answer：First, I need to determine the cost price of one pen. Let's assume the cost of one pen is ( C ).Next, calculate the total cost price for 100 pens:[text{Total Cost Price} = 100 times C = 100C]The trader gains the cost of 30 pens, which means the gain is:[text{Gain} = 30C]The selling price of 100 pens is the total cost price plus the gain:[text{Selling Price} = 100C + 30C = 130C]To find the gain percentage, use the formula:[text{Gain Percentage} = left( frac{text{Gain}}{text{Cost Price}} right) times 100]Substituting the values:[text{Gain Percentage} = left( frac{30C}{100C} right) times 100 = 30%]

answer：Alright, so I have this problem here where there's a natural number on a computer screen, and it's divisible by 7. There's a cursor between two adjacent digits, and I need to prove that there's some digit I can insert into that cursor any number of times, and the resulting number will still be divisible by 7. Hmm, okay, let me try to wrap my head around this.First, let me try to visualize the problem. Suppose the number is something like 1603, and the cursor is between the 6 and the 0. So it's like 1 6|0 3. I need to find a digit, say 'a', such that if I insert 'a' between the 6 and the 0 any number of times, the new number will still be divisible by 7. So inserting once would make it 16a03, inserting twice would make it 16aa03, and so on. Got it.Since the original number is divisible by 7, let's denote it as N. So N is divisible by 7. Let's think about how numbers work. When you insert a digit into a number, you're effectively splitting the number into two parts and inserting the digit in between. So, if I have a number like ABCD, and I insert a digit 'a' between B and C, it becomes ABaCD.Now, mathematically, how does inserting a digit affect the number's divisibility? Let's break it down. Suppose the original number is N = A * 10^k + B, where A is the part before the cursor and B is the part after the cursor. When we insert a digit 'a' between A and B, the new number becomes N' = A * 10^{k+1} + a * 10^k + B.Wait, is that right? Let me check. If A is the part before the cursor and B is the part after, then inserting 'a' would shift B one place to the right, so it's A * 10^{k+1} + a * 10^k + B. Yeah, that seems correct.Since N is divisible by 7, we have N = 7 * m for some integer m. So, A * 10^k + B = 7 * m. We need to find a digit 'a' such that A * 10^{k+1} + a * 10^k + B is also divisible by 7.Let me write that out:A * 10^{k+1} + a * 10^k + B ≡ 0 mod 7.But we know that A * 10^k + B ≡ 0 mod 7, so let's substitute that in:(A * 10^k + B) * 10 + a * 10^k ≡ 0 mod 7.Which simplifies to:0 * 10 + a * 10^k ≡ 0 mod 7.So, a * 10^k ≡ 0 mod 7.Hmm, that seems too straightforward. But wait, 10 and 7 are coprime, right? So 10^k is also coprime with 7. That means that for a * 10^k ≡ 0 mod 7, we must have a ≡ 0 mod 7. But a is a single digit, so a must be 0 or 7. But 0 is a digit, so inserting 0 any number of times would keep the number divisible by 7.Wait, is that the case? Let me test with an example. Suppose N is 1603, which is divisible by 7 (1603 ÷ 7 = 229). If I insert 0 between 6 and 0, I get 16003. Is 16003 divisible by 7? Let's check: 16003 ÷ 7 = 2286.142..., which is not an integer. Hmm, that's a problem.So, my initial reasoning must be flawed. Maybe I made a mistake in the algebra. Let me go back.Starting again: N = A * 10^k + B ≡ 0 mod 7.After inserting 'a', the new number N' = A * 10^{k+1} + a * 10^k + B.So, N' = 10 * (A * 10^k + B) - 9 * B + a * 10^k.Wait, that might not be helpful. Alternatively, let's express N' in terms of N.N' = A * 10^{k+1} + a * 10^k + B = 10 * (A * 10^k) + a * 10^k + B.But N = A * 10^k + B, so N' = 10 * (A * 10^k) + a * 10^k + B = 10*(N - B) + a * 10^k + B = 10N - 10B + a * 10^k + B = 10N - 9B + a * 10^k.So, N' = 10N - 9B + a * 10^k.Since N is divisible by 7, 10N is also divisible by 7. So, N' ≡ -9B + a * 10^k mod 7.We need N' ≡ 0 mod 7, so:-9B + a * 10^k ≡ 0 mod 7.Which simplifies to:a * 10^k ≡ 9B mod 7.But 10 ≡ 3 mod 7, so 10^k ≡ 3^k mod 7. Therefore:a * 3^k ≡ 9B mod 7.Simplify 9 mod 7 is 2, so:a * 3^k ≡ 2B mod 7.So, we have:a ≡ (2B) * (3^k)^{-1} mod 7.Since 3 and 7 are coprime, 3^k has an inverse modulo 7. So, for each k, we can find the inverse of 3^k mod 7, multiply it by 2B, and that gives us the required a.But a has to be a digit, so between 0 and 9. So, for each k, we can find such an a.Wait, but k is the number of digits after the cursor, right? So, depending on where the cursor is, k can vary. But the problem states that the cursor is between two adjacent digits, so k is fixed for a given number.Therefore, for a given number N, with a cursor between two digits, k is fixed, so we can compute a as (2B) * (3^k)^{-1} mod 7, and that gives us the digit a that we can insert any number of times.Wait, but inserting a multiple times, how does that affect the number? Let's say we insert a once, then again, etc. Each time we insert a, we're effectively doing the same operation: inserting a digit between A and B.But if we insert a multiple times, the number becomes A followed by multiple a's followed by B. So, for example, inserting a twice would be A a a B.Let me see if the same reasoning applies. Suppose we insert a once, getting N' = A a B, which is divisible by 7. If we insert a again, we get N'' = A a a B. Is N'' also divisible by 7?Let's apply the same logic. N' = A a B is divisible by 7. Now, inserting another a between A and a B would give N'' = A a a B. Let's express N'' in terms of N'.N'' = A * 10^{k+1} + a * 10^{k} + a * 10^{k-1} + B.Wait, that seems complicated. Maybe a better approach is to consider the general case where we insert m copies of a between A and B, resulting in N_m = A followed by m a's followed by B.Expressed mathematically, N_m = A * 10^{k + m} + a * (10^{k + m - 1} + 10^{k + m - 2} + ... + 10^{k}) + B.That's a geometric series for the a's. The sum of the geometric series is a * (10^{k} * (10^{m} - 1)/9).So, N_m = A * 10^{k + m} + a * (10^{k} * (10^{m} - 1)/9) + B.But since N = A * 10^k + B is divisible by 7, let's see if N_m is also divisible by 7.Let me write N_m as:N_m = 10^m * (A * 10^k + B) + a * (10^{k} * (10^{m} - 1)/9).Since A * 10^k + B = N is divisible by 7, 10^m * N is also divisible by 7. So, the first term is divisible by 7. Therefore, for N_m to be divisible by 7, the second term must also be divisible by 7.So, we need:a * (10^{k} * (10^{m} - 1)/9) ≡ 0 mod 7.But 10 and 7 are coprime, so 10^k is coprime with 7. Also, 10^m - 1 is a number like 9, 99, 999, etc., which is divisible by 9. So, (10^m - 1)/9 is an integer, specifically a number consisting of m 1's.Therefore, the term becomes a * 10^k * (111...1) with m 1's.We need this to be divisible by 7. Since 10^k is coprime with 7, and (111...1) is some integer, we need a such that a * (111...1) ≡ 0 mod 7.But (111...1) mod 7 cycles every 6 digits because 10^6 ≡ 1 mod 7. So, depending on m, (111...1) mod 7 can be different.Wait, this seems complicated. Maybe there's a simpler way.Going back to the initial step, we had:a * 3^k ≡ 2B mod 7.So, for each k, we can find a digit a that satisfies this congruence. Once we have such an a, inserting it any number of times will maintain the divisibility by 7.But how does inserting multiple a's affect the number? Let's think recursively.Suppose we have N_1 = A a B, which is divisible by 7. Now, inserting another a gives N_2 = A a a B. We can write N_2 as N_1 with an additional a inserted before B. But actually, it's inserting a between A and a B, so it's more like A a a B.Wait, maybe it's better to think of it as N_2 = A * 10^{k+2} + a * 10^{k+1} + a * 10^k + B.But N_1 = A * 10^{k+1} + a * 10^k + B.So, N_2 = 10 * N_1 - 9 * B + a * 10^{k+1}.Wait, that might not be helpful. Alternatively, let's express N_2 in terms of N_1.N_2 = A * 10^{k+2} + a * 10^{k+1} + a * 10^k + B = 10 * (A * 10^{k+1} + a * 10^k) + a * 10^k + B.But A * 10^{k+1} + a * 10^k = N_1 - B.So, N_2 = 10 * (N_1 - B) + a * 10^k + B = 10N_1 - 10B + a * 10^k + B = 10N_1 - 9B + a * 10^k.But N_1 is divisible by 7, so 10N_1 is divisible by 7. Therefore, N_2 ≡ -9B + a * 10^k mod 7.But from the initial condition, we had a * 3^k ≡ 2B mod 7. So, a * 10^k ≡ 2B mod 7 because 10^k ≡ 3^k mod 7.Therefore, -9B + a * 10^k ≡ -9B + 2B ≡ -7B ≡ 0 mod 7.So, N_2 ≡ 0 mod 7. Therefore, N_2 is divisible by 7.Similarly, if we insert another a, we can show that N_3 is divisible by 7, and so on. Therefore, by induction, inserting a any number of times will result in a number divisible by 7.So, the key is to find such an a that satisfies a * 3^k ≡ 2B mod 7. Since 3 and 7 are coprime, 3^k has an inverse modulo 7. Therefore, for each k, there exists a unique a modulo 7 that satisfies this equation. Since a is a digit (0-9), we can always find such an a.For example, let's take the number 1603 again. Let's compute k and B.Wait, in 1603, if the cursor is between 6 and 0, then A is 16 and B is 03. So, A = 16, B = 3, and k is the number of digits in B, which is 2.So, k = 2. Then, 3^k = 9 ≡ 2 mod 7. So, we have:a * 2 ≡ 2 * 3 mod 7.Wait, 2B is 2*3=6. So,a * 2 ≡ 6 mod 7.Therefore, a ≡ 3 mod 7. So, a can be 3 or 10, but since it's a digit, a=3.Let me test this. Inserting 3 between 6 and 0 in 1603 gives 16303. Is 16303 divisible by 7? Let's check: 16303 ÷ 7 = 2329, which is an integer. Good.Inserting another 3 gives 163303. Let's check: 163303 ÷ 7 = 23329, which is also an integer. Nice, it works.Another example: Let's take N = 105, which is divisible by 7. Suppose the cursor is between 1 and 0, so A=1, B=05, k=2.Compute a:3^k = 9 ≡ 2 mod 7.2B = 2*5=10 ≡ 3 mod 7.So, a * 2 ≡ 3 mod 7.Multiplying both sides by the inverse of 2 mod 7, which is 4, since 2*4=8≡1 mod7.Thus, a ≡ 3*4=12≡5 mod7.So, a=5.Inserting 5 between 1 and 0 gives 1505. Is 1505 divisible by 7? 1505 ÷7=215, yes.Inserting another 5 gives 15505. 15505 ÷7=2215, which is also divisible by 7.Great, it works.So, in general, for any number N divisible by 7, with a cursor between two digits, we can compute k (the number of digits after the cursor), compute B (the number formed by the digits after the cursor), and then solve for a in the congruence a * 3^k ≡ 2B mod7. Since 3^k is invertible mod7, there exists a unique solution for a modulo7, and since a is a digit, we can always find such an a. Therefore, inserting this a any number of times will keep the number divisible by7.I think that's the gist of it. The key steps are expressing the new number after insertion in terms of the original number, using modular arithmetic to find the required digit a, and then showing that inserting a multiple times maintains the divisibility by7 through induction or recursive reasoning.

answer：Okay, so I have this problem where numbers from 1 to n are arranged randomly around a circle. I need to prove that the sum of the absolute differences between adjacent numbers is at least 2n - 2. Hmm, let me try to wrap my head around this.First, let me visualize the problem. Imagine a circle with numbers 1, 2, 3, ..., n placed around it in some order. Each number has two neighbors, and I need to calculate the absolute difference between each pair of neighbors and then sum all those differences. The claim is that no matter how I arrange these numbers, this total sum will always be at least 2n - 2.Alright, so maybe I should start by considering a simple case, like when n is small, say n=3 or n=4. That might give me some intuition.Let's take n=3. The numbers are 1, 2, 3. If I arrange them around a circle, there are two possible arrangements: 1-2-3-1 or 1-3-2-1. Let's compute the sum for both.First arrangement: 1-2-3-1Differences: |2-1| = 1, |3-2| = 1, |1-3| = 2. Sum: 1 + 1 + 2 = 4.Second arrangement: 1-3-2-1Differences: |3-1| = 2, |2-3| = 1, |1-2| = 1. Sum: 2 + 1 + 1 = 4.So in both cases, the sum is 4. Now, 2n - 2 for n=3 is 4, so it matches. That's a good sign.Let's try n=4. Numbers are 1, 2, 3, 4. There are more arrangements, but let me pick a couple.First arrangement: 1-2-3-4-1Differences: |2-1|=1, |3-2|=1, |4-3|=1, |1-4|=3. Sum: 1+1+1+3=6.Second arrangement: 1-3-2-4-1Differences: |3-1|=2, |2-3|=1, |4-2|=2, |1-4|=3. Sum: 2+1+2+3=8.Third arrangement: 1-4-2-3-1Differences: |4-1|=3, |2-4|=2, |3-2|=1, |1-3|=2. Sum: 3+2+1+2=8.Wait, 2n - 2 for n=4 is 6. So in the first arrangement, the sum is exactly 6, which is the minimum. In the other arrangements, the sum is higher. So it seems that the minimal sum is indeed 2n - 2, and other arrangements can have higher sums.Okay, so for n=3 and n=4, the minimal sum is achieved when the numbers are arranged in order, either increasing or decreasing. So maybe the minimal sum occurs when the numbers are arranged in a way that minimizes the jumps between consecutive numbers.But wait, in the circle, the arrangement is cyclic, so if you arrange them in order, you have a jump from n back to 1, which is a big difference of n-1. Hmm, in the n=4 case, that jump was 3, which is indeed n-1.So perhaps the minimal sum is achieved when the numbers are arranged in order, but because it's a circle, you have that one big jump. But in the n=3 case, arranging them in order gives a jump of 2, which is n-1 as well.Wait, but in the n=4 case, arranging them in order gives a sum of 6, which is exactly 2n - 2. So maybe the minimal sum is indeed 2n - 2, and that occurs when the numbers are arranged in order, either clockwise or counterclockwise.But in the n=4 case, arranging them as 1-2-3-4-1 gives a sum of 6, which is 2n - 2. Similarly, arranging them as 4-3-2-1-4 would also give the same sum.So perhaps, in general, arranging the numbers in order around the circle gives the minimal sum, which is 2n - 2, and any other arrangement would result in a higher sum.But how can I prove this for any n?Maybe I can think about the total variation around the circle. Since the numbers are arranged in a circle, the sum of the differences must account for the fact that you have to go from 1 to n and back to 1, which introduces a large difference.Alternatively, maybe I can model this as a traveling salesman problem, where the cities are the numbers, and the distances are the absolute differences. But I'm not sure if that's helpful here.Wait, perhaps another approach is to consider the contributions of each number to the total sum. Each number is adjacent to two others, so each number contributes twice to the total sum, once for each neighbor.But the absolute differences are |a - b| and |a - c| for a number a with neighbors b and c. So each number a contributes |a - b| + |a - c| to the total sum.Hmm, but I'm not sure how to use that directly.Alternatively, maybe I can consider the total sum as the sum over all edges of |i - j|, where i and j are adjacent numbers on the circle.Since the numbers are arranged in a circle, each number has exactly two neighbors, so the total number of edges is n.Wait, but the sum is over n edges, each contributing |i - j|.I need to find the minimal possible sum of |i - j| over all cyclic arrangements of 1 to n.Hmm, so maybe I can model this as finding a cyclic permutation of 1 to n that minimizes the sum of adjacent differences.But I need to show that the minimal sum is 2n - 2.Wait, in the n=3 case, the minimal sum is 4, which is 2*3 - 2 = 4.In the n=4 case, minimal sum is 6, which is 2*4 - 2 = 6.So maybe for general n, the minimal sum is 2n - 2.But how can I prove that?Perhaps I can use induction. Let me try that.Base case: n=2. Numbers are 1 and 2 arranged around a circle. The sum is |1-2| + |2-1| = 1 + 1 = 2. 2n - 2 = 2*2 - 2 = 2. So it holds.Assume it holds for n=k, i.e., the minimal sum is 2k - 2.Now, consider n=k+1. How can I arrange the numbers 1 to k+1 around a circle to minimize the sum?Wait, but I'm not sure how to extend the induction step here. Maybe another approach.Alternatively, maybe I can consider the total sum of all possible adjacent differences and find a lower bound.Wait, but the sum depends on the arrangement, so I need to find the minimal possible sum.Alternatively, perhaps I can think about the problem in terms of the maximum and minimum numbers.In any arrangement, the number 1 and the number n must be adjacent to some numbers. The difference |n - x| and |1 - y|, where x and y are the neighbors of n and 1, respectively.But since the numbers are arranged in a circle, 1 and n must be separated by some number of steps. The minimal total difference would occur when 1 and n are as close as possible, but in a circle, they can be adjacent, but that would create a large difference of n-1.Wait, but in the minimal sum arrangement, 1 and n are not adjacent. Instead, they are separated by one number, so that the differences |n - (n-1)| and |1 - 2| are both 1, and the jump from n-1 to 1 is n-2, which is still large, but maybe that's not the case.Wait, no, in the minimal arrangement, the numbers are arranged in order, so 1 is next to 2 and n, and n is next to n-1 and 1. So the differences are |2 - 1| = 1, |n - (n-1)| = 1, and |1 - n| = n-1. So the total sum includes two differences of 1 and one difference of n-1, but wait, in a circle with n numbers, each number has two neighbors, so the total number of differences is n.Wait, no, in the case of n=3, arranging them in order gives differences of 1, 1, and 2, which sums to 4. For n=4, arranging them in order gives differences of 1, 1, 1, and 3, which sums to 6.Wait, so in general, arranging the numbers in order gives n-1 differences of 1 and one difference of n-1, so the total sum is (n-1)*1 + (n-1) = 2n - 2.Ah, that's the key! So when arranging the numbers in order, you have n-1 adjacent pairs with a difference of 1, and one pair with a difference of n-1, giving a total sum of 2n - 2.Now, to show that this is indeed the minimal sum, I need to argue that any other arrangement would result in a larger sum.Suppose that in some arrangement, the numbers are not in order. Then, there must be at least one place where a larger jump occurs, which would increase the total sum.Wait, but how can I formalize this?Maybe I can consider that in any arrangement, the numbers must increase and decrease around the circle. The total increase and decrease must balance out because it's a circle.But the sum of absolute differences is related to the total variation.Wait, another idea: the sum of absolute differences around the circle must be at least the total distance traveled when going from 1 to n and back.Wait, that might not be precise.Alternatively, perhaps I can use the fact that the sum of absolute differences is minimized when the arrangement is as "smooth" as possible, i.e., in order.But I need a more rigorous approach.Wait, maybe I can use the concept of a traveling salesman tour on a line, where the minimal tour is just going from 1 to n and back, but in a circle, it's a bit different.Alternatively, perhaps I can use the triangle inequality.Wait, consider that the sum of absolute differences around the circle must be at least the sum of the differences when going from 1 to n and back.Wait, but that might not directly apply.Alternatively, perhaps I can consider that in any arrangement, the numbers must traverse from 1 to n and back, so the total sum must account for that.Wait, maybe another approach: consider that in any circular arrangement, the sum of the differences is equal to twice the sum of the differences from 1 to n, but that doesn't seem right.Wait, perhaps I can think of the arrangement as a permutation, and the sum of absolute differences as the sum over adjacent transpositions.But I'm not sure.Wait, another idea: the minimal sum occurs when the numbers are arranged in order, either increasing or decreasing, because any deviation from this order would introduce larger jumps, thus increasing the total sum.So, if I can show that any non-sequential arrangement would result in a larger sum, then the minimal sum is indeed 2n - 2.Alternatively, perhaps I can use the concept of inversions. In a sorted array, there are no inversions, but in a circular arrangement, it's a bit different.Wait, maybe I can use the fact that the sum of absolute differences is minimized when the permutation is as close to the identity permutation as possible.But in a circle, the identity permutation is just one possible arrangement, but there are others.Wait, perhaps I can use the concept of linear extensions. If I break the circle at some point, say between n and 1, then the arrangement becomes a linear arrangement from 1 to n, and the sum of differences is minimized when it's in order.But since it's a circle, breaking it anywhere would give a linear arrangement, but the minimal sum would still be achieved when the numbers are in order.Wait, maybe I can formalize this.Suppose I fix the position of 1. Then, to minimize the sum, the numbers should increase by 1 as we move clockwise around the circle. This would result in the minimal sum, as any deviation would require a larger jump somewhere.Therefore, the minimal sum is achieved when the numbers are arranged in order, giving a sum of 2n - 2.But I need to make this argument more rigorous.Alternatively, perhaps I can use the concept of the total variation. The total variation around the circle must be at least the difference between the maximum and minimum values, but since it's a circle, it's a bit more involved.Wait, another approach: consider that in any arrangement, the sum of the absolute differences must be at least the sum of the differences when going from 1 to n and back, which would be 2(n - 1).Wait, that seems promising.Let me think: if I start at 1, go to n, and then back to 1, the total difference would be |n - 1| + |1 - n| = 2(n - 1). But in the circle, this path is just one possible path, but the actual arrangement might have a different path.Wait, but in the minimal arrangement, the numbers are arranged in order, so the path from 1 to n is through 2, 3, ..., n-1, n, and then back to 1. The sum of differences along this path is (n - 1) + (n - 1) = 2n - 2.Wait, but in reality, the circle has n edges, each contributing a difference, so the total sum is the sum of all these edges.But in the minimal arrangement, n-1 of these edges have a difference of 1, and one edge has a difference of n-1, so the total sum is (n - 1)*1 + (n - 1) = 2n - 2.Now, suppose that in some other arrangement, the numbers are not in order. Then, there must be at least one edge where the difference is greater than 1, which would make the total sum larger than 2n - 2.Wait, but how can I be sure that the total sum can't be smaller?Alternatively, perhaps I can use the concept of the minimal spanning tree or something similar, but I'm not sure.Wait, another idea: consider that the sum of absolute differences is equivalent to the total length of the edges in the circular arrangement. To minimize this total length, the numbers should be arranged in order, as any deviation would require longer edges.Therefore, the minimal total length is achieved when the numbers are arranged in order, giving a sum of 2n - 2.But I need to make this more precise.Alternatively, perhaps I can use the concept of the minimal sum of adjacent differences in a circular arrangement.Wait, I recall that in a linear arrangement, the minimal sum of adjacent differences is achieved when the numbers are sorted, giving a sum of n - 1. But in a circular arrangement, it's different because the first and last elements are also adjacent.So, in a circular arrangement, the minimal sum would be the minimal sum for a linear arrangement plus the difference between the first and last elements.In the minimal linear arrangement, the sum is n - 1, and the difference between the first and last elements is n - 1, so the total sum for the circle would be (n - 1) + (n - 1) = 2n - 2.Therefore, the minimal sum for the circular arrangement is 2n - 2.That seems to make sense.So, to summarize, in a linear arrangement, the minimal sum of adjacent differences is n - 1, achieved when the numbers are sorted. In a circular arrangement, we have an additional edge connecting the first and last elements, which adds another n - 1 to the sum, resulting in a total minimal sum of 2n - 2.Therefore, the sum of the absolute differences between adjacent numbers arranged in a circle is at least 2n - 2.

answer：Okay, so I have this problem with two parts. It involves finding equations of lines based on given conditions. Let me try to figure this out step by step.First, the problem mentions two lines, l₁: x - 2y + 3 = 0 and l₂: 2x + 3y - 8 = 0. Their intersection point is M. I need to find the equation of another line l that passes through M and is parallel to l₃: x + 3y + 1 = 0 for part (1). Then, for part (2), I need another line l that passes through M and has a distance of 1 from point P(0,4).Starting with part (1). I think the first step is to find the coordinates of point M, the intersection of l₁ and l₂. To do that, I need to solve the system of equations given by l₁ and l₂.So, let me write down the equations:1. x - 2y + 3 = 02. 2x + 3y - 8 = 0I can solve this system using substitution or elimination. Maybe elimination is easier here. Let me try to eliminate one variable. Let's see, if I multiply the first equation by 2, I'll have:1. 2x - 4y + 6 = 02. 2x + 3y - 8 = 0Now, subtract the second equation from the first:(2x - 4y + 6) - (2x + 3y - 8) = 0 - 0Simplify:2x - 4y + 6 - 2x - 3y + 8 = 0The 2x terms cancel out:-4y - 3y + 6 + 8 = 0Combine like terms:-7y + 14 = 0So, -7y = -14Divide both sides by -7:y = 2Now that I have y, I can plug it back into one of the original equations to find x. Let's use the first equation:x - 2y + 3 = 0Substitute y = 2:x - 2(2) + 3 = 0x - 4 + 3 = 0x - 1 = 0x = 1So, the intersection point M is (1, 2). Got that.Now, part (1) asks for the equation of line l that passes through M and is parallel to l₃: x + 3y + 1 = 0.I remember that parallel lines have the same slope. So, I need to find the slope of l₃ and then use that to write the equation of line l.First, let me find the slope of l₃. The equation is x + 3y + 1 = 0. Let me rewrite it in slope-intercept form (y = mx + b) to find the slope.Starting with x + 3y + 1 = 0.Subtract x and 1 from both sides:3y = -x - 1Divide both sides by 3:y = (-1/3)x - 1/3So, the slope (m) of l₃ is -1/3.Since line l is parallel to l₃, it will have the same slope, -1/3.Now, line l passes through point M(1, 2). I can use the point-slope form of a line to write its equation.Point-slope form is: y - y₁ = m(x - x₁)Where (x₁, y₁) is a point on the line and m is the slope.Plugging in the values:y - 2 = (-1/3)(x - 1)Let me simplify this equation.First, distribute the slope:y - 2 = (-1/3)x + 1/3Now, add 2 to both sides to get y by itself:y = (-1/3)x + 1/3 + 2Convert 2 to thirds to add it to 1/3:2 = 6/3So, y = (-1/3)x + 1/3 + 6/3Combine the constants:y = (-1/3)x + 7/3Hmm, but the problem might prefer the equation in standard form (Ax + By + C = 0). Let me rearrange it.Starting from y = (-1/3)x + 7/3Multiply both sides by 3 to eliminate fractions:3y = -x + 7Bring all terms to one side:x + 3y - 7 = 0So, the equation of line l is x + 3y - 7 = 0. That seems right.Let me double-check. The slope is -1/3, which matches l₃, and it passes through (1,2). Plugging in x=1, y=2:1 + 3(2) - 7 = 1 + 6 - 7 = 0. Yes, that works.Alright, part (1) seems done.Moving on to part (2). I need to find the equation of line l that passes through M(1,2) and has a distance of 1 from point P(0,4).Hmm, distance from a point to a line. I remember the formula for the distance from a point (x₀, y₀) to the line Ax + By + C = 0 is |Ax₀ + By₀ + C| / sqrt(A² + B²).So, in this case, the line l passes through M(1,2), so it has some equation, and the distance from P(0,4) to this line is 1.But I don't know the equation of line l yet. I need to find it.Let me think about how to approach this. Since the line passes through M(1,2), I can write its equation in point-slope form, but I don't know the slope. Alternatively, I can write it in the form y = kx + b, but since it passes through (1,2), I can express b in terms of k.Alternatively, maybe it's better to represent the line in general form: Ax + By + C = 0, knowing that it passes through (1,2). So, substituting (1,2) into the equation:A(1) + B(2) + C = 0 => A + 2B + C = 0So, C = -A - 2BTherefore, the equation of line l can be written as Ax + By - A - 2B = 0.But this might complicate things. Maybe it's better to use the slope-intercept form.Let me try that.Let me denote the slope of line l as k. Then, the equation of line l is:y - 2 = k(x - 1)Which can be rewritten as:y = kx - k + 2Or, in standard form:kx - y + (2 - k) = 0So, A = k, B = -1, C = 2 - kNow, the distance from P(0,4) to this line is 1.Using the distance formula:|A*0 + B*4 + C| / sqrt(A² + B²) = 1Substituting A, B, C:|k*0 + (-1)*4 + (2 - k)| / sqrt(k² + (-1)²) = 1Simplify numerator:|0 - 4 + 2 - k| = |-2 - k|Denominator:sqrt(k² + 1)So, the equation becomes:| -2 - k | / sqrt(k² + 1) = 1Which can be written as:|k + 2| / sqrt(k² + 1) = 1Because |-2 - k| is the same as |k + 2|So, |k + 2| = sqrt(k² + 1)To solve this equation, I can square both sides to eliminate the square root and absolute value.So, (k + 2)² = (sqrt(k² + 1))²Which simplifies to:k² + 4k + 4 = k² + 1Subtract k² from both sides:4k + 4 = 1Subtract 4 from both sides:4k = -3Divide by 4:k = -3/4So, the slope k is -3/4.Therefore, the equation of line l is:y - 2 = (-3/4)(x - 1)Let me write this in standard form.Multiply both sides by 4 to eliminate fractions:4(y - 2) = -3(x - 1)Expand both sides:4y - 8 = -3x + 3Bring all terms to one side:3x + 4y - 8 - 3 = 0Wait, that would be:3x + 4y - 11 = 0Wait, let me check:Starting from 4(y - 2) = -3(x - 1)4y - 8 = -3x + 3Bring all terms to the left:3x + 4y - 8 - 3 = 0Which is 3x + 4y - 11 = 0Yes, that's correct.But wait, I should also consider the case when the line is vertical because I assumed the slope exists (i.e., the line isn't vertical). So, if the line is vertical, its equation would be x = 1, since it passes through M(1,2).Let me check if x = 1 is a possible solution.The distance from P(0,4) to the line x = 1 is |0 - 1| / sqrt(1² + 0²) = 1 / 1 = 1.Which is exactly the distance we need. So, x = 1 is also a valid solution.Therefore, there are two possible lines: one with slope -3/4, which is 3x + 4y - 11 = 0, and the vertical line x = 1.So, for part (2), the equations are x = 1 and 3x + 4y - 11 = 0.Let me recap:For part (1), I found the intersection point M(1,2) by solving the system of l₁ and l₂. Then, since the line l is parallel to l₃, it has the same slope, which I found to be -1/3. Using point-slope form, I derived the equation x + 3y - 7 = 0.For part (2), I considered the general equation of the line passing through M(1,2) with slope k, converted it to standard form, and used the distance formula from point P(0,4) to set up an equation. Solving for k gave me k = -3/4, leading to the equation 3x + 4y - 11 = 0. Additionally, I considered the vertical line case, which turned out to be x = 1, also satisfying the distance condition.I think that's thorough. I don't see any mistakes in my calculations, but let me double-check the distance for both lines.First, for 3x + 4y - 11 = 0 and point P(0,4):Distance = |3*0 + 4*4 - 11| / sqrt(3² + 4²) = |0 + 16 - 11| / 5 = |5| / 5 = 1. Correct.For x = 1 and point P(0,4):Distance is |0 - 1| / sqrt(1 + 0) = 1 / 1 = 1. Correct.So both lines satisfy the distance condition.Therefore, my answers should be correct.**Final Answer**(1) boxed{x + 3y - 7 = 0}(2) boxed{x = 1 text{ or } 3x + 4y - 11 = 0}

answer：Okay, so I have this problem about arithmetic and geometric sequences. Let me try to understand what it's asking.First, there's an arithmetic sequence {a_n} where the first term a₁ is 1, and the common difference d is greater than 0. Then, it says that a₂, a₅, and a₁₄ are the second, third, and fourth terms of a geometric sequence {b_n}. I need to find the general term formulas for both sequences and then find the sum S_n of the first n terms of the sequence {a_n + b_n}.Alright, let's start with part (1). I need to find the general term for {a_n} and {b_n}.Since {a_n} is an arithmetic sequence with a₁ = 1 and common difference d, the general term is a_n = a₁ + (n-1)d. So, a_n = 1 + (n-1)d. That's straightforward.Now, the tricky part is that a₂, a₅, and a₁₄ form a geometric sequence. Let me write down what those terms are.a₂ = 1 + (2-1)d = 1 + da₅ = 1 + (5-1)d = 1 + 4da₁₄ = 1 + (14-1)d = 1 + 13dSo, in the geometric sequence {b_n}, these correspond to the second, third, and fourth terms. That means:b₂ = a₂ = 1 + db₃ = a₅ = 1 + 4db₄ = a₁₄ = 1 + 13dIn a geometric sequence, each term is the previous term multiplied by a common ratio r. So, b₃ = b₂ * r and b₄ = b₃ * r. Therefore, we can write:b₃ = b₂ * r => 1 + 4d = (1 + d) * rb₄ = b₃ * r => 1 + 13d = (1 + 4d) * rSo now we have two equations:1) 1 + 4d = (1 + d) * r2) 1 + 13d = (1 + 4d) * rI can use these two equations to solve for r and d. Let me substitute r from the first equation into the second equation.From equation 1: r = (1 + 4d)/(1 + d)Substitute into equation 2:1 + 13d = (1 + 4d) * [(1 + 4d)/(1 + d)]So, 1 + 13d = (1 + 4d)^2 / (1 + d)Multiply both sides by (1 + d) to eliminate the denominator:(1 + 13d)(1 + d) = (1 + 4d)^2Let me expand both sides.Left side: (1 + 13d)(1 + d) = 1*1 + 1*d + 13d*1 + 13d*d = 1 + d + 13d + 13d² = 1 + 14d + 13d²Right side: (1 + 4d)^2 = 1^2 + 2*1*4d + (4d)^2 = 1 + 8d + 16d²So, setting left side equal to right side:1 + 14d + 13d² = 1 + 8d + 16d²Subtract 1 from both sides:14d + 13d² = 8d + 16d²Bring all terms to one side:14d + 13d² - 8d - 16d² = 0Simplify:(14d - 8d) + (13d² - 16d²) = 06d - 3d² = 0Factor out 3d:3d(2 - d) = 0So, 3d = 0 => d = 0, or 2 - d = 0 => d = 2But the problem says d > 0, so d = 2.Okay, so the common difference d is 2. Now, let's find the general term for {a_n}.a_n = 1 + (n - 1)*2 = 1 + 2n - 2 = 2n - 1So, a_n = 2n - 1.Now, let's find the general term for the geometric sequence {b_n}.We know that b₂ = a₂ = 1 + d = 1 + 2 = 3b₃ = a₅ = 1 + 4d = 1 + 8 = 9b₄ = a₁₄ = 1 + 13d = 1 + 26 = 27So, the terms of {b_n} are 3, 9, 27,... which is a geometric sequence with first term b₁ = ?Wait, we have b₂ = 3, b₃ = 9, b₄ = 27.So, the common ratio r is b₃ / b₂ = 9 / 3 = 3.So, the common ratio r = 3.Since b₂ = 3, then b₁ = b₂ / r = 3 / 3 = 1.Therefore, the general term for {b_n} is b_n = b₁ * r^{n - 1} = 1 * 3^{n - 1} = 3^{n - 1}.So, b_n = 3^{n - 1}.Alright, that takes care of part (1). Now, part (2) asks for the sum S_n of the first n terms of the sequence {a_n + b_n}.So, S_n = sum_{k=1}^{n} (a_k + b_k) = sum_{k=1}^{n} a_k + sum_{k=1}^{n} b_kWe can compute each sum separately.First, sum_{k=1}^{n} a_k. Since {a_n} is an arithmetic sequence with a₁ = 1 and common difference d = 2, the sum is:Sum_a = n/2 * [2a₁ + (n - 1)d] = n/2 * [2*1 + (n - 1)*2] = n/2 * [2 + 2n - 2] = n/2 * [2n] = n^2So, Sum_a = n².Next, sum_{k=1}^{n} b_k. Since {b_n} is a geometric sequence with b₁ = 1 and common ratio r = 3, the sum is:Sum_b = b₁ * (r^n - 1)/(r - 1) = 1 * (3^n - 1)/(3 - 1) = (3^n - 1)/2So, Sum_b = (3^n - 1)/2.Therefore, the total sum S_n = Sum_a + Sum_b = n² + (3^n - 1)/2.Let me write that as S_n = n² + (3^n - 1)/2.I think that's the answer. Let me just check if I did everything correctly.For part (1), we found d = 2, so a_n = 2n - 1. Then, the geometric sequence terms are 3, 9, 27,... which gives r = 3, so b_n = 3^{n - 1}. That seems correct.For part (2), the sum of the arithmetic sequence is n², which makes sense because the sum of the first n odd numbers is n². The sum of the geometric sequence is (3^n - 1)/2, which is correct because the sum of a geometric series is (r^n - 1)/(r - 1). So, adding them together, S_n = n² + (3^n - 1)/2.Yeah, that seems right. I don't think I made any mistakes here.

answer：To determine the distance between the centers of two externally tangent circles, we start by noting their radii. Circle O has a radius of 8 units, and circle O' has a radius of 3 units. When two circles are externally tangent, the distance between their centers is equal to the sum of their radii. This is because the circles touch at exactly one point, and the line connecting their centers passes through this point of tangency.Therefore, we add the radii of the two circles to find the distance between their centers:[OO' = R + r = 8 + 3 = 11]Thus, the distance between the centers of the two circles is 11 units.