answer：First, I need to identify the given information and what is being asked. The problem states that the width of a rectangle is 2 centimeters longer than its length, and the perimeter of the rectangle is 16 centimeters. I need to find the length of the width.I'll start by assigning variables to the unknown quantities. Let ( L ) represent the length of the rectangle in centimeters. Since the width is 2 centimeters longer than the length, I can express the width as ( W = L + 2 ).Next, I'll recall the formula for the perimeter of a rectangle, which is ( P = 2L + 2W ). Given that the perimeter is 16 centimeters, I can set up the equation:[ 16 = 2L + 2W ]Now, I'll substitute the expression for ( W ) from the first equation into the perimeter equation:[ 16 = 2L + 2(L + 2) ]I'll simplify the equation by expanding and combining like terms:[ 16 = 2L + 2L + 4 ][ 16 = 4L + 4 ]To isolate ( L ), I'll subtract 4 from both sides of the equation:[ 12 = 4L ]Then, I'll divide both sides by 4 to solve for ( L ):[ L = 3 ]Now that I have the value of ( L ), I'll substitute it back into the expression for ( W ) to find the width:[ W = 3 + 2 ][ W = 5 ]Therefore, the length of the width is 5 centimeters.

answer：Okay, so I have this problem about an arithmetic sequence. Let me try to understand what it's asking. It says that in the arithmetic sequence {a_n}, the sum of the third term and the seventh term is 16, and the sum of the first ten terms is 85. I need to find the common difference of the sequence.Alright, first, let me recall what an arithmetic sequence is. It's a sequence where each term after the first is obtained by adding a constant difference to the preceding term. So, if the first term is a₁ and the common difference is d, then the nth term, a_n, can be written as a₁ + (n-1)d.Given that, let's break down the information provided.First, a₃ + a₇ = 16. Let me write expressions for a₃ and a₇ using the formula for the nth term.a₃ = a₁ + 2d (since n=3, so 3-1=2)a₇ = a₁ + 6d (since n=7, so 7-1=6)Adding these together: a₃ + a₇ = (a₁ + 2d) + (a₁ + 6d) = 2a₁ + 8d = 16.So, equation one is 2a₁ + 8d = 16.Next, the sum of the first ten terms, S₁₀, is 85. The formula for the sum of the first n terms of an arithmetic sequence is S_n = n/2 * (2a₁ + (n-1)d). Let's apply that for n=10.S₁₀ = 10/2 * (2a₁ + 9d) = 5*(2a₁ + 9d) = 10a₁ + 45d = 85.So, equation two is 10a₁ + 45d = 85.Now, I have a system of two equations:1. 2a₁ + 8d = 162. 10a₁ + 45d = 85I need to solve this system to find the value of d, the common difference.Let me see. Maybe I can solve equation one for a₁ and substitute into equation two.From equation one: 2a₁ + 8d = 16.Let me subtract 8d from both sides: 2a₁ = 16 - 8d.Then, divide both sides by 2: a₁ = (16 - 8d)/2 = 8 - 4d.So, a₁ = 8 - 4d.Now, substitute this expression for a₁ into equation two.Equation two is 10a₁ + 45d = 85.Substituting a₁: 10*(8 - 4d) + 45d = 85.Let me compute 10*(8 - 4d): that's 80 - 40d.So, equation becomes: 80 - 40d + 45d = 85.Combine like terms: (-40d + 45d) = 5d.So, 80 + 5d = 85.Subtract 80 from both sides: 5d = 5.Divide both sides by 5: d = 1.So, the common difference d is 1.Wait, let me double-check my calculations to make sure I didn't make a mistake.Starting with equation one: 2a₁ + 8d = 16. If d=1, then 2a₁ + 8*1 = 16 => 2a₁ + 8 = 16 => 2a₁ = 8 => a₁ = 4.Then, equation two: 10a₁ + 45d = 85. Plugging in a₁=4 and d=1: 10*4 + 45*1 = 40 + 45 = 85. Yep, that checks out.Also, checking a₃ + a₇: a₁=4, d=1.a₃ = 4 + 2*1 = 6a₇ = 4 + 6*1 = 10a₃ + a₇ = 6 + 10 = 16. Perfect, that's correct.So, everything seems to add up. Therefore, the common difference is indeed 1.I think that's solid. I don't see any mistakes in my steps, and the checks confirm the solution. So, yeah, d=1 is the answer.**Final Answer**The common difference of the arithmetic sequence is boxed{1}.

answer：Alright, so I've got this inequality to prove: [ x_1! cdot x_2! cdot ldots cdot x_n! geq left( leftlfloor frac{x_1 + x_2 + cdots + x_n}{n} rightrfloor ! right)^n ]where ( x_1, x_2, ldots, x_n ) are non-negative integers, and ( leftlfloor y rightrfloor ) is the floor function, meaning the largest integer not exceeding ( y ). I also need to figure out when equality holds.Okay, let's break this down. First, I need to understand what this inequality is saying. On the left side, we have the product of factorials of each ( x_i ). On the right side, we're taking the floor of the average of all ( x_i )s, then taking the factorial of that floor value, and raising it to the power of ( n ).So, essentially, the inequality is comparing the product of individual factorials to the nth power of the factorial of the average (rounded down). I need to show that the product of the factorials is always at least as big as this nth power.Let me think about some simple cases to get an intuition.**Case 1: All ( x_i ) are equal.**Suppose ( x_1 = x_2 = ldots = x_n = k ). Then the left side is ( (k!)^n ). The average is ( k ), so the right side is ( (k!)^n ). So in this case, equality holds. That's good to know.**Case 2: Some ( x_i ) are different.**Let's take ( n = 2 ) for simplicity. Suppose ( x_1 = 3 ) and ( x_2 = 1 ). Then the left side is ( 3! cdot 1! = 6 cdot 1 = 6 ). The average is ( (3 + 1)/2 = 2 ), so the right side is ( (2!)^2 = 2^2 = 4 ). Indeed, ( 6 geq 4 ).Another example: ( x_1 = 4 ), ( x_2 = 2 ). Left side: ( 4! cdot 2! = 24 cdot 2 = 48 ). Average: ( 3 ), right side: ( 3!^2 = 36 ). So ( 48 geq 36 ).Wait, so when the numbers are unequal, the product seems to be larger. That suggests that the product of factorials is minimized when all ( x_i ) are equal. So maybe the inequality is a consequence of some sort of convexity or inequality principle.**Thinking about Inequalities:**I recall that for positive numbers, the product is minimized when the numbers are as equal as possible, given a fixed sum. This is related to the AM-GM inequality, but here we're dealing with factorials, which are multiplicative functions.Factorials grow very rapidly, so perhaps the product of factorials is minimized when the ( x_i ) are as equal as possible.**Formalizing the Idea:**Let me denote ( S = x_1 + x_2 + ldots + x_n ). Then, the average is ( frac{S}{n} ), and we're taking the floor of that, which is ( q = leftlfloor frac{S}{n} rightrfloor ).So, the right side is ( (q!)^n ). The left side is the product of ( x_i! ).I need to show that ( prod_{i=1}^n x_i! geq (q!)^n ).Given that ( q ) is the floor of the average, ( q leq frac{S}{n} < q + 1 ). So, ( S = nq + r ) where ( 0 leq r < n ).This means that ( r ) of the ( x_i )s are ( q + 1 ) and the remaining ( n - r ) are ( q ).Wait, but in the problem, the ( x_i ) can be any non-negative integers, not necessarily partitioned into ( q ) and ( q + 1 ).But perhaps I can consider that the minimal product occurs when the ( x_i ) are as equal as possible, i.e., when ( r ) of them are ( q + 1 ) and the rest are ( q ).So, if I can show that the product ( prod_{i=1}^n x_i! ) is minimized when the ( x_i ) are as equal as possible, then the inequality would follow.**Using the Concept of Majorization:**I remember that in mathematics, majorization is a preorder on vectors that compares how "spread out" one vector is compared to another. If one vector majorizes another, then certain inequalities hold, especially for Schur-convex functions.Factorials are increasing functions, so perhaps the product of factorials is Schur-convex. If that's the case, then the minimal product occurs when the vector is as equal as possible.But I'm not entirely sure about the Schur-convexity of the product of factorials. Maybe I need a different approach.**Looking for a Direct Proof:**Alternatively, I can try to use induction on ( n ) or on the sum ( S ).**Base Case: ( n = 1 ).**If ( n = 1 ), then the inequality becomes ( x_1! geq (x_1!)^1 ), which is equality. So, the base case holds.**Inductive Step:**Assume the inequality holds for ( n - 1 ). Now, consider ( n ).But I'm not sure how to apply the inductive hypothesis here because the right side involves the floor of the average, which complicates things.**Alternative Approach: Using the AM-GM Inequality for Factorials:**Wait, perhaps I can relate the product of factorials to the factorial of the sum.I know that ( (x_1 + x_2 + ldots + x_n)! ) is greater than or equal to the product ( x_1! x_2! ldots x_n! ) when the ( x_i ) are arranged in a certain way, but I'm not sure if that's directly applicable here.Alternatively, maybe I can use Stirling's approximation for factorials, which approximates ( n! ) as ( n^n e^{-n} sqrt{2pi n} ). But since we're dealing with integers and exact inequalities, Stirling's formula might not give the precise result needed.**Considering the Function ( f(k) = ln(k!) ):**Since ( ln(k!) ) is a convex function, by Jensen's inequality, we have:[ frac{1}{n} sum_{i=1}^n ln(x_i!) geq lnleft( leftlfloor frac{S}{n} rightrfloor ! right) ]Multiplying both sides by ( n ):[ sum_{i=1}^n ln(x_i!) geq n lnleft( leftlfloor frac{S}{n} rightrfloor ! right) ]Exponentiating both sides:[ prod_{i=1}^n x_i! geq left( leftlfloor frac{S}{n} rightrfloor ! right)^n ]Which is exactly the inequality we need to prove!Wait, does this hold? Let me verify.**Jensen's Inequality:**Jensen's inequality states that for a convex function ( f ), the following holds:[ frac{1}{n} sum_{i=1}^n f(x_i) geq fleft( frac{1}{n} sum_{i=1}^n x_i right) ]In our case, ( f(k) = ln(k!) ), which is convex because the second difference is positive.But in our case, we have ( leftlfloor frac{S}{n} rightrfloor ) instead of ( frac{S}{n} ). So, since ( leftlfloor frac{S}{n} rightrfloor leq frac{S}{n} ), and ( f ) is increasing, we have:[ fleft( leftlfloor frac{S}{n} rightrfloor right) leq fleft( frac{S}{n} right) ]Therefore,[ frac{1}{n} sum_{i=1}^n f(x_i) geq fleft( frac{S}{n} right) geq fleft( leftlfloor frac{S}{n} rightrfloor right) ]So,[ frac{1}{n} sum_{i=1}^n f(x_i) geq fleft( leftlfloor frac{S}{n} rightrfloor right) ]Multiplying both sides by ( n ):[ sum_{i=1}^n f(x_i) geq n fleft( leftlfloor frac{S}{n} rightrfloor right) ]Which translates back to:[ prod_{i=1}^n x_i! geq left( leftlfloor frac{S}{n} rightrfloor ! right)^n ]So, this proves the inequality using Jensen's inequality.**When Does Equality Hold?**Equality in Jensen's inequality holds when all ( x_i ) are equal, or when the function is linear. Since ( f(k) = ln(k!) ) is strictly convex, equality holds only when all ( x_i ) are equal.Therefore, equality holds if and only if ( x_1 = x_2 = ldots = x_n ).**Conclusion:**Using Jensen's inequality with the convex function ( f(k) = ln(k!) ), we've shown that the product of the factorials is at least the nth power of the factorial of the floor of the average. Equality occurs precisely when all ( x_i ) are equal.**Final Answer**The inequality holds, and equality occurs if and only if all ( x_i ) are equal. Thus, the final answer is boxed{text{Equality holds when all } x_i text{ are equal}}.

answer：Alright, so I'm trying to tackle these three geometry problems about trihedral angles and tetrahedrons. Let me start by understanding each part step by step.**Part (a): Prove that the planes passing through each edge of a trihedral angle perpendicularly to the opposite face intersect at a single line (the altitude line of the trihedral angle).**Okay, a trihedral angle is like the corner of a pyramid where three faces meet. So, imagine the tip of a pyramid where three triangular faces come together. Each edge of this trihedral angle is where two faces meet. The problem is asking about planes that pass through each of these edges and are perpendicular to the opposite face.Let me visualize this. If I have a trihedral angle with edges SA, SB, and SC, each meeting at point S. The opposite face to edge SA would be the face formed by SB and SC, right? So, a plane passing through SA and perpendicular to the face SBC. Similarly, planes through SB perpendicular to face SAC and through SC perpendicular to face SAB.I need to show that these three planes intersect along a single line, which is the altitude line. The altitude line from S should be perpendicular to the base ABC. So, maybe if I can show that all three planes intersect along this altitude line, that would prove it.Let me think about how two planes intersect. If two planes are both perpendicular to the same line, they should intersect along a line that's also perpendicular to that line. Wait, but in this case, each plane is perpendicular to a different face. Hmm.Maybe I can use the fact that each plane is perpendicular to a face, so their intersection should be a line that's perpendicular to all three faces. But in a trihedral angle, the three faces meet at a single point, so the only line that can be perpendicular to all three is the altitude from that point.Alternatively, maybe I can use coordinates. Let me assign coordinates to the trihedral angle. Let’s place point S at the origin (0,0,0). Let’s say the edges SA, SB, and SC are along the x, y, and z-axes respectively. So, points A, B, and C can be at (a,0,0), (0,b,0), and (0,0,c).Now, the face opposite to SA is the face SBC, which lies in the y-z plane. A plane passing through SA (the x-axis) and perpendicular to the y-z plane would be the x-y plane or the x-z plane? Wait, no. A plane perpendicular to the y-z plane (which is the face SBC) would have a normal vector in the direction of the x-axis. So, the plane equation would be something like y = k or z = k? Wait, no. If a plane is perpendicular to the y-z plane, its normal vector should be along the x-axis. So, the plane equation would be x = constant. But it has to pass through SA, which is the x-axis. So, the plane would be x = 0, but that's just the y-z plane itself, which is the face SBC. That doesn't make sense.Wait, maybe I'm misunderstanding. The plane passes through edge SA and is perpendicular to the opposite face SBC. So, the plane contains SA and is perpendicular to face SBC. Since face SBC is in the y-z plane, a plane perpendicular to it would have a normal vector in the y-z plane. Hmm, this is getting confusing.Maybe a better approach is to think about the normal vectors. The plane through SA and perpendicular to face SBC. The face SBC has a normal vector that can be found by the cross product of vectors SB and SC. So, vectors SB = (0,b,0) and SC = (0,0,c). Their cross product is (b*c, 0, 0), which is along the x-axis. So, the normal vector of face SBC is along the x-axis.Therefore, a plane perpendicular to face SBC must have a normal vector that is parallel to the x-axis. So, the plane through SA (the x-axis) and perpendicular to face SBC would have a normal vector along the x-axis. But any plane containing the x-axis with a normal vector along the x-axis would just be the x-y plane or x-z plane? Wait, no. If the normal vector is along the x-axis, the plane would be of the form y = k or z = k, but it has to contain the x-axis. So, the only plane containing the x-axis and with normal vector along the x-axis is the x-y plane or x-z plane? Wait, no, because the x-y plane has normal vector along z, and x-z plane has normal vector along y.I'm getting confused here. Maybe I need to think differently. If the plane is perpendicular to face SBC, which has normal vector along x, then the plane must be parallel to the x-axis. Wait, no, being perpendicular to a plane means that the plane's normal vector is parallel to the original plane's normal vector. So, if face SBC has normal vector along x, then a plane perpendicular to face SBC would have a normal vector that's orthogonal to x, meaning in the y-z plane.But the plane we're considering passes through edge SA, which is along the x-axis. So, the plane contains the x-axis and has a normal vector in the y-z plane. That means the plane can be represented as a linear combination of y and z. For example, y = m z, but it has to contain the x-axis, so when y=0 and z=0, x can be anything. So, the plane equation would be y = m z.Similarly, the plane through SB perpendicular to face SAC. Face SAC is in the x-z plane, so its normal vector is along the y-axis. Therefore, the plane through SB (the y-axis) and perpendicular to face SAC must have a normal vector orthogonal to y, meaning in the x-z plane. So, its equation would be something like x = n z.And the plane through SC perpendicular to face SAB. Face SAB is in the x-y plane, so its normal vector is along the z-axis. Therefore, the plane through SC (the z-axis) and perpendicular to face SAB must have a normal vector orthogonal to z, meaning in the x-y plane. So, its equation would be something like x = p y.Now, we have three planes:1. y = m z (through SA)2. x = n z (through SB)3. x = p y (through SC)We need to find where these three planes intersect. Let's solve these equations together.From equation 3: x = p y.From equation 2: x = n z. So, p y = n z => z = (p/n) y.From equation 1: y = m z = m*(p/n) y => y = (m p / n) y.If y ≠ 0, then 1 = (m p / n). So, m p = n.But this is a condition on the constants m, n, p. However, in reality, these planes are defined by the geometry of the trihedral angle, so m, n, p are determined by the angles between the edges.But regardless, the intersection of these three planes is a line. Because each pair of planes intersects along a line, and all three lines must coincide if the planes are consistent.Alternatively, since each plane is constructed to be perpendicular to a face and pass through an edge, their intersection should be the line that is perpendicular to all three faces, which is the altitude from S.Therefore, all three planes intersect along the altitude line from S.**Part (b): If two altitude lines of a tetrahedron intersect, then the other two also intersect.**Alright, a tetrahedron has four vertices, and each vertex has an altitude line, which is the line from the vertex perpendicular to the opposite face.So, if two of these altitude lines intersect, we need to show that the other two must also intersect.Let me think about the properties of tetrahedrons and their altitudes. In a general tetrahedron, the altitudes don't necessarily intersect at a single point unless it's an orthocentric tetrahedron.But here, the problem is saying that if two altitudes intersect, then the other two must also intersect. So, maybe it's implying that if two altitudes intersect, the tetrahedron must be orthocentric, meaning all four altitudes intersect at a single point.Wait, but the problem says "the other two also intersect," not necessarily at the same point. Hmm.Let me consider two altitude lines intersecting. Suppose altitude from vertex A intersects altitude from vertex B at some point O. Then, we need to show that altitudes from C and D also intersect.But in a tetrahedron, the altitudes are lines in 3D space. If two of them intersect, does that force the other two to intersect as well?Maybe we can use the concept of planes. If two altitudes intersect, they lie on a common plane. Since each altitude is perpendicular to a face, this common plane must be related to the faces opposite to A and B.Alternatively, perhaps we can use duality or some reciprocal properties.Wait, another approach: In a tetrahedron, if two altitudes intersect, then the tetrahedron is orthocentric, meaning all four altitudes intersect at a single point. So, if two altitudes intersect, then all four must intersect, hence the other two also intersect.But I need to verify this. Is it true that if two altitudes of a tetrahedron intersect, then all four altitudes intersect?I think yes, because the intersection point would have to satisfy the condition of being the orthocenter for all four faces. So, if two altitudes intersect, they define a point that is the orthocenter, and thus the other two altitudes must pass through this point as well.Therefore, if two altitude lines intersect, the other two must also intersect at the same point.**Part (c): If three altitude lines of a tetrahedron intersect at a single point, then the fourth line also passes through this point; in this case, the altitude lines coincide with the altitudes, and the tetrahedron is orthogonal.**This seems similar to part (b). If three altitudes intersect at a single point, then the fourth must also pass through that point, making the tetrahedron orthogonal, meaning all edges are perpendicular to each other.Wait, in a tetrahedron, being orthogonal usually means that all three edges meeting at a vertex are mutually perpendicular. So, if the tetrahedron is orthogonal, then all its altitudes coincide with its edges, and they all intersect at the vertex.But the problem says that if three altitude lines intersect at a single point, then the fourth also passes through this point, and the tetrahedron is orthogonal.So, if three altitudes intersect at a point, that point must be the orthocenter, and the fourth altitude must also pass through it. Moreover, the tetrahedron must be orthogonal, meaning all edges are perpendicular.This makes sense because in an orthogonal tetrahedron, all altitudes coincide with the edges, and they all meet at the right angles vertex.So, if three altitudes intersect, it forces the tetrahedron to have all edges perpendicular, making it orthogonal, and thus the fourth altitude must also pass through the intersection point.**Summary:**For part (a), by considering the planes through each edge and their perpendicularity to the opposite faces, we can show that they intersect along the altitude line.For parts (b) and (c), the key idea is that if two or three altitudes intersect, it imposes a condition on the tetrahedron making it orthocentric or orthogonal, respectively, ensuring that all altitudes intersect at a single point.

answer：Alright, so I have this equation here: 1/a minus 1/b equals 1/37. I need to find all integer solutions for a and b. Hmm, okay, let's see. I remember that when dealing with equations involving reciprocals, sometimes it's helpful to combine the fractions or find a common denominator. Let me try that.First, I'll write the equation again:1/a - 1/b = 1/37To combine the fractions on the left side, I need a common denominator, which would be ab. So, rewriting the left side:(b - a)/ab = 1/37Okay, so now I have (b - a)/ab equals 1/37. Maybe I can cross-multiply to get rid of the fractions. Let's try that:37(b - a) = abAlright, so expanding the left side:37b - 37a = abHmm, now I have an equation with both a and b terms. I want to rearrange this to make it easier to solve for a and b. Let's move all terms to one side:ab - 37b + 37a = 0Hmm, this looks a bit messy. Maybe I can factor this somehow. Let me see if I can factor by grouping. I'll try to group the terms:ab - 37b + 37a - 37*37 = -37*37Wait, why did I do that? I think I added and subtracted 37 squared to complete the rectangle or something. Let me check:So, starting from ab - 37b + 37a = 0, I added -37*37 to both sides to get:ab - 37b + 37a - 37*37 = -37*37Now, let's factor the left side. Looking at ab - 37b + 37a - 37*37, I can factor out b from the first two terms and 37 from the last two:b(a - 37) + 37(a - 37) = -37*37Oh, wait, that's better. So, factoring out (a - 37):(a - 37)(b + 37) = -37*37Which simplifies to:(a - 37)(b + 37) = -1369Wait, but 37 squared is 1369, so that's correct. Now, I need to find integer solutions for a and b such that (a - 37) times (b + 37) equals -1369.Since 1369 is 37 squared, its positive factors are 1, 37, and 1369. But since the product is negative, one of the factors must be positive and the other negative.So, let's list all possible pairs of integers (d, e) such that d * e = -1369. Remembering that both d and e can be positive or negative.The pairs would be:1. (1, -1369)2. (-1, 1369)3. (37, -37)4. (-37, 37)5. (1369, -1)6. (-1369, 1)Okay, so these are all the possible pairs. Now, for each pair, I can set (a - 37) equal to d and (b + 37) equal to e, and solve for a and b.Let's go through each pair:1. d = 1, e = -1369So,a - 37 = 1 => a = 38b + 37 = -1369 => b = -1369 - 37 = -1406So, one solution is (38, -1406)2. d = -1, e = 1369a - 37 = -1 => a = 36b + 37 = 1369 => b = 1369 - 37 = 1332So, another solution is (36, 1332)3. d = 37, e = -37a - 37 = 37 => a = 74b + 37 = -37 => b = -37 - 37 = -74So, another solution is (74, -74)4. d = -37, e = 37a - 37 = -37 => a = 0b + 37 = 37 => b = 0Wait, a = 0 and b = 0? But in the original equation, we have 1/a and 1/b, which would be undefined if a or b is zero. So, this solution is invalid because division by zero is not allowed. So, we have to discard this pair.5. d = 1369, e = -1a - 37 = 1369 => a = 1369 + 37 = 1406b + 37 = -1 => b = -1 - 37 = -38So, another solution is (1406, -38)6. d = -1369, e = 1a - 37 = -1369 => a = -1369 + 37 = -1332b + 37 = 1 => b = 1 - 37 = -36So, another solution is (-1332, -36)Okay, so compiling all the valid solutions:1. (38, -1406)2. (36, 1332)3. (74, -74)4. (1406, -38)5. (-1332, -36)Wait, but earlier I thought (74, -74) was a solution, but let me check if that works in the original equation.Plugging a = 74 and b = -74 into 1/a - 1/b:1/74 - 1/(-74) = 1/74 + 1/74 = 2/74 = 1/37Yes, that works. So, it is a valid solution.Similarly, let's check another one, say (36, 1332):1/36 - 1/1332Let me compute this:1/36 is approximately 0.027777...1/1332 is approximately 0.000751...Subtracting them: 0.027777... - 0.000751... ≈ 0.027026...Now, 1/37 is approximately 0.027027..., which is very close, so that works.Similarly, let's check (38, -1406):1/38 - 1/(-1406) = 1/38 + 1/1406Calculating:1/38 ≈ 0.026315...1/1406 ≈ 0.000711...Adding them: ≈ 0.027026..., which is 1/37. So, that works too.Alright, so all these solutions seem to check out.But wait, the problem asks for integer solutions, so negative integers are allowed unless specified otherwise. So, all these pairs are valid.However, sometimes problems might specify positive integers, but since it's not mentioned here, I think all these are acceptable.But just to be thorough, let me check if there are any other factor pairs I might have missed.Earlier, I considered all pairs where d and e are integers such that d * e = -1369. Since 1369 is 37 squared, its only positive factors are 1, 37, and 1369. Therefore, the negative factors would be -1, -37, -1369.So, I think I have considered all possible factor pairs.Therefore, the integer solutions are:(38, -1406), (36, 1332), (74, -74), (1406, -38), (-1332, -36)Wait, but I also had the pair (a, b) = (0, 0), but that's invalid because division by zero is undefined. So, that's not a valid solution.Is there any other pair? Let me think. Since 1369 is a square number, its factor pairs are limited, so I think I have covered all possibilities.Therefore, the integer solutions are the five pairs I listed above.But wait, let me double-check if I missed any factor pairs. For example, could there be a factor pair where one is positive and the other is negative, but not just the ones I listed?Well, since 1369 is positive, and we're looking for a product of -1369, one factor has to be positive and the other negative. So, the possible pairs are indeed the ones I listed: (1, -1369), (-1, 1369), (37, -37), (-37, 37), (1369, -1), (-1369, 1). So, I think I have all of them.Therefore, the integer solutions are:(38, -1406), (36, 1332), (74, -74), (1406, -38), (-1332, -36)But wait, let me check if (74, -74) is the only solution where both a and b are non-zero. Yes, because in the other solutions, one is positive and the other is negative.But just to make sure, let me see if there's any other way to factor this.Alternatively, starting from the equation:(a - 37)(b + 37) = -1369We can think of this as (a - 37) and (b + 37) being integer divisors of -1369.Since -1369 is negative, one of the terms must be positive and the other negative.So, the possible pairs are:(1, -1369), (-1, 1369), (37, -37), (-37, 37), (1369, -1), (-1369, 1)Which is exactly what I considered earlier.Therefore, the solutions are as above.So, in conclusion, the integer solutions (a, b) are:(38, -1406), (36, 1332), (74, -74), (1406, -38), (-1332, -36)But wait, let me check if (74, -74) is indeed a solution:1/74 - 1/(-74) = 1/74 + 1/74 = 2/74 = 1/37Yes, that's correct.Similarly, checking (36, 1332):1/36 - 1/1332 = (37 - 1)/1332 = 36/1332 = 1/37Yes, that works.And (38, -1406):1/38 - 1/(-1406) = 1/38 + 1/1406 = (37 + 1)/1406 = 38/1406 = 1/37Yes, correct.Similarly, (1406, -38):1/1406 - 1/(-38) = 1/1406 + 1/38 = (1 + 37)/1406 = 38/1406 = 1/37Yes, that works.And (-1332, -36):1/(-1332) - 1/(-36) = -1/1332 + 1/36 = (-1 + 37)/1332 = 36/1332 = 1/37Yes, that works too.So, all these solutions are valid.Therefore, the integer solutions are:(38, -1406), (36, 1332), (74, -74), (1406, -38), (-1332, -36)I think that's all of them.

answer：First, I need to understand the given unit conversions. I know that 1 mile is equal to 8 furlongs, and 1 furlong is equal to 50 rods.To find out how many rods are in one mile, I can multiply the number of furlongs in a mile by the number of rods in a furlong. This means multiplying 8 furlongs by 50 rods per furlong.Calculating this gives me 8 multiplied by 50, which equals 400 rods. Therefore, there are 400 rods in one mile.