💡First, I recognize that (0.overline{3}) is a repeating decimal, which can be expressed as a fraction. Let (x = 0.overline{3}).To eliminate the repeating decimal, I multiply both sides of the equation by 10:[10x = 3.overline{3}]Next, I subtract the original equation from this new equation to isolate (x):[10x - x = 3.overline{3} - 0.overline{3}][9x = 3]Solving for (x), I find:[x = frac{3}{9} = frac{1}{3}]Now, I multiply this fraction by 8:[frac{1}{3} times 8 = frac{8}{3}]Finally, I express (frac{8}{3}) as a mixed number:[frac{8}{3} = 2 frac{2}{3}]Thus, the product of (0.overline{3}) and 8 is (2frac{2}{3}).

💡Alright, so I'm trying to figure out this problem where we have an inequality involving x and some fraction with a in the numerator and x raised to the power of n in the denominator. The problem gives some examples and then asks for the general case where x + a/x^n is greater than or equal to n + 1, and we need to find what a is in terms of n.First, let me look at the examples they provided to understand the pattern. The first example is x + 1/x ≥ 2. That makes sense because I remember from algebra that for any positive x, the expression x + 1/x is always greater than or equal to 2. This is derived from the AM-GM inequality, which states that the arithmetic mean is greater than or equal to the geometric mean. In this case, the arithmetic mean of x and 1/x is (x + 1/x)/2, and the geometric mean is sqrt(x * 1/x) = 1. So, (x + 1/x)/2 ≥ 1, which simplifies to x + 1/x ≥ 2.The second example they gave is x + 4/x² ≥ 3. Hmm, okay, so they split x into two equal parts: x/2 + x/2, and then added 4/x². So, it's like they're using three terms instead of two. Then, applying the AM-GM inequality again, the arithmetic mean of these three terms is (x/2 + x/2 + 4/x²)/3, and the geometric mean is the cube root of (x/2 * x/2 * 4/x²). Let me compute that:The geometric mean is the cube root of (x/2 * x/2 * 4/x²). Multiplying those together: (x * x * 4) / (2 * 2 * x²) = (4x²) / (4x²) = 1. So, the cube root of 1 is 1. Therefore, the arithmetic mean is greater than or equal to 1, which means (x/2 + x/2 + 4/x²)/3 ≥ 1, so multiplying both sides by 3 gives x + 4/x² ≥ 3.Okay, so in the first example, we had two terms: x and 1/x, and a was 1. In the second example, we had three terms: x/2, x/2, and 4/x², and a was 4. So, it seems like when we split x into n equal parts, we get n terms of x/n, and then we have another term a/x^n. Then, applying the AM-GM inequality, we get that the sum is greater than or equal to n + 1.So, generalizing this, if we have x + a/x^n, we can split x into n equal parts, each being x/n, and then add the term a/x^n. So, we have n terms of x/n and one term of a/x^n, making a total of n + 1 terms. Applying the AM-GM inequality, the arithmetic mean of these n + 1 terms is greater than or equal to the geometric mean.Let me write that out:(x/n + x/n + ... + x/n + a/x^n) / (n + 1) ≥ ( (x/n)^n * (a/x^n) )^(1/(n + 1))Multiplying both sides by (n + 1):x + a/x^n ≥ (n + 1) * ( (x/n)^n * (a/x^n) )^(1/(n + 1))Now, simplifying the right-hand side:First, let's compute (x/n)^n * (a/x^n):(x/n)^n = x^n / n^nMultiplying that by a/x^n gives:(x^n / n^n) * (a / x^n) = a / n^nSo, the geometric mean is (a / n^n)^(1/(n + 1))Therefore, the inequality becomes:x + a/x^n ≥ (n + 1) * (a / n^n)^(1/(n + 1))But according to the problem statement, we want this to be equal to n + 1. So, we have:(n + 1) * (a / n^n)^(1/(n + 1)) = n + 1Dividing both sides by (n + 1):(a / n^n)^(1/(n + 1)) = 1Raising both sides to the power of (n + 1):a / n^n = 1^(n + 1) = 1Therefore, a = n^nSo, the value of a is n raised to the power of n.Wait, let me double-check that. If a = n^n, then in the first example, where n = 1, a should be 1^1 = 1, which matches the first example. In the second example, n = 2, so a = 2^2 = 4, which also matches the second example. So, that seems consistent.Therefore, the general conclusion is that a = n^n.

💡Alright, so I have this function f(x) = ln x + a/x, and I need to tackle two parts. Let me start with part (I): finding the monotonic intervals when a is less than zero.First, I remember that to find monotonic intervals, I need to look at the derivative of the function. So, let me compute f'(x). The derivative of ln x is 1/x, and the derivative of a/x is -a/x². So putting that together, f'(x) = 1/x - a/x². Hmm, maybe I can combine these terms over a common denominator. That would be (x - a)/x².Now, since a is less than zero, let's think about what this means for f'(x). The denominator x² is always positive because any real number squared is positive. So the sign of f'(x) depends on the numerator, which is x - a. Since a is negative, subtracting a negative is like adding a positive. So x - a is the same as x + |a|, right? Because a is negative, |a| is positive.So, x + |a| is always positive for x > 0 because both x and |a| are positive. Therefore, the numerator is always positive, and since the denominator is always positive, f'(x) is always positive. That means the function is monotonically increasing on its entire domain, which is (0, ∞).Okay, that seems straightforward. So for part (I), the function is increasing everywhere when a is negative.Moving on to part (II): If the minimum value of f(x) on the interval [1, e] is 3/2, find the value of a.Hmm, okay. So I need to find 'a' such that the minimum of f(x) on [1, e] is 3/2. Let's think about how to approach this.First, I know that the minimum of a function on a closed interval can occur either at a critical point inside the interval or at one of the endpoints. So, I need to find the critical points by setting f'(x) = 0 and then evaluate f(x) at those critical points and at the endpoints x=1 and x=e.From part (I), we have f'(x) = (x - a)/x². Setting this equal to zero gives x - a = 0, so x = a. Therefore, the critical point is at x = a.Now, since we're looking at the interval [1, e], the critical point x = a will only be inside the interval if a is between 1 and e. So, I need to consider different cases based on the value of 'a'.Case 1: a < 1. In this case, the critical point x = a is less than 1, so it's outside our interval. Therefore, the function is either increasing or decreasing throughout the entire interval [1, e]. Let's check the sign of f'(x) on [1, e].Since a < 1, and x is in [1, e], x - a is positive because x >= 1 and a < 1. So, f'(x) is positive on [1, e], meaning the function is increasing on this interval. Therefore, the minimum occurs at x=1.So, f(1) = ln 1 + a/1 = 0 + a = a. According to the problem, the minimum is 3/2, so a = 3/2. But wait, in this case, we assumed a < 1, but 3/2 is greater than 1. That's a contradiction. So, this case doesn't hold.Case 2: a = 1. Then, the critical point is at x=1, which is the endpoint. So, let's see the behavior of f'(x) around x=1. For x > 1, f'(x) = (x - 1)/x², which is positive. So, the function is increasing on (1, e]. Therefore, the minimum is at x=1, which is f(1) = 1. But the problem states the minimum is 3/2, so this case also doesn't hold.Case 3: 1 < a < e. Here, the critical point x = a is inside the interval [1, e]. So, we need to check the behavior of f(x) around this critical point. Let's see:For x < a, f'(x) = (x - a)/x². Since x < a, the numerator is negative, so f'(x) is negative. Therefore, the function is decreasing on [1, a).For x > a, f'(x) = (x - a)/x². Since x > a, the numerator is positive, so f'(x) is positive. Therefore, the function is increasing on (a, e].So, the function decreases from x=1 to x=a and then increases from x=a to x=e. Therefore, the minimum occurs at x=a.So, f(a) = ln a + a/a = ln a + 1. According to the problem, this minimum is 3/2. So, ln a + 1 = 3/2. Solving for a:ln a = 3/2 - 1 = 1/2Therefore, a = e^(1/2) = sqrt(e). Now, we need to check if this a is within our assumed range of 1 < a < e. Since sqrt(e) is approximately 1.648, which is between 1 and e (approximately 2.718), this is valid.Case 4: a = e. Then, the critical point is at x=e, which is the endpoint. Let's check the behavior of f'(x) around x=e. For x < e, f'(x) = (x - e)/x². Since x < e, the numerator is negative, so f'(x) is negative. Therefore, the function is decreasing on [1, e). So, the minimum is at x=e, which is f(e) = ln e + e/e = 1 + 1 = 2. But the problem states the minimum is 3/2, so this case doesn't hold.Case 5: a > e. Here, the critical point x=a is greater than e, so it's outside our interval. Therefore, on [1, e], f'(x) = (x - a)/x². Since a > e and x <= e, x - a is negative. So, f'(x) is negative on [1, e], meaning the function is decreasing throughout the interval. Therefore, the minimum occurs at x=e, which is f(e) = 2. Again, this contradicts the given minimum of 3/2.So, the only valid case is Case 3, where a = sqrt(e). Therefore, the value of a is sqrt(e).Wait, let me double-check my calculations. In Case 3, I set f(a) = 3/2, which led to ln a + 1 = 3/2, so ln a = 1/2, hence a = e^(1/2). That seems correct. And since e^(1/2) is approximately 1.648, which is indeed between 1 and e, this makes sense.I think that's solid. So, the answer for part (II) is a = sqrt(e).**Final Answer**The value of ( a ) is (boxed{sqrt{e}}).

💡Okay, so I have this problem here where a point C is randomly chosen on a line segment AB that's 12 cm long. Then, they construct a rectangle using AC and BC as adjacent sides. I need to find the probability that the area of this rectangle is greater than 20 square centimeters. Hmm, let me try to break this down step by step.First, let me visualize the problem. We have a line segment AB that's 12 cm long. Point C is somewhere on this segment, dividing it into two parts: AC and BC. Since C is chosen randomly, the lengths of AC and BC can vary, but their sum will always be 12 cm. So, if I let AC be some length x, then BC would naturally be 12 - x.Now, the rectangle is constructed with AC and BC as adjacent sides. That means the sides of the rectangle are x and 12 - x. Therefore, the area of the rectangle, which I'll call S, should be the product of these two lengths. So, S = x * (12 - x). That makes sense because the area of a rectangle is length times width.The problem asks for the probability that this area S is greater than 20 square centimeters. So, I need to find the probability that x*(12 - x) > 20. Let me write that down as an inequality:x*(12 - x) > 20Hmm, okay. Let me expand this inequality to make it easier to solve. Multiplying x by (12 - x) gives:12x - x² > 20Now, let me rearrange this inequality to standard quadratic form. I'll subtract 20 from both sides:12x - x² - 20 > 0Which simplifies to:-x² + 12x - 20 > 0Hmm, quadratic inequalities can be tricky, but I think I remember that we can solve them by first finding the roots of the corresponding quadratic equation and then determining the intervals where the inequality holds.So, let me set the quadratic expression equal to zero:-x² + 12x - 20 = 0I can multiply both sides by -1 to make it a bit easier to work with, but I have to remember that multiplying by a negative number reverses the inequality signs. So, multiplying both sides by -1 gives:x² - 12x + 20 = 0Now, I need to solve this quadratic equation. I can use the quadratic formula, which is:x = [ -b ± sqrt(b² - 4ac) ] / (2a)In this equation, a = 1, b = -12, and c = 20. Plugging these values into the formula:x = [ 12 ± sqrt( (-12)² - 4*1*20 ) ] / (2*1)x = [ 12 ± sqrt(144 - 80) ] / 2x = [ 12 ± sqrt(64) ] / 2x = [ 12 ± 8 ] / 2So, the two solutions are:x = (12 + 8)/2 = 20/2 = 10x = (12 - 8)/2 = 4/2 = 2Therefore, the roots are x = 2 and x = 10. These are the points where the quadratic expression equals zero. Now, since the quadratic coefficient is positive (the coefficient of x² is 1, which is positive), the parabola opens upwards. That means the expression x² - 12x + 20 is positive outside the interval [2, 10] and negative inside the interval [2, 10].But wait, in our original inequality after multiplying by -1, we had:x² - 12x + 20 > 0Which corresponds to the regions where the quadratic is positive, which is x < 2 or x > 10. However, remember that we had multiplied the inequality by -1 earlier, so we need to reverse the inequality sign. Let me think about this carefully.Wait, actually, let me backtrack a bit. The original inequality after rearrangement was:-x² + 12x - 20 > 0Which is equivalent to:x² - 12x + 20 < 0Because I multiplied both sides by -1, which flips the inequality. So, the inequality x² - 12x + 20 < 0 is true between the roots, that is, for x between 2 and 10.Therefore, the solution to the inequality x*(12 - x) > 20 is 2 < x < 10.So, x must be greater than 2 cm and less than 10 cm for the area of the rectangle to be greater than 20 square centimeters.Now, since point C is chosen randomly along the 12 cm segment AB, the probability we're looking for is the length of the interval where x satisfies 2 < x < 10 divided by the total length of AB, which is 12 cm.So, the favorable interval is from 2 cm to 10 cm, which is a length of 10 - 2 = 8 cm.Therefore, the probability P is:P = (Length of favorable interval) / (Total length of AB) = 8 / 12Simplifying this fraction by dividing numerator and denominator by 4:P = 2 / 3So, the probability that the area of the rectangle is greater than 20 square centimeters is 2/3.Wait, let me double-check my steps to make sure I didn't make any mistakes.1. I set AC = x, so BC = 12 - x. Correct.2. Area S = x*(12 - x). Correct.3. Set up inequality x*(12 - x) > 20. Correct.4. Expanded to 12x - x² > 20. Correct.5. Rearranged to -x² + 12x - 20 > 0. Correct.6. Multiplied by -1 to get x² - 12x + 20 < 0. Remembered to flip the inequality. Correct.7. Solved quadratic equation x² - 12x + 20 = 0 using quadratic formula. Got roots at x = 2 and x = 10. Correct.8. Since quadratic opens upwards, the inequality x² - 12x + 20 < 0 is satisfied between the roots, so 2 < x < 10. Correct.9. Calculated favorable interval length as 8 cm. Correct.10. Divided by total length 12 cm to get probability 8/12 = 2/3. Correct.Everything seems to check out. So, I think my answer is correct.**Final Answer**The probability is boxed{dfrac{2}{3}}.

💡Alright, so I need to prove that the number of positive divisors of a positive integer ( n ) doesn't exceed ( 2sqrt{n} ). Hmm, okay. Let me think about this step by step.First, I know that a divisor of ( n ) is a number that divides ( n ) without leaving a remainder. For example, the divisors of 6 are 1, 2, 3, and 6. So, the number of divisors here is 4.Now, the problem is asking me to show that no matter what positive integer ( n ) I choose, the number of its positive divisors can't be more than ( 2sqrt{n} ). That seems interesting because ( sqrt{n} ) grows much slower than ( n ) itself, so this is a relatively tight bound.Let me consider how divisors come in pairs. For any divisor ( d ) of ( n ), there is another divisor ( frac{n}{d} ). So, if ( d ) is a divisor, then ( frac{n}{d} ) is also a divisor. This is because ( d times frac{n}{d} = n ), which means both ( d ) and ( frac{n}{d} ) divide ( n ).Now, if I list out all the divisors of ( n ), I can pair them up such that each pair multiplies to ( n ). For example, for ( n = 6 ), the pairs are (1, 6) and (2, 3). Each pair multiplies to 6.Here's an idea: if I can show that each pair has at least one divisor that is less than or equal to ( sqrt{n} ), then the number of such pairs would be limited by ( sqrt{n} ), and hence the total number of divisors would be limited by ( 2sqrt{n} ).Let me test this idea with an example. Take ( n = 16 ). The divisors are 1, 2, 4, 8, 16. The pairs are (1, 16), (2, 8), and (4, 4). Now, ( sqrt{16} = 4 ). In each pair, one number is less than or equal to 4, and the other is greater than or equal to 4. So, the number of pairs is 3, and the total number of divisors is 5, which is less than ( 2 times 4 = 8 ).Wait, but in this case, the number of divisors is 5, which is less than 8. So, the bound holds. But I need to make sure this works for any ( n ).Another example: ( n = 12 ). Divisors are 1, 2, 3, 4, 6, 12. Pairs are (1, 12), (2, 6), (3, 4). ( sqrt{12} approx 3.464 ). Each pair has one number less than or equal to 3.464 and the other greater than or equal to 3.464. The number of pairs is 3, and the total number of divisors is 6, which is less than ( 2 times 3.464 approx 6.928 ). So, again, the bound holds.But wait, in this case, the number of divisors is exactly 6, which is close to ( 2sqrt{12} ). So, the bound is tight in some cases.Now, let's think about how to generalize this. For any positive integer ( n ), the number of positive divisors is equal to the number of such pairs. Each pair consists of a divisor ( d ) and ( frac{n}{d} ). Now, if ( d leq sqrt{n} ), then ( frac{n}{d} geq sqrt{n} ). Conversely, if ( d > sqrt{n} ), then ( frac{n}{d} < sqrt{n} ).Therefore, every divisor ( d ) can be paired with another divisor ( frac{n}{d} ), and exactly one of the pair is less than or equal to ( sqrt{n} ). So, the number of divisors less than or equal to ( sqrt{n} ) is equal to the number of divisors greater than or equal to ( sqrt{n} ).If ( n ) is not a perfect square, then ( sqrt{n} ) is not an integer, and all divisors less than ( sqrt{n} ) pair with divisors greater than ( sqrt{n} ). Hence, the total number of divisors is twice the number of divisors less than ( sqrt{n} ).If ( n ) is a perfect square, then ( sqrt{n} ) is an integer and is a divisor of ( n ). In this case, the number of divisors less than ( sqrt{n} ) is equal to the number of divisors greater than ( sqrt{n} ), and ( sqrt{n} ) itself is counted once. Therefore, the total number of divisors is ( 2 times ) (number of divisors less than ( sqrt{n} )) ( + 1 ).In both cases, whether ( n ) is a perfect square or not, the number of divisors is at most ( 2sqrt{n} ). Because the number of divisors less than or equal to ( sqrt{n} ) is at most ( sqrt{n} ), and hence the total number of divisors is at most ( 2sqrt{n} ).Wait, but in the case of a perfect square, the total number of divisors is ( 2k + 1 ), where ( k ) is the number of divisors less than ( sqrt{n} ). So, if ( k leq sqrt{n} ), then ( 2k + 1 leq 2sqrt{n} + 1 ). But the problem states that the number of divisors does not exceed ( 2sqrt{n} ). So, is there a mistake here?Let me think again. If ( n ) is a perfect square, say ( n = m^2 ), then ( m ) is a divisor, and the number of divisors less than ( m ) is equal to the number of divisors greater than ( m ). So, the total number of divisors is ( 2k + 1 ), where ( k ) is the number of divisors less than ( m ). Now, ( k ) is at most ( m - 1 ), because ( m ) itself is a divisor.But ( m = sqrt{n} ), so ( k leq sqrt{n} - 1 ). Therefore, the total number of divisors is ( 2k + 1 leq 2(sqrt{n} - 1) + 1 = 2sqrt{n} - 1 ), which is still less than ( 2sqrt{n} ).So, in both cases, whether ( n ) is a perfect square or not, the number of divisors does not exceed ( 2sqrt{n} ).Wait, but in the case of ( n = 1 ), which is a perfect square, the number of divisors is 1, and ( 2sqrt{1} = 2 ), so 1 ≤ 2, which holds.Another test case: ( n = 4 ). Divisors are 1, 2, 4. Number of divisors is 3. ( 2sqrt{4} = 4 ). So, 3 ≤ 4, which holds.Another test case: ( n = 9 ). Divisors are 1, 3, 9. Number of divisors is 3. ( 2sqrt{9} = 6 ). So, 3 ≤ 6, which holds.Okay, so it seems to hold for perfect squares as well.Therefore, the reasoning is as follows:1. Every divisor ( d ) of ( n ) can be paired with ( frac{n}{d} ).2. In each pair, one divisor is ≤ ( sqrt{n} ) and the other is ≥ ( sqrt{n} ).3. The number of divisors ≤ ( sqrt{n} ) is at most ( sqrt{n} ).4. Therefore, the total number of divisors is at most ( 2sqrt{n} ).This seems to cover all cases, whether ( n ) is a perfect square or not.I think this makes sense. So, the key idea is that divisors come in pairs, and each pair has one element ≤ ( sqrt{n} ) and the other ≥ ( sqrt{n} ). Therefore, the number of such pairs is limited by ( sqrt{n} ), leading to the total number of divisors being at most ( 2sqrt{n} ).I don't see any flaws in this reasoning. It seems solid. Maybe I should try to write it more formally to make sure.Let me try to formalize the proof:**Proof:**Let ( n ) be a positive integer. Consider the set of positive divisors of ( n ). For each divisor ( d ) of ( n ), we can pair it with ( frac{n}{d} ), which is also a divisor of ( n ).Now, observe that for any divisor ( d ), either ( d leq sqrt{n} ) or ( d geq sqrt{n} ). If ( d leq sqrt{n} ), then ( frac{n}{d} geq sqrt{n} ), and vice versa.Therefore, the set of divisors can be partitioned into pairs ( (d, frac{n}{d}) ) where one element is ≤ ( sqrt{n} ) and the other is ≥ ( sqrt{n} ).If ( n ) is not a perfect square, then ( sqrt{n} ) is not an integer, and all pairs consist of distinct divisors. Hence, the number of such pairs is equal to the number of divisors ≤ ( sqrt{n} ), which is at most ( sqrt{n} ). Therefore, the total number of divisors is at most ( 2sqrt{n} ).If ( n ) is a perfect square, then ( sqrt{n} ) is an integer and is a divisor of ( n ). In this case, the number of divisors less than ( sqrt{n} ) is equal to the number of divisors greater than ( sqrt{n} ), and ( sqrt{n} ) itself is counted once. Therefore, the total number of divisors is ( 2k + 1 ), where ( k ) is the number of divisors less than ( sqrt{n} ). Since ( k leq sqrt{n} - 1 ), the total number of divisors is ( 2k + 1 leq 2(sqrt{n} - 1) + 1 = 2sqrt{n} - 1 ), which is still less than ( 2sqrt{n} ).Thus, in both cases, whether ( n ) is a perfect square or not, the number of positive divisors of ( n ) does not exceed ( 2sqrt{n} ).**Conclusion:**Therefore, we have proven that for any positive integer ( n ), the number of positive divisors of ( n ) is at most ( 2sqrt{n} ).

💡Okay, so I have this problem here about a rectangular solid, which I think is like a box shape, right? It says that the lengths of three edges that share a vertex are 3, 4, and 5. I need to find the surface area of the circumscribed sphere around this solid. The options are A: 200π, B: 50π, C: 200, and D: 50.Hmm, let me visualize this. A rectangular solid has length, width, and height, and all the edges meet at a vertex. So, if the edges are 3, 4, and 5, that means the solid has dimensions 3x4x5. Now, a circumscribed sphere around this solid would touch all the vertices of the solid, right? So, the sphere is perfectly surrounding the box.I remember that for a rectangular solid, the diameter of the circumscribed sphere is equal to the space diagonal of the solid. The space diagonal is the longest diagonal that goes from one corner of the box to the opposite corner, passing through the interior of the box. So, if I can find the space diagonal, that will give me the diameter of the sphere, and then I can find the radius, which is half of that.The formula for the space diagonal (d) of a rectangular solid with sides a, b, c is d = √(a² + b² + c²). Let me plug in the values: a = 3, b = 4, c = 5.Calculating the squares: 3² = 9, 4² = 16, 5² = 25. Adding them up: 9 + 16 + 25 = 50. So, the space diagonal is √50. Simplifying that, √50 is 5√2. So, the diameter of the sphere is 5√2. Therefore, the radius (r) is half of that, which is (5√2)/2.Now, I need to find the surface area of the sphere. The formula for the surface area (S) of a sphere is S = 4πr². Let me plug in the radius I found: r = (5√2)/2.Calculating r²: [(5√2)/2]² = (25 * 2)/4 = 50/4 = 12.5. So, r² is 12.5.Now, plug that into the surface area formula: S = 4π * 12.5 = 50π. So, the surface area is 50π.Looking back at the options, that's option B: 50π. So, I think that's the answer.Wait, let me double-check my steps to make sure I didn't make a mistake. First, I found the space diagonal correctly using the formula √(a² + b² + c²). Plugging in 3, 4, 5, I got √50, which is correct. Then, the radius is half of that, so (5√2)/2. Squaring that gives 12.5, and multiplying by 4π gives 50π. Yeah, that seems right.I think I might have confused diameter and radius at first, but I caught myself and corrected it. So, I'm pretty confident now that the surface area is 50π, which is option B.**Final Answer**boxed{B}