💡To determine the value of ( a ) such that the point ( A(3, a) ) lies on the line ( 2x + y - 7 = 0 ), follow these steps:1. **Understand the Problem**: The point ( A(3, a) ) must satisfy the equation of the line ( 2x + y - 7 = 0 ).2. **Substitute the Coordinates into the Line Equation**: [ 2(3) + a - 7 = 0 ] 3. **Simplify the Equation**: [ 6 + a - 7 = 0 ] [ a - 1 = 0 ] 4. **Solve for ( a )**: [ a = 1 ]

💡To determine how many of the 300 smallest positive integers are congruent to 5 modulo 7, we can follow these steps:1. **Understand the Congruence Condition**: An integer ( x ) is congruent to 5 modulo 7 if it can be expressed in the form: [ x = 7n + 5 ] where ( n ) is a non-negative integer.2. **Set Up the Inequality**: We need ( x ) to be within the range of the first 300 positive integers: [ 1 leq 7n + 5 leq 300 ]3. **Solve for ( n )**: - Subtract 5 from all parts of the inequality: [ -4 leq 7n leq 295 ] - Divide by 7: [ -frac{4}{7} leq n leq frac{295}{7} ] - Calculate ( frac{295}{7} approx 42.142857 ). The largest integer ( n ) satisfying this is 42.4. **Count the Valid Values of ( n )**: Since ( n ) starts from 0 (as ( 7 cdot 0 + 5 = 5 ) is the smallest number satisfying the condition and still within the positive range), the total number of valid ( n ) is: [ 42 - 0 + 1 = 43 ]5. **Conclusion**: There are 43 integers from 1 to 300 inclusive that are congruent to 5 modulo 7.

💡First, I need to determine how many 10-meter segments fit into the 225-meter yard. I'll divide the total length by the distance between consecutive trees.225 meters divided by 10 meters equals 22.5 segments.Since we can't have half a segment, I'll consider only the whole number, which is 22 segments.Each segment represents the space between two trees. Therefore, the number of trees is one more than the number of segments.So, 22 segments plus 1 additional tree equals 23 trees in total.

💡Okay, so I have this problem with two parts. Let me try to tackle them one by one. First, I need to prove that the sequence {n/a_n - 1} is a geometric progression and find an expression for a_n. Then, I have to use mathematical induction to prove an inequality involving the product of a_1 to a_n. Hmm, sounds a bit involved, but let's break it down.Starting with part 1. I know that a_1 is given as 5/4. The recursive formula is a_n = (5n a_{n-1}) / (4a_{n-1} + n - 1) for n ≥ 2. I need to look at the sequence {n/a_n - 1} and show it's a geometric progression. A geometric progression has a common ratio between consecutive terms, so if I can show that the ratio of (n/a_n - 1) to ((n-1)/a_{n-1} - 1) is constant, that would do it.Let me write down what n/a_n - 1 is. From the recursive formula:a_n = (5n a_{n-1}) / (4a_{n-1} + n - 1)So, taking reciprocal:1/a_n = (4a_{n-1} + n - 1) / (5n a_{n-1})Multiply both sides by n:n/a_n = (4a_{n-1} + n - 1) / (5a_{n-1})Simplify:n/a_n = (4a_{n-1})/(5a_{n-1}) + (n - 1)/(5a_{n-1})Which simplifies to:n/a_n = 4/5 + (n - 1)/(5a_{n-1})So, subtracting 1 from both sides:n/a_n - 1 = 4/5 - 1 + (n - 1)/(5a_{n-1})Which is:n/a_n - 1 = -1/5 + (n - 1)/(5a_{n-1})Hmm, that seems a bit messy. Wait, maybe I can factor out 1/5:n/a_n - 1 = (1/5)( (n - 1)/a_{n-1} - 1 )Oh, that's better! So, this shows that each term is 1/5 times the previous term. That means the ratio between consecutive terms is 1/5, which is constant. So, the sequence {n/a_n - 1} is indeed a geometric progression with common ratio 1/5.Now, to find the expression for a_n, I can use the fact that it's a geometric progression. The first term is when n=1: 1/a_1 - 1. Since a_1 = 5/4, 1/a_1 = 4/5, so 4/5 - 1 = -1/5. So, the first term is -1/5, and the common ratio is 1/5.Therefore, the general term is:(n/a_n - 1) = (-1/5) * (1/5)^{n - 1} = - (1/5)^nSo, solving for a_n:n/a_n = 1 - (1/5)^nTherefore, a_n = n / (1 - (1/5)^n )Let me check this for n=1: a_1 should be 5/4. Plugging in n=1: 1 / (1 - 1/5) = 1 / (4/5) = 5/4. Correct. For n=2, let's compute a_2 using the recursive formula:a_2 = (5*2*a_1)/(4a_1 + 2 - 1) = (10*(5/4))/(4*(5/4) + 1) = (50/4)/(5 + 1) = (12.5)/6 ≈ 2.0833Using the expression: a_2 = 2 / (1 - 1/25) = 2 / (24/25) = 2*(25/24) = 50/24 ≈ 2.0833. Correct. Seems good.So, part 1 is done. Now, part 2: Use induction to prove that a_1*a_2*...*a_n < n! / (1 - 1/5 - 1/5^2 - ... - 1/5^n) for n ≥ 2.First, let me understand the right-hand side. The denominator is 1 minus a finite geometric series. The sum of 1/5 + 1/5^2 + ... + 1/5^n is (1/5)(1 - (1/5)^n)/(1 - 1/5) = (1 - (1/5)^n)/4. So, 1 - sum = 1 - (1 - (1/5)^n)/4 = (4 - 1 + (1/5)^n)/4 = (3 + (1/5)^n)/4. Wait, no, let me compute it correctly.Wait, sum from k=1 to n of 1/5^k is (1/5)(1 - (1/5)^n)/(1 - 1/5) ) = (1 - (1/5)^n)/4. So, 1 - sum = 1 - (1 - (1/5)^n)/4 = (4 - 1 + (1/5)^n)/4 = (3 + (1/5)^n)/4. Hmm, but I'm not sure if that's necessary for the induction. Maybe I can keep it as 1 - sum.So, the inequality is:a_1 a_2 ... a_n < n! / (1 - 1/5 - 1/5^2 - ... - 1/5^n )Let me denote S_n = 1 - 1/5 - 1/5^2 - ... - 1/5^n. So, the inequality is a_1 a_2 ... a_n < n! / S_n.I need to prove this by induction. Let's start with the base case, n=2.Base case: n=2.Compute a_1 a_2: a_1 = 5/4, a_2 = 50/24 ≈ 2.0833. So, product is (5/4)*(50/24) = (250)/96 ≈ 2.6042.Compute RHS: 2! / (1 - 1/5 - 1/25) = 2 / (1 - 0.2 - 0.04) = 2 / (0.76) ≈ 2.6316.So, 2.6042 < 2.6316. True.Now, assume for n=k, the inequality holds: a_1 a_2 ... a_k < k! / S_k.Need to show for n=k+1: a_1 a_2 ... a_k a_{k+1} < (k+1)! / S_{k+1}.From the induction hypothesis, a_1...a_k < k! / S_k.So, a_1...a_k a_{k+1} < (k! / S_k) * a_{k+1}.We need to show that (k! / S_k) * a_{k+1} ≤ (k+1)! / S_{k+1}.Simplify:(k! / S_k) * a_{k+1} ≤ (k+1)! / S_{k+1}Divide both sides by k!:a_{k+1} / S_k ≤ (k+1) / S_{k+1}So, need to show that a_{k+1} / S_k ≤ (k+1)/S_{k+1}From part 1, a_{k+1} = (k+1)/(1 - (1/5)^{k+1})So, plug that in:[(k+1)/(1 - (1/5)^{k+1})] / S_k ≤ (k+1)/S_{k+1}Simplify:1/(1 - (1/5)^{k+1}) / S_k ≤ 1/S_{k+1}Multiply both sides by S_k S_{k+1}:S_{k+1} ≤ S_k (1 - (1/5)^{k+1})But S_{k+1} = 1 - 1/5 - ... - 1/5^{k+1} = S_k - 1/5^{k+1}So, S_{k+1} = S_k - 1/5^{k+1}So, the inequality becomes:S_k - 1/5^{k+1} ≤ S_k (1 - (1/5)^{k+1})Expand RHS:S_k - S_k (1/5)^{k+1}So, inequality:S_k - 1/5^{k+1} ≤ S_k - S_k (1/5)^{k+1}Subtract S_k from both sides:-1/5^{k+1} ≤ - S_k (1/5)^{k+1}Multiply both sides by -1 (inequality reverses):1/5^{k+1} ≥ S_k (1/5)^{k+1}Divide both sides by (1/5)^{k+1} (positive, so inequality remains):1 ≥ S_kBut S_k = 1 - (1/5 + ... + 1/5^k) < 1, since all terms are positive. So, 1 ≥ S_k is true.Therefore, the inequality holds, so the inductive step is proven.Thus, by induction, the inequality holds for all n ≥ 2.Wait, but I need to make sure that all steps are valid. When I multiplied both sides by S_k S_{k+1}, I need to ensure they are positive, which they are since S_n = 1 - sum of positive terms, but since sum is less than 1, S_n is positive. Also, when dividing by (1/5)^{k+1}, it's positive, so inequality direction remains.Yes, seems solid.So, overall, part 2 is proven by induction.**Final Answer**1. The expression for ( a_n ) is (boxed{a_n = frac{n}{1 - left( frac{1}{5} right)^n}}).2. The inequality is proven and the result is (boxed{a_1 cdot a_2 cdots a_n < frac{n!}{1 - frac{1}{5} - frac{1}{5^2} - ldots - frac{1}{5^n}}}).

💡Alright, so I'm trying to figure out this problem where Calvin and Hobbes play a game. Hobbes picks a family of subsets, and Calvin tries to collect all elements of one of those subsets. The goal is to find the largest possible size of this family where Hobbes still has a winning strategy. Hmm, okay, let's break this down step by step.First, the game setup: Calvin and Hobbes take turns picking numbers from 1 to 2020, with Calvin going first. They each end up with 1010 numbers. Calvin wins if his set includes all elements of some subset in the family (mathcal{F}) that Hobbes chose. If Calvin doesn't manage that, Hobbes wins.So, Hobbes wants to pick a family (mathcal{F}) such that no matter how Calvin picks his numbers, he can't cover any of the subsets in (mathcal{F}). But Hobbes also wants this family to be as large as possible. Interesting.I remember something about strategies where players can block each other from completing certain sets. Maybe something like pairing numbers or using some kind of mirroring strategy? Let me think.If Hobbes pairs the numbers, like ({1,2}, {3,4}, ldots, {2019,2020}), that's 1010 pairs. Then, whenever Calvin picks a number from a pair, Hobbes picks the other number in that pair. This way, Calvin can never have both numbers from any pair. So, if Hobbes ensures that Calvin can't have both elements of any pair, then Calvin can't have any subset that requires both elements of a pair.Wait, so if Hobbes does this, then Calvin's set will always miss at least one element from each pair. Therefore, any subset in (mathcal{F}) that requires both elements of any pair can't be completed by Calvin. So, Hobbes can include all subsets that require both elements of at least one pair in (mathcal{F}), and Calvin can't complete any of them.But how many such subsets are there? Let's see. The total number of subsets of ({1,2,ldots,2020}) is (2^{2020}). But we need to count the subsets that include both elements of at least one pair.Alternatively, maybe it's easier to count the subsets that don't include both elements of any pair and subtract that from the total. The subsets that don't include both elements of any pair are exactly the subsets where for each pair, we can choose none, one, or the other, but not both. So for each pair, there are 3 choices: neither, first, or second. Therefore, the number of such subsets is (3^{1010}).So, the number of subsets that include both elements of at least one pair is (2^{2020} - 3^{1010}). But wait, is that the family (mathcal{F}) that Hobbes can pick? Because if Hobbes picks all subsets that include both elements of at least one pair, then Calvin can't complete any of them because Hobbes is blocking the second element of any pair Calvin picks.But wait, the problem says that Calvin wins if he has all elements of some member of (mathcal{F}). So, if (mathcal{F}) consists of all subsets that include both elements of at least one pair, then Calvin can't complete any of them because Hobbes is always blocking the second element. Therefore, Hobbes can pick this family (mathcal{F}) and have a winning strategy.But the question is asking for the largest possible size of such a family. So, is (2^{2020} - 3^{1010}) the largest possible? Or is there a larger family?Wait, actually, I think I might have made a mistake here. Because if Hobbes picks all subsets that include both elements of at least one pair, that's a huge family, but maybe there's a larger family where Hobbes can still block Calvin.Alternatively, maybe the maximum family is actually (4^{1010} - 3^{1010}). Let me think about that.If we consider each pair as a separate entity, then for each pair, there are four possibilities: neither element is chosen, only the first, only the second, or both. But Calvin can't choose both because Hobbes is blocking. So, for each pair, Calvin can have at most one element. Therefore, the number of subsets that Calvin can have is (3^{1010}), as for each pair, he can choose none, first, or second.But the total number of subsets is (2^{2020}), which is equal to (4^{1010}). So, the number of subsets that Calvin cannot have is (4^{1010} - 3^{1010}). Therefore, the family (mathcal{F}) can be as large as (4^{1010} - 3^{1010}), and Hobbes can ensure that Calvin doesn't have any of these subsets by using the pairing strategy.Wait, so that makes sense. Because for each pair, Calvin can't have both, so the subsets that require both elements of any pair are exactly the ones that Calvin can't have. And the number of such subsets is (4^{1010} - 3^{1010}).But let me double-check this. If we think of each pair independently, the number of subsets that include both elements of a specific pair is (2^{2020 - 2}), because the other 2018 elements can be anything. But since there are 1010 pairs, we might be tempted to multiply by 1010, but that would overcount subsets that include both elements of multiple pairs.So, actually, using the principle of inclusion-exclusion, the number of subsets that include both elements of at least one pair is equal to the total number of subsets minus the number of subsets that include at most one element from each pair, which is (2^{2020} - 3^{1010}). But (2^{2020}) is (4^{1010}), so it's the same as (4^{1010} - 3^{1010}).Therefore, the largest possible size of the family (mathcal{F}) is (4^{1010} - 3^{1010}).But let me think if there's a way to make (mathcal{F}) even larger. Suppose Hobbes doesn't just block pairs but uses a more sophisticated strategy. Maybe he can block more subsets by using a different pairing or something else. But I don't think so because the pairing strategy is already quite efficient in blocking all subsets that require both elements of any pair.Alternatively, maybe there's a way to have a family larger than (4^{1010} - 3^{1010}), but I can't think of how. Because any larger family would include subsets that don't require both elements of any pair, and those subsets could potentially be completed by Calvin.Wait, actually, no. Because if the family includes subsets that don't require both elements of any pair, then Calvin could potentially complete them. So, to ensure that Calvin can't complete any subset in (mathcal{F}), (mathcal{F}) must consist only of subsets that require both elements of at least one pair. Therefore, the maximum size is indeed (4^{1010} - 3^{1010}).So, I think that's the answer. The largest possible size of the family (mathcal{F}) is (4^{1010} - 3^{1010}).

💡Alright, so I've got this equation to solve: 81 equals 4 times 16 to the power of (x minus 2). Hmm, okay. Let me write that down to make it clearer: 81 = 4 * 16^(x-2). I need to find the value of x that makes this equation true.First, I think I should try to isolate the exponential part, which is 16^(x-2). To do that, I can divide both sides of the equation by 4. Let me see, 81 divided by 4 is... 20.25. So, that simplifies the equation to 20.25 = 16^(x-2). Okay, that looks a bit simpler.Now, I have 20.25 equals 16 raised to the power of (x-2). I remember that when dealing with exponential equations, it's helpful to express both sides with the same base if possible. Let me think about the numbers here: 16 is 2 to the 4th power, right? So, 16 is 2^4. And 20.25, well, that's a bit trickier. Let me see if I can express 20.25 as a power of 4 or something similar.Wait, 20.25 is actually 81 divided by 4, which is 81/4. Hmm, 81 is 3^4, and 4 is 2^2. So, 81/4 is (3^4)/(2^2). Maybe that's useful? I'm not sure yet. Alternatively, 20.25 is 4.5 squared because 4.5 times 4.5 is 20.25. So, 20.25 is (9/2)^2. Hmm, maybe that's helpful.Let me write down what I have so far:20.25 = 16^(x-2)Expressed as powers:(9/2)^2 = (2^4)^(x-2)Hmm, okay, so on the left side, I have (9/2)^2, and on the right side, I have (2^4)^(x-2). Maybe I can take the logarithm of both sides to solve for x. But before I do that, let me see if I can express both sides with the same base.Wait, 9 is 3^2, so 9/2 is (3^2)/2. So, (9/2)^2 is (3^2 / 2)^2, which is 3^4 / 2^2. So, 20.25 is 3^4 / 2^2. On the other side, 16^(x-2) is (2^4)^(x-2), which is 2^(4(x-2)). So, now I have:3^4 / 2^2 = 2^(4(x-2))Hmm, so I have 3^4 divided by 2^2 equals 2 raised to some power. I don't see an obvious way to express 3^4 in terms of 2's, so maybe taking logarithms is the way to go.Let me take the natural logarithm of both sides. So, ln(3^4 / 2^2) = ln(2^(4(x-2))). Using logarithm properties, ln(a/b) is ln(a) - ln(b), and ln(a^c) is c*ln(a). So, applying that:ln(3^4) - ln(2^2) = 4(x-2) * ln(2)Which simplifies to:4*ln(3) - 2*ln(2) = 4(x-2)*ln(2)Now, I can divide both sides by ln(2) to make it simpler:(4*ln(3) - 2*ln(2)) / ln(2) = 4(x-2)Let me compute the left side:First, 4*ln(3) divided by ln(2) is 4*(ln(3)/ln(2)), which is 4*log base 2 of 3. Similarly, 2*ln(2)/ln(2) is just 2. So, the left side becomes:4*log2(3) - 2 = 4(x-2)Now, let's solve for x. First, divide both sides by 4:log2(3) - 0.5 = x - 2Then, add 2 to both sides:log2(3) - 0.5 + 2 = xSimplify:log2(3) + 1.5 = xSo, x is equal to log base 2 of 3 plus 1.5. Alternatively, 1.5 is 3/2, so x = log2(3) + 3/2.Wait, let me double-check my steps to make sure I didn't make a mistake. Starting from:4*ln(3) - 2*ln(2) = 4(x-2)*ln(2)Divide both sides by ln(2):4*(ln(3)/ln(2)) - 2 = 4(x-2)Which is 4*log2(3) - 2 = 4x - 8Then, add 8 to both sides:4*log2(3) + 6 = 4xDivide both sides by 4:log2(3) + 1.5 = xYes, that seems correct. So, x is log base 2 of 3 plus 1.5.Alternatively, I could express this in terms of logarithms with a different base if needed, but I think this is a valid solution.Just to make sure, let me plug this value back into the original equation to verify.Original equation: 81 = 4*16^(x-2)Compute 16^(x-2):x = log2(3) + 1.5x - 2 = log2(3) + 1.5 - 2 = log2(3) - 0.5So, 16^(x-2) = 16^(log2(3) - 0.5)16 is 2^4, so 16^(log2(3) - 0.5) = (2^4)^(log2(3) - 0.5) = 2^(4*(log2(3) - 0.5)) = 2^(4*log2(3) - 2)Simplify:2^(4*log2(3)) * 2^(-2) = (2^log2(3))^4 * (1/4) = (3)^4 * (1/4) = 81 * (1/4) = 81/4 = 20.25So, 4*16^(x-2) = 4*20.25 = 81, which matches the left side of the equation. Therefore, the solution is correct.So, x equals log base 2 of 3 plus 1.5, or in fractional form, log2(3) + 3/2.