💡Alright, so I've got this equation to solve: (leftlfloor frac{x}{2} rightrfloor + leftlfloor frac{x}{4} rightrfloor = x). I remember that the floor function, denoted by (lfloor x rfloor), gives the greatest integer less than or equal to (x). So, for example, (lfloor 3.7 rfloor = 3) and (lfloor -1.2 rfloor = -2). Okay, so the equation involves floor functions of (x/2) and (x/4), and their sum equals (x). I need to find all real numbers (x) that satisfy this equation. Hmm, floor functions can be tricky because they're piecewise constant functions, meaning they stay constant over intervals and then jump by 1 at certain points. So, maybe I should consider different intervals where the floor functions are constant and solve the equation within each interval.Let me think about how to approach this. Since the equation involves both (x/2) and (x/4), it might be helpful to express everything in terms of a common variable or to look for a substitution that simplifies the equation. Maybe I can let (y = x/4), which would make (x/2 = 2y). Then the equation becomes (lfloor 2y rfloor + lfloor y rfloor = 4y). That seems a bit simpler, but I'm not sure yet. Let me write that down:Let (y = frac{x}{4}), so (x = 4y). Then the equation becomes:[lfloor 2y rfloor + lfloor y rfloor = 4y]Okay, so now I have an equation in terms of (y). Maybe I can analyze this equation by considering the fractional part of (y). Let me recall that any real number (y) can be written as (n + r), where (n) is an integer and (0 leq r < 1) is the fractional part. So, let's express (y) as (n + r), where (n) is an integer and (0 leq r < 1).Substituting (y = n + r) into the equation:[lfloor 2(n + r) rfloor + lfloor n + r rfloor = 4(n + r)]Simplify the left side:[lfloor 2n + 2r rfloor + lfloor n + r rfloor]Since (n) is an integer, (lfloor 2n + 2r rfloor = 2n + lfloor 2r rfloor) and (lfloor n + r rfloor = n + lfloor r rfloor). But since (0 leq r < 1), (lfloor r rfloor = 0). Therefore, the left side becomes:[2n + lfloor 2r rfloor + n + 0 = 3n + lfloor 2r rfloor]The right side is:[4n + 4r]So, the equation simplifies to:[3n + lfloor 2r rfloor = 4n + 4r]Let me rearrange this equation:[lfloor 2r rfloor = n + 4r]Hmm, this is interesting. Now, (lfloor 2r rfloor) is an integer because the floor function always returns an integer. On the right side, (n) is an integer, and (4r) is a real number between 0 (inclusive) and 4 (exclusive). So, (n + 4r) must also be an integer because it's equal to (lfloor 2r rfloor).Let me denote (k = lfloor 2r rfloor), which is an integer. Then:[k = n + 4r]But (k) is an integer, and (n) is an integer, so (4r) must also be an integer. Let me denote (4r = m), where (m) is an integer. Since (0 leq r < 1), (0 leq m < 4). Therefore, (m) can be 0, 1, 2, or 3.So, (r = frac{m}{4}), where (m = 0, 1, 2, 3). Now, let's substitute back into the equation (k = n + 4r):[k = n + m]But (k = lfloor 2r rfloor = lfloor frac{2m}{4} rfloor = lfloor frac{m}{2} rfloor). So:[lfloor frac{m}{2} rfloor = n + m]Wait, that seems a bit off. Let me double-check. Since (r = frac{m}{4}), then (2r = frac{m}{2}). Therefore, (lfloor 2r rfloor = lfloor frac{m}{2} rfloor). So, the equation becomes:[lfloor frac{m}{2} rfloor = n + m]But (n) is an integer, and (m) is 0, 1, 2, or 3. Let's consider each possible value of (m) and see what (n) must be.Case 1: (m = 0)Then, (lfloor frac{0}{2} rfloor = 0), so:[0 = n + 0 implies n = 0]So, (y = n + r = 0 + 0 = 0), which means (x = 4y = 0).Case 2: (m = 1)Then, (lfloor frac{1}{2} rfloor = 0), so:[0 = n + 1 implies n = -1]So, (y = n + r = -1 + frac{1}{4} = -frac{3}{4}), which means (x = 4y = -3).Case 3: (m = 2)Then, (lfloor frac{2}{2} rfloor = 1), so:[1 = n + 2 implies n = -1]So, (y = n + r = -1 + frac{2}{4} = -frac{1}{2}), which means (x = 4y = -2).Case 4: (m = 3)Then, (lfloor frac{3}{2} rfloor = 1), so:[1 = n + 3 implies n = -2]So, (y = n + r = -2 + frac{3}{4} = -frac{5}{4}), which means (x = 4y = -5).So, the possible solutions are (x = 0, -3, -2, -5). Let me verify each of these to make sure they satisfy the original equation.Verification for (x = 0):[leftlfloor frac{0}{2} rightrfloor + leftlfloor frac{0}{4} rightrfloor = 0 + 0 = 0 = x]So, it works.Verification for (x = -3):[leftlfloor frac{-3}{2} rightrfloor + leftlfloor frac{-3}{4} rightrfloor = leftlfloor -1.5 rightrfloor + leftlfloor -0.75 rightrfloor = -2 + (-1) = -3 = x]So, it works.Verification for (x = -2):[leftlfloor frac{-2}{2} rightrfloor + leftlfloor frac{-2}{4} rightrfloor = leftlfloor -1 rightrfloor + leftlfloor -0.5 rightrfloor = -1 + (-1) = -2 = x]So, it works.Verification for (x = -5):[leftlfloor frac{-5}{2} rightrfloor + leftlfloor frac{-5}{4} rightrfloor = leftlfloor -2.5 rightrfloor + leftlfloor -1.25 rightrfloor = -3 + (-2) = -5 = x]So, it works.Therefore, all four solutions satisfy the original equation. I think that's all the possible solutions because we've considered all possible values of (m) and found corresponding (n) and (x). So, the solutions are (x = 0, -3, -2, -5).**Final Answer**The solutions are (boxed{0}), (boxed{-3}), (boxed{-2}), and (boxed{-5}).

💡First, identify the total number of trouser-jacket combinations. Roberto has 5 pairs of trousers, with 3 that can be paired with any of the 3 jackets and 2 that can only be paired with 2 specific jackets.Calculate the combinations for the 3 unrestricted trousers:3 trousers × 3 jackets = 9 combinations.Calculate the combinations for the 2 restricted trousers:2 trousers × 2 jackets = 4 combinations.Add both combinations to get the total trouser-jacket combinations:9 + 4 = 13 combinations.Finally, multiply the total trouser-jacket combinations by the number of shirts to find the total number of outfits:13 × 7 shirts = 91 outfits.

💡First, I need to determine the work rates of Model X and Model Y computers. Model X completes the task in 72 minutes, so its rate is 1/72 tasks per minute. Model Y completes the task in 36 minutes, so its rate is 1/36 tasks per minute.Since the company uses the same number of each model, let’s denote the number of Model X computers as X and the number of Model Y computers as Y, with X = Y.The combined work rate of X Model X computers and X Model Y computers is calculated by adding their individual rates:X * (1/72) + X * (1/36).To simplify, I find a common denominator for the fractions:X * (1/72 + 2/72) = X * (3/72) = X * (1/24).This means the combined work rate is X/24 tasks per minute. Since the task needs to be completed in 1 minute, the combined rate must equal 1 task per minute:X/24 = 1.Solving for X, I multiply both sides by 24:X = 24.Therefore, 24 Model X computers were used.

💡Okay, so I have this problem here asking how many distinct trees with exactly 7 vertices there are. I remember that in graph theory, a tree is a connected graph without any cycles. It also mentions that a tree with n vertices has exactly n-1 edges. So, for 7 vertices, that would be 6 edges.First, I need to recall what makes trees distinct. I think it has to do with their structure, meaning the way the vertices are connected. Two trees are considered different if their structures aren't identical, even if you can relabel the vertices to make them look the same. So, it's about the unlabeled trees, not the labeled ones.I remember something called Cayley's formula, which gives the number of labeled trees on n vertices. Cayley's formula says that the number of labeled trees is n^(n-2). For n=7, that would be 7^(7-2) = 7^5. Calculating that, 7^5 is 16807. But wait, that's the number of labeled trees, right? The problem is asking for distinct trees, which I think refers to unlabeled trees.So, I need to find the number of unlabeled trees with 7 vertices. I don't remember the exact number, but I think there's a known sequence for the number of unlabeled trees. Maybe I can look it up or recall some values.I think the number of unlabeled trees increases as n increases, but not as rapidly as labeled trees. For small n, like n=1, there's only 1 tree. For n=2, also 1 tree. For n=3, there's 1 tree as well. Wait, is that right? No, for n=3, there are actually 2 trees: one where all three vertices are connected in a line, and another where one vertex is connected to the other two. Hmm, maybe I was wrong before.Let me try to list them out for small n to get a sense:- n=1: 1 tree (just a single vertex)- n=2: 1 tree (two vertices connected by an edge)- n=3: 2 trees (a path of three vertices and a star with one central vertex connected to the other two)- n=4: Let's see, I think there are 3 trees. One is a path of four vertices, another is a star with one central vertex connected to the other three, and the third is a tree where one vertex is connected to two others, and one of those is connected to the fourth vertex.- n=5: I think there are 7 trees. It starts getting more complicated here.- n=6: I believe there are 11 trees.- n=7: So, following this pattern, I think it might be 12 or 13 trees.Wait, but I'm not sure. Maybe I should look for a formula or a way to calculate it. I remember that the number of unlabeled trees can be found using something called the "Catalan numbers," but I'm not entirely sure if that's correct. Let me think.Catalan numbers count the number of binary trees, among other things, but I don't think they directly give the number of unlabeled trees. Maybe it's related to integer partitions or something else.Alternatively, I recall that the number of unlabeled trees can be found using the formula involving the number of labeled trees divided by the number of automorphisms, but that seems complicated.Wait, maybe I can use the fact that the number of unlabeled trees grows roughly exponentially, but I need the exact number for n=7. I think I remember that the number of unlabeled trees for n=7 is 11. Let me try to verify that.If n=1: 1n=2: 1n=3: 2n=4: 3n=5: 7n=6: 11n=7: 12Hmm, so for n=6, it's 11, and for n=7, it's 12. But I'm not entirely sure. Maybe it's 11 for n=7? I think I need to double-check.Alternatively, I can think about the possible structures of trees with 7 vertices. Let's try to enumerate them.First, the simplest tree is a straight line: 1-2-3-4-5-6-7. That's one tree.Then, trees where one vertex is connected to multiple branches. For example, a star tree where one central vertex is connected to all others: 1 connected to 2,3,4,5,6,7. That's another tree.Then, trees with two branches. For example, one vertex connected to two other vertices, each of which is connected to more vertices. Let's see:- One central vertex connected to two vertices, each of which is connected to two more vertices: 1-2-3-4 and 1-5-6-7. Wait, that might not be unique.Actually, it's getting complicated. Maybe I should look for a pattern or a formula.I think the number of unlabeled trees for n=7 is 11. So, the answer is 11, which is option C.

💡Okay, so I have this problem here where I need to figure out how many different denominators are possible when expressing the repeating decimal (0.overline{abc}) as a fraction in its lowest terms. Here, (a), (b), and (c) are digits, and they aren't all nine or all zero. Hmm, let me try to break this down step by step.First, I remember that any repeating decimal can be expressed as a fraction. Specifically, for a three-digit repeating decimal like (0.overline{abc}), the fraction form is (frac{abc}{999}). So, if (abc) is a three-digit number, say 123, then the decimal would be (0.overline{123} = frac{123}{999}). That makes sense because 999 is the number with three 9s, corresponding to the three repeating digits.Now, the problem mentions that this fraction should be in its lowest terms. That means I need to simplify (frac{abc}{999}) by dividing both the numerator and the denominator by their greatest common divisor (GCD). So, the simplified fraction will have a denominator of (frac{999}{text{GCD}(abc, 999)}). Therefore, the possible denominators depend on the factors of 999.Let me factorize 999 to find its prime factors. I know that 999 divided by 9 is 111, and 111 divided by 3 is 37. So, 999 breaks down into prime factors as (3^3 times 37). That means the factors of 999 are all the products of these primes raised to different powers. Specifically, the exponents for 3 can be 0, 1, 2, or 3, and for 37, they can be 0 or 1. So, the factors are:- (3^0 times 37^0 = 1)- (3^1 times 37^0 = 3)- (3^2 times 37^0 = 9)- (3^3 times 37^0 = 27)- (3^0 times 37^1 = 37)- (3^1 times 37^1 = 111)- (3^2 times 37^1 = 333)- (3^3 times 37^1 = 999)So, the factors of 999 are 1, 3, 9, 27, 37, 111, 333, and 999. That gives us 8 factors in total.But wait, the problem states that (a), (b), and (c) are not all nine and not all zero. So, (abc) can't be 999 or 000. If (abc) were 999, then the fraction would be (frac{999}{999} = 1), which is an integer, not a fraction with denominator 1. Similarly, if (abc) were 000, the fraction would be 0, which is also not a fraction with denominator 1. So, we need to exclude these cases.But hold on, when we simplify (frac{abc}{999}), the denominator becomes (frac{999}{text{GCD}(abc, 999)}). So, if (abc) is 999, the GCD would be 999, making the denominator 1. Similarly, if (abc) is 000, the GCD would be undefined because 0 can't be in the denominator. But since the problem specifies that (a), (b), and (c) are not all zero, we don't have to worry about that case.Therefore, the denominator can be any factor of 999 except for 1. That leaves us with 3, 9, 27, 37, 111, 333, and 999. So, that's 7 different denominators.But let me double-check to make sure I haven't missed anything. The key here is that (abc) can be any three-digit number from 001 to 998, excluding 999. Each of these will have a GCD with 999 that is a factor of 999. Therefore, when we divide 999 by each possible GCD, we get the denominators. Since the GCD can't be 999 (because (abc) isn't 999), the denominator can't be 1. So, all other factors of 999 are possible denominators.Let me consider an example. If (abc = 111), then (text{GCD}(111, 999) = 111), so the denominator becomes (frac{999}{111} = 9). Similarly, if (abc = 123), which is co-prime with 999 (since 123 factors into 3 × 41, and 41 isn't a factor of 999), the GCD would be 3, so the denominator would be (frac{999}{3} = 333). Another example: if (abc = 333), then (text{GCD}(333, 999) = 333), so the denominator is (frac{999}{333} = 3). Wait, that seems contradictory because 333 is a factor of 999, so dividing 999 by 333 gives 3. So, depending on the GCD, the denominator can be any of the factors except 1.Wait, but in this case, if (abc = 333), the simplified fraction is (frac{333}{999} = frac{1}{3}), so the denominator is 3. Similarly, if (abc = 111), the simplified fraction is (frac{111}{999} = frac{1}{9}), so denominator is 9. If (abc = 27), then (text{GCD}(27, 999) = 27), so denominator is (frac{999}{27} = 37). Wait, so 37 is also a possible denominator. For example, if (abc = 37), then (text{GCD}(37, 999) = 37), so the denominator is (frac{999}{37} = 27). Hmm, so 27 is also a possible denominator. Similarly, if (abc = 9), (text{GCD}(9, 999) = 9), so denominator is (frac{999}{9} = 111). Wait, so 111 is also a possible denominator. For example, if (abc = 9), then the fraction is (frac{9}{999} = frac{1}{111}). Similarly, if (abc = 3), (text{GCD}(3, 999) = 3), so denominator is (frac{999}{3} = 333). Wait, so 333 is also a possible denominator. For example, if (abc = 3), the fraction is (frac{3}{999} = frac{1}{333}). And finally, if (abc = 1), (text{GCD}(1, 999) = 1), so denominator is 999. But wait, (abc = 001) is allowed since they are not all zero. So, (frac{001}{999}) simplifies to (frac{1}{999}), so denominator is 999.So, putting it all together, the possible denominators are 3, 9, 27, 37, 111, 333, and 999. That's 7 different denominators. But just to make sure, let me think if there's any case where the denominator could be something else. Since 999 only has those factors, and we've accounted for all except 1, I think 7 is the correct number. Wait, but what about if (abc) is a multiple of 3 and 37? For example, (abc = 111), which is 3 × 37. Then, (text{GCD}(111, 999) = 111), so denominator is 9. Similarly, if (abc = 333), which is 3 × 111, (text{GCD}(333, 999) = 333), so denominator is 3. Another example: (abc = 222), which is 2 × 111. (text{GCD}(222, 999)). Let's see, 222 factors into 2 × 3 × 37, and 999 is 3³ × 37. So, the GCD is 3 × 37 = 111. Therefore, denominator is 999 / 111 = 9. Wait, so even if (abc) has multiple factors, as long as they are common with 999, the GCD will be the product of the lowest powers of the common primes. So, the denominator will always be a factor of 999. Therefore, the denominators possible are exactly the factors of 999 except for 1. Since 999 has 8 factors, excluding 1 gives us 7 denominators. I think that's solid. I can't think of any other factors or cases where the denominator would be something else. So, the answer should be 7.

💡To solve for ( m - n ) given the polynomial ( -2x^{2} + mx + nx^{2} - 5x - 1 + 4x ), and knowing its value is independent of ( x ), we proceed as follows:First, we simplify the given polynomial:[-2x^{2} + mx + nx^{2} - 5x - 1 + 4x = (-2 + n)x^{2} + (m - 5 + 4)x - 1 = (-2 + n)x^{2} + (m - 1)x - 1.]For the polynomial's value to be independent of ( x ), the coefficients of ( x^2 ) and ( x ) must be zero. This leads to the system of equations:[-2 + n = 0,][m - 1 = 0.]Solving these equations gives us:[n = 2,][m = 1.]Therefore, to find ( m - n ), we calculate:[m - n = 1 - 2 = -1.]Thus, the value of ( m - n ) is (boxed{-1}).