answer：Alright, so I have this problem where I need to prove that if the sum of f(S) over all subsets S with an odd number of elements is 1/2, then at least one of the a_k must be 1/2. Hmm, okay. Let me try to unpack this step by step.First, let me understand what f(S) represents. For any subset S, f(S) is the product of a_i for i in S multiplied by the product of (1 - a_j) for j not in S. So, it's like a probability of selecting exactly the elements in S, where each element is chosen with probability a_i and not chosen with probability (1 - a_i). That makes sense.So, if I think of each a_i as the probability of selecting element i, then f(S) is the probability of selecting exactly the subset S. Now, the sum over all subsets S with an odd number of elements of f(S) is given as 1/2. So, this is the probability that a randomly selected subset (with probabilities defined by the a_i's) has an odd number of elements.Wait, so if I have n coins, each with probability a_i of landing heads, then f(S) is the probability of getting exactly the heads in S and tails elsewhere. Then, the sum over all subsets S with an odd number of elements is the probability that the number of heads is odd. So, the problem is saying that the probability of getting an odd number of heads is 1/2, and we need to show that one of the a_i's must be 1/2.Okay, that's a nice interpretation. So, maybe I can use some probability theory or generating functions to approach this.Let me recall that the probability generating function for the number of heads is the product over all i of (1 - a_i + a_i x). If I plug in x = -1, I get the difference between the probability of even and odd number of heads. Specifically, the generating function evaluated at x = -1 is equal to the sum over all subsets S of (-1)^{|S|} f(S). So, that would be the difference between the probability of even and odd subsets.Let me write that down:G(-1) = sum_{S} (-1)^{|S|} f(S) = sum_{|S| even} f(S) - sum_{|S| odd} f(S).But we know that sum_{|S| odd} f(S) = 1/2, and since the total probability is 1, sum_{|S| even} f(S) must also be 1/2. Therefore, G(-1) = 1/2 - 1/2 = 0.So, G(-1) = 0. But G(x) is the product over all i of (1 - a_i + a_i x). Therefore, G(-1) = product_{i=1}^n (1 - a_i + a_i (-1)) = product_{i=1}^n (1 - 2 a_i).So, product_{i=1}^n (1 - 2 a_i) = 0.But the product of these terms is zero only if at least one of the terms is zero. That is, for some k, 1 - 2 a_k = 0, which implies a_k = 1/2.Therefore, there must be some k such that a_k = 1/2.Wait, that seems straightforward. Let me check if I missed anything.So, G(x) is the generating function for the number of heads, which is a standard approach in probability. Evaluating it at x = -1 gives the difference between the probabilities of even and odd numbers of heads. Since the sum of probabilities for odd subsets is 1/2, the difference is zero, which implies the product of (1 - 2 a_i) is zero. Therefore, at least one a_i must be 1/2.Yes, that seems correct. I think I got it.**Final Answer**boxed{a_k = frac{1}{2} text{ for some } k}

answer：Alright, so I have this problem here where the terminal side of angle α passes through the point (3, -4). I need to find the value of sin α + cos α. Hmm, okay. Let me try to remember how to approach this.First, I recall that when dealing with angles in standard position (meaning their vertex is at the origin and they start along the positive x-axis), the terminal side passing through a specific point gives us information about the sine and cosine of that angle. Specifically, if the terminal side passes through the point (x, y), then:- sin α = y / r- cos α = x / rwhere r is the distance from the origin to the point (x, y). So, r is like the hypotenuse of a right triangle formed by the x and y coordinates.Alright, so in this case, the point is (3, -4). That means x = 3 and y = -4. I need to find r first. To find r, I can use the Pythagorean theorem, right? Because r is the hypotenuse of a right triangle with legs of length 3 and 4.So, r = √(x² + y²) = √(3² + (-4)²) = √(9 + 16) = √25 = 5.Okay, so r is 5. Got that down.Now, let's compute sin α and cos α.Starting with sin α: sin α = y / r = (-4) / 5 = -4/5.Then, cos α: cos α = x / r = 3 / 5.So, sin α is -4/5 and cos α is 3/5.Now, the question asks for sin α + cos α. So, let's add those two values together.sin α + cos α = (-4/5) + (3/5) = (-4 + 3)/5 = (-1)/5 = -1/5.Hmm, so the sum is -1/5. Let me check the answer choices.A: ±1/5B: ±7/5C: -1/5D: 7/5So, -1/5 is option C. That seems straightforward, but let me make sure I didn't make any mistakes.Wait, why is option A ±1/5? Shouldn't it be just -1/5? Because both sin and cos are determined by the specific point (3, -4). The angle α is in a specific quadrant, right? Since x is positive and y is negative, the point is in the fourth quadrant. In the fourth quadrant, sine is negative and cosine is positive, which matches our calculations: sin α is negative and cos α is positive.So, sin α + cos α is definitely negative because -4/5 is more negative than 3/5 is positive. So, adding them gives a negative number, which is -1/5. So, option C is correct.But wait, why is option A ±1/5? Is there a possibility that the angle could be in another quadrant where the sum could be positive? Hmm, no, because the terminal side passes through (3, -4), which is fixed in the fourth quadrant. So, the angle α is uniquely determined by that point, and thus sin α + cos α is uniquely determined as -1/5.Therefore, I think the answer is definitely C: -1/5.Just to double-check, let me visualize the point (3, -4). It's in the fourth quadrant, 3 units to the right on the x-axis and 4 units down on the y-axis. The distance from the origin is 5, as we calculated. So, sin is opposite over hypotenuse, which is -4/5, and cos is adjacent over hypotenuse, which is 3/5. Adding those gives -1/5.Yep, that seems right. I don't see any mistakes in my reasoning. So, I'm confident that the correct answer is C: -1/5.**Final Answer**boxed{C}

answer：Okay, so I need to solve this problem about the school buying books for the "Campus Reading Festival." There are two parts to the problem. Let me start with part (1).First, the problem says that the unit price of type A books is 10 more than the unit price of type B books. So, if I let the price of type B books be, say, ( y ) dollars, then the price of type A books would be ( y + 10 ) dollars. That makes sense.Next, it says that purchasing 3 type A books and 2 type B books costs a total of 130. So, I can write an equation for that. The cost of 3 type A books would be ( 3 times (y + 10) ) and the cost of 2 type B books would be ( 2 times y ). Adding those together should equal 130.So, the equation is:[ 3(y + 10) + 2y = 130 ]Let me simplify this equation step by step. First, distribute the 3 into ( y + 10 ):[ 3y + 30 + 2y = 130 ]Now, combine like terms:[ 3y + 2y + 30 = 130 ][ 5y + 30 = 130 ]Subtract 30 from both sides:[ 5y = 100 ]Divide both sides by 5:[ y = 20 ]So, the unit price of type B books is 20. Then, the unit price of type A books is ( y + 10 = 20 + 10 = 30 ) dollars.Alright, that seems straightforward. So, part (1) is done. Type A books are 30 each, and type B books are 20 each.Now, moving on to part (2). The school plans to purchase a total of 40 books, and the total budget does not exceed 980. I need to find the maximum number of type A books they can purchase.Let me denote the number of type A books as ( a ). Then, the number of type B books would be ( 40 - a ), since the total is 40 books.The total cost for type A books would be ( 30a ) dollars, and for type B books, it would be ( 20(40 - a) ) dollars. The sum of these should be less than or equal to 980.So, the inequality is:[ 30a + 20(40 - a) leq 980 ]Let me simplify this inequality step by step. First, distribute the 20:[ 30a + 800 - 20a leq 980 ]Combine like terms:[ 10a + 800 leq 980 ]Subtract 800 from both sides:[ 10a leq 180 ]Divide both sides by 10:[ a leq 18 ]So, the maximum number of type A books they can purchase is 18.Wait, let me double-check my calculations to make sure I didn't make any mistakes.Starting with the equation:[ 30a + 20(40 - a) leq 980 ][ 30a + 800 - 20a leq 980 ][ 10a + 800 leq 980 ][ 10a leq 180 ][ a leq 18 ]Yes, that seems correct. So, the school can buy at most 18 type A books without exceeding the budget.Just to be thorough, let me calculate the total cost if they buy 18 type A books and 22 type B books.Cost of type A: ( 18 times 30 = 540 ) dollarsCost of type B: ( 22 times 20 = 440 ) dollarsTotal cost: ( 540 + 440 = 980 ) dollarsPerfect, that's exactly the budget. So, 18 type A books is the maximum they can purchase without going over 980.I think that's all for this problem. It wasn't too bad once I set up the equations properly. Just needed to translate the words into mathematical expressions and solve step by step.

answer：First, I need to determine the number of years between 2008 and 2011. This includes the years 2008, 2009, 2010, and 2011, totaling four years.Next, I'll identify which of these years are leap years. A leap year occurs every four years and has 366 days instead of the usual 365. Among the years in question, 2008 is a leap year, while 2009, 2010, and 2011 are not.Now, I'll calculate the total number of days. The three non-leap years contribute 365 days each, and the one leap year contributes 366 days.Adding these together: (3 × 365) + 366 = 1095 + 366 = 1461 days.Therefore, the total number of days from 2008 through 2011 is 1461 days.

answer：Okay, so I have this problem here with two propositions, p and q. Let me try to understand each one step by step.First, proposition p says: "The solution set of the inequality ( a^x > 1 ) (where ( a > 0 ) and ( a neq 1 )) is ( { x | x < 0 } )." Hmm, okay. So, I remember that exponential functions can be increasing or decreasing depending on the base ( a ). If ( a > 1 ), then ( a^x ) is an increasing function. If ( 0 < a < 1 ), it's a decreasing function.So, for ( a^x > 1 ), if ( a > 1 ), then since it's increasing, ( a^x > 1 ) when ( x > 0 ). On the other hand, if ( 0 < a < 1 ), since it's decreasing, ( a^x > 1 ) when ( x < 0 ). So, the solution set depends on whether ( a ) is greater than 1 or between 0 and 1.Proposition p claims that the solution set is ( x < 0 ). That would only be true if ( 0 < a < 1 ). If ( a > 1 ), the solution set would be ( x > 0 ), which is different. So, p is true when ( 0 < a < 1 ) and false otherwise.Now, moving on to proposition q: "The domain of the function ( y = lg(x^2 - x + a) ) is ( mathbb{R} )." Okay, so the domain of a logarithmic function is where the argument is positive. So, ( x^2 - x + a > 0 ) for all real numbers x.To ensure that ( x^2 - x + a > 0 ) for all x, the quadratic must be always positive. That means the quadratic has no real roots and opens upwards. The quadratic ( x^2 - x + a ) opens upwards because the coefficient of ( x^2 ) is positive. So, we just need to ensure it has no real roots.The discriminant of the quadratic ( x^2 - x + a ) is ( D = (-1)^2 - 4(1)(a) = 1 - 4a ). For the quadratic to have no real roots, the discriminant must be negative. So, ( 1 - 4a < 0 ), which simplifies to ( a > frac{1}{4} ).Therefore, proposition q is true when ( a > frac{1}{4} ) and false otherwise.Now, the problem states that ( p lor q ) is true and ( p land q ) is false. Let me recall what these logical connectives mean.- ( p lor q ) is true if at least one of p or q is true.- ( p land q ) is true only if both p and q are true.Given that ( p land q ) is false, it means that p and q cannot both be true at the same time. So, either p is false, or q is false, or both are false.But since ( p lor q ) is true, at least one of p or q must be true. So, combining these two conditions, we have that exactly one of p or q is true, and the other is false.So, let's break it down into two cases:1. Case 1: p is true and q is false.2. Case 2: p is false and q is true.Let's analyze each case.**Case 1: p is true and q is false.**From earlier, p is true when ( 0 < a < 1 ). q is false when ( a leq frac{1}{4} ). So, for both conditions to hold, ( a ) must satisfy both ( 0 < a < 1 ) and ( a leq frac{1}{4} ). The overlap of these intervals is ( 0 < a leq frac{1}{4} ).**Case 2: p is false and q is true.**p is false when ( a leq 0 ) or ( a geq 1 ). However, since ( a > 0 ) is given in the problem (from proposition p's condition), we can disregard ( a leq 0 ). So, p is false when ( a geq 1 ). q is true when ( a > frac{1}{4} ). So, for both conditions to hold, ( a ) must satisfy both ( a geq 1 ) and ( a > frac{1}{4} ). The overlap here is simply ( a geq 1 ).Now, combining both cases, the possible values of ( a ) are ( 0 < a leq frac{1}{4} ) and ( a geq 1 ). So, the range of ( a ) is the union of these two intervals.Therefore, the range of the real number ( a ) is ( (0, frac{1}{4}] cup [1, +infty) ).Wait, hold on, in the second case, when ( a geq 1 ), is ( q ) necessarily true? Let me double-check.If ( a geq 1 ), then ( a > frac{1}{4} ) is automatically satisfied, so ( q ) is true. So, yes, in that case, p is false and q is true, which satisfies the conditions.Similarly, in the first case, when ( 0 < a leq frac{1}{4} ), p is true (since ( 0 < a < 1 )) and q is false (since ( a leq frac{1}{4} )), which also satisfies the conditions.So, putting it all together, the range of ( a ) is ( (0, frac{1}{4}] cup [1, +infty) ).But wait, in the initial analysis, I considered ( a geq 1 ) as part of the solution. However, let me check if ( a = 1 ) is allowed. From proposition p, ( a neq 1 ), so ( a = 1 ) is excluded. So, actually, the interval should be ( a > 1 ), not ( a geq 1 ).Therefore, the correct range is ( (0, frac{1}{4}] cup (1, +infty) ).Yes, that makes sense because ( a = 1 ) is excluded from the original condition for proposition p.So, final answer: ( (0, frac{1}{4}] cup (1, +infty) ).**Final Answer**The range of the real number ( a ) is boxed{(0, dfrac{1}{4}] cup (1, +infty)}.

answer：Okay, so I have this problem where I need to prove that for positive integers ( a ) and ( b ), the expressions ( an^2 + b ) and ( a(n+1)^2 + b ) can only both be perfect squares for finitely many integers ( n ). Hmm, interesting. Let me try to break this down step by step.First, let me understand what the problem is asking. We have two quadratic expressions in terms of ( n ): ( an^2 + b ) and ( a(n+1)^2 + b ). The claim is that there are only finitely many integers ( n ) such that both of these expressions are perfect squares. So, my goal is to show that beyond a certain point, as ( n ) grows, it's impossible for both ( an^2 + b ) and ( a(n+1)^2 + b ) to be squares simultaneously.Let me denote ( an^2 + b = x^2 ) and ( a(n+1)^2 + b = y^2 ), where ( x ) and ( y ) are integers. So, I have two equations:1. ( an^2 + b = x^2 )2. ( a(n+1)^2 + b = y^2 )I need to find the number of integer solutions ( n ) such that both equations hold. The problem states that this number is finite.Let me subtract the first equation from the second to eliminate ( b ):( a(n+1)^2 + b - (an^2 + b) = y^2 - x^2 )Simplifying the left side:( a(n^2 + 2n + 1) + b - an^2 - b = a(2n + 1) )So, ( a(2n + 1) = y^2 - x^2 )I can factor the right side as a difference of squares:( a(2n + 1) = (y - x)(y + x) )So, now I have:( (y - x)(y + x) = a(2n + 1) )This equation tells me that the product of ( y - x ) and ( y + x ) is equal to ( a ) times an odd integer ( 2n + 1 ). Since ( a ) and ( 2n + 1 ) are positive integers, and ( y > x ) (since ( y^2 > x^2 )), both ( y - x ) and ( y + x ) are positive integers.Let me denote ( d_1 = y - x ) and ( d_2 = y + x ). Then, ( d_1 times d_2 = a(2n + 1) ). Also, since ( d_1 ) and ( d_2 ) are factors of ( a(2n + 1) ), they must satisfy ( d_1 < d_2 ) and ( d_1 times d_2 = a(2n + 1) ).Additionally, since ( d_1 = y - x ) and ( d_2 = y + x ), adding these two equations gives:( d_1 + d_2 = 2y )And subtracting them gives:( d_2 - d_1 = 2x )So, both ( d_1 + d_2 ) and ( d_2 - d_1 ) must be even, which implies that ( d_1 ) and ( d_2 ) are both even or both odd. Since ( a(2n + 1) ) is the product, and ( 2n + 1 ) is odd, the factors ( d_1 ) and ( d_2 ) must also be odd if ( a ) is odd, or one even and one odd if ( a ) is even. Hmm, this might complicate things a bit.But perhaps I can consider the ratio between ( d_2 ) and ( d_1 ). Let me denote ( k = frac{d_2}{d_1} ). Then, ( d_2 = k d_1 ), and so:( d_1 times k d_1 = a(2n + 1) )Which simplifies to:( k d_1^2 = a(2n + 1) )So, ( 2n + 1 = frac{k d_1^2}{a} )Since ( 2n + 1 ) must be an integer, ( a ) must divide ( k d_1^2 ). Therefore, ( k d_1^2 ) must be a multiple of ( a ).But ( k ) is a rational number since ( d_2 ) and ( d_1 ) are integers. Let me assume that ( k ) is an integer. Wait, is that necessarily true? If ( d_1 ) and ( d_2 ) are integers, then ( k = frac{d_2}{d_1} ) could be a rational number, but not necessarily an integer. Hmm, perhaps I need to adjust my approach.Alternatively, let me consider that ( d_1 ) and ( d_2 ) are divisors of ( a(2n + 1) ). Since ( d_1 ) and ( d_2 ) are factors, and their product is ( a(2n + 1) ), each of them must be a factor of ( a(2n + 1) ). But since ( a ) and ( 2n + 1 ) are coprime? Wait, not necessarily. If ( a ) is even, then ( 2n + 1 ) is odd, so they are coprime. If ( a ) is odd, then ( 2n + 1 ) is also odd, so they might share common factors.Wait, perhaps I should consider the greatest common divisor (gcd) of ( d_1 ) and ( d_2 ). Since ( d_1 ) and ( d_2 ) are both factors of ( a(2n + 1) ), their gcd must divide ( a(2n + 1) ). Also, since ( d_1 ) and ( d_2 ) are such that ( d_2 - d_1 = 2x ) and ( d_1 + d_2 = 2y ), their gcd must divide both ( 2x ) and ( 2y ). Therefore, the gcd of ( d_1 ) and ( d_2 ) must divide ( 2 times gcd(x, y) ).This is getting a bit abstract. Maybe I should try to express ( x ) and ( y ) in terms of ( d_1 ) and ( d_2 ). From earlier, we have:( y = frac{d_1 + d_2}{2} )( x = frac{d_2 - d_1}{2} )So, substituting back into the first equation ( an^2 + b = x^2 ):( an^2 + b = left( frac{d_2 - d_1}{2} right)^2 )Similarly, the second equation is already used to get the relation between ( d_1 ), ( d_2 ), and ( n ). So, perhaps I can express ( n ) in terms of ( d_1 ) and ( d_2 ).From ( 2n + 1 = frac{d_1 d_2}{a} ), we have:( n = frac{d_1 d_2}{2a} - frac{1}{2} )So, substituting this into the expression for ( x^2 ):( aleft( frac{d_1 d_2}{2a} - frac{1}{2} right)^2 + b = left( frac{d_2 - d_1}{2} right)^2 )Let me expand the left-hand side:First, compute ( left( frac{d_1 d_2}{2a} - frac{1}{2} right)^2 ):( left( frac{d_1 d_2 - a}{2a} right)^2 = frac{(d_1 d_2 - a)^2}{4a^2} )So, multiplying by ( a ):( a times frac{(d_1 d_2 - a)^2}{4a^2} = frac{(d_1 d_2 - a)^2}{4a} )Adding ( b ):( frac{(d_1 d_2 - a)^2}{4a} + b )Set this equal to ( left( frac{d_2 - d_1}{2} right)^2 ):( frac{(d_1 d_2 - a)^2}{4a} + b = frac{(d_2 - d_1)^2}{4} )Multiply both sides by 4a to eliminate denominators:( (d_1 d_2 - a)^2 + 4ab = a(d_2 - d_1)^2 )Let me expand both sides:Left side: ( (d_1 d_2 - a)^2 + 4ab )= ( d_1^2 d_2^2 - 2a d_1 d_2 + a^2 + 4ab )Right side: ( a(d_2 - d_1)^2 )= ( a(d_2^2 - 2d_1 d_2 + d_1^2) )= ( a d_2^2 - 2a d_1 d_2 + a d_1^2 )So, setting left side equal to right side:( d_1^2 d_2^2 - 2a d_1 d_2 + a^2 + 4ab = a d_2^2 - 2a d_1 d_2 + a d_1^2 )Let me bring all terms to the left side:( d_1^2 d_2^2 - 2a d_1 d_2 + a^2 + 4ab - a d_2^2 + 2a d_1 d_2 - a d_1^2 = 0 )Simplify term by term:- ( d_1^2 d_2^2 )- ( -2a d_1 d_2 + 2a d_1 d_2 = 0 )- ( a^2 + 4ab )- ( -a d_2^2 - a d_1^2 )So, combining:( d_1^2 d_2^2 + a^2 + 4ab - a d_2^2 - a d_1^2 = 0 )Let me factor out ( a ) from the last three terms:( d_1^2 d_2^2 + a(a + 4b) - a(d_1^2 + d_2^2) = 0 )Hmm, not sure if that helps. Maybe rearrange terms:( d_1^2 d_2^2 - a d_1^2 - a d_2^2 + a^2 + 4ab = 0 )Let me factor ( d_1^2 ) and ( d_2^2 ):( d_1^2(d_2^2 - a) - a d_2^2 + a^2 + 4ab = 0 )Alternatively, perhaps factor differently:( d_1^2 d_2^2 - a d_1^2 - a d_2^2 + a^2 + 4ab = 0 )This seems complicated. Maybe I can factor this expression as a quadratic in terms of ( d_1^2 ) or ( d_2^2 ). Let me try to consider it as a quadratic in ( d_1^2 ):Let me denote ( u = d_1^2 ), then the equation becomes:( u d_2^2 - a u - a d_2^2 + a^2 + 4ab = 0 )This is linear in ( u ):( u(d_2^2 - a) - a d_2^2 + a^2 + 4ab = 0 )Solving for ( u ):( u = frac{a d_2^2 - a^2 - 4ab}{d_2^2 - a} )Simplify numerator:( a(d_2^2 - a - 4b) )So,( u = frac{a(d_2^2 - a - 4b)}{d_2^2 - a} )Which can be written as:( u = a left( 1 - frac{a + 4b}{d_2^2 - a} right) )Since ( u = d_1^2 ) must be a positive integer, the expression ( frac{a + 4b}{d_2^2 - a} ) must be rational, and in fact, an integer because ( u ) is an integer.Let me denote ( t = d_2^2 - a ). Then,( u = a left( 1 - frac{a + 4b}{t} right) )Which implies:( u = a - frac{a(a + 4b)}{t} )Since ( u ) must be positive, ( t ) must be a positive divisor of ( a(a + 4b) ). So, ( t ) divides ( a(a + 4b) ), and ( t = d_2^2 - a ).Therefore, ( d_2^2 = t + a ), and since ( t ) divides ( a(a + 4b) ), ( d_2^2 ) must be equal to ( a + t ), where ( t ) is a positive divisor of ( a(a + 4b) ).Given that ( a ) and ( b ) are fixed positive integers, the number of positive divisors of ( a(a + 4b) ) is finite. Therefore, the number of possible ( t ) is finite, which in turn implies that the number of possible ( d_2 ) is finite.For each such ( d_2 ), we can compute ( u = d_1^2 ) as:( u = a - frac{a(a + 4b)}{t} )But since ( u ) must be a perfect square, this imposes a condition on ( t ). Specifically, ( a - frac{a(a + 4b)}{t} ) must be a perfect square. Therefore, for each divisor ( t ) of ( a(a + 4b) ), we can check if ( a - frac{a(a + 4b)}{t} ) is a perfect square.Since there are only finitely many such ( t ), there are only finitely many possible ( d_1 ) and ( d_2 ), which correspond to finitely many solutions ( n ).Therefore, the number of integers ( n ) for which both ( an^2 + b ) and ( a(n+1)^2 + b ) are perfect squares is finite.Wait, let me double-check if I missed something. I assumed that ( d_1 ) and ( d_2 ) are positive integers, which they are because ( y > x ) and ( y, x ) are positive integers. Also, I considered ( t ) as a positive divisor, which is correct because ( d_2^2 = t + a ) must be greater than ( a ), so ( t ) is positive.Another point to consider is whether different ( t ) can lead to the same ( n ). But since each ( t ) gives a specific ( d_2 ), and from ( d_2 ) we can compute ( n ) as ( n = frac{d_1 d_2}{2a} - frac{1}{2} ), each ( t ) would lead to a unique ( n ) or possibly a few, but still finite.Therefore, my conclusion is that there are only finitely many such ( n ), which proves the original statement.**Final Answer**boxed{text{Only finitely many integers } n text{ satisfy the condition.}}