answer：Okay, so I have this problem where I need to determine if the function ( f(x) = x^n ) is a solution to the functional inequality ( f(x) + f(1 - x) > 1 ). If it is, I need to find the domain ( D ); if not, I have to explain why. Alright, let me break this down step by step.First, I know that ( f(x) = x^n ) where ( n ) is a positive integer. The inequality given is ( f(x) + f(1 - x) > 1 ). So, substituting ( f(x) ) into the inequality, it becomes ( x^n + (1 - x)^n > 1 ). That seems straightforward.I think I should analyze this inequality for different values of ( n ). Maybe starting with small values of ( n ) like 1, 2, 3, etc., to see if there's a pattern or a general rule.Let's start with ( n = 1 ). Then the inequality becomes ( x + (1 - x) > 1 ). Simplifying that, ( x + 1 - x = 1 ), so the inequality is ( 1 > 1 ), which is not true. So, for ( n = 1 ), there are no solutions. That means ( f(x) = x ) is not a solution to the inequality.Moving on to ( n = 2 ). The inequality becomes ( x^2 + (1 - x)^2 > 1 ). Let me expand that: ( x^2 + 1 - 2x + x^2 > 1 ). Combining like terms, that's ( 2x^2 - 2x + 1 > 1 ). Subtracting 1 from both sides, we get ( 2x^2 - 2x > 0 ). Factoring out 2x, it becomes ( 2x(x - 1) > 0 ).To solve this inequality, I can consider the critical points where the expression equals zero, which are ( x = 0 ) and ( x = 1 ). These points divide the real line into three intervals: ( (-infty, 0) ), ( (0, 1) ), and ( (1, infty) ). I'll test each interval.1. For ( x < 0 ), say ( x = -1 ): ( 2(-1)(-1 - 1) = 2(-1)(-2) = 4 > 0 ). So, the inequality holds here.2. For ( 0 < x < 1 ), say ( x = 0.5 ): ( 2(0.5)(0.5 - 1) = 2(0.5)(-0.5) = -0.5 < 0 ). So, the inequality does not hold here.3. For ( x > 1 ), say ( x = 2 ): ( 2(2)(2 - 1) = 2(2)(1) = 4 > 0 ). So, the inequality holds here.Therefore, for ( n = 2 ), the solutions are ( x in (-infty, 0) cup (1, infty) ). So, ( f(x) = x^2 ) is a solution with this domain.Hmm, interesting. Maybe this pattern holds for higher values of ( n ). Let me test ( n = 3 ).For ( n = 3 ), the inequality becomes ( x^3 + (1 - x)^3 > 1 ). Let me expand ( (1 - x)^3 ): ( 1 - 3x + 3x^2 - x^3 ). So, adding ( x^3 ), we get ( x^3 + 1 - 3x + 3x^2 - x^3 = 1 - 3x + 3x^2 ). Therefore, the inequality simplifies to ( 1 - 3x + 3x^2 > 1 ). Subtracting 1 from both sides, ( -3x + 3x^2 > 0 ). Factoring out 3x, we get ( 3x(x - 1) > 0 ).This is similar to the ( n = 2 ) case. The critical points are ( x = 0 ) and ( x = 1 ). Testing intervals:1. ( x < 0 ): Let ( x = -1 ): ( 3(-1)(-1 - 1) = 3(-1)(-2) = 6 > 0 ). Inequality holds.2. ( 0 < x < 1 ): Let ( x = 0.5 ): ( 3(0.5)(0.5 - 1) = 3(0.5)(-0.5) = -0.75 < 0 ). Inequality doesn't hold.3. ( x > 1 ): Let ( x = 2 ): ( 3(2)(2 - 1) = 3(2)(1) = 6 > 0 ). Inequality holds.So, for ( n = 3 ), the solution is also ( x in (-infty, 0) cup (1, infty) ). Interesting, same as ( n = 2 ).Wait, maybe this is a general pattern. Let me try ( n = 4 ).For ( n = 4 ), the inequality is ( x^4 + (1 - x)^4 > 1 ). Let me expand ( (1 - x)^4 ): ( 1 - 4x + 6x^2 - 4x^3 + x^4 ). Adding ( x^4 ), we get ( x^4 + 1 - 4x + 6x^2 - 4x^3 + x^4 = 2x^4 - 4x^3 + 6x^2 - 4x + 1 ). So, the inequality is ( 2x^4 - 4x^3 + 6x^2 - 4x + 1 > 1 ). Subtracting 1, we get ( 2x^4 - 4x^3 + 6x^2 - 4x > 0 ).Let me factor this expression. Maybe factor out a 2x: ( 2x(x^3 - 2x^2 + 3x - 2) > 0 ). Hmm, let's see if I can factor the cubic polynomial ( x^3 - 2x^2 + 3x - 2 ). Trying rational roots, possible roots are 1 and 2.Testing ( x = 1 ): ( 1 - 2 + 3 - 2 = 0 ). So, ( x = 1 ) is a root. Therefore, we can factor it as ( (x - 1)(x^2 - x + 2) ). So, the expression becomes ( 2x(x - 1)(x^2 - x + 2) > 0 ).Now, ( x^2 - x + 2 ) has discriminant ( (-1)^2 - 4(1)(2) = 1 - 8 = -7 ), which is negative. So, ( x^2 - x + 2 ) is always positive. Therefore, the sign of the expression depends on ( 2x(x - 1) ).So, ( 2x(x - 1) > 0 ). The critical points are ( x = 0 ) and ( x = 1 ). Testing intervals:1. ( x < 0 ): Let ( x = -1 ): ( 2(-1)(-1 - 1) = 2(-1)(-2) = 4 > 0 ). Inequality holds.2. ( 0 < x < 1 ): Let ( x = 0.5 ): ( 2(0.5)(0.5 - 1) = 2(0.5)(-0.5) = -0.5 < 0 ). Inequality doesn't hold.3. ( x > 1 ): Let ( x = 2 ): ( 2(2)(2 - 1) = 2(2)(1) = 4 > 0 ). Inequality holds.So, again, the solution is ( x in (-infty, 0) cup (1, infty) ). It seems like for ( n geq 2 ), the solution is the same. Wait, is this always the case? Let me think about ( n = 5 ) just to be sure.For ( n = 5 ), the inequality is ( x^5 + (1 - x)^5 > 1 ). Let's expand ( (1 - x)^5 ): ( 1 - 5x + 10x^2 - 10x^3 + 5x^4 - x^5 ). Adding ( x^5 ), we get ( x^5 + 1 - 5x + 10x^2 - 10x^3 + 5x^4 - x^5 = 1 - 5x + 10x^2 - 10x^3 + 5x^4 ). So, the inequality becomes ( 1 - 5x + 10x^2 - 10x^3 + 5x^4 > 1 ). Subtracting 1, we get ( -5x + 10x^2 - 10x^3 + 5x^4 > 0 ).Factor out a 5x: ( 5x(-1 + 2x - 2x^2 + x^3) > 0 ). Let me factor the cubic polynomial ( x^3 - 2x^2 + 2x - 1 ). Trying ( x = 1 ): ( 1 - 2 + 2 - 1 = 0 ). So, ( x = 1 ) is a root. Therefore, factor it as ( (x - 1)(x^2 - x + 1) ). So, the expression becomes ( 5x(x - 1)(x^2 - x + 1) > 0 ).Again, ( x^2 - x + 1 ) has discriminant ( (-1)^2 - 4(1)(1) = 1 - 4 = -3 ), which is negative, so it's always positive. Thus, the sign depends on ( 5x(x - 1) ).So, ( 5x(x - 1) > 0 ). Critical points at ( x = 0 ) and ( x = 1 ). Testing intervals:1. ( x < 0 ): ( x = -1 ): ( 5(-1)(-1 - 1) = 5(-1)(-2) = 10 > 0 ). Inequality holds.2. ( 0 < x < 1 ): ( x = 0.5 ): ( 5(0.5)(0.5 - 1) = 5(0.5)(-0.5) = -1.25 < 0 ). Inequality doesn't hold.3. ( x > 1 ): ( x = 2 ): ( 5(2)(2 - 1) = 5(2)(1) = 10 > 0 ). Inequality holds.Same result again. So, it seems consistent that for ( n geq 2 ), the solution is ( x in (-infty, 0) cup (1, infty) ).But wait, why is this happening? Is there a general reason why for ( n geq 2 ), the inequality ( x^n + (1 - x)^n > 1 ) holds only when ( x < 0 ) or ( x > 1 )?Let me think about the behavior of the function ( f(x) = x^n + (1 - x)^n ). For ( n geq 2 ), both ( x^n ) and ( (1 - x)^n ) are convex functions. Their sum is also convex. The minimum of this function occurs at some point between 0 and 1.Wait, actually, let's consider the function ( f(x) = x^n + (1 - x)^n ). Let's compute its derivative to find critical points.( f'(x) = n x^{n-1} - n (1 - x)^{n-1} ).Setting ( f'(x) = 0 ):( n x^{n-1} = n (1 - x)^{n-1} )Divide both sides by ( n ):( x^{n-1} = (1 - x)^{n-1} )Taking the (n-1)th root (since n-1 is positive):( x = 1 - x )So, ( 2x = 1 ) => ( x = 0.5 ).Therefore, the function ( f(x) ) has a critical point at ( x = 0.5 ). Let's check the second derivative to see if it's a minimum or maximum.( f''(x) = n(n - 1)x^{n-2} + n(n - 1)(1 - x)^{n-2} ).Since ( n geq 2 ), ( f''(x) ) is positive for all ( x ) except possibly at the endpoints. Therefore, ( x = 0.5 ) is a point of local minimum.So, the minimum value of ( f(x) ) occurs at ( x = 0.5 ). Let's compute ( f(0.5) ):( f(0.5) = (0.5)^n + (0.5)^n = 2 times (0.5)^n = 2^{1 - n} ).So, the minimum value is ( 2^{1 - n} ). For ( n geq 2 ), ( 2^{1 - n} leq 0.5 ). Therefore, the function ( f(x) ) reaches its minimum at ( x = 0.5 ) with a value less than or equal to 0.5, which is less than 1. But wait, our inequality is ( f(x) > 1 ). So, even though the function has a minimum below 1, we need to find where it exceeds 1.Since ( f(x) ) is convex and has a minimum at ( x = 0.5 ), it will tend to infinity as ( x ) approaches ( pm infty ). Therefore, the function will cross the value 1 at some points on either side of the minimum.But in our earlier cases, for ( n = 2, 3, 4, 5 ), the solutions were ( x in (-infty, 0) cup (1, infty) ). So, maybe the function ( f(x) = x^n + (1 - x)^n ) is greater than 1 only when ( x ) is outside the interval [0, 1].Wait, let me test ( x = 2 ) for ( n = 2 ): ( 2^2 + (1 - 2)^2 = 4 + 1 = 5 > 1 ). Similarly, ( x = -1 ): ( (-1)^2 + (1 - (-1))^2 = 1 + 4 = 5 > 1 ). So, yes, outside [0, 1], the function exceeds 1.But what about ( x = 0.5 ): ( (0.5)^2 + (0.5)^2 = 0.25 + 0.25 = 0.5 < 1 ). So, the function is below 1 in [0, 1].Therefore, for ( n geq 2 ), ( f(x) + f(1 - x) > 1 ) holds when ( x ) is outside the interval [0, 1], i.e., ( x in (-infty, 0) cup (1, infty) ).But wait, is this always true? Let me think about ( n = 0 ). Wait, ( n ) is a positive integer, so ( n geq 1 ). We already saw that for ( n = 1 ), it doesn't hold anywhere, and for ( n geq 2 ), it holds outside [0, 1].Is there a way to generalize this without testing each ( n )?Let me consider the function ( f(x) = x^n + (1 - x)^n ). For ( x ) in [0, 1], both ( x ) and ( 1 - x ) are between 0 and 1. Since ( n geq 2 ), raising them to the power ( n ) will make them smaller or equal to 1. Therefore, ( x^n leq x ) and ( (1 - x)^n leq 1 - x ) for ( x in [0, 1] ). So, ( f(x) = x^n + (1 - x)^n leq x + (1 - x) = 1 ). Hence, ( f(x) leq 1 ) for ( x in [0, 1] ).But our inequality is ( f(x) > 1 ). Therefore, the inequality cannot hold for any ( x ) in [0, 1]. So, the solutions must lie outside [0, 1], i.e., ( x in (-infty, 0) cup (1, infty) ).Moreover, for ( x > 1 ), ( x^n ) grows rapidly, and ( (1 - x)^n = (x - 1)^n ) also grows, so their sum will definitely exceed 1. Similarly, for ( x < 0 ), ( x^n ) (if ( n ) is even) is positive and large, and ( (1 - x)^n ) is also positive and large, so their sum exceeds 1.Wait, but what if ( n ) is odd? For ( x < 0 ), ( x^n ) is negative if ( n ) is odd. However, ( (1 - x)^n ) when ( x < 0 ) becomes ( (1 - x)^n = (1 + |x|)^n ), which is positive and large. So, even if ( x^n ) is negative, ( (1 - x)^n ) is positive and dominates, making the sum positive and greater than 1.Let me test ( n = 3 ) with ( x = -1 ): ( (-1)^3 + (1 - (-1))^3 = -1 + 8 = 7 > 1 ). So, yes, even for odd ( n ), the sum is greater than 1 when ( x < 0 ).Therefore, regardless of whether ( n ) is even or odd (as long as ( n geq 2 )), the inequality ( x^n + (1 - x)^n > 1 ) holds for ( x in (-infty, 0) cup (1, infty) ).So, putting it all together:- For ( n = 1 ), there are no solutions.- For ( n geq 2 ), the solutions are ( x in (-infty, 0) cup (1, infty) ).Therefore, ( f(x) = x^n ) is a solution to the inequality ( f(x) + f(1 - x) > 1 ) if and only if ( n geq 2 ), and the domain ( D ) is ( (-infty, 0) cup (1, infty) ).I think that covers all cases. I don't see any mistakes in my reasoning, but let me just recap:1. For ( n = 1 ), the inequality simplifies to ( 1 > 1 ), which is false.2. For ( n geq 2 ), the function ( f(x) = x^n + (1 - x)^n ) has a minimum at ( x = 0.5 ) with value ( 2^{1 - n} leq 0.5 ), which is less than 1. However, as ( x ) moves away from [0, 1], the function grows without bound, thus exceeding 1. Therefore, the solutions are outside [0, 1].Yes, that makes sense. So, the final answer is that ( f(x) = x^n ) is a solution for ( n geq 2 ) with domain ( D = (-infty, 0) cup (1, infty) ).

answer：Okay, so I have this problem where I need to find the remainder when a certain number n is divided by 17. The number n is defined as the sum of the products of all subsets of size 25 from the set O, which consists of odd numbers between 0 and 100. Hmm, that sounds a bit complicated, but let me try to break it down step by step.First, let me understand the set O. It's the set of odd numbers between 0 and 100. So, that would be 1, 3, 5, ..., up to 99. Let me count how many numbers are in O. Since 100 divided by 2 is 50, there are 50 odd numbers between 0 and 100. So, O has 50 elements.Next, T is the set of all subsets of O that have exactly 25 elements. So, T is a collection of all possible ways to choose 25 odd numbers from the 50 available. For each subset S in T, P(S) is the product of the elements in S. Then, n is the sum of all these products. So, n is like adding up the products of every possible combination of 25 odd numbers from O.Now, I need to find the remainder when n is divided by 17. That is, compute n mod 17. Since n is a huge number, directly computing it isn't feasible, so I need a smarter approach.I remember that in combinatorics, sums of products of subsets can be related to polynomial coefficients. Specifically, if I consider the polynomial whose roots are the elements of O, then the sum of the products of all subsets of size k is the coefficient of x^{50 - k} in the expanded polynomial. Wait, let me think about that again.Actually, if I have a polynomial where each term is (x + a_i) for each element a_i in O, then expanding this polynomial will give me terms where the coefficient of x^{50 - k} is the sum of all products of k elements from O. So, in this case, since I need the sum of products of 25 elements, I need the coefficient of x^{25} in the expansion of the product (x + 1)(x + 3)(x + 5)...(x + 99).But wait, in the problem, P(S) is the product of the elements, not the product of (x + a_i). So, actually, if I set x = 1, then the product (1 + a_i) would give me the sum of all subsets. But in this case, I need the sum of products of subsets of size exactly 25. So, maybe I need to consider the generating function where each term is (1 + a_i x), and then the coefficient of x^{25} would be the sum of products of 25 elements.Yes, that makes sense. So, let me define the generating function as:G(x) = (1 + 1x)(1 + 3x)(1 + 5x)...(1 + 99x)Then, the coefficient of x^{25} in G(x) is exactly n, which is the sum we're looking for. So, n is the coefficient of x^{25} in G(x).Now, I need to compute this coefficient modulo 17. Computing such a large polynomial directly is not practical, so I need to find a way to simplify this modulo 17.First, let's note that 17 is a prime number, which might be helpful. Also, since we're working modulo 17, we can reduce each coefficient modulo 17 as we go along.But before that, let me see if there's a pattern or a property that can help me simplify G(x) modulo 17.Looking at the elements of O, which are 1, 3, 5, ..., 99, I can note that modulo 17, these numbers repeat every 17*2 = 34 numbers because 17 is prime and 2 is the step between consecutive odd numbers. Wait, actually, the odd numbers modulo 17 cycle every 17 numbers, but since we're dealing with odd numbers, the cycle would be every 17*2 = 34 numbers.But let me check: 17 is prime, so the multiplicative group modulo 17 has order 16. Hmm, not sure if that's directly helpful.Alternatively, perhaps I can pair the terms in G(x) in a way that simplifies the product modulo 17.Wait, another approach: Since we're dealing with modulo 17, and the generating function is a product of linear terms, perhaps I can consider the product modulo 17 and see if it can be expressed in a simpler form.Let me consider the set O modulo 17. Each element of O is an odd number between 1 and 99, so modulo 17, they can be represented as 1, 3, 5, ..., 15, 0, 2, 4, ..., 16, etc., but since we're only considering odd numbers, modulo 17, they will cover all residues from 1 to 16, but only the odd ones.Wait, 17 is odd, so the residues modulo 17 for odd numbers would be 1, 3, 5, 7, 9, 11, 13, 15, 0, 2, 4, 6, 8, 10, 12, 14, 16. But since we're only considering odd numbers, the residues would be 1, 3, 5, 7, 9, 11, 13, 15, and then 17 is equivalent to 0, but 17 is not in O since O goes up to 99. Wait, actually, 17 is included in O because 17 is odd and between 0 and 100. So, 17 is in O, and 17 modulo 17 is 0.So, in O, modulo 17, we have the residues: 1, 3, 5, 7, 9, 11, 13, 15, 0, 2, 4, 6, 8, 10, 12, 14, 16, and then repeats.But since O has 50 elements, which is more than 17, the residues modulo 17 will repeat multiple times.Wait, let me think differently. Since we're working modulo 17, perhaps the generating function can be expressed as a product over residues modulo 17, each raised to the number of times they appear in O.But that might complicate things. Alternatively, perhaps I can note that the product (1 + a_i x) modulo 17 can be simplified by grouping terms with the same residue modulo 17.But let me first see how many times each residue appears in O modulo 17.Since O has 50 elements, and 50 divided by 17 is approximately 2 with a remainder of 16. So, each residue from 1 to 16 modulo 17 will appear either 3 or 2 times in O.Wait, let me calculate it precisely.The number of elements in O is 50. Since 50 = 2*17 + 16, so each residue from 1 to 16 modulo 17 will appear 3 times for the first 16 residues and 2 times for the remaining 16 residues? Wait, that doesn't add up.Wait, actually, 50 divided by 17 is 2 with a remainder of 16. So, each residue from 1 to 16 will appear 2 times, and then the first 16 residues will appear one more time, making it 3 times for residues 1 to 16.Wait, let me check: 17 residues (0 to 16), but since we're only considering odd numbers, the residues modulo 17 will be the odd residues: 1, 3, 5, 7, 9, 11, 13, 15, and then 0, 2, 4, 6, 8, 10, 12, 14, 16. Wait, no, 0 is even? No, 0 is even, but in our case, 17 is included, which is odd, so 17 modulo 17 is 0, which is even? Wait, no, 0 is neither even nor odd, but in our case, 17 is odd, so 17 is included in O, and 17 modulo 17 is 0.Wait, perhaps I'm overcomplicating. Let me list the residues of O modulo 17.Each element in O is an odd number from 1 to 99. So, modulo 17, these numbers will cycle through 1, 3, 5, 7, 9, 11, 13, 15, 0, 2, 4, 6, 8, 10, 12, 14, 16, and then repeat.Since 17 is the modulus, each residue from 0 to 16 will appear approximately 50 / 17 ≈ 2.94 times. So, some residues will appear 3 times, and others 2 times.Specifically, since 17*2 = 34, and 50 - 34 = 16, so the first 16 residues will appear 3 times, and the remaining 1 residue (which is 17, but 17 is 0 modulo 17) will appear 2 times.Wait, but 17 is included in O, so 0 modulo 17 appears once (from 17), and then 17 + 34 = 51, which is also in O, so 51 modulo 17 is 0. Similarly, 85 is in O, so 85 modulo 17 is 0. So, 0 modulo 17 appears 3 times: 17, 51, 85.Wait, so 0 modulo 17 appears 3 times, and the other residues from 1 to 16 modulo 17 will appear either 3 or 2 times.Wait, let me count:From 1 to 99, stepping by 2, we have 50 numbers.Each residue modulo 17 repeats every 34 numbers (since 17*2=34). So, in 50 numbers, how many full cycles of 34 are there? 50 / 34 ≈ 1.47, so one full cycle and a partial cycle.In each full cycle of 34 numbers, each residue modulo 17 appears exactly 2 times (since 34 / 17 = 2). Then, in the partial cycle of 16 numbers (50 - 34 = 16), each residue from 1 to 16 modulo 17 will appear once more.Therefore, residues 1 to 16 modulo 17 will appear 2 + 1 = 3 times, and residue 0 modulo 17 will appear 2 times (from 17 and 51) plus one more time from 85, making it 3 times as well.Wait, but 17, 51, 85 are all in O, so 0 modulo 17 appears 3 times.So, in total, each residue from 0 to 16 modulo 17 appears 3 times in O.Wait, but 17*3 = 51, which is more than 50. Hmm, maybe I made a mistake.Wait, 50 elements in O. Each residue modulo 17 appears floor(50 / 17) = 2 times, with a remainder of 50 - 2*17 = 16. So, the first 16 residues will have an extra occurrence, making them 3 times, and the remaining residue (which is 17, i.e., 0 modulo 17) will have 2 times.But wait, 0 modulo 17 is already included in the residues, so perhaps the first 16 residues (1 to 16) will appear 3 times, and residue 0 will appear 2 times.But in O, 17 is included, which is 0 modulo 17, and 51 and 85 are also included, which are also 0 modulo 17. So, 0 modulo 17 appears 3 times.Wait, so maybe each residue from 0 to 16 appears 3 times? But 17*3 = 51, which is more than 50. So, that can't be.Wait, perhaps I need to think differently. Let me list the numbers in O and their residues modulo 17.O = {1, 3, 5, 7, 9, 11, 13, 15, 17, 19, 21, 23, 25, 27, 29, 31, 33, 35, 37, 39, 41, 43, 45, 47, 49, 51, 53, 55, 57, 59, 61, 63, 65, 67, 69, 71, 73, 75, 77, 79, 81, 83, 85, 87, 89, 91, 93, 95, 97, 99}Now, let's compute each number modulo 17:1 mod17=13 mod17=35 mod17=57 mod17=79 mod17=911 mod17=1113 mod17=1315 mod17=1517 mod17=019 mod17=221 mod17=423 mod17=625 mod17=827 mod17=1029 mod17=1231 mod17=1433 mod17=1635 mod17=1 (35-34=1)37 mod17=3 (37-34=3)39 mod17=5 (39-34=5)41 mod17=7 (41-34=7)43 mod17=9 (43-34=9)45 mod17=11 (45-34=11)47 mod17=13 (47-34=13)49 mod17=15 (49-34=15)51 mod17=0 (51-51=0)53 mod17=2 (53-51=2)55 mod17=4 (55-51=4)57 mod17=6 (57-51=6)59 mod17=8 (59-51=8)61 mod17=10 (61-51=10)63 mod17=12 (63-51=12)65 mod17=14 (65-51=14)67 mod17=16 (67-51=16)69 mod17=1 (69-68=1)71 mod17=3 (71-68=3)73 mod17=5 (73-68=5)75 mod17=7 (75-68=7)77 mod17=9 (77-68=9)79 mod17=11 (79-68=11)81 mod17=13 (81-68=13)83 mod17=15 (83-68=15)85 mod17=0 (85-85=0)87 mod17=2 (87-85=2)89 mod17=4 (89-85=4)91 mod17=6 (91-85=6)93 mod17=8 (93-85=8)95 mod17=10 (95-85=10)97 mod17=12 (97-85=12)99 mod17=14 (99-85=14)So, now let's count how many times each residue appears:Residue 0: 17, 51, 85 → 3 timesResidue 1: 1, 35, 69 → 3 timesResidue 2: 19, 53, 87 → 3 timesResidue 3: 3, 37, 71 → 3 timesResidue 4: 21, 55, 89 → 3 timesResidue 5: 5, 39, 73 → 3 timesResidue 6: 23, 57, 91 → 3 timesResidue 7: 7, 41, 75 → 3 timesResidue 8: 25, 59, 93 → 3 timesResidue 9: 9, 43, 77 → 3 timesResidue 10: 27, 61, 95 → 3 timesResidue 11: 11, 45, 79 → 3 timesResidue 12: 13, 63, 97 → 3 timesResidue 13: 15, 47, 81 → 3 timesResidue 14: 17 is 0, but 31, 65, 99: 31, 65, 99 → 3 timesWait, 31 mod17=14, 65 mod17=14, 99 mod17=14. So, residue 14 appears 3 times.Residue 15: 15, 49, 83 → 3 timesResidue 16: 33, 67 → Wait, 33 mod17=16, 67 mod17=16, and 91 mod17=6, so only 33 and 67. Wait, 33, 67, and 91? No, 91 mod17=6, so only 33 and 67. Wait, but earlier I thought residue 16 appears 3 times, but according to this, it's only 2 times.Wait, let me recount:Residue 16: 33, 67, and 101? But 101 is beyond 99, so only 33 and 67. So, residue 16 appears 2 times.Wait, but earlier I thought each residue appears 3 times, but residue 16 appears only 2 times.Wait, let me check the list again:Looking at the residues:- 0: 3 times- 1: 3 times- 2: 3 times- 3: 3 times- 4: 3 times- 5: 3 times- 6: 3 times- 7: 3 times- 8: 3 times- 9: 3 times- 10: 3 times- 11: 3 times- 12: 3 times- 13: 3 times- 14: 3 times- 15: 3 times- 16: 2 timesWait, so residue 16 appears only 2 times. That's because 17*3=51, and 50 elements, so one residue is short by one.So, in total, residues 0 to 15 appear 3 times each, and residue 16 appears 2 times.Wait, but in our list above, residue 16 is 33, 67, and 101, but 101 is not in O, so only 33 and 67, which are 2 times.Therefore, in G(x), the generating function, we can write it as:G(x) = (1 + 1x)^3 * (1 + 3x)^3 * (1 + 5x)^3 * ... * (1 + 15x)^3 * (1 + 16x)^2But wait, actually, each residue from 0 to 16 is represented by (1 + a_i x), where a_i is the residue. But since residue 0 appears 3 times, it would be (1 + 0x)^3 = 1^3 = 1. So, it doesn't affect the product.Similarly, residues 1 to 15 appear 3 times, and residue 16 appears 2 times.Wait, but in our case, the generating function is the product over all elements in O, so each element is represented as (1 + a_i x). Therefore, modulo 17, each (1 + a_i x) can be considered as (1 + (a_i mod17) x).Therefore, G(x) modulo 17 is:G(x) ≡ [ (1 + 1x)^3 * (1 + 3x)^3 * (1 + 5x)^3 * ... * (1 + 15x)^3 * (1 + 16x)^2 ] mod17But since 16 ≡ -1 mod17, we can write (1 + 16x) ≡ (1 - x) mod17.So, G(x) ≡ [ (1 + x)^3 * (1 + 3x)^3 * (1 + 5x)^3 * ... * (1 + 15x)^3 * (1 - x)^2 ] mod17Now, that's a bit complicated, but maybe we can find a pattern or a way to simplify this product.Alternatively, perhaps we can note that the product (1 + x)(1 + 3x)...(1 + 15x) modulo 17 is related to the polynomial (x^16 - 1) due to properties of finite fields.Wait, let me recall that in the field GF(17), the multiplicative group is cyclic of order 16. So, the product of (x - a) for all non-zero a in GF(17) is x^16 - 1.But in our case, we have (1 + a_i x) terms, which are similar but not exactly the same.Wait, perhaps I can make a substitution. Let me set y = x, then each term (1 + a_i x) can be written as (1 + a_i y). If I consider the product over all a_i in GF(17)* (the multiplicative group), then the product would be (1 + y)^{16} - 1, but I'm not sure.Wait, actually, in GF(17), the product of (1 + a_i y) for all a_i in GF(17)* is equal to (y^{16} - 1), because the product of (x - a) for all a in GF(17)* is x^{16} - 1. But in our case, it's (1 + a_i y), which is similar to (y - (-1/a_i)).Wait, maybe I'm overcomplicating. Let me think differently.Since we have the product of (1 + a_i x) for a_i in O, and modulo 17, each a_i is equivalent to some residue from 0 to 16. But since 0 doesn't contribute anything (as (1 + 0x) = 1), we can ignore it.So, the product becomes:G(x) ≡ [ (1 + x)^3 * (1 + 3x)^3 * (1 + 5x)^3 * ... * (1 + 15x)^3 * (1 - x)^2 ] mod17Now, let's note that 3, 5, ..., 15 are all the odd residues modulo 17 except 1 and 16 (which is -1). So, the product can be written as:G(x) ≡ (1 + x)^3 * (1 - x)^2 * [ product_{k=1}^{8} (1 + (2k+1)x)^3 ] mod17Wait, but 2k+1 for k=1 to 8 gives 3,5,...,17, but 17 is 0, so maybe not.Alternatively, perhaps I can pair terms to simplify.Wait, another idea: Since the multiplicative group modulo 17 is cyclic of order 16, and the product of all non-zero elements is -1 (Wilson's theorem). But I'm not sure how that helps here.Alternatively, perhaps I can consider that the product of (1 + a_i x) for a_i in GF(17)* is equal to (1 - x^{16}) / (1 - x), but I'm not sure.Wait, actually, in GF(17)[x], the product of (x - a) for all a in GF(17)* is x^{16} - 1. So, if I set x = 1/y, then the product becomes (1/y)^{16} - 1 = (1 - y^{16}) / y^{16}. Hmm, not sure.Alternatively, perhaps I can consider the generating function as:G(x) ≡ [ product_{a in GF(17)*} (1 + a x)^{3} ] * (1 - x)^2 mod17But since the product over a in GF(17)* of (1 + a x) is equal to (1 - x^{16}) / (1 - x), because:product_{a in GF(17)*} (x - a) = x^{16} - 1So, replacing x with -1/x, we get:product_{a in GF(17)*} (-1/x - a) = (-1/x)^{16} - 1 = (1/x^{16}) - 1But that's equal to product_{a in GF(17)*} (-1 - a x)/x = (-1)^{16} product_{a in GF(17)*} (1 + a x) / x^{16} = product_{a in GF(17)*} (1 + a x) / x^{16} = (1/x^{16} - 1)Therefore, product_{a in GF(17)*} (1 + a x) = (1 - x^{16}) / (1 - x)Wait, that seems promising.So, product_{a in GF(17)*} (1 + a x) = (1 - x^{16}) / (1 - x)Therefore, [ product_{a in GF(17)*} (1 + a x) ]^3 = [ (1 - x^{16}) / (1 - x) ]^3So, G(x) ≡ [ (1 - x^{16})^3 / (1 - x)^3 ] * (1 - x)^2 mod17Simplifying, G(x) ≡ (1 - x^{16})^3 / (1 - x) mod17Which is equal to (1 - x^{16})^3 * (1 - x)^{-1} mod17Now, (1 - x)^{-1} is the generating function for the geometric series: 1 + x + x^2 + x^3 + ... So, modulo 17, it's 1 + x + x^2 + ... + x^{16} + x^{17} + ..., but since we're working modulo 17, higher powers can be reduced.But perhaps we can write (1 - x^{16})^3 as 1 - 3x^{16} + 3x^{32} - x^{48}So, G(x) ≡ (1 - 3x^{16} + 3x^{32} - x^{48}) * (1 + x + x^2 + ... + x^{16} + x^{17} + ...) mod17But since we're looking for the coefficient of x^{25}, we can ignore terms beyond x^{25}.So, let's write:G(x) ≡ (1 - 3x^{16} + 3x^{32} - x^{48}) * (1 + x + x^2 + ... + x^{25} + ...) mod17But since we're only interested in the coefficient of x^{25}, we can truncate the series at x^{25}.So, let's denote S = 1 + x + x^2 + ... + x^{25}Then, G(x) ≡ (1 - 3x^{16} + 3x^{32} - x^{48}) * S mod17But x^{32} and x^{48} are beyond x^{25}, so when multiplied by S, their contributions to x^{25} would come from terms where x^{32} * x^{-7} and x^{48} * x^{-23}, but since S only goes up to x^{25}, those terms would not contribute. Therefore, we can ignore the x^{32} and x^{48} terms when considering the coefficient of x^{25}.Thus, G(x) ≡ (1 - 3x^{16}) * S mod17So, the coefficient of x^{25} in G(x) is equal to the coefficient of x^{25} in (1 - 3x^{16}) * S.Which is equal to the coefficient of x^{25} in S minus 3 times the coefficient of x^{9} in S.Because when you multiply by (1 - 3x^{16}), the coefficient of x^{25} comes from 1 * x^{25} and -3x^{16} * x^{9}.The coefficient of x^{25} in S is 1 (since S is 1 + x + x^2 + ... + x^{25}).The coefficient of x^{9} in S is also 1.Therefore, the coefficient of x^{25} in G(x) is 1 - 3*1 = -2.But we're working modulo 17, so -2 ≡ 15 mod17.Therefore, the remainder when n is divided by 17 is 15.Wait, let me double-check that.We have G(x) ≡ (1 - 3x^{16}) * S mod17Coefficient of x^{25} is 1 (from 1 * x^{25}) minus 3*(coefficient of x^{9} in S), which is 3*1=3.So, 1 - 3 = -2 ≡ 15 mod17.Yes, that seems correct.So, the final answer is 15.

answer：Okay, so I have this logarithmic equation: log p plus 2 log q equals log of (2p plus q). Hmm, I need to find p in terms of q. Alright, let's break this down step by step.First, I remember some logarithmic properties. One of them is that if you have log a plus log b, it's equal to log(ab). Also, coefficients in front of the log can be turned into exponents. So, 2 log q is the same as log(q squared). Let me write that down:log p + 2 log q = log(2p + q)So, that becomes:log p + log(q²) = log(2p + q)Now, using the property that log a + log b = log(ab), I can combine the left side:log(p * q²) = log(2p + q)Since the logs are equal, their arguments must be equal too. So, I can set the inside parts equal:p * q² = 2p + qAlright, now I have an equation without logs. Let me rearrange this to solve for p. I'll bring all the terms with p to one side:p * q² - 2p = qFactor out p from the left side:p(q² - 2) = qNow, to solve for p, I'll divide both sides by (q² - 2):p = q / (q² - 2)Hmm, let me check if this makes sense. If I plug this back into the original equation, does it hold true?Let me substitute p = q / (q² - 2) into the left side of the original equation:log(p) + 2 log(q) = log(q / (q² - 2)) + 2 log(q)That's equal to log(q) - log(q² - 2) + 2 log(q) = 3 log(q) - log(q² - 2)On the right side, log(2p + q):2p + q = 2*(q / (q² - 2)) + q = (2q)/(q² - 2) + q = (2q + q(q² - 2))/(q² - 2) = (2q + q³ - 2q)/(q² - 2) = q³/(q² - 2)So, log(2p + q) = log(q³/(q² - 2)) = log(q³) - log(q² - 2) = 3 log(q) - log(q² - 2)Which matches the left side. So, it seems correct.I should also consider the domain of the original logarithmic equation. The arguments of the logs must be positive. So, p > 0, q > 0, and 2p + q > 0. Since p is expressed in terms of q, I need to ensure that p is positive. So, q / (q² - 2) > 0.This means that q and (q² - 2) must have the same sign. Since q > 0, then (q² - 2) must also be positive. So, q² - 2 > 0, which implies q² > 2, so q > sqrt(2). So, q must be greater than sqrt(2) for p to be positive.But the problem doesn't specify any constraints on q, so I guess we just go with the solution as long as q > sqrt(2). Anyway, the question is just asking for p in terms of q, so p = q / (q² - 2) is the answer.Looking at the options, that's option B.

answer：To determine the greatest whole number less than 200 with an odd number of positive factors, I start by recalling that a number has an odd number of factors if and only if it is a perfect square. This is because factors come in pairs, but in the case of perfect squares, one of the factors is repeated, resulting in an odd count.Next, I identify the largest perfect square under 200. Calculating the squares of consecutive integers:- (13^2 = 169)- (14^2 = 196)- (15^2 = 225) (which exceeds 200)Since 196 is the largest perfect square under 200, it is the number with an odd number of positive factors that meets the criteria.

answer：Alright, so I've got this geometry problem here, and I need to figure out which statement is incorrect. Let's see, the options are A, B, C, and D. I'll go through each one step by step to understand what they're saying and whether they're correct or not.Starting with **Statement A**: "Two parallel lines have equal alternate interior angles." Hmm, okay, I remember from my geometry class that when two parallel lines are cut by a transversal, the alternate interior angles are equal. So, if I imagine two parallel lines and a transversal crossing them, the angles that are on opposite sides of the transversal but inside the two lines are equal. Yeah, that sounds right. So, Statement A seems correct.Moving on to **Statement B**: "Two parallel lines have equal corresponding angles." Alright, corresponding angles are the angles that are in the same position relative to the transversal and the two lines. I think they're also equal when the lines are parallel. Wait, isn't that similar to alternate interior angles? Or is it different? Let me visualize it. If I have two parallel lines and a transversal, the corresponding angles are like the ones in the same corner at each intersection. Yeah, those should be equal too. So, Statement B also seems correct.Next is **Statement C**: "Equal corresponding angles imply two parallel lines." Okay, this is kind of the converse of Statement B. If the corresponding angles are equal, does that mean the lines are parallel? I think that's one of the theorems we learned. If two lines are cut by a transversal and the corresponding angles are equal, then the lines must be parallel. Yeah, that makes sense. So, Statement C is correct as well.Finally, **Statement D**: "Two lines parallel to the same line are parallel." Hmm, this is about the transitive property of parallel lines, right? If line A is parallel to line B, and line C is parallel to line B, then line A should be parallel to line C. I think that's true in Euclidean geometry. So, Statement D seems correct too.Wait a minute, all four statements seem correct based on what I remember. But the question is asking which one is incorrect. Did I miss something? Maybe I need to think more carefully about each statement.Let's revisit **Statement B**: "Two parallel lines have equal corresponding angles." I thought this was correct because corresponding angles are equal when lines are parallel. But maybe there's a catch here. Are there situations where corresponding angles aren't equal even if the lines are parallel? Or perhaps the statement is too broad?Thinking about it, corresponding angles are defined when two lines are cut by a transversal. If the lines are parallel, then the corresponding angles are equal. But if the lines aren't parallel, the corresponding angles aren't equal. So, the statement is correct in the context where the lines are parallel. Maybe the issue is with the wording? It says "two parallel lines have equal corresponding angles," which is true, but perhaps it's not the most precise way to state it.Alternatively, maybe the problem is with **Statement D**: "Two lines parallel to the same line are parallel." I thought this was true, but is there a scenario where this isn't the case? In Euclidean geometry, yes, two lines parallel to the same line are parallel to each other. But in non-Euclidean geometries, like spherical or hyperbolic, this might not hold. However, I think the question is assuming Euclidean geometry since it's the standard unless specified otherwise.Hmm, I'm still not sure. Maybe I need to check an example. Let's say we have three lines: line A, line B, and line C. If line A is parallel to line B, and line C is also parallel to line B, then line A should be parallel to line C. That seems to hold true. So, Statement D is correct.Going back to **Statement B**, maybe the issue is that it's not specifying that the lines are cut by a transversal. The statement just says "two parallel lines have equal corresponding angles," but corresponding angles are defined in the context of a transversal. So, without mentioning the transversal, the statement might be incomplete or misleading. That could be the reason why it's considered incorrect.Alternatively, perhaps the problem is that corresponding angles are not necessarily equal unless the lines are parallel. But the statement says "two parallel lines have equal corresponding angles," which is correct. So, I'm still confused about why Statement B would be incorrect.Wait, maybe the issue is with the term "corresponding angles." In some contexts, corresponding angles can refer to angles in similar triangles, not necessarily in the context of parallel lines. If that's the case, then the statement might be mixing concepts. But I think in the context of parallel lines and transversals, corresponding angles are well-defined.I'm going in circles here. Let me try to summarize:- **Statement A**: Correct.- **Statement B**: Seems correct, but maybe incomplete or misleading.- **Statement C**: Correct.- **Statement D**: Correct.If all statements are correct, then the question is flawed. But since the question asks for the incorrect statement, I must have made a mistake in my reasoning. Perhaps Statement B is incorrect because it doesn't specify that the corresponding angles are formed by a transversal. Without that, the statement is too vague or incorrect.Alternatively, maybe Statement B is incorrect because corresponding angles are not necessarily equal unless the lines are parallel, but the statement says "two parallel lines have equal corresponding angles," which is correct. So, I'm still unsure.Given that, I'll go with the initial thought that Statement B might be considered incorrect due to incomplete wording, even though the content is correct.

answer：Okay, so I need to find all quadratic polynomials ( P(x) ) such that for all ( x, y geq 0 ), the inequality ( P(x+y) geq P(x) + P(y) ) holds. Hmm, quadratic polynomials, so they have the form ( P(x) = ax^2 + bx + c ), right? Let me write that down.So, ( P(x) = ax^2 + bx + c ). Now, the inequality given is ( P(x+y) geq P(x) + P(y) ) for all ( x, y geq 0 ). Let me substitute ( x+y ) into the polynomial.Calculating ( P(x+y) ):[P(x+y) = a(x+y)^2 + b(x+y) + c = a(x^2 + 2xy + y^2) + b(x + y) + c = ax^2 + 2axy + ay^2 + bx + by + c]Now, let's compute ( P(x) + P(y) ):[P(x) + P(y) = (ax^2 + bx + c) + (ay^2 + by + c) = ax^2 + ay^2 + bx + by + 2c]So, the inequality ( P(x+y) geq P(x) + P(y) ) becomes:[ax^2 + 2axy + ay^2 + bx + by + c geq ax^2 + ay^2 + bx + by + 2c]Let me subtract ( ax^2 + ay^2 + bx + by ) from both sides to simplify:[2axy + c geq 2c]Which simplifies to:[2axy geq c]Wait, so ( 2axy geq c ) for all ( x, y geq 0 ). Hmm, this seems a bit tricky. Let me think about this. Since ( x ) and ( y ) can be any non-negative real numbers, including very small ones, how can ( 2axy ) always be greater than or equal to ( c )?If ( c ) is positive, then for very small ( x ) and ( y ), ( 2axy ) would be very small, possibly less than ( c ), which would violate the inequality. So, maybe ( c ) has to be non-positive? Let me test that.If ( c leq 0 ), then ( 2axy geq c ) because ( 2axy ) is non-negative (since ( a ) is a coefficient of ( x^2 ), which for a quadratic polynomial, ( a ) can be positive or negative, but let's see) and ( c ) is non-positive. So, ( 2axy geq c ) would hold because a non-negative number is always greater than or equal to a non-positive number.But wait, is that the only condition? Let me check if ( a ) has any restrictions. If ( a ) is positive, then ( 2axy ) is positive, which is good because ( c ) is non-positive. If ( a ) is negative, then ( 2axy ) is negative, which might not necessarily satisfy ( 2axy geq c ) because ( c ) is also non-positive. Hmm, that might complicate things.Wait, let me go back. The original polynomial is quadratic, so ( a ) cannot be zero. So, ( a ) is either positive or negative. Let me consider both cases.Case 1: ( a > 0 ).In this case, ( 2axy ) is non-negative because ( x ) and ( y ) are non-negative. So, ( 2axy geq c ) must hold for all ( x, y geq 0 ). Since ( c ) is a constant, and ( 2axy ) can be made as small as we like by choosing small ( x ) and ( y ), the only way for ( 2axy geq c ) to hold for all ( x, y geq 0 ) is if ( c leq 0 ). Because if ( c ) were positive, then for sufficiently small ( x ) and ( y ), ( 2axy ) would be less than ( c ), violating the inequality.Case 2: ( a < 0 ).Here, ( 2axy ) is non-positive because ( a ) is negative and ( x, y ) are non-negative. So, ( 2axy geq c ) becomes a non-positive number being greater than or equal to ( c ). If ( c ) is positive, then ( 2axy ) (which is non-positive) cannot be greater than or equal to a positive ( c ). So, ( c ) must be non-positive as well.But wait, even if ( c ) is non-positive, ( 2axy ) is non-positive, so we have a non-positive number on the left and a non-positive number on the right. So, the inequality ( 2axy geq c ) would require that ( 2axy ) is greater than or equal to ( c ). But since both are non-positive, this would mean that ( 2axy ) is less negative than ( c ), or ( c ) is more negative than ( 2axy ). But since ( x ) and ( y ) can be any non-negative numbers, including very large ones, ( 2axy ) can become very negative (since ( a ) is negative). So, ( 2axy ) can be less than ( c ) if ( c ) is not sufficiently negative.Wait, this is getting complicated. Let me think again.If ( a < 0 ), then ( 2axy ) can be made arbitrarily negative by choosing large ( x ) and ( y ). So, unless ( c ) is also negative infinity, which is impossible, the inequality ( 2axy geq c ) cannot hold for all ( x, y geq 0 ). Therefore, ( a ) cannot be negative because it would violate the inequality for large ( x ) and ( y ).Therefore, ( a ) must be positive. So, ( a > 0 ) and ( c leq 0 ).But wait, let me check if there are any other conditions. The polynomial is quadratic, so ( a neq 0 ). We've established ( a > 0 ) and ( c leq 0 ). Is that all?Wait, let's go back to the inequality ( 2axy geq c ). Since ( a > 0 ) and ( c leq 0 ), ( 2axy ) is non-negative and ( c ) is non-positive. So, ( 2axy geq c ) is always true because a non-negative number is always greater than or equal to a non-positive number. So, as long as ( a > 0 ) and ( c leq 0 ), the inequality holds for all ( x, y geq 0 ).But wait, is that the only condition? Let me test with specific values.Let me choose ( x = y = 1 ). Then, ( P(2) = a(4) + b(2) + c ). ( P(1) + P(1) = (a + b + c) + (a + b + c) = 2a + 2b + 2c ).So, the inequality becomes:[4a + 2b + c geq 2a + 2b + 2c]Simplify:[2a - c geq 0]Which is:[2a geq c]But since ( c leq 0 ), and ( a > 0 ), ( 2a geq c ) is automatically satisfied because ( 2a ) is positive and ( c ) is non-positive. So, this doesn't add any new condition.What about another test case? Let me choose ( x = 0 ) and ( y = 0 ). Then, ( P(0) = c ). ( P(0) + P(0) = 2c ). So, the inequality is ( c geq 2c ), which simplifies to ( c leq 0 ). So, that's consistent with our earlier conclusion.Another test case: ( x = 1 ), ( y = 0 ). Then, ( P(1) = a + b + c ). ( P(1) + P(0) = (a + b + c) + c = a + b + 2c ). The inequality is ( a + b + c geq a + b + 2c ), which simplifies to ( c geq 2c ), so ( c leq 0 ). Again, consistent.Wait, so all these test cases just confirm that ( c leq 0 ) and ( a > 0 ). Is there any other condition?Let me think about the coefficient ( b ). In the original inequality, after simplifying, we only had ( 2axy geq c ). The term involving ( b ) canceled out. So, does ( b ) have any restrictions?Wait, let me check again. When I expanded ( P(x+y) ) and ( P(x) + P(y) ), the ( bx ) and ( by ) terms canceled out. So, the coefficient ( b ) doesn't affect the inequality ( 2axy geq c ). Therefore, ( b ) can be any real number? Hmm, that seems odd.Wait, but let me think about the behavior of the polynomial. If ( b ) is very large and positive, does it affect the inequality? Let me see.Suppose ( b ) is very large and positive. Then, ( P(x) = ax^2 + bx + c ) would have a very steep linear term. But in the inequality ( P(x+y) geq P(x) + P(y) ), the linear terms ( bx ) and ( by ) cancel out when subtracting ( P(x) + P(y) ) from ( P(x+y) ). So, the linear term doesn't influence the inequality. Therefore, ( b ) can indeed be any real number, positive or negative, without violating the inequality.Wait, but let me test with a specific example. Let me choose ( a = 1 ), ( b = 100 ), ( c = -1 ). So, ( P(x) = x^2 + 100x - 1 ). Let's check the inequality for ( x = 1 ), ( y = 1 ).Compute ( P(2) = 4 + 200 - 1 = 203 ). Compute ( P(1) + P(1) = (1 + 100 - 1) + (1 + 100 - 1) = 100 + 100 = 200 ). So, ( 203 geq 200 ), which holds.Another test: ( x = 2 ), ( y = 3 ). ( P(5) = 25 + 500 - 1 = 524 ). ( P(2) + P(3) = (4 + 200 - 1) + (9 + 300 - 1) = 203 + 308 = 511 ). So, ( 524 geq 511 ), which holds.What if ( b ) is negative? Let me choose ( a = 1 ), ( b = -100 ), ( c = -1 ). So, ( P(x) = x^2 - 100x - 1 ). Let's test ( x = 1 ), ( y = 1 ).( P(2) = 4 - 200 - 1 = -197 ). ( P(1) + P(1) = (1 - 100 - 1) + (1 - 100 - 1) = (-100) + (-100) = -200 ). So, ( -197 geq -200 ), which holds.Another test: ( x = 2 ), ( y = 3 ). ( P(5) = 25 - 500 - 1 = -476 ). ( P(2) + P(3) = (4 - 200 - 1) + (9 - 300 - 1) = (-197) + (-292) = -489 ). So, ( -476 geq -489 ), which holds.Hmm, so even with a large negative ( b ), the inequality still holds. Therefore, it seems that ( b ) can indeed be any real number, positive or negative, as long as ( a > 0 ) and ( c leq 0 ).Wait, but let me think about the behavior as ( x ) and ( y ) become very large. For example, let ( x = y = t ), where ( t ) is a large positive number. Then, ( P(2t) = a(4t^2) + b(2t) + c = 4a t^2 + 2b t + c ). ( P(t) + P(t) = 2(a t^2 + b t + c) = 2a t^2 + 2b t + 2c ). The inequality becomes:[4a t^2 + 2b t + c geq 2a t^2 + 2b t + 2c]Simplify:[2a t^2 - c geq 0]Since ( a > 0 ) and ( c leq 0 ), as ( t ) becomes very large, ( 2a t^2 ) dominates, making the left side positive. So, the inequality holds for large ( t ).What about when ( t ) is very small? Let ( t ) approach 0. Then, ( P(2t) approx 4a t^2 + 2b t + c ) and ( P(t) + P(t) approx 2a t^2 + 2b t + 2c ). The inequality becomes:[4a t^2 + 2b t + c geq 2a t^2 + 2b t + 2c]Simplify:[2a t^2 - c geq 0]As ( t ) approaches 0, this reduces to ( -c geq 0 ), which is ( c leq 0 ), consistent with our earlier conclusion.Therefore, it seems that the only conditions required are ( a > 0 ) and ( c leq 0 ), with no restrictions on ( b ).Wait, but let me think again. Is there any case where ( b ) could cause a problem? For example, if ( b ) is extremely large, positive or negative, could that affect the inequality?Wait, in the inequality ( 2axy geq c ), ( b ) doesn't appear because it cancels out. So, regardless of the value of ( b ), as long as ( a > 0 ) and ( c leq 0 ), the inequality holds. Therefore, ( b ) can be any real number.So, putting it all together, the quadratic polynomials ( P(x) = ax^2 + bx + c ) that satisfy ( P(x+y) geq P(x) + P(y) ) for all ( x, y geq 0 ) must have ( a > 0 ) and ( c leq 0 ), with ( b ) being any real number.Wait, but let me check one more time. Suppose ( a = 1 ), ( b = 0 ), ( c = 0 ). Then, ( P(x) = x^2 ). Let's test ( x = 1 ), ( y = 1 ). ( P(2) = 4 ), ( P(1) + P(1) = 1 + 1 = 2 ). So, ( 4 geq 2 ), which holds. Another test: ( x = 2 ), ( y = 3 ). ( P(5) = 25 ), ( P(2) + P(3) = 4 + 9 = 13 ). ( 25 geq 13 ), holds.Now, if ( c = 0 ), it's allowed because ( c leq 0 ). If ( c ) is negative, say ( c = -1 ), then ( P(x) = x^2 + bx - 1 ). Testing ( x = 1 ), ( y = 1 ): ( P(2) = 4 + 2b - 1 = 3 + 2b ). ( P(1) + P(1) = (1 + b - 1) + (1 + b - 1) = 2b ). So, ( 3 + 2b geq 2b ), which simplifies to ( 3 geq 0 ), which is always true.Another test with ( c = -1 ): ( x = 2 ), ( y = 3 ). ( P(5) = 25 + 5b - 1 = 24 + 5b ). ( P(2) + P(3) = (4 + 2b - 1) + (9 + 3b - 1) = (3 + 2b) + (8 + 3b) = 11 + 5b ). So, ( 24 + 5b geq 11 + 5b ), which simplifies to ( 24 geq 11 ), which is true.Therefore, it seems that as long as ( a > 0 ) and ( c leq 0 ), the inequality holds for all ( x, y geq 0 ), regardless of the value of ( b ).Wait, but let me consider a case where ( a > 0 ), ( c leq 0 ), but ( b ) is such that the polynomial might not be convex or something. Wait, quadratic polynomials are always convex or concave depending on the sign of ( a ). Since ( a > 0 ), the polynomial is convex. But convexity alone doesn't necessarily imply the inequality ( P(x+y) geq P(x) + P(y) ). However, in our case, the inequality reduces to ( 2axy geq c ), which is satisfied as long as ( a > 0 ) and ( c leq 0 ).Wait, but let me think about the function's behavior. If ( P(x) ) is convex, then ( P(x+y) geq P(x) + P(y) - P(0) ), but in our case, the inequality is ( P(x+y) geq P(x) + P(y) ). So, it's a stronger condition than convexity.But in our analysis, we found that as long as ( a > 0 ) and ( c leq 0 ), the inequality holds. So, maybe that's the answer.Wait, but let me think about the case when ( c = 0 ). Then, ( P(x) = ax^2 + bx ). The inequality becomes ( a(x+y)^2 + b(x+y) geq ax^2 + bx + ay^2 + by ). Simplifying, ( 2axy geq 0 ), which is true since ( a > 0 ) and ( x, y geq 0 ).If ( c < 0 ), then ( 2axy geq c ) is automatically true because ( 2axy geq 0 ) and ( c leq 0 ).Therefore, the conditions are indeed ( a > 0 ) and ( c leq 0 ), with no restrictions on ( b ).Wait, but let me check if ( b ) can be anything. Suppose ( b ) is extremely large negative, say ( b = -1000 ), ( a = 1 ), ( c = -1 ). Then, ( P(x) = x^2 - 1000x - 1 ). Let's test ( x = 1 ), ( y = 1 ). ( P(2) = 4 - 2000 - 1 = -1997 ). ( P(1) + P(1) = (1 - 1000 - 1) + (1 - 1000 - 1) = (-1000) + (-1000) = -2000 ). So, ( -1997 geq -2000 ), which holds.Another test: ( x = 1000 ), ( y = 1000 ). ( P(2000) = (2000)^2 - 1000*2000 - 1 = 4,000,000 - 2,000,000 - 1 = 2,000,000 - 1 = 1,999,999 ). ( P(1000) + P(1000) = (1,000,000 - 1,000,000 - 1) + (1,000,000 - 1,000,000 - 1) = (-1) + (-1) = -2 ). So, ( 1,999,999 geq -2 ), which holds.Wait, but in this case, ( P(2000) ) is much larger than ( P(1000) + P(1000) ). So, even with a large negative ( b ), the inequality holds because the quadratic term dominates for large ( x ) and ( y ).Therefore, I think my conclusion is correct. The quadratic polynomials ( P(x) = ax^2 + bx + c ) satisfy ( P(x+y) geq P(x) + P(y) ) for all ( x, y geq 0 ) if and only if ( a > 0 ) and ( c leq 0 ), with ( b ) being any real number.Wait, but let me think about the case when ( c = 0 ). Then, ( P(x) = ax^2 + bx ). The inequality becomes ( a(x+y)^2 + b(x+y) geq ax^2 + bx + ay^2 + by ). Simplifying, ( 2axy geq 0 ), which is true because ( a > 0 ) and ( x, y geq 0 ).If ( c < 0 ), then ( 2axy geq c ) is automatically true because ( 2axy geq 0 ) and ( c leq 0 ).Therefore, the only conditions are ( a > 0 ) and ( c leq 0 ), with no restrictions on ( b ).Wait, but let me think about the case when ( a = 0 ). Then, the polynomial becomes linear: ( P(x) = bx + c ). But the problem specifies quadratic polynomials, so ( a neq 0 ). Therefore, ( a ) must be positive.In summary, the quadratic polynomials satisfying the given inequality are those where the coefficient of ( x^2 ) is positive, and the constant term is non-positive, with no restrictions on the linear coefficient.