answer：Joy has 30 thin rods, each of integer lengths from 1 cm to 30 cm. She places rods of lengths 3 cm, 7 cm, and 15 cm on a table and wants to choose a fourth rod to form a quadrilateral with positive area. To determine how many of the remaining rods can be chosen, we need to ensure that the sum of any three sides is greater than the fourth side.First, we identify the lengths of the rods already placed on the table: 3 cm, 7 cm, and 15 cm. Let the length of the fourth rod be ( d ).To form a quadrilateral, the following inequalities must hold:1. ( 3 + 7 + 15 > d )2. ( 3 + 7 + d > 15 )3. ( 3 + 15 + d > 7 )4. ( 7 + 15 + d > 3 )Simplifying these inequalities:1. ( 25 > d ) or ( d < 25 )2. ( 10 + d > 15 ) or ( d > 5 )3. ( 18 + d > 7 ) (always true since ( d ) is positive)4. ( 22 + d > 3 ) (always true since ( d ) is positive)Thus, the valid range for ( d ) is ( 5 < d < 25 ). The integer values of ( d ) that satisfy this are 6, 7, 8, ..., 24. This gives us 19 possible values.However, we must exclude the rods already used: 7 cm and 15 cm. Therefore, we subtract 2 from 19, resulting in 17 valid choices.Thus, Joy can choose from (boxed{17}) different rods to form a quadrilateral with positive area.

answer：Okay, so I need to find the decimal equivalent of the fraction 5/8. Hmm, I remember that fractions can be converted to decimals by dividing the numerator by the denominator. So, in this case, I need to divide 5 by 8. Let me think about how to do that step by step.First, I know that 8 goes into 5 zero times, so I would write a 0. and then proceed with the division. To make it easier, I can add a decimal point and some zeros to the right of 5, turning it into 5.000. Now, I can start dividing 8 into 50.8 goes into 50 six times because 8 times 6 is 48. I subtract 48 from 50 and get 2. Then, I bring down the next 0, making it 20. 8 goes into 20 two times because 8 times 2 is 16. Subtracting 16 from 20 leaves me with 4. I bring down another 0, making it 40.Now, 8 goes into 40 exactly five times because 8 times 5 is 40. Subtracting 40 from 40 leaves no remainder. So, putting it all together, I have 0.625 as the decimal equivalent of 5/8.Wait, let me double-check that. If I multiply 0.625 by 8, do I get back to 5? Let's see: 0.625 times 8 is 5. Yes, that checks out. So, I'm confident that 5/8 as a decimal is 0.625.Looking at the options provided:A) 0.625B) 0.8C) 0.5D) 1.25E) 0.05The correct answer should be A) 0.625.

answer：First, I need to determine the total number of possible combinations for the letter lock. Since there are 3 rings and each ring has 6 different letters, the total number of combinations is calculated by multiplying the number of letters on each ring: 6 × 6 × 6 = 216.Next, I recognize that only one specific combination will successfully open the lock. Therefore, the number of unsuccessful attempts is the total number of combinations minus the one successful attempt: 216 - 1 = 215.Thus, the maximum number of distinct unsuccessful attempts to open the lock is 215.

answer：Alright, so I have this problem where α and β are acute angles. That means both α and β are between 0 and 90 degrees, or in radians, between 0 and π/2. The problem gives me two cosine values: cos(α + β) = 12/13 and cos(2α + β) = 3/5. I need to find the value of cos α. The options are A: 56/65, B: 16/65, C: either 56/65 or 16/65, and D: None of the above.Okay, let's start by recalling some trigonometric identities. Since we have cosines of sums of angles, maybe I can use the cosine addition formula. The formula for cos(A - B) is cos A cos B + sin A sin B. Wait, but in this case, I have cos(α + β) and cos(2α + β). Hmm, maybe I can express cos α in terms of these.Let me think. If I let θ = α + β, then 2α + β would be θ + α. So, cos(2α + β) is cos(θ + α). Maybe I can write that as cos θ cos α - sin θ sin α. But I know cos θ is 12/13, so I can find sin θ using the Pythagorean identity. Since θ is α + β and both α and β are acute, θ must be between 0 and π, but more specifically, since both are acute, θ is between 0 and π/2 + π/2 = π, but actually, since α and β are each less than π/2, their sum is less than π. But given that cos(α + β) is positive (12/13), θ must be in the first quadrant, so θ is between 0 and π/2.So, sin θ = sqrt(1 - cos²θ) = sqrt(1 - (144/169)) = sqrt(25/169) = 5/13.Similarly, for cos(2α + β) = 3/5, which is cos(θ + α). Let's denote φ = 2α + β, so cos φ = 3/5. Since φ = θ + α, and θ is between 0 and π/2, and α is between 0 and π/2, φ is between θ and θ + π/2. But since θ is already between 0 and π/2, φ could be up to π. However, cos φ is positive (3/5), so φ must be in the first quadrant as well, meaning φ is between 0 and π/2. Wait, but if θ is already up to π/2, adding α (which is up to π/2) could make φ up to π. But since cos φ is positive, φ must be less than π/2. So, φ is between 0 and π/2.Therefore, sin φ = sqrt(1 - cos²φ) = sqrt(1 - 9/25) = sqrt(16/25) = 4/5.Now, going back to the expression for cos(θ + α). Using the cosine addition formula:cos(θ + α) = cos θ cos α - sin θ sin α.We know cos(θ + α) is 3/5, cos θ is 12/13, sin θ is 5/13. So plugging in:3/5 = (12/13) cos α - (5/13) sin α.So, we have the equation:(12/13) cos α - (5/13) sin α = 3/5.Let me write that as:12 cos α - 5 sin α = (3/5) * 13.Calculating the right side: 3/5 * 13 = 39/5.So, 12 cos α - 5 sin α = 39/5.Hmm, that's one equation with two variables, cos α and sin α. But I know that cos²α + sin²α = 1. So, maybe I can set up a system of equations.Let me denote x = cos α and y = sin α. Then, we have:12x - 5y = 39/5,andx² + y² = 1.So, two equations:1) 12x - 5y = 39/5,2) x² + y² = 1.Let me solve equation 1 for y:12x - 5y = 39/5,Multiply both sides by 5 to eliminate the denominator:60x - 25y = 39,Then, rearranged:-25y = -60x + 39,Divide both sides by -25:y = (60x - 39)/25.So, y = (60x - 39)/25.Now, substitute this into equation 2:x² + [(60x - 39)/25]^2 = 1.Let me compute [(60x - 39)/25]^2:= (60x - 39)^2 / 625= (3600x² - 2*60x*39 + 39²) / 625= (3600x² - 4680x + 1521) / 625.So, equation 2 becomes:x² + (3600x² - 4680x + 1521)/625 = 1.Multiply both sides by 625 to eliminate the denominator:625x² + 3600x² - 4680x + 1521 = 625.Combine like terms:(625x² + 3600x²) = 4225x²,So, 4225x² - 4680x + 1521 = 625.Subtract 625 from both sides:4225x² - 4680x + 1521 - 625 = 0,Which simplifies to:4225x² - 4680x + 896 = 0.Hmm, that's a quadratic in x. Let me see if I can simplify it.First, let's check if all coefficients are divisible by a common factor. 4225, 4680, 896.4225: 4225 ÷ 5 = 845, 845 ÷ 5 = 169, which is 13². So, 4225 = 5² * 13².4680 ÷ 5 = 936, 936 ÷ 5 = 187.2, not integer. 4680 ÷ 13 = 360, which is integer. So, 4680 = 13 * 360.896 ÷ 13 is approximately 68.92, not integer. So, no common factor besides 1. So, we have to solve the quadratic as it is.Quadratic equation: 4225x² - 4680x + 896 = 0.Let me write it as:4225x² - 4680x + 896 = 0.To solve for x, we can use the quadratic formula:x = [4680 ± sqrt(4680² - 4*4225*896)] / (2*4225).First, compute the discriminant D:D = 4680² - 4*4225*896.Compute 4680²:4680 * 4680. Let's compute 468 * 468 first.468 * 468:Compute 400*400 = 160000,400*68 = 27200,68*400 = 27200,68*68 = 4624.So, (400 + 68)^2 = 400² + 2*400*68 + 68² = 160000 + 54400 + 4624 = 160000 + 54400 = 214400 + 4624 = 219,024.So, 4680² = (468*10)² = 468² * 100 = 219024 * 100 = 21,902,400.Now, compute 4*4225*896:First, 4*4225 = 16,900.16,900 * 896.Let me compute 16,900 * 800 = 13,520,000,16,900 * 96 = ?Compute 16,900 * 90 = 1,521,000,16,900 * 6 = 101,400,So, 1,521,000 + 101,400 = 1,622,400.So, total 13,520,000 + 1,622,400 = 15,142,400.So, D = 21,902,400 - 15,142,400 = 6,760,000.sqrt(D) = sqrt(6,760,000). Let's see, 2600² = 6,760,000, because 26² = 676, so 2600² = 676*10,000 = 6,760,000.So, sqrt(D) = 2600.Therefore, x = [4680 ± 2600] / (2*4225).Compute numerator:First, 4680 + 2600 = 7280,4680 - 2600 = 2080.Denominator: 2*4225 = 8450.So, x = 7280 / 8450 or x = 2080 / 8450.Simplify both fractions.First, 7280 / 8450.Divide numerator and denominator by 10: 728 / 845.Check if 728 and 845 have a common factor.728 ÷ 13 = 56, because 13*56 = 728.845 ÷ 13 = 65, because 13*65 = 845.So, 728 / 845 = 56 / 65.Similarly, 2080 / 8450.Divide numerator and denominator by 10: 208 / 845.Check if 208 and 845 have a common factor.208 ÷ 13 = 16, because 13*16 = 208.845 ÷ 13 = 65, as before.So, 208 / 845 = 16 / 65.Therefore, x = 56/65 or x = 16/65.So, cos α is either 56/65 or 16/65.Wait, but we need to check if both solutions are valid because sometimes when we square equations, we can introduce extraneous solutions.So, let's check both possibilities.First, let's take x = 56/65.Then, y = (60x - 39)/25.Compute 60x: 60*(56/65) = (60*56)/65 = 3360/65.3360 ÷ 65: 65*51 = 3315, so 3360 - 3315 = 45, so 51 + 45/65 = 51 + 9/13 = 51.6923.Wait, but let me compute it as fractions:60*(56/65) = (60/65)*56 = (12/13)*56 = (12*56)/13 = 672/13.Then, 672/13 - 39 = 672/13 - 507/13 = (672 - 507)/13 = 165/13.So, y = (165/13)/25 = (165)/(13*25) = 165/325 = 33/65.So, y = 33/65.Now, check if x² + y² = 1:(56/65)² + (33/65)² = (3136 + 1089)/4225 = 4225/4225 = 1. So, that's valid.Now, check the other solution, x = 16/65.Compute y = (60x - 39)/25.60x = 60*(16/65) = 960/65 = 192/13.192/13 - 39 = 192/13 - 507/13 = (192 - 507)/13 = (-315)/13.So, y = (-315/13)/25 = (-315)/(13*25) = (-63)/65.So, y = -63/65.Now, check x² + y²:(16/65)² + (-63/65)² = (256 + 3969)/4225 = 4225/4225 = 1. So, that's also valid.But wait, α is an acute angle, so sin α must be positive because sine is positive in the first quadrant. So, y = sin α = -63/65 would imply that sin α is negative, which contradicts α being acute. Therefore, y must be positive, so we discard the solution x = 16/65 because it gives y negative.Therefore, the only valid solution is x = 56/65.So, cos α = 56/65.Wait, but let me double-check. If x = 16/65, then y = -63/65, which would mean sin α is negative, which isn't possible because α is acute. So, yes, we must discard x = 16/65.Therefore, the answer is 56/65, which is option A.But wait, let me think again. When I solved for y, I got y = (60x - 39)/25. If x = 16/65, then 60x = 60*(16/65) = 960/65 ≈ 14.769, and 14.769 - 39 = -24.231, which is negative, so y is negative. Since α is acute, sin α must be positive, so this solution is invalid.Therefore, only x = 56/65 is valid.Wait, but in the initial problem, the options are A: 56/65, B: 16/65, C: either, D: none. So, since only 56/65 is valid, the answer is A.But wait, in my earlier steps, I considered φ = 2α + β, which is θ + α. Since θ is α + β, and both α and β are acute, θ is less than π/2 + π/2 = π, but since cos θ is positive, θ is less than π/2. Similarly, φ = θ + α, which is less than π/2 + π/2 = π, but since cos φ is positive, φ is less than π/2. So, φ is between 0 and π/2, which means α must be such that θ + α < π/2. But θ is α + β, so α + β + α < π/2, which implies 2α + β < π/2. But wait, we were given that cos(2α + β) = 3/5, which is positive, so 2α + β is indeed less than π/2. Therefore, α must be such that 2α + β < π/2.But since α and β are both positive, this imposes a restriction on α. So, if 2α + β < π/2, and β is positive, then α must be less than π/4, because if α were π/4, then 2α would be π/2, and adding β (which is positive) would make it more than π/2, which contradicts. Therefore, α must be less than π/4, so cos α must be greater than cos(π/4) = √2/2 ≈ 0.707. Now, 56/65 ≈ 0.8615, which is greater than √2/2, and 16/65 ≈ 0.246, which is less than √2/2. But since α is less than π/4, cos α must be greater than √2/2, so 56/65 is valid, and 16/65 is invalid.Therefore, only 56/65 is the valid solution.Wait, but earlier I thought that 16/65 would give a negative sine, which is invalid, so that's another reason to discard it. So, yes, only 56/65 is valid.Therefore, the answer is A: 56/65.But wait, let me check the calculations again to make sure I didn't make any mistakes.Starting from:cos(α + β) = 12/13,cos(2α + β) = 3/5.We set θ = α + β,So, cos θ = 12/13,sin θ = 5/13,Then, cos(θ + α) = 3/5,Which is cos(2α + β) = 3/5.Using the cosine addition formula:cos(θ + α) = cos θ cos α - sin θ sin α = 3/5.So,(12/13) cos α - (5/13) sin α = 3/5.Multiply both sides by 13:12 cos α - 5 sin α = 39/5.Then, I set x = cos α, y = sin α,So,12x - 5y = 39/5,andx² + y² = 1.Solving for y:y = (60x - 39)/25.Substitute into x² + y² = 1,x² + [(60x - 39)/25]^2 = 1,Which led to quadratic equation,4225x² - 4680x + 896 = 0,Solutions x = 56/65 and x = 16/65.Checked validity:x = 56/65 gives y = 33/65, which is positive, valid.x = 16/65 gives y = -63/65, which is negative, invalid.Therefore, only x = 56/65 is valid.Hence, the answer is A: 56/65.Yes, that seems correct.

answer：Okay, so I have this trigonometry problem here: Given that sin α minus 3 cos α equals zero, I need to find the value of cos 2α plus tan α. Hmm, let me try to figure this out step by step.First, the equation given is sin α - 3 cos α = 0. I think I can rearrange this equation to find a relationship between sin α and cos α. If I move the 3 cos α to the other side, I get sin α = 3 cos α. That seems useful because it relates sine and cosine, which might help me find tan α.Wait, tan α is sin α over cos α, right? So if sin α equals 3 cos α, then tan α would be (3 cos α) / cos α. Oh, that simplifies to 3. So tan α is 3. That was straightforward.Now, I need to find cos 2α. I remember there are a few formulas for cos 2α. One of them is cos² α - sin² α, another is 1 - 2 sin² α, and another is 2 cos² α - 1. Maybe I can use one of these identities. Since I already know tan α, which is 3, perhaps I can express cos 2α in terms of tan α.I think there's a formula that relates cos 2α directly to tan α. Let me recall... Oh, yes! It's (1 - tan² α) / (1 + tan² α). So, substituting tan α = 3 into this formula, I get (1 - 3²) / (1 + 3²). Calculating that, 3 squared is 9, so it becomes (1 - 9) / (1 + 9) which is (-8)/10. Simplifying that, it's -4/5. So cos 2α is -4/5.Now, I need to find cos 2α plus tan α. I have cos 2α as -4/5 and tan α as 3. So adding them together: -4/5 + 3. To add these, I should express 3 as a fraction with denominator 5. That would be 15/5. So, -4/5 + 15/5 equals (15 - 4)/5, which is 11/5.Wait, let me double-check my steps to make sure I didn't make any mistakes. Starting with sin α - 3 cos α = 0, I got sin α = 3 cos α, which leads to tan α = 3. Then, using the identity for cos 2α in terms of tan α, I substituted and got -4/5. Adding that to tan α, which is 3, converted to fifths, gives 11/5. That seems correct.Is there another way to verify this? Maybe by finding sin α and cos α explicitly. If tan α = 3, then I can think of a right triangle where the opposite side is 3 and the adjacent side is 1, so the hypotenuse would be sqrt(1² + 3²) = sqrt(10). Therefore, sin α is 3/sqrt(10) and cos α is 1/sqrt(10).Let me compute cos 2α using the identity cos² α - sin² α. So, cos² α is (1/sqrt(10))² = 1/10, and sin² α is (3/sqrt(10))² = 9/10. Therefore, cos 2α = 1/10 - 9/10 = -8/10 = -4/5. That matches what I got earlier.Then, tan α is 3, so adding cos 2α + tan α is -4/5 + 3, which is indeed 11/5. Okay, that confirms my answer.I think I'm confident that the value of cos 2α + tan α is 11/5.

answer：Okay, so I have this problem where I need to find out if there are 2018 pairs of natural numbers (x, y) that satisfy four specific conditions. Let me try to understand each condition one by one and see how I can approach this.First, condition 1 says that in each pair, x and y are different. That means for every pair (x, y), x ≠ y. Okay, that's straightforward.Condition 2 mentions that in each successive pair, the number x is 1 greater than the number x in the previous pair. So if I have a pair (x1, y1), the next pair should be (x1 + 1, y2). Similarly, condition 3 says that in each successive pair, the number y is 1 greater than the number y in the previous pair. So, following the same logic, if I have (x1, y1), the next pair is (x1 + 1, y1 + 1).Condition 4 is a bit more involved: in each pair, x is divisible by y. So, for every pair (x, y), x must be a multiple of y. That means x/y should be an integer.Alright, so putting it all together, I need to find 2018 such pairs where each x is one more than the previous x, each y is one more than the previous y, x and y are different in each pair, and x is divisible by y in every pair.Let me think about how to construct such pairs. Since x and y are both increasing by 1 each time, maybe I can express x and y in terms of a starting point and then increment them step by step.Suppose I start with some initial pair (x1, y1). Then the next pair would be (x1 + 1, y1 + 1), the one after that would be (x1 + 2, y1 + 2), and so on, up to (x1 + 2017, y1 + 2017). So, in general, the nth pair would be (x1 + n - 1, y1 + n - 1) for n from 1 to 2018.Now, the key is to choose x1 and y1 such that for every n, (x1 + n - 1) is divisible by (y1 + n - 1). Also, x and y must be different in each pair, but since we're starting with x1 ≠ y1 and both are increasing by 1 each time, as long as x1 ≠ y1, all subsequent pairs will also have x ≠ y.So, the main challenge is to find x1 and y1 such that for every k from 0 to 2017, (x1 + k) is divisible by (y1 + k). That is, (x1 + k) mod (y1 + k) = 0 for all k in 0 to 2017.Hmm, that seems tricky. Maybe I can think of it differently. If I set y1 to be 1, then y increases by 1 each time, so y would be 1, 2, 3, ..., 2018. Then, x needs to be divisible by each of these y's. So, x1 needs to be divisible by 1, x1 + 1 needs to be divisible by 2, x1 + 2 needs to be divisible by 3, and so on, up to x1 + 2017 needs to be divisible by 2018.Wait, that sounds like x1 needs to satisfy a series of congruences. Specifically, x1 ≡ 0 mod 1, x1 ≡ -1 mod 2, x1 ≡ -2 mod 3, ..., x1 ≡ -2017 mod 2018.But mod 1 is trivial because every number is congruent to 0 mod 1. So, the real constraints start from mod 2 onwards.This seems like a system of congruences that can be solved using the Chinese Remainder Theorem (CRT). The CRT tells us that if the moduli are pairwise coprime, then there exists a solution. However, in this case, the moduli are 2, 3, ..., 2018, which are not all pairwise coprime. So, I need to check if the system is solvable.Alternatively, maybe I can choose x1 to be a multiple of the least common multiple (LCM) of the numbers from 1 to 2018. That way, x1 would be divisible by each y1 + k, but I need to adjust it so that x1 + k is divisible by y1 + k.Wait, if I set x1 to be the LCM of 1 to 2018, then x1 is divisible by each y1 + k, but x1 + k would not necessarily be divisible by y1 + k. Hmm, that might not work.Alternatively, maybe I can set x1 = LCM(1 to 2018) - 1. Then, x1 + 1 = LCM(1 to 2018), which is divisible by 2, x1 + 2 = LCM(1 to 2018) + 1, which needs to be divisible by 3, and so on. But I'm not sure if that would work.Wait, let's think about it differently. If I set y1 = 1, then y increases by 1 each time, so y becomes 1, 2, 3, ..., 2018. Then, x needs to be such that x + k is divisible by k + 1 for k from 0 to 2017.So, x needs to satisfy x ≡ -k mod (k + 1) for each k from 0 to 2017.This is a system of congruences where x ≡ 0 mod 1, x ≡ -1 mod 2, x ≡ -2 mod 3, ..., x ≡ -2017 mod 2018.This system can be solved using the Chinese Remainder Theorem if the moduli are pairwise coprime. However, since the moduli are 1, 2, 3, ..., 2018, which are not all pairwise coprime, the CRT doesn't directly apply. But I can still try to find a solution.Alternatively, I can think of x as being congruent to -k modulo (k + 1) for each k. That means x ≡ -k mod (k + 1), which can be rewritten as x ≡ 1 mod (k + 1) because -k ≡ 1 mod (k + 1).Wait, that's interesting. So, x ≡ 1 mod 2, x ≡ 1 mod 3, x ≡ 1 mod 4, ..., x ≡ 1 mod 2018.If that's the case, then x ≡ 1 mod LCM(2, 3, 4, ..., 2018). So, x can be written as x = LCM(2, 3, 4, ..., 2018) * m + 1 for some integer m.But LCM(2, 3, 4, ..., 2018) is a huge number, and x needs to be a natural number, so m can be 0, 1, 2, etc. If m = 0, then x = 1, but then x + k = 1 + k, which needs to be divisible by k + 1, which it is, because 1 + k is divisible by k + 1. Wait, that's trivially true because any number is divisible by itself.Wait a minute, if x = 1, then x + k = 1 + k, which is divisible by k + 1. So, x = 1 satisfies all the congruences. Therefore, x1 = 1, y1 = 1.But wait, condition 1 says that in each pair, x and y are different. So, if x1 = 1 and y1 = 1, then x1 = y1, which violates condition 1.Ah, that's a problem. So, I need to choose x1 and y1 such that x1 ≠ y1, and then x increases by 1 each time, y increases by 1 each time, and x is divisible by y in each pair.So, maybe I can choose y1 to be something else. Let's say y1 = 2. Then y would be 2, 3, 4, ..., 2019. Then, x needs to be such that x + k is divisible by (2 + k) for k from 0 to 2017.So, x ≡ -k mod (2 + k) for k from 0 to 2017.Again, this is a system of congruences. Let's see if we can find a pattern or a way to solve this.Alternatively, maybe I can choose x1 to be a multiple of y1, and then ensure that x1 + k is a multiple of y1 + k for each k.Wait, if x1 is a multiple of y1, say x1 = y1 * m for some integer m, then x1 + k = y1 * m + k. We need y1 * m + k to be divisible by y1 + k.So, (y1 * m + k) / (y1 + k) should be an integer.Let me denote y1 as a, so x1 = a * m.Then, (a * m + k) / (a + k) should be an integer for each k from 0 to 2017.Hmm, that seems complicated. Maybe I can choose a = 1, but then x1 = m, and x1 + k = m + k needs to be divisible by 1 + k. But 1 + k divides m + k, which is similar to the previous case, but we saw that x1 = 1 works, but x1 = y1, which is not allowed.Alternatively, maybe I can choose a = 2018! (2018 factorial). Then, y1 = 2018!, and x1 can be set to 2018! + 1. Then, x1 + k = 2018! + 1 + k, and y1 + k = 2018! + k.Now, we need to check if 2018! + 1 + k is divisible by 2018! + k.Wait, 2018! + 1 + k = (2018! + k) + 1. So, (2018! + k) + 1 divided by (2018! + k) is 1 + 1/(2018! + k), which is not an integer because 1/(2018! + k) is less than 1 and not an integer.So, that doesn't work.Wait, maybe I need to adjust x1 differently. If I set x1 = (2018! + 1) * something, but I'm not sure.Alternatively, maybe I can set x1 = LCM(1, 2, 3, ..., 2018) + 1. Then, x1 + k = LCM(1, 2, 3, ..., 2018) + 1 + k. Since LCM(1, 2, 3, ..., 2018) is divisible by all numbers from 1 to 2018, then LCM + 1 is congruent to 1 modulo each of these numbers. So, LCM + 1 + k is congruent to 1 + k modulo (k + 1). Wait, 1 + k modulo (k + 1) is 0, because 1 + k = (k + 1). So, LCM + 1 + k is divisible by k + 1.Wait, that seems promising. Let me check.If x1 = LCM(1, 2, 3, ..., 2018) + 1, then x1 + k = LCM + 1 + k. Since LCM is divisible by (k + 1), then LCM + 1 + k ≡ 0 + 1 + k ≡ (k + 1) ≡ 0 mod (k + 1). So, x1 + k is divisible by k + 1.Therefore, if I set y1 = 1, then y increases by 1 each time, so y becomes 1, 2, 3, ..., 2018. And x1 = LCM + 1, so x increases by 1 each time, so x becomes LCM + 1, LCM + 2, ..., LCM + 2018.But wait, in this case, x1 = LCM + 1 and y1 = 1, so x1 ≠ y1, which satisfies condition 1. Then, each subsequent pair has x increasing by 1 and y increasing by 1, so x and y are always different because x is always greater than y (since LCM is a huge number, much larger than 2018).Therefore, this construction satisfies all four conditions.So, the pairs would be:(x1, y1) = (LCM(1, 2, ..., 2018) + 1, 1)(x2, y2) = (LCM(1, 2, ..., 2018) + 2, 2)...(x2018, y2018) = (LCM(1, 2, ..., 2018) + 2018, 2018)Each x is divisible by y, x and y are different, and each x and y increase by 1 from the previous pair.Therefore, such pairs do exist.I think this approach works because by choosing x1 as LCM + 1, we ensure that x1 + k is divisible by k + 1 for each k from 0 to 2017. This satisfies condition 4, and since x and y are constructed to increase by 1 each time and start with x1 ≠ y1, all other conditions are satisfied.So, the answer is yes, there are 2018 such pairs.