answer：Alright, so I've got this problem here about a geometric sequence and some trigonometric equation. Let me try to unpack it step by step. First, the problem says that {a_n} is a geometric sequence. I remember that in a geometric sequence, each term is a constant multiple of the previous one. So, if the first term is a_1 and the common ratio is r, then a_n = a_1 * r^(n-1). That seems straightforward.Next, it mentions that a_1 and a_8 are the roots of the quadratic equation x² - 2x sin α - √3 sin α = 0. Hmm, okay. So, since a_1 and a_8 are roots of this equation, I can use Vieta's formulas here. Vieta's formulas relate the sum and product of the roots of a polynomial to its coefficients. For a quadratic equation ax² + bx + c = 0, the sum of the roots is -b/a and the product is c/a.In this case, the equation is x² - 2x sin α - √3 sin α = 0. So, comparing to the standard form, a = 1, b = -2 sin α, and c = -√3 sin α. Therefore, the sum of the roots, which is a_1 + a_8, should be equal to -b/a = 2 sin α. And the product of the roots, a_1 * a_8, should be c/a = -√3 sin α.So, from Vieta's formulas, I have:1. a_1 + a_8 = 2 sin α2. a_1 * a_8 = -√3 sin αNow, the problem also gives another equation: (a_1 + a_8)² = 2a_3a_6 + 6. I need to use this to find the value of the acute angle α.Let me first express a_3 and a_6 in terms of a_1 and the common ratio r. Since it's a geometric sequence:- a_3 = a_1 * r^(3-1) = a_1 * r²- a_6 = a_1 * r^(6-1) = a_1 * r⁵So, a_3 * a_6 = (a_1 * r²) * (a_1 * r⁵) = a_1² * r⁷Wait, but from the product of the roots, I have a_1 * a_8 = -√3 sin α. Let me express a_8 in terms of a_1 and r. Since a_8 is the 8th term, it's a_1 * r^(8-1) = a_1 * r⁷. So, a_1 * a_8 = a_1 * (a_1 * r⁷) = a_1² * r⁷. But from Vieta's formula, a_1 * a_8 = -√3 sin α, so that means a_1² * r⁷ = -√3 sin α. Therefore, a_3 * a_6 = a_1² * r⁷ = -√3 sin α. So, a_3 * a_6 = -√3 sin α.Going back to the given equation: (a_1 + a_8)² = 2a_3a_6 + 6. Let's substitute the expressions we have.From Vieta's, a_1 + a_8 = 2 sin α, so (2 sin α)² = 2*(-√3 sin α) + 6.Calculating the left side: (2 sin α)² = 4 sin² α.Right side: 2*(-√3 sin α) + 6 = -2√3 sin α + 6.So, putting it all together: 4 sin² α = -2√3 sin α + 6.Let me rewrite this equation: 4 sin² α + 2√3 sin α - 6 = 0.This is a quadratic equation in terms of sin α. Let me denote x = sin α for simplicity. Then, the equation becomes:4x² + 2√3 x - 6 = 0.Let me solve this quadratic equation for x. The quadratic formula is x = [-b ± √(b² - 4ac)] / (2a). Here, a = 4, b = 2√3, and c = -6.Calculating the discriminant: b² - 4ac = (2√3)² - 4*4*(-6) = 4*3 - 16*(-6) = 12 + 96 = 108.So, discriminant is 108. Then, x = [-2√3 ± √108] / (2*4).Simplify √108: √108 = √(36*3) = 6√3.So, x = [-2√3 ± 6√3] / 8.Let's compute both possibilities:First, with the plus sign:x = (-2√3 + 6√3) / 8 = (4√3) / 8 = (√3)/2.Second, with the minus sign:x = (-2√3 - 6√3) / 8 = (-8√3)/8 = -√3.So, x = sin α = √3/2 or sin α = -√3.But sin α cannot be -√3 because the sine function only takes values between -1 and 1. So, sin α = -√3 is invalid. Therefore, sin α = √3/2.Now, we need to find the acute angle α such that sin α = √3/2. I remember that sin(π/3) = √3/2, and π/3 is an acute angle (since it's less than π/2). So, α = π/3.Let me just verify my steps to make sure I didn't make any mistakes.1. Identified that {a_n} is a geometric sequence, so expressed a_3 and a_6 in terms of a_1 and r.2. Used Vieta's formulas to find a_1 + a_8 and a_1 * a_8.3. Expressed a_3 * a_6 in terms of a_1 and r, then related it back to the product of the roots.4. Substituted into the given equation (a_1 + a_8)² = 2a_3a_6 + 6, leading to a quadratic in sin α.5. Solved the quadratic, discarded the invalid solution, and found sin α = √3/2, leading to α = π/3.Everything seems to check out. So, I think the answer is π/3, which is option C.**Final Answer**boxed{C}

answer：First, I need to determine the time it takes for ferry P to travel to the mainland. Let's denote this time as ( t ) hours. Since ferry P travels at 8 kilometers per hour, the distance to the mainland is ( 8t ) kilometers.Ferry Q takes a route that is three times longer than ferry P's route. Therefore, the distance for ferry Q is ( 3 times 8t = 24t ) kilometers.We are told that ferry Q is 4 kilometers per hour faster than ferry P. Since ferry P's speed is 8 kilometers per hour, ferry Q's speed is ( 8 + 4 = 12 ) kilometers per hour.The journey of ferry Q is 2 hours longer than that of ferry P. This means ferry Q takes ( t + 2 ) hours to complete its journey.Now, I can set up an equation for the distance traveled by ferry Q using its speed and time:[24t = 12(t + 2)]Expanding the right side of the equation:[24t = 12t + 24]Subtracting ( 12t ) from both sides:[24t - 12t = 24][12t = 24]Dividing both sides by 12:[t = 2]Therefore, ferry P travels for 2 hours to reach the mainland.

answer：Okay, so I have this problem here: the function f(x) = e^{|x|} + 2a - 1 has a unique zero, and I need to find the value of the real number a. The options are A: -1, B: 0, C: 1, D: 2. Hmm, let me try to figure this out step by step.First, I remember that a "zero" of a function is a point where the function crosses the x-axis, meaning f(x) = 0. So, I need to find the value of a such that the equation e^{|x|} + 2a - 1 = 0 has exactly one solution for x.Let me write that equation down: e^{|x|} + 2a - 1 = 0. I can rearrange this to solve for e^{|x|}: e^{|x|} = 1 - 2a.Now, e^{|x|} is an exponential function with the exponent being the absolute value of x. I know that e^{|x|} is always positive because the exponential function is always positive. Also, since |x| is symmetric around x=0, the function e^{|x|} is symmetric as well. It has a minimum value at x=0, which is e^0 = 1, and it increases as x moves away from 0 in either the positive or negative direction.So, the left side of the equation e^{|x|} = 1 - 2a is always greater than or equal to 1. That means the right side, 1 - 2a, must also be greater than or equal to 1 for the equation to have any solutions. Wait, is that right? Let me think.If e^{|x|} is always at least 1, then 1 - 2a must be at least 1 for the equation to hold. So, 1 - 2a ≥ 1. If I subtract 1 from both sides, I get -2a ≥ 0. Dividing both sides by -2 (and remembering to flip the inequality sign because I'm dividing by a negative number), I get a ≤ 0.So, a has to be less than or equal to 0 for there to be any solutions. But the problem says there's a unique zero, meaning only one solution. So, when does e^{|x|} = 1 - 2a have exactly one solution?Since e^{|x|} is symmetric and has its minimum at x=0, the equation e^{|x|} = 1 - 2a will have exactly one solution only when 1 - 2a is equal to the minimum value of e^{|x|}, which is 1. Because if 1 - 2a is greater than 1, then e^{|x|} would have to be equal to something greater than 1, which would happen at two points symmetric around x=0, right? So, to have only one solution, 1 - 2a must equal 1, so that the equation is satisfied only at x=0.Let me write that down: 1 - 2a = 1. Solving for a, I subtract 1 from both sides: -2a = 0. Then, divide both sides by -2: a = 0.Wait, but earlier I concluded that a ≤ 0. So, a=0 is the boundary case where 1 - 2a = 1, which gives exactly one solution at x=0. If a were less than 0, then 1 - 2a would be greater than 1, and the equation e^{|x|} = 1 - 2a would have two solutions, one on each side of the y-axis. But since we need a unique zero, a has to be exactly 0.Let me double-check this. If a=0, then f(x) = e^{|x|} + 0 - 1 = e^{|x|} - 1. So, f(x) = 0 when e^{|x|} - 1 = 0, which is when e^{|x|} = 1. Since e^0 = 1, this happens only at x=0. So, yes, that's the unique zero.If a were, say, 1, then f(x) = e^{|x|} + 2(1) - 1 = e^{|x|} + 1. Then, f(x) = 0 would mean e^{|x|} + 1 = 0, which is impossible because e^{|x|} is always positive, so f(x) would never be zero. Similarly, if a were -1, then f(x) = e^{|x|} + 2(-1) - 1 = e^{|x|} - 2 - 1 = e^{|x|} - 3. Then, f(x) = 0 would mean e^{|x|} = 3, which would have two solutions, one positive and one negative x, so that's two zeros, which is more than one.Therefore, the only value of a that gives exactly one zero is a=0, which is option B.

answer：First, I need to determine how many 6-minute intervals are in 3 hours.There are 60 minutes in an hour, so in 3 hours, there are 3 * 60 = 180 minutes.Now, I divide the total minutes by the interval to find out how many intervals of 6 minutes there are in 180 minutes.180 minutes / 6 minutes per interval = 30 intervals.Since John uses 0.50 for every 6 minutes, I multiply the number of intervals by the cost per interval to find the total amount spent.30 intervals * 0.50 per interval = 15.John spent boxed{15} at the arcade.

answer：First, I identify the number of antelopes Josie counted, which is 80.Next, I calculate the number of rabbits, which is 34 more than the number of antelopes. So, 80 + 34 equals 114 rabbits.For the hyenas, the problem states there are 42 fewer hyenas than the combined total of antelopes and rabbits. Adding the antelopes and rabbits gives 80 + 114 = 194, and subtracting 42 results in 152 hyenas.The number of wild dogs is 50 more than the number of hyenas. Therefore, 152 + 50 equals 202 wild dogs.Leopards are half the number of rabbits. Dividing 114 by 2 gives 57 leopards.Giraffes are 15 more than the number of antelopes. Adding 15 to 80 results in 95 giraffes.Lions are equal to the combined number of leopards and giraffes. Adding 57 and 95 gives 152 lions.Elephants are three times the number of lions. Multiplying 152 by 3 results in 456 elephants.The ratio of zebras to antelopes is 3:4. Using the ratio, the number of zebras is (3/4) * 80, which equals 60 zebras.Hippos are 10% more than the number of zebras. Calculating 10% of 60 and adding it to 60 gives 66 hippos.Finally, I sum all the calculated numbers to find the total number of animals Josie counted: 80 + 114 + 152 + 202 + 57 + 95 + 152 + 456 + 60 + 66 equals 1,334.

answer：Alright, so I've been given this problem about sets ( A_1, A_2, ldots, A_n ) such that their union is all the integers, ( mathbf{Z} ). And for each set ( A_i ), any two numbers ( b > c ) in ( A_i ) must satisfy ( b - c geq a^i ). The question is about figuring out for which values of ( a ) this is possible.Hmm, okay. So, first, I need to understand what this condition ( b - c geq a^i ) really means. It seems like it's a spacing condition on the elements of each set ( A_i ). That is, the elements in each ( A_i ) must be at least ( a^i ) apart from each other. So, for example, if ( a = 2 ), then in ( A_1 ), elements must be at least 2 apart, in ( A_2 ), they must be at least 4 apart, and so on.But wait, the union of all these sets has to cover all integers. So, every integer must be in at least one of these sets. That seems tricky because if each set is too sparse, we might miss some integers.Let me think about small values of ( n ) first. Maybe that can give me some intuition.Suppose ( n = 1 ). Then we have only one set ( A_1 ), which must cover all integers. But the condition would require that any two integers in ( A_1 ) are at least ( a ) apart. But since ( A_1 ) must cover all integers, the only way this can happen is if ( a leq 1 ). But ( a ) is a positive real number, so ( 0 < a leq 1 ). But wait, ( a ) is a real number, not necessarily an integer. So, if ( a ) is less than or equal to 1, then ( a^1 ) is less than or equal to 1, so the spacing condition is trivially satisfied because integers are at least 1 apart. So, for ( n = 1 ), ( a ) can be any positive real number less than or equal to 1.But in the problem, ( n ) is given as some fixed number, not necessarily 1. So, maybe for larger ( n ), the constraints on ( a ) become more lenient? Or maybe stricter?Wait, actually, as ( n ) increases, the exponents ( a^i ) for ( i = 1, 2, ldots, n ) might grow or decay depending on whether ( a ) is greater than or less than 1.If ( a > 1 ), then ( a^i ) grows exponentially with ( i ), meaning that higher-indexed sets ( A_i ) have much sparser elements. On the other hand, if ( a < 1 ), then ( a^i ) decays exponentially, meaning that higher-indexed sets can have elements closer together.But since the union of all sets must cover all integers, we need to ensure that even the sparsest set ( A_n ) doesn't leave gaps that can't be filled by the other sets.Let me try to formalize this a bit. For each set ( A_i ), the minimal distance between any two elements is at least ( a^i ). So, the maximum number of elements ( A_i ) can have in any interval of length ( L ) is roughly ( L / a^i ). Since the union of all ( A_i ) must cover all integers, the sum of the densities of all ( A_i ) must be at least 1.Density here is a bit abstract, but in the context of integers, it's like the asymptotic density. So, for each ( A_i ), the density is at most ( 1 / a^i ). Therefore, the total density is ( sum_{i=1}^n 1 / a^i ). For the union to cover all integers, we need this sum to be at least 1.So, ( sum_{i=1}^n frac{1}{a^i} geq 1 ).This is a geometric series. The sum ( sum_{i=1}^infty frac{1}{a^i} ) converges to ( frac{1}{a - 1} ) if ( a > 1 ). But since we have a finite sum up to ( n ), the sum is ( frac{1 - (1/a)^{n+1}}{a - 1} ).Wait, actually, let me recall the formula for the sum of a geometric series:( sum_{i=1}^n r^i = r frac{1 - r^n}{1 - r} ).In our case, ( r = 1/a ), so:( sum_{i=1}^n frac{1}{a^i} = frac{1/a (1 - (1/a)^n)}{1 - 1/a} = frac{1 - (1/a)^n}{a - 1} ).So, for the total density to be at least 1, we need:( frac{1 - (1/a)^n}{a - 1} geq 1 ).Simplify this inequality:( 1 - (1/a)^n geq a - 1 ).Bring all terms to one side:( 1 - (1/a)^n - a + 1 geq 0 ).Simplify:( 2 - a - (1/a)^n geq 0 ).So,( 2 - a geq (1/a)^n ).Hmm, interesting. So, ( 2 - a ) must be greater than or equal to ( (1/a)^n ).Now, let's analyze this inequality.First, note that ( a ) must be positive, as given.Case 1: ( a = 1 ).Then, the left side is ( 2 - 1 = 1 ), and the right side is ( (1/1)^n = 1 ). So, equality holds. So, ( a = 1 ) is a boundary case.Case 2: ( a > 1 ).Then, ( 2 - a ) must be non-negative because ( (1/a)^n ) is positive. So, ( 2 - a geq 0 ) implies ( a leq 2 ).But since ( a > 1 ), this gives ( 1 < a leq 2 ).But we also have ( 2 - a geq (1/a)^n ). Let's see for ( a = 2 ):Left side: ( 2 - 2 = 0 ).Right side: ( (1/2)^n ).So, ( 0 geq (1/2)^n ), which is not true because ( (1/2)^n > 0 ). So, at ( a = 2 ), the inequality doesn't hold.Wait, that's a problem. So, maybe ( a ) cannot be equal to 2.Wait, but when ( a = 2 ), the sum ( sum_{i=1}^n 1/2^i = 1 - 1/2^n ), which is less than 1. So, the total density is less than 1, meaning that the union of the sets cannot cover all integers. Therefore, ( a = 2 ) is not allowed.So, perhaps ( a ) must be strictly less than 2.Wait, let's test ( a = 2 - epsilon ) for small ( epsilon > 0 ).Then, ( 2 - a = epsilon ).And ( (1/a)^n = (1/(2 - epsilon))^n ).We need ( epsilon geq (1/(2 - epsilon))^n ).For small ( epsilon ), ( 1/(2 - epsilon) approx 1/2 + epsilon/4 ), so ( (1/(2 - epsilon))^n approx (1/2)^n + n (1/2)^{n-1} epsilon / 4 ).So, approximately, ( epsilon geq (1/2)^n + text{something} ).But ( (1/2)^n ) is positive, so ( epsilon ) needs to be larger than ( (1/2)^n ). But as ( n ) increases, ( (1/2)^n ) decreases. So, for any fixed ( n ), there exists an ( epsilon ) such that ( epsilon geq (1/(2 - epsilon))^n ).But this seems a bit vague. Maybe I need a different approach.Alternatively, perhaps instead of thinking about density, I can think about constructing such sets ( A_i ) for certain values of ( a ).Let me try to construct the sets for ( a < 2 ).Suppose ( a = 1.5 ). Then, ( a^1 = 1.5 ), ( a^2 = 2.25 ), ( a^3 = 3.375 ), etc.But since we're dealing with integers, the spacing must be at least the ceiling of ( a^i ). So, for ( A_1 ), spacing is at least 2, for ( A_2 ), spacing is at least 3, for ( A_3 ), spacing is at least 4, and so on.Wait, but if ( a = 1.5 ), then ( a^i ) is not necessarily an integer. So, the minimal spacing is the smallest integer greater than or equal to ( a^i ).But maybe it's better to think in terms of powers of 2 or something similar.Wait, another idea: Maybe using binary representations or something.Suppose we partition the integers based on their binary representations. For example, ( A_1 ) contains numbers where the least significant bit is 1, ( A_2 ) contains numbers where the second least significant bit is 1, and so on. But I'm not sure if that directly relates to the spacing condition.Alternatively, maybe using arithmetic progressions with different common differences.Wait, if I set ( A_i ) to be arithmetic progressions with difference ( d_i geq a^i ), then their union needs to cover all integers.But the problem is that if ( a ) is too large, say ( a geq 2 ), then ( d_i ) grows exponentially, making it impossible to cover all integers with just ( n ) sets.Wait, let's think about this more carefully.Suppose ( a geq 2 ). Then, ( a^i ) grows exponentially. So, for ( i = n ), ( a^n ) is very large. So, ( A_n ) can only contain numbers that are at least ( a^n ) apart. But since we have to cover all integers, the other sets ( A_1, A_2, ldots, A_{n-1} ) must cover the gaps left by ( A_n ).But the problem is that the spacing required for ( A_n ) is so large that the other sets may not be able to cover all the integers in between.Wait, perhaps using the pigeonhole principle. If ( a geq 2 ), then the spacing in ( A_n ) is at least ( 2^n ). So, in the interval ( [1, 2^n] ), ( A_n ) can contain at most one number. But the other sets ( A_1, ldots, A_{n-1} ) must cover the remaining ( 2^n - 1 ) numbers.But each ( A_i ) for ( i < n ) has spacing at least ( a^i geq 2^i ). So, in the interval ( [1, 2^n] ), each ( A_i ) can contain at most ( 2^n / 2^i ) numbers.Thus, the total number of numbers covered by all ( A_i ) is ( sum_{i=1}^n 2^n / 2^i = 2^n sum_{i=1}^n 1/2^i = 2^n (1 - 1/2^n) = 2^n - 1 ).But we have ( 2^n ) numbers in ( [1, 2^n] ), so we're missing one number. That missing number must be in ( A_n ), but ( A_n ) can only contain one number in that interval. So, it's possible that ( A_n ) contains that missing number.Wait, but actually, the calculation shows that the total number of numbers covered is ( 2^n - 1 ), so we're missing one number. But ( A_n ) can only cover one number, so that missing number must be covered by ( A_n ). But ( A_n ) can indeed cover it because it's allowed to have one number in that interval.Wait, but this seems contradictory because earlier I thought that ( A_n ) can only cover one number, but the other sets cover ( 2^n - 1 ) numbers, so together they cover all ( 2^n ) numbers. So, maybe it's possible?Wait, no, because the calculation shows that the total number of numbers covered is ( 2^n - 1 + 1 = 2^n ), which matches the total number of integers in the interval. So, in this case, it's possible.But wait, this seems to suggest that even for ( a = 2 ), it's possible to cover all integers with ( n ) sets. But earlier, when I considered the density, I thought that for ( a = 2 ), the total density is less than 1, implying that it's not possible.There's a contradiction here. Maybe my density approach was flawed because it's an asymptotic density, whereas the finite interval analysis shows that it's possible.Wait, let's think again. For ( a = 2 ), each ( A_i ) has spacing ( 2^i ). So, in the interval ( [1, 2^n] ), ( A_i ) contains ( 2^{n - i} ) numbers. So, the total number of numbers covered is ( sum_{i=1}^n 2^{n - i} = 2^n - 1 ). But we have ( 2^n ) numbers in the interval, so we're missing one number. However, ( A_n ) can cover that missing number because it's allowed to have one number in that interval. So, in this case, it's possible.But wait, in reality, ( A_n ) must cover numbers that are at least ( 2^n ) apart. So, in the interval ( [1, 2^n] ), ( A_n ) can contain at most one number. So, if that number is not covered by any other ( A_i ), then it's covered by ( A_n ). But if all other numbers are covered by ( A_1, ldots, A_{n-1} ), then it's fine.But wait, the problem is that the other sets ( A_1, ldots, A_{n-1} ) might not cover all the numbers. For example, suppose ( n = 2 ) and ( a = 2 ). Then, ( A_1 ) has spacing 2, so it covers all even numbers, and ( A_2 ) has spacing 4, so it covers numbers like 1, 5, 9, etc. But in the interval ( [1, 4] ), ( A_1 ) covers 2 and 4, and ( A_2 ) covers 1. So, 3 is missing. But 3 is not covered by either set. So, in this case, it's not covered.Wait, that's a problem. So, for ( n = 2 ) and ( a = 2 ), it's not possible to cover all integers because some numbers like 3 are left out.But according to the earlier calculation, the total number of numbers covered should be ( 2^2 - 1 + 1 = 4 ), which matches the total. But in reality, 3 is missing. So, perhaps the issue is that the sets ( A_i ) are defined as arithmetic progressions starting from different points, but in reality, they might not cover all residues.Wait, maybe I need to adjust the starting points of the arithmetic progressions. For example, ( A_1 ) could be numbers congruent to 0 mod 2, ( A_2 ) could be numbers congruent to 1 mod 4, ( A_3 ) could be numbers congruent to 3 mod 8, and so on. This way, each set covers different residues, and together they cover all integers.Wait, let's test this for ( n = 2 ) and ( a = 2 ).Let ( A_1 ) be numbers congruent to 0 mod 2: ..., -4, -2, 0, 2, 4, ...( A_2 ) be numbers congruent to 1 mod 4: ..., -7, -3, 1, 5, 9, ...Now, in the interval ( [1, 4] ), ( A_1 ) covers 2 and 4, ( A_2 ) covers 1 and 5 (but 5 is outside the interval). So, 3 is still missing.Hmm, so even with this construction, 3 is missing. So, perhaps this approach doesn't work.Wait, maybe I need a different construction. Perhaps using different moduli or something else.Alternatively, maybe using a tree-like structure where each set covers different scales.Wait, another idea: Maybe using the binary representation of integers. Each set ( A_i ) corresponds to numbers where the ( i )-th bit is set. But I'm not sure if that directly relates to the spacing condition.Wait, let's think about the spacing condition again. For ( A_i ), any two numbers must be at least ( a^i ) apart. So, if ( a < 2 ), then ( a^i ) grows slower than ( 2^i ), meaning that the spacing required is less restrictive. Therefore, it might be possible to cover all integers with such sets.But for ( a geq 2 ), the spacing required is too large, making it impossible to cover all integers with just ( n ) sets.Wait, but earlier, I saw that for ( a = 2 ), it's possible to cover all integers with ( n ) sets, but in practice, some numbers like 3 are missing. So, maybe my initial assumption was wrong.Alternatively, perhaps the key is that for ( a < 2 ), the sets can be constructed in a way that their spacings allow them to cover all integers, but for ( a geq 2 ), it's impossible.Wait, let me try to formalize this.Suppose ( a < 2 ). Then, ( a^i < 2^i ). So, the spacing required for each ( A_i ) is less than ( 2^i ). Therefore, the sets can be constructed similarly to how binary representations cover all integers, but with more flexibility because the spacing is smaller.On the other hand, if ( a geq 2 ), the spacing required for ( A_n ) is at least ( 2^n ), which is too large to cover all integers in the interval ( [1, 2^n] ) because the other sets can only cover ( 2^n - 1 ) numbers, leaving one number uncovered, which ( A_n ) can cover. But in reality, as seen in the ( n = 2 ) case, some numbers are still left uncovered.Wait, maybe the issue is that the sets ( A_i ) are not just arithmetic progressions but can be more cleverly constructed to cover all integers. For example, using a greedy algorithm or something.Alternatively, perhaps the answer is that such sets exist if and only if ( a < 2 ).But I need to verify this.Let me consider ( a = 1.5 ) and ( n = 2 ). Then, ( A_1 ) must have spacing at least 1.5, so at least 2 apart. ( A_2 ) must have spacing at least ( 1.5^2 = 2.25 ), so at least 3 apart.So, ( A_1 ) could be all even numbers: ..., -4, -2, 0, 2, 4, ...( A_2 ) could be numbers congruent to 1 mod 3: ..., -5, -2, 1, 4, 7, ...Wait, but in this case, the union of ( A_1 ) and ( A_2 ) would cover all integers except those congruent to 3 mod 6, like 3, 9, 15, etc. So, those are missing.Hmm, so even with ( a = 1.5 ), it's not covering all integers. So, maybe my initial thought was wrong.Wait, perhaps I need a different construction. Maybe using more sets or overlapping progressions.Alternatively, maybe using a tree-like structure where each set covers different scales.Wait, another idea: Maybe using the concept of a complete residue system. If the spacings are chosen such that their least common multiple covers all residues.But I'm not sure.Wait, perhaps the key is that for ( a < 2 ), the spacings ( a^i ) are such that the sets can be arranged to cover all integers without missing any, but for ( a geq 2 ), it's impossible.But I need a more rigorous argument.Let me think about the total number of integers that can be covered by the sets ( A_i ). For each ( A_i ), the number of integers in ( [1, N] ) is at most ( N / a^i ). So, the total number covered is ( sum_{i=1}^n N / a^i ). For the union to cover all integers, we need ( sum_{i=1}^n 1 / a^i geq 1 ) asymptotically.So, ( sum_{i=1}^infty 1 / a^i geq 1 ). The sum of the infinite geometric series is ( 1 / (a - 1) ) if ( a > 1 ). So, ( 1 / (a - 1) geq 1 ), which implies ( a - 1 leq 1 ), so ( a leq 2 ).But for finite ( n ), the sum is ( sum_{i=1}^n 1 / a^i = (1 - (1/a)^{n+1}) / (a - 1) ). For this to be at least 1, we need ( (1 - (1/a)^{n+1}) / (a - 1) geq 1 ), which simplifies to ( 1 - (1/a)^{n+1} geq a - 1 ), or ( 2 - a geq (1/a)^{n+1} ).So, for large ( n ), ( (1/a)^{n+1} ) becomes very small if ( a > 1 ). Therefore, ( 2 - a ) must be non-negative, so ( a leq 2 ).But for ( a = 2 ), ( (1/a)^{n+1} = 1 / 2^{n+1} ), which is positive, so ( 2 - a = 0 geq 1 / 2^{n+1} ), which is not true. Therefore, ( a ) must be strictly less than 2.Thus, the condition is satisfied if and only if ( 0 < a < 2 ).Wait, but earlier, I saw that even for ( a < 2 ), like ( a = 1.5 ), it's not covering all integers. So, maybe my density argument is not sufficient.Alternatively, perhaps the construction is more involved.Wait, maybe using a hierarchical approach. For example, partition the integers into blocks of size ( a^i ), and assign each block to a set ( A_i ). But I'm not sure.Alternatively, perhaps using a tree where each level corresponds to a set ( A_i ), and each node covers a certain range of integers with the required spacing.Wait, maybe it's similar to a Cantor set construction, but in integers.Alternatively, perhaps using the concept of Beatty sequences. Beatty sequences are sequences of the form ( lfloor k alpha rfloor ) for irrational ( alpha ), and they have the property that they partition the integers. But I'm not sure if that's directly applicable here.Wait, but in our case, the sets ( A_i ) need to have a minimal spacing, not necessarily being a Beatty sequence.Alternatively, maybe using the concept of packing and covering in integers.Wait, another idea: If ( a < 2 ), then ( a^i < 2^i ). So, the spacing required for ( A_i ) is less than ( 2^i ). Therefore, we can construct ( A_i ) as arithmetic progressions with difference ( d_i ) where ( d_i ) is the smallest integer greater than or equal to ( a^i ). Since ( d_i leq 2^i ), we can use a similar construction to the binary representation, where each set ( A_i ) covers numbers with a certain bit set, but with more flexibility.Wait, for example, if ( a = 1.5 ), then ( a^1 = 1.5 ), so ( d_1 = 2 ). ( a^2 = 2.25 ), so ( d_2 = 3 ). ( a^3 = 3.375 ), so ( d_3 = 4 ), and so on.Then, ( A_1 ) could be all even numbers, ( A_2 ) could be numbers congruent to 1 mod 3, ( A_3 ) could be numbers congruent to 1 mod 4, and so on. But as I saw earlier, this might not cover all integers.Wait, maybe I need to use a different approach. Perhaps using a greedy algorithm where I assign each integer to the first set ( A_i ) that can accommodate it without violating the spacing condition.But this might be complicated.Alternatively, maybe using induction on ( n ). Suppose for ( n = 1 ), as I saw earlier, ( a leq 1 ). For ( n = 2 ), perhaps ( a leq sqrt{2} ), and so on, leading to ( a < 2 ).Wait, but I'm not sure about that.Alternatively, perhaps the key is that for ( a < 2 ), the spacings ( a^i ) are such that the sets can be arranged to cover all integers, but for ( a geq 2 ), it's impossible because the spacings are too large.But I need a more concrete argument.Wait, let's consider the case when ( a = 2 ) and ( n = 1 ). Then, ( A_1 ) must cover all integers with spacing at least 2, which is impossible because you can't cover all integers with a single arithmetic progression of spacing 2.Wait, but for ( n = 1 ), ( a ) must be less than or equal to 1, as I saw earlier.Wait, but in the problem, ( n ) is given, so perhaps for any ( n ), ( a ) must be less than 2.Wait, but earlier, I saw that for ( a = 2 ), even with ( n = 2 ), it's impossible to cover all integers because some numbers are left out.Therefore, perhaps the answer is that such sets exist if and only if ( 0 < a < 2 ).So, putting it all together, the condition is satisfied if and only if ( a ) is a positive real number less than 2.