only 99 numbers. Hmm, maybe I need to adjust.Wait, perhaps I should consider that ( n ) must satisfy both ( n geq frac{5}{3}(m + 100) ) and ( n leq 2m ). Let's compute these for each ( m ).For ( m = 500 ):[ frac{5}{3}(600) = 1000 leq n leq 1000 ]So, ( n = 1000 ). But as before, ( n = 1000 ) gives only 99 integers between 500 and 600. So, maybe ( m ) starts at 501?Wait, let's check ( m = 501 ):[ frac{5}{3}(601) ≈ 1001.666 leq n leq 1002 ]So, ( n ) can be 1002.Let me check ( n = 1002 ):( 0.5n = 501 ), ( 0.6n = 601.2 ). So, the integers between 501 and 601.2 are 502 to 601, which is 99 numbers. Still not 100.Wait, maybe I need to adjust the inequalities. Perhaps the number of integers is ( lfloor 0.6n rfloor - lceil 0.5n rceil + 1 ). Because if the endpoints are integers, they might be excluded or included.Wait, the problem says "between ( 0.5n ) and ( 0.6n )", which typically means excluding the endpoints. So, if ( 0.5n ) and ( 0.6n ) are integers, they are not included. So, the number of integers is ( lfloor 0.6n - 1 rfloor - lceil 0.5n + 1 rceil + 1 ). Hmm, this is getting complicated.Alternatively, maybe it's better to think in terms of the interval length. The number of integers between ( a ) and ( b ) is approximately ( b - a ), but adjusted for the fractional parts.Given that ( 0.6n - 0.5n = 0.1n ), and we need exactly 100 integers in this interval. So, ( 0.1n ) should be just over 100, but not so much that it includes 101.So, ( 100 < 0.1n < 101 ), which gives ( 1000 < n < 1010 ). But since ( n ) is an integer, ( n ) can be from 1001 to 1009. But wait, earlier when I tried ( n = 1000 ), it gave 99 numbers, so maybe ( n ) needs to be such that ( 0.1n ) is just over 100, meaning ( n ) is just over 1000.But let's test ( n = 1001 ):( 0.5n = 500.5 ), ( 0.6n = 600.6 ). The integers between 500.5 and 600.6 are 501 to 600, which is 100 numbers. Perfect.Similarly, ( n = 1009 ):( 0.5n = 504.5 ), ( 0.6n = 605.4 ). The integers between 504.5 and 605.4 are 505 to 605, which is 101 numbers. Wait, that's too many.Wait, so ( n = 1009 ) gives 101 numbers. So, maybe the upper limit is lower.Wait, let's check ( n = 1005 ):( 0.5n = 502.5 ), ( 0.6n = 603 ). The integers between 502.5 and 603 are 503 to 602, which is 100 numbers.Wait, ( n = 1005 ) gives 100 numbers. What about ( n = 1010 ):( 0.5n = 505 ), ( 0.6n = 606 ). The integers between 505 and 606 are 506 to 605, which is 100 numbers.Wait, so ( n = 1010 ) also gives 100 numbers. But earlier, ( n = 1009 ) gave 101 numbers. So, maybe the upper limit is 1010.Wait, let me check ( n = 1010 ):( 0.5n = 505 ), ( 0.6n = 606 ). The integers between 505 and 606 are 506 to 605, which is 100 numbers.What about ( n = 1011 ):( 0.5n = 505.5 ), ( 0.6n = 606.6 ). The integers between 505.5 and 606.6 are 506 to 606, which is 101 numbers. So, too many.So, the upper limit is 1010.Now, for the lower limit, let's see. When does ( n ) start giving exactly 100 numbers.We saw that ( n = 1000 ) gives 99 numbers. ( n = 1001 ) gives 100 numbers. So, the smallest ( n ) is 1001.Wait, but earlier, I thought ( m ) starts at 500, but when ( m = 500 ), ( n = 1000 ) gives 99 numbers. So, maybe the smallest ( n ) is 1001.But wait, let's check ( n = 997 ):( 0.5n = 498.5 ), ( 0.6n = 598.2 ). The integers between 498.5 and 598.2 are 499 to 598, which is 100 numbers. So, ( n = 997 ) also gives 100 numbers.Wait, so maybe the lower limit is lower than 1001.Let me check ( n = 997 ):( 0.5n = 498.5 ), ( 0.6n = 598.2 ). The integers between 498.5 and 598.2 are 499 to 598, which is 100 numbers.What about ( n = 996 ):( 0.5n = 498 ), ( 0.6n = 597.6 ). The integers between 498 and 597.6 are 499 to 597, which is 99 numbers. So, ( n = 996 ) gives 99 numbers.So, ( n = 997 ) is the smallest ( n ) that gives exactly 100 numbers.Similarly, ( n = 1010 ) is the largest ( n ) that gives exactly 100 numbers, as ( n = 1011 ) gives 101 numbers.Therefore, the smallest ( n ) is 997, and the largest ( n ) is 1010.