answer：Okay, so I've got this problem here: A three-digit number had the sum of its digits subtracted from it, resulting in 261. I need to find the second digit of the original number. Hmm, let me try to figure this out step by step.First, let's break down what's given. We have a three-digit number. Let's represent this number as ABC, where A is the hundreds digit, B is the tens digit, and C is the units digit. So, the number itself can be expressed as 100A + 10B + C.The problem says that if we subtract the sum of its digits from the number, we get 261. The sum of the digits would be A + B + C. So, mathematically, this can be written as:100A + 10B + C - (A + B + C) = 261Let me simplify this equation. First, distribute the subtraction:100A + 10B + C - A - B - C = 261Now, combine like terms. The C terms cancel out:100A - A + 10B - B + C - C = 261Which simplifies to:99A + 9B = 261Hmm, okay, so we have 99A + 9B = 261. I notice that both terms on the left side are divisible by 9, so let's factor that out:9(11A + B) = 261Now, divide both sides by 9 to simplify further:11A + B = 29Alright, so now we have 11A + B = 29. Since A and B are digits, A can be from 1 to 9 (because it's a three-digit number), and B can be from 0 to 9.Let me think about possible values for A and B that satisfy this equation. Let's solve for B:B = 29 - 11ASince B has to be a single digit (0-9), let's see which value of A makes B fall within that range.Let's try A = 2:B = 29 - 11*2 = 29 - 22 = 7Okay, B = 7, which is a valid digit. Let's check if this works.So, the original number would be 100*2 + 10*7 + C = 200 + 70 + C = 270 + C.Now, subtract the sum of the digits: 2 + 7 + C = 9 + C.So, 270 + C - (9 + C) = 270 + C - 9 - C = 261.Yes, that works out perfectly. So, regardless of what C is, the result is 261. But the problem only asks for the second digit, which is B, and we found that B = 7.Just to be thorough, let's check if there are any other possible values for A.If A = 3:B = 29 - 11*3 = 29 - 33 = -4That's not a valid digit, so A can't be 3.If A = 1:B = 29 - 11*1 = 18That's also not a valid digit since B has to be less than 10.So, the only valid solution is A = 2 and B = 7. Therefore, the second digit of the original number is 7.

answer：Alright, so I have this problem about a triangular pyramid, which is also called a tetrahedron. The problem says that any triangular cross-section of this pyramid has an area smaller than at least one of its faces. Hmm, okay, I need to prove that. Let me try to break this down.First, let me visualize a tetrahedron. It has four triangular faces, and each face is a triangle. A cross-section, in this case, would be another triangle formed by cutting through the pyramid with a plane. So, if I imagine slicing the pyramid with a knife, the shape of the slice would be a triangle. The claim is that this triangle's area is smaller than at least one of the original faces of the pyramid.Okay, so maybe I can start by considering different types of cross-sections. There are cross-sections that pass through edges of the pyramid and those that pass through vertices. Maybe I should handle these cases separately.Let's think about a cross-section that passes through an edge. Suppose I have a tetrahedron ABCD, and I make a cross-section that includes edge CD. So, the cross-section would be triangle RCD, where R is some point on another edge of the pyramid, say between A and B. Now, the area of triangle RCD should be compared to the areas of the faces ACD and BCD.Wait, how do I compare these areas? Maybe I can use some properties of triangles or areas in three-dimensional space. I recall that the area of a triangle can be calculated using vectors or coordinates, but I'm not sure if that's the right approach here.Alternatively, maybe I can use some geometric intuition. If R is between A and B, then the height of triangle RCD from R to CD should be less than the maximum of the heights from A and B to CD. That makes sense because R is somewhere in between A and B, so its height can't be greater than both.So, if the height of RCD is less than or equal to the maximum of the heights of ACD and BCD, then the area of RCD should be less than or equal to the maximum of the areas of ACD and BCD. But the problem says "smaller than at least one of its faces," so if it's equal to one, that's still okay because it's not greater than both.Okay, that seems to handle the case where the cross-section passes through an edge. Now, what about a cross-section that passes through a vertex? Let's say I have a cross-section that goes through vertex A and some other points on the edges. So, the cross-section would be triangle APR, where P and R are points on other edges.Hmm, how do I compare the area of APR to the areas of the faces? Maybe I can use a similar approach as before, considering the heights from A to the opposite edges or something like that. But I'm not entirely sure.Wait, maybe I need to think about projections or something related to the orientation of the cross-section relative to the faces. If I project the cross-section onto one of the faces, the area might shrink, which could help me compare them.Alternatively, perhaps I can use the fact that the cross-section is contained within the pyramid, so its area can't exceed the area of any face it's parallel to or something like that. But I'm not sure if that's accurate.I'm getting a bit stuck here. Maybe I should try to formalize this a bit more. Let's assign coordinates to the vertices of the tetrahedron to make things more concrete. Suppose I place vertex A at (0,0,0), B at (1,0,0), C at (0,1,0), and D at (0,0,1). Then, any cross-section can be represented by a plane equation, and I can find the intersection points with the edges to define the cross-sectional triangle.Once I have the coordinates of the cross-sectional triangle, I can calculate its area using the cross product formula. Then, I can compare this area to the areas of the original faces, which I can also calculate.But this seems like a lot of work, and I'm not sure if it's the most efficient way to approach the problem. Maybe there's a more geometric or algebraic proof that doesn't require coordinates.Wait, another idea: perhaps I can use the concept of convexity. A tetrahedron is a convex polyhedron, so any cross-section should also be a convex polygon, in this case, a triangle. Maybe the area of this triangle is constrained by the areas of the faces in some way.I'm still not entirely sure how to make this precise, though. Maybe I need to think about the relationship between the cross-section and the faces in terms of their positions and orientations.Another thought: if I consider the cross-section as a linear transformation of one of the faces, then the area might be scaled by some factor. But I'm not sure if that's applicable here.Hmm, I'm going in circles a bit. Maybe I should look for some inequalities or theorems related to areas of cross-sections in convex polyhedra. I recall that there's something called the isoperimetric inequality, but I don't think that's directly applicable here.Wait, perhaps I can use the fact that the cross-section lies entirely within the tetrahedron, so its area can't exceed the area of any face it intersects. But I need to make this more precise.Let me try to formalize this. Suppose I have a cross-section triangle T inside the tetrahedron. If T intersects a face F of the tetrahedron, then T lies entirely within F or partially outside. But since the tetrahedron is convex, T must lie entirely within F if it's a cross-section. Wait, no, that's not necessarily true because the cross-section could intersect multiple faces.Actually, a cross-section of a tetrahedron will intersect three edges, forming a triangle that doesn't necessarily lie entirely within any single face. So, I need to compare the area of this triangle to the areas of the four faces.But how? Maybe I can use some kind of averaging argument. If the cross-section is smaller than all faces, then it's certainly smaller than at least one. Wait, no, the problem says it's smaller than at least one, not necessarily all.So, perhaps I can assume, for contradiction, that the cross-section is larger than all four faces and reach a contradiction. But I'm not sure how to proceed with that.Alternatively, maybe I can use the fact that the cross-section is a linear combination of the faces in some way, and thus its area is bounded by the maximum area of the faces.Wait, I think I'm overcomplicating this. Let's go back to the initial idea with heights. If I can show that the height of the cross-section from some vertex is less than the height of at least one face, then the area would be smaller.But I need to make sure that this holds for any cross-section. Maybe I can use some kind of pigeonhole principle argument. Since the cross-section is determined by three points on the edges, at least one of these points must be closer to a vertex than the corresponding point on a face.Hmm, that might work. Let me try to make this more precise.Suppose I have a cross-section triangle with vertices on three different edges of the tetrahedron. For each edge, the point where the cross-section intersects the edge divides the edge into two segments. If I consider the ratios of these segments, at least one of them must be less than or equal to the corresponding ratio on a face.Wait, I'm not sure. Maybe I need to think about the barycentric coordinates or something like that.Alternatively, maybe I can use the fact that the cross-section is affine equivalent to one of the faces, but scaled down. So, its area must be smaller.But I'm not entirely confident about that. Maybe I need to look for a more straightforward geometric argument.Wait, another idea: consider the areas of the cross-section and the faces as functions of the position of the cutting plane. Maybe I can show that the area of the cross-section reaches its maximum when it coincides with one of the faces, and thus, any other cross-section must have a smaller area.But I'm not sure if that's true. Actually, I think the maximum area cross-section of a tetrahedron is indeed one of its faces, so any other cross-section must have a smaller area.Yes, that makes sense. Because if you move the cutting plane away from a face, the cross-sectional area decreases. So, the maximum possible area of a cross-section is equal to the area of one of the faces, and any other cross-section must have a smaller area.Therefore, any triangular cross-section must have an area smaller than at least one of the faces, specifically the one it's not coinciding with.Okay, I think that's the key idea. So, to summarize, the maximum area of a cross-section in a tetrahedron is achieved when the cross-section coincides with one of the faces. Therefore, any other cross-section must have a smaller area, meaning it's smaller than at least one face.I think that's a solid argument. Maybe I can make it more rigorous by considering the continuity of the area as the cutting plane moves and showing that the area is maximized when the plane coincides with a face.Yeah, that should do it. So, in conclusion, any triangular cross-section of a tetrahedron has an area smaller than at least one of its faces because the maximum cross-sectional area is achieved by one of the faces themselves.

answer：Okay, so I have this equation: log base 2 of (x + 3) minus log base 2 of (x + 1) equals 1. Hmm, I need to find the value of x, which is denoted as a in the problem. Alright, let me think about how to approach this.First, I remember that when you subtract two logarithms with the same base, it's equivalent to the logarithm of the quotient of their arguments. So, log base 2 of (x + 3) minus log base 2 of (x + 1) is the same as log base 2 of ((x + 3)/(x + 1)). That simplifies the equation a bit.So, rewriting the equation, I get:log₂((x + 3)/(x + 1)) = 1Now, I need to solve for x. I know that if log base b of something equals c, then that something is equal to b raised to the power of c. Applying that here, since the base is 2 and the result is 1, I can rewrite the equation as:(x + 3)/(x + 1) = 2¹Which simplifies to:(x + 3)/(x + 1) = 2Okay, so now I have a rational equation. To solve for x, I can cross-multiply to eliminate the denominator. Multiplying both sides by (x + 1) gives:x + 3 = 2(x + 1)Let me expand the right side:x + 3 = 2x + 2Now, I want to get all the x terms on one side and the constants on the other. Subtracting x from both sides:3 = x + 2Then, subtracting 2 from both sides:1 = xSo, x equals 1. But wait, I should check if this solution is valid in the original equation because sometimes when dealing with logarithms, you might get extraneous solutions.Let me plug x = 1 back into the original logarithmic expressions to make sure they are defined and that the equation holds.First, checking log₂(x + 3):x + 3 = 1 + 3 = 4, which is positive, so log₂(4) is defined.Next, checking log₂(x + 1):x + 1 = 1 + 1 = 2, which is also positive, so log₂(2) is defined.Now, let's verify the original equation:log₂(4) - log₂(2) = ?We know that log₂(4) is 2 because 2² = 4, and log₂(2) is 1 because 2¹ = 2. So:2 - 1 = 1Which matches the right side of the equation. Therefore, x = 1 is indeed a valid solution.I think that's it. I didn't encounter any issues during the solving process, and the solution checks out when plugged back into the original equation. So, the value of a is 1.

answer：Alright, so I have this problem here about a city applying for the title of "National Sanitary Civilized City." They need to inspect 200 restaurants, and they're starting with a random selection of 5 restaurants using a random number table. The task is to figure out the second number among these five when starting from the fifth column of the given random number table.First, let me understand the setup. The restaurants are numbered from 001 to 200, which makes sense because they need to cover all 200 restaurants uniquely. The random number table provided is a sequence of two-digit numbers: 84, 42, 17, 53, 31, 57, 24, 55, 06, 88, 77, 04, 74, 47, 67, 21, 76, 33, 50, 25, 83, 92, 12, 06, 76.The instruction says to start from the fifth column and read the numbers sequentially. Hmm, okay. So, I need to figure out what the fifth column is in this table. Let me list out the numbers with their positions to visualize the columns better.Let's assume that each two-digit number occupies a column. So, the first column is 84, the second is 42, the third is 17, the fourth is 53, the fifth is 31, the sixth is 57, and so on. Wait, is that correct? Or is each digit a column? Hmm, the problem says "starting from the fifth column," but it's not entirely clear whether each two-digit number is a column or each single digit is a column.Looking back at the problem statement: "starting from the fifth column and reading the numbers sequentially." Since the random number table is given as two-digit numbers separated by spaces, it's more likely that each two-digit number is a column. So, the first column is 84, second is 42, third is 17, fourth is 53, fifth is 31, sixth is 57, and so on.So, starting from the fifth column, which is 31, and then reading sequentially. That means the first number we pick is 31. Then the next is 57, followed by 24, 55, 06, 88, 77, 04, 74, 47, 67, 21, 76, 33, 50, 25, 83, 92, 12, 06, 76.But wait, we need to select 5 restaurants. So, starting from the fifth column, which is 31, we read the numbers one by one until we get five valid restaurant numbers.However, each restaurant is numbered from 001 to 200, so each number must be a three-digit number. The random number table provides two-digit numbers, so perhaps we need to combine them or interpret them differently.Wait, maybe each two-digit number is part of a larger sequence. Let me think. If we have 84, 42, 17, 53, 31, 57, 24, 55, 06, 88, 77, 04, 74, 47, 67, 21, 76, 33, 50, 25, 83, 92, 12, 06, 76, these are all two-digit numbers. To get three-digit numbers for the restaurant IDs, we might need to read three consecutive two-digit numbers and combine them into a six-digit number, but that seems complicated.Alternatively, perhaps each two-digit number is part of a sequence where we read them as individual digits, making three-digit numbers by taking three consecutive digits. Let me try that approach.So, the random number table is: 84 42 17 53 31 57 24 55 06 88 77 04 74 47 67 21 76 33 50 25 83 92 12 06 76.If I write them out without spaces: 84421753315724550688770474476721763350258392120676.Now, starting from the fifth column. Wait, but if each two-digit number is a column, then the fifth column is 31. So, starting from 31, and reading sequentially.But if we need three-digit numbers, perhaps we need to take three digits at a time starting from the fifth column. Let me clarify.Alternatively, maybe each two-digit number is treated as part of a three-digit number by overlapping. For example, starting from 31, the next number is 57, so combining 31 and 57 gives 3157, but that's four digits. Hmm, not helpful.Wait, perhaps the numbers are read as individual digits. So, the random number table is a sequence of digits: 8,4,4,2,1,7,5,3,3,1,5,7,2,4,5,5,0,6,8,8,7,7,0,4,7,4,4,7,6,7,2,1,7,6,3,3,5,0,2,5,8,3,9,2,1,2,0,6,7,6.But that seems too long. Wait, maybe each two-digit number is a single entry, so 84 is one, 42 is another, etc. So, starting from the fifth column, which is 31, and then reading the next numbers: 57, 24, 55, 06, 88, 77, 04, 74, 47, 67, 21, 76, 33, 50, 25, 83, 92, 12, 06, 76.But each restaurant is numbered from 001 to 200, so each number must be between 001 and 200. The two-digit numbers in the table are from 04 to 92, so they are all within 00 to 99. So, perhaps we can treat each two-digit number as the last two digits of a three-digit number, assuming the first digit is 0. So, 31 becomes 031, 57 becomes 057, 24 becomes 024, etc.But wait, 031 is 31, which is within 001-200. Similarly, 057 is 57, which is also within range. So, perhaps that's the approach.So, starting from the fifth column, which is 31, we take that as 031. Then the next number is 57, which is 057, then 24 as 024, then 55 as 055, then 06 as 006, then 88 as 088, and so on.But we need to select 5 restaurants. So, starting from 31, we take 031, then 057, then 024, then 055, then 006. So, the five numbers are 031, 057, 024, 055, 006.But wait, the problem says "starting from the fifth column and reading the numbers sequentially." So, the fifth column is 31, which is the first number. Then the next numbers are 57, 24, 55, 06. So, the five numbers are 31, 57, 24, 55, 06.But these are two-digit numbers. Since the restaurants are numbered from 001 to 200, we need to convert these into three-digit numbers. So, 31 becomes 031, 57 becomes 057, 24 becomes 024, 55 becomes 055, 06 becomes 006.So, the five numbers are 031, 057, 024, 055, 006. Therefore, the second number among these five is 057.Wait, but the answer given was 068. Hmm, that doesn't match. Maybe I made a mistake in interpreting the columns.Let me try another approach. Perhaps the columns are individual digits, not two-digit numbers. So, the random number table is a sequence of digits: 8,4,4,2,1,7,5,3,3,1,5,7,2,4,5,5,0,6,8,8,7,7,0,4,7,4,4,7,6,7,2,1,7,6,3,3,5,0,2,5,8,3,9,2,1,2,0,6,7,6.If each digit is a column, then the fifth column is the fifth digit, which is 1 (since the first digit is 8, second is 4, third is 4, fourth is 2, fifth is 1). So, starting from the fifth column, which is 1, and reading sequentially, we get the digits: 1,7,5,3,3,1,5,7,2,4,5,5,0,6,8,8,7,7,0,4,7,4,4,7,6,7,2,1,7,6,3,3,5,0,2,5,8,3,9,2,1,2,0,6,7,6.Now, to form three-digit numbers, we can take three digits at a time: 175, 331, 572, 455, 068, 877, 047, 447, 672, 176, 335, 025, 839, 212, 067, 6.But we need to stop when we have five valid numbers within 001-200. So, starting from 175, which is 175, that's within range. Next is 331, which is 331, but that's above 200, so we skip it. Next is 572, which is way above 200, skip. Next is 455, also above 200, skip. Next is 068, which is 68, within range. Next is 877, too high. Next is 047, which is 47, within range. Next is 447, too high. Next is 672, too high. Next is 176, which is 176, within range. Next is 335, too high. Next is 025, which is 25, within range. Next is 839, too high. Next is 212, which is 212, above 200, skip. Next is 067, which is 67, within range. Next is 6, which is incomplete, so we might stop here.So, the valid numbers we get are: 175, 068, 047, 176, 025, 067. But we need only five. So, the first five valid numbers are 175, 068, 047, 176, 025. Therefore, the second number is 068.Ah, that makes sense now. So, the initial approach was incorrect because I treated the two-digit numbers as columns, but actually, each digit is a column. Therefore, starting from the fifth column (the fifth digit), we read the digits sequentially and form three-digit numbers, skipping those above 200. The second valid number is 068.So, the second number among the five selected is 068.

answer：Okay, so I have this problem: convert ( sqrt{2} e^{13pi i/4} ) to rectangular form. Hmm, rectangular form is like the standard a + bi form, right? So, I need to figure out what the real part and the imaginary part are from this exponential form.First, I remember that Euler's formula says that ( e^{itheta} = cos(theta) + isin(theta) ). So, in this case, the expression is ( sqrt{2} e^{13pi i/4} ), which should be equal to ( sqrt{2} [cos(13pi/4) + isin(13pi/4)] ). Okay, that makes sense.But wait, the angle here is ( 13pi/4 ). That seems a bit large. I think angles in the exponential form are usually given in terms of multiples of ( 2pi ), so maybe I can simplify this angle by subtracting multiples of ( 2pi ) to find an equivalent angle between 0 and ( 2pi ).Let me calculate how many times ( 2pi ) goes into ( 13pi/4 ). Well, ( 2pi ) is the same as ( 8pi/4 ). So, if I subtract ( 8pi/4 ) from ( 13pi/4 ), I get ( 5pi/4 ). So, ( 13pi/4 ) is the same as ( 5pi/4 ) when considering the unit circle because angles are periodic with period ( 2pi ).Alright, so now I can rewrite the original expression as ( sqrt{2} e^{5pi i/4} ), which is ( sqrt{2} [cos(5pi/4) + isin(5pi/4)] ). Now, I need to find the values of ( cos(5pi/4) ) and ( sin(5pi/4) ).I remember that ( 5pi/4 ) is in the third quadrant of the unit circle, where both cosine and sine are negative. The reference angle for ( 5pi/4 ) is ( 5pi/4 - pi = pi/4 ). So, the cosine and sine of ( 5pi/4 ) should be the same as those of ( pi/4 ), but negative.I know that ( cos(pi/4) = sin(pi/4) = sqrt{2}/2 ). Therefore, ( cos(5pi/4) = -sqrt{2}/2 ) and ( sin(5pi/4) = -sqrt{2}/2 ).Substituting these back into the expression, I get:( sqrt{2} [ -sqrt{2}/2 + i(-sqrt{2}/2) ] ).Now, I need to multiply ( sqrt{2} ) with each term inside the brackets. Let's do that step by step.First, ( sqrt{2} times (-sqrt{2}/2) ). Multiplying the radicals: ( sqrt{2} times sqrt{2} = 2 ). So, it becomes ( -2/2 = -1 ).Next, ( sqrt{2} times (-sqrt{2}/2) ) for the imaginary part. Similarly, this is also ( -2/2 = -1 ).Putting it all together, the expression simplifies to ( -1 - i ).Wait, let me double-check that. So, ( sqrt{2} times (-sqrt{2}/2) = -(sqrt{2} times sqrt{2}) / 2 = -2/2 = -1 ). Yep, that's correct for both the real and imaginary parts.So, after simplifying, the rectangular form of ( sqrt{2} e^{13pi i/4} ) is ( -1 - i ).I think that makes sense. The angle ( 5pi/4 ) is in the third quadrant, so both the real and imaginary parts should be negative, which they are. Also, the magnitude is ( sqrt{2} ), and when we multiply it by ( sqrt{2}/2 ), we get 1, which matches the result.Just to make sure, let's think about the magnitude. The original expression has a magnitude of ( sqrt{2} ), and in rectangular form, the magnitude should be ( sqrt{(-1)^2 + (-1)^2} = sqrt{1 + 1} = sqrt{2} ). That checks out.So, I feel confident that the rectangular form is indeed ( -1 - i ).**Final Answer**The rectangular form of ( sqrt{2} e^{13pi i/4} ) is boxed{-1 - i}.

answer：Okay, so I have this problem here: "The arithmetic mean of an even number of consecutive even integers is z. Find the sum of the smallest and largest of these integers in terms of z." Hmm, let me try to wrap my head around this.First, let's break down what's given. We have an even number of consecutive even integers. So, for example, if we have four numbers, they might be 2, 4, 6, 8. Or if we have six numbers, maybe 4, 6, 8, 10, 12, 14. The key points here are that the number of integers is even, and each integer is consecutive even numbers.The arithmetic mean of these integers is given as z. The arithmetic mean is just the average, right? So if I have a set of numbers, I add them all up and divide by how many there are. In this case, since it's an even number of integers, let's say there are 2n integers. So, n is some positive integer, and the total number of integers is 2n.Now, the problem is asking for the sum of the smallest and largest of these integers in terms of z. So, if I can find expressions for the smallest and largest integers, add them together, and express that sum in terms of z, that should be the answer.Let me think about how to model this. Let's denote the smallest integer as b. Since they are consecutive even integers, the next one would be b + 2, then b + 4, and so on. So, the sequence would be: b, b + 2, b + 4, ..., up to the 2n-th term.What's the largest integer in this sequence? Well, the first term is b, the second is b + 2, so the k-th term is b + 2(k - 1). Therefore, the 2n-th term would be b + 2(2n - 1) = b + 4n - 2. So, the largest integer is b + 4n - 2.So, the smallest integer is b, and the largest is b + 4n - 2. Therefore, their sum would be b + (b + 4n - 2) = 2b + 4n - 2. Hmm, okay, that's the sum we need to find in terms of z.Now, let's relate this to the arithmetic mean z. The arithmetic mean is the sum of all the integers divided by the number of integers. So, the sum of all the integers is z multiplied by the number of integers, which is 2n. So, sum = 2n * z.But we can also express the sum of the integers in another way. Since these are consecutive even integers, they form an arithmetic sequence. The sum of an arithmetic sequence can be calculated as the number of terms multiplied by the average of the first and last terms. So, sum = number of terms * (first term + last term)/2.In this case, the number of terms is 2n, the first term is b, and the last term is b + 4n - 2. So, sum = 2n * (b + (b + 4n - 2))/2 = 2n * (2b + 4n - 2)/2 = 2n * (b + 2n - 1).But we also know that the sum is equal to 2n * z. So, setting these equal:2n * (b + 2n - 1) = 2n * zWe can divide both sides by 2n (assuming n ≠ 0, which it isn't because we have a positive number of integers):b + 2n - 1 = zSo, b + 2n - 1 = z. Therefore, b = z - 2n + 1.Wait, hold on. Let me check that algebra again. If b + 2n - 1 = z, then b = z - 2n + 1. Hmm, that seems correct.But I'm trying to find the sum of the smallest and largest integers, which is 2b + 4n - 2. Let's substitute b from the equation above into this sum.So, sum = 2b + 4n - 2 = 2(z - 2n + 1) + 4n - 2Let's expand that:= 2z - 4n + 2 + 4n - 2Now, let's combine like terms:-4n + 4n cancels out, and +2 -2 also cancels out.So, we're left with 2z.Wait, that's interesting. So, the sum of the smallest and largest integers is 2z.But let me think again. Is this possible? Let's test it with an example.Suppose we have four consecutive even integers: 2, 4, 6, 8.Their arithmetic mean is (2 + 4 + 6 + 8)/4 = 20/4 = 5. So, z = 5.The smallest integer is 2, the largest is 8. Their sum is 10, which is 2z = 10. That checks out.Another example: six consecutive even integers: 4, 6, 8, 10, 12, 14.Arithmetic mean is (4 + 6 + 8 + 10 + 12 + 14)/6 = 54/6 = 9. So, z = 9.Smallest is 4, largest is 14. Their sum is 18, which is 2z = 18. That also checks out.Wait, so regardless of the number of integers, as long as it's an even number of consecutive even integers, the sum of the smallest and largest is twice the arithmetic mean. That's a neat result.So, in general, for any even number of consecutive even integers, the sum of the smallest and largest is 2z.Therefore, the answer should be 2z.But let me try to think about why this happens. In an arithmetic sequence, the average is equal to the average of the first and last terms. Since we have an even number of terms, the average is still the average of the first and last terms. Therefore, the sum of the first and last terms is twice the average, which is 2z.Ah, that makes sense. So, in any arithmetic sequence, whether it's even or odd number of terms, the average is the average of the first and last terms. Therefore, the sum of the first and last terms is twice the average. So, in this case, since we have an arithmetic sequence of consecutive even integers, the same logic applies.Therefore, regardless of the number of terms, as long as it's an arithmetic sequence, the sum of the first and last terms is twice the average. So, in this problem, since the average is z, the sum is 2z.So, that's why the answer is 2z.I think that's solid. I've tested it with examples, and the reasoning makes sense. So, I'm confident that the sum of the smallest and largest integers is 2z.