💡Okay, so I have this problem where I need to show that ( n ) is an even number given certain conditions. Let me try to understand the problem step by step.First, the problem defines ( Omega(N) ) for a positive integer ( N > 1 ) with its unique prime factorization ( N = p_1^{alpha_1} p_2^{alpha_2} dotsb p_k^{alpha_k} ). The function ( Omega(N) ) is the sum of the exponents in the prime factorization, so ( Omega(N) = alpha_1 + alpha_2 + dotsb + alpha_k ). Got it.Next, we have a polynomial ( p(x) = (x + a_1)(x + a_2) dotsb (x + a_n) ) where each ( a_i ) is a positive integer. The condition given is that for all positive integers ( k ), ( Omega(P(k)) ) is even. Here, ( P(k) ) would be ( (k + a_1)(k + a_2) dotsb (k + a_n) ). So, the product of these linear terms evaluated at any positive integer ( k ) has an even number of prime factors when you count their exponents.I need to show that ( n ), the number of terms in the product, is even. Hmm, okay. So, the key is that no matter what ( k ) I choose, the total number of prime factors (counting multiplicities) of ( P(k) ) is even. That must impose some condition on ( n ).Let me think about how ( Omega ) behaves. Since ( Omega ) is additive over multiplication, ( Omega(P(k)) = Omega(k + a_1) + Omega(k + a_2) + dotsb + Omega(k + a_n) ). So, the sum of the ( Omega ) values of each ( k + a_i ) must be even for every ( k ).Now, each ( Omega(k + a_i) ) can be either even or odd. If I have a sum of numbers, each of which is either even or odd, the total sum is even if and only if there are an even number of odd terms in the sum. So, for the sum ( Omega(k + a_1) + Omega(k + a_2) + dotsb + Omega(k + a_n) ) to be even for all ( k ), the number of terms where ( Omega(k + a_i) ) is odd must be even for every ( k ).This seems tricky because ( k ) can be any positive integer. So, I need to ensure that no matter what ( k ) is, the number of ( a_i ) such that ( Omega(k + a_i) ) is odd is even. How can I ensure that?Maybe I can think about specific choices of ( k ) that might reveal something about the ( a_i ). For example, if I choose ( k = 1 ), then ( P(1) = (1 + a_1)(1 + a_2) dotsb (1 + a_n) ). The ( Omega ) of this product is even. Similarly, for ( k = 2 ), ( P(2) = (2 + a_1)(2 + a_2) dotsb (2 + a_n) ), and ( Omega(P(2)) ) is even.But how does this help me? Maybe I need to consider the parity of ( Omega(k + a_i) ). Let's denote ( omega_i(k) = Omega(k + a_i) mod 2 ). So, ( omega_i(k) ) is 0 if ( Omega(k + a_i) ) is even and 1 if it's odd. Then, the condition is that ( sum_{i=1}^n omega_i(k) equiv 0 mod 2 ) for all ( k ).So, I have a system where for every ( k ), the sum of these ( omega_i(k) ) is 0 modulo 2. I need to find out what this implies about ( n ).Perhaps I can consider the behavior of ( omega_i(k) ) as a function of ( k ). For each ( a_i ), ( omega_i(k) ) is the parity of the number of prime factors (with multiplicity) of ( k + a_i ). This is a bit abstract, but maybe I can think about how often ( k + a_i ) is a prime or a prime power.Wait, if ( k + a_i ) is a prime, then ( Omega(k + a_i) = 1 ), which is odd. If ( k + a_i ) is a prime square, then ( Omega(k + a_i) = 2 ), which is even. Similarly, higher powers would have even ( Omega ) if the exponent is even and odd if it's odd.But this seems complicated because ( k + a_i ) can be any number depending on ( k ). Maybe instead of looking at specific ( k ), I can consider the behavior as ( k ) varies.If I fix ( a_i ), then ( k + a_i ) can be any integer greater than ( a_i ). So, for each ( a_i ), ( omega_i(k) ) alternates between 0 and 1 depending on whether ( k + a_i ) has an even or odd number of prime factors.But how can the sum of these ( omega_i(k) ) always be even? It seems like each ( omega_i(k) ) is a function that can be 0 or 1, and their sum must always be even. The only way this can happen is if the number of functions ( omega_i(k) ) that can be 1 is even. But wait, that might not necessarily be the case because ( omega_i(k) ) can vary with ( k ).Alternatively, maybe I can think about the polynomial ( p(x) ) and its properties. Since ( p(x) ) is a product of linear terms, it's a polynomial of degree ( n ). The condition is about the number of prime factors of ( p(k) ) for all ( k ).But I'm not sure how to connect the degree of the polynomial to the parity of the number of prime factors. Maybe I need a different approach.Let me consider specific cases. Suppose ( n = 1 ). Then ( p(x) = x + a_1 ). Then ( Omega(p(k)) = Omega(k + a_1) ). For this to be even for all ( k ), ( k + a_1 ) must always have an even number of prime factors. But that's impossible because, for example, if ( k + a_1 ) is prime, ( Omega(k + a_1) = 1 ), which is odd. So ( n = 1 ) is impossible.Similarly, if ( n = 2 ), then ( p(x) = (x + a_1)(x + a_2) ). Then ( Omega(p(k)) = Omega(k + a_1) + Omega(k + a_2) ). For this sum to be even for all ( k ), the number of times ( Omega(k + a_i) ) is odd must be even. So, either both ( Omega(k + a_1) ) and ( Omega(k + a_2) ) are even, or both are odd.Is it possible for ( Omega(k + a_1) ) and ( Omega(k + a_2) ) to always have the same parity? That would require that ( k + a_1 ) and ( k + a_2 ) have the same parity in their number of prime factors for all ( k ).Wait, but ( k + a_1 ) and ( k + a_2 ) differ by ( a_2 - a_1 ). So, if ( a_2 - a_1 ) is even, then ( k + a_1 ) and ( k + a_2 ) have the same parity. If ( a_2 - a_1 ) is odd, then they have opposite parity.But ( Omega ) depends on the number of prime factors, not just the number itself. So, even if ( k + a_1 ) and ( k + a_2 ) have different parities, their ( Omega ) values could still have the same parity.Hmm, this is getting complicated. Maybe I need to think about the fact that ( Omega ) is a completely additive function. So, the sum ( Omega(P(k)) ) is the sum of ( Omega(k + a_i) ).If I can show that each ( Omega(k + a_i) ) is odd for an even number of ( i ), then their sum would be even. But how can I ensure that?Wait, maybe I can use the fact that the sum must be even for all ( k ). So, if I consider the sum modulo 2, it's always 0. That means the sum of the parities of ( Omega(k + a_i) ) is 0 modulo 2 for all ( k ).This is similar to saying that the sum of certain functions (each being the parity of ( Omega(k + a_i) )) is always 0 modulo 2. Maybe I can think of this as a linear algebra problem over the field ( mathbb{F}_2 ).Each ( omega_i(k) ) is a function from ( mathbb{N} ) to ( mathbb{F}_2 ), and their sum is the zero function. So, the functions ( omega_i ) must be linearly dependent in such a way that their sum is zero.But I'm not sure if this approach is helpful. Maybe I need a different angle.Let me think about the behavior of ( Omega ) for different numbers. For example, if ( k + a_i ) is a prime, ( Omega(k + a_i) = 1 ). If it's a square of a prime, ( Omega(k + a_i) = 2 ). If it's a product of two distinct primes, ( Omega(k + a_i) = 2 ). If it's a cube of a prime, ( Omega(k + a_i) = 3 ), and so on.So, ( Omega(k + a_i) ) is odd if ( k + a_i ) is a prime power with an odd exponent or a product of an odd number of primes. Wait, no, ( Omega ) counts the total number of prime factors with multiplicity. So, it's odd if the total count is odd, regardless of the exponents.So, for example, ( 12 = 2^2 times 3 ) has ( Omega(12) = 3 ), which is odd. ( 16 = 2^4 ) has ( Omega(16) = 4 ), which is even.So, ( Omega(k + a_i) ) is odd if ( k + a_i ) has an odd number of prime factors when counted with multiplicity.Now, if I can find a way to ensure that the number of ( a_i ) such that ( k + a_i ) has an odd number of prime factors is even for all ( k ), then the sum ( Omega(P(k)) ) would be even.But how can I ensure that? It seems like a very strong condition because ( k ) can be any positive integer, so ( k + a_i ) can be any integer greater than ( a_i ).Maybe I can consider the fact that for each ( a_i ), the function ( omega_i(k) = Omega(k + a_i) mod 2 ) is periodic or has some regularity. But I'm not sure.Alternatively, perhaps I can use the fact that the sum of these parities must be zero for all ( k ). So, if I consider the sum as a function, it's identically zero. Maybe I can use generating functions or something.Wait, another idea: suppose I fix ( k ) and vary ( a_i ). But no, ( a_i ) are fixed, and ( k ) varies.Wait, maybe I can consider the difference between two different ( k ) values. For example, take ( k ) and ( k + 1 ). Then, ( P(k + 1) = (k + 1 + a_1)(k + 1 + a_2) dotsb (k + 1 + a_n) ). The difference between ( P(k + 1) ) and ( P(k) ) is some polynomial, but I'm not sure if that helps.Alternatively, maybe I can consider the fact that for ( k ) large enough, ( k + a_i ) can be prime or composite. But again, not sure.Wait, perhaps I can use the fact that the sum ( sum_{i=1}^n omega_i(k) equiv 0 mod 2 ) for all ( k ). So, this is a linear equation in ( mathbb{F}_2 ) that must hold for all ( k ). The only way this can happen is if each ( omega_i(k) ) is the same function, but that might not necessarily be the case.Alternatively, maybe the functions ( omega_i(k) ) must be such that their sum is zero for all ( k ), which would imply that they are linearly dependent in a very strong way.Wait, perhaps I can think about the fact that if ( n ) is odd, then it's impossible for the sum to always be even. Let me test this idea.Suppose ( n ) is odd. Then, the sum of ( n ) terms, each of which is 0 or 1, would be odd if an odd number of them are 1. But the condition requires that the sum is always even. So, if ( n ) is odd, it's impossible for the sum to always be even because you can have configurations where an odd number of ( omega_i(k) ) are 1, leading to an odd sum.Wait, but that's not necessarily true because ( omega_i(k) ) can vary with ( k ). So, even if ( n ) is odd, maybe for each ( k ), the number of ( omega_i(k) ) that are 1 is even, making the sum even. But is that possible?I think it's not possible because if ( n ) is odd, you can't have the sum of an odd number of variables each being 0 or 1 to always be even. Because if you have an odd number of variables, the sum can be even or odd depending on the number of 1s. So, to have the sum always even, you need an even number of variables that can be 1.Wait, that makes sense. If you have an odd number of terms, you can't have the sum always even because you can have cases where an odd number of terms are 1, making the sum odd. Therefore, ( n ) must be even.Wait, but is that rigorous? Let me think again.Suppose ( n ) is odd. Then, consider the sum ( S(k) = sum_{i=1}^n omega_i(k) mod 2 ). For ( S(k) ) to be 0 for all ( k ), the number of ( omega_i(k) ) that are 1 must be even for every ( k ).But if ( n ) is odd, then the sum of an odd number of bits (each 0 or 1) can be either even or odd. However, the condition requires it to always be even. So, if ( n ) is odd, it's impossible for the sum to always be even because you can have configurations where an odd number of ( omega_i(k) ) are 1, leading to an odd sum.Therefore, ( n ) must be even.Wait, but I'm not sure if this is entirely correct because ( omega_i(k) ) are not arbitrary; they are specific functions based on the number of prime factors. Maybe there's a way for the sum to always be even even if ( n ) is odd.But I think the key point is that if ( n ) is odd, you can choose ( k ) such that an odd number of ( omega_i(k) ) are 1, making the sum odd, which contradicts the condition. Therefore, ( n ) must be even.So, putting it all together, the number of terms ( n ) must be even to ensure that the sum ( Omega(P(k)) ) is always even for all positive integers ( k ).

💡First, determine the number of shirts Avery donates, which is 4.Next, calculate the number of pants by doubling the number of shirts: 2 * 4 = 8 pants.Then, find the number of shorts by taking half of the number of pants: 8 / 2 = 4 shorts.Finally, add up all the pieces of clothing: 4 shirts + 8 pants + 4 shorts = 16 pieces of clothes.

💡Alright, so I have this problem here where I need to solve an inequality using a custom operation defined as ( x oplus y = x(1 - y) ). The inequality given is ( (x - 1) oplus (x + 2) < 0 ). Hmm, okay, let me try to unpack this step by step.First, I need to understand what the operation ( oplus ) does. It takes two real numbers, ( x ) and ( y ), and returns ( x ) multiplied by ( (1 - y) ). So, it's like a modified multiplication where the second operand is subtracted from 1 before multiplying. Got it.Now, applying this operation to ( (x - 1) ) and ( (x + 2) ), I substitute into the definition:( (x - 1) oplus (x + 2) = (x - 1)(1 - (x + 2)) ).Let me simplify the expression inside the parentheses first:( 1 - (x + 2) = 1 - x - 2 = -x - 1 ).So, now the expression becomes:( (x - 1)(-x - 1) ).Hmm, that looks a bit messy. Maybe I can factor out a negative sign to make it simpler. Let me try that:( (x - 1)(-x - 1) = - (x - 1)(x + 1) ).Wait, is that right? Let me check:( (x - 1)(-x - 1) = (x - 1)(-1)(x + 1) = - (x - 1)(x + 1) ). Yeah, that seems correct.So, now the inequality ( (x - 1) oplus (x + 2) < 0 ) becomes:( - (x - 1)(x + 1) < 0 ).Hmm, if I multiply both sides of the inequality by -1, I remember that this will reverse the inequality sign. So, let's do that:( (x - 1)(x + 1) > 0 ).Alright, now I have a product of two terms, ( (x - 1) ) and ( (x + 1) ), and I need this product to be greater than zero. To solve this inequality, I can analyze the sign of each term in different intervals determined by the roots of the equation ( (x - 1)(x + 1) = 0 ). The roots are ( x = 1 ) and ( x = -1 ). These roots divide the real number line into three intervals:1. ( x < -1 )2. ( -1 < x < 1 )3. ( x > 1 )Now, I'll test each interval to see where the product ( (x - 1)(x + 1) ) is positive.**Interval 1: ( x < -1 )**Let's pick a test point, say ( x = -2 ).- ( x - 1 = -2 - 1 = -3 ) (negative)- ( x + 1 = -2 + 1 = -1 ) (negative)The product of two negatives is positive. So, ( (x - 1)(x + 1) > 0 ) in this interval.**Interval 2: ( -1 < x < 1 )**Let's choose ( x = 0 ).- ( x - 1 = 0 - 1 = -1 ) (negative)- ( x + 1 = 0 + 1 = 1 ) (positive)The product of a negative and a positive is negative. So, ( (x - 1)(x + 1) < 0 ) in this interval.**Interval 3: ( x > 1 )**Let's take ( x = 2 ).- ( x - 1 = 2 - 1 = 1 ) (positive)- ( x + 1 = 2 + 1 = 3 ) (positive)The product of two positives is positive. So, ( (x - 1)(x + 1) > 0 ) in this interval.Putting this all together, the inequality ( (x - 1)(x + 1) > 0 ) holds true when ( x < -1 ) or ( x > 1 ).But wait, let me double-check my steps to make sure I didn't make a mistake. Starting from the operation definition, substitution, simplification, and then solving the inequality. Everything seems to follow logically. The critical points are correctly identified, and testing each interval gives consistent results.So, the solution set for the inequality ( (x - 1) oplus (x + 2) < 0 ) is all real numbers less than -1 or greater than 1. Looking back at the options provided:A: ( -1 < x < 1 ) B: ( -2 < x < 1 ) C: ( x < -1 ) or ( x > 1 ) D: ( x < -2 ) or ( x > 1 )The correct choice is option C, which matches our solution.**Final Answer**boxed{C}

💡First, expand the complex number ( z = (a - i)^2 ) to obtain its standard form.[z = (a - i)^2 = a^2 - 2ai + i^2 = a^2 - 2ai - 1]This simplifies to:[z = (a^2 - 1) - 2ai]Since ( z ) is a pure imaginary number, the real part must be zero:[a^2 - 1 = 0]Solving this equation gives:[a^2 = 1 implies a = pm 1]Additionally, the imaginary part must not be zero:[-2a neq 0 implies a neq 0]Therefore, the real number ( a ) is ( pm 1 ).

💡First, I need to determine the number of days it takes for both A and B to complete the task individually. I know that B completes 60% of the task, which takes B 21.6 days to finish. A leaves 12 days before the task is completed, so A works for 9.6 days.Next, I'll calculate A's work rate. Since A completes 40% of the task in 9.6 days, I can find out how many days A would take to complete the entire task alone by setting up the equation 0.4 * y = 9.6, where y is the number of days A would take alone. Solving for y gives me y = 24 days.Now, I know that B would take 36 days to complete the task alone. To find out how many more days B takes than A, I subtract the number of days A takes from the number of days B takes: 36 - 24 = 12 days.Therefore, B takes 12 more days than A to finish the task.

💡To determine the number of digits in the expression (3^{15} times 5^{10}), I can use logarithms to simplify the calculation.First, I'll take the logarithm base 10 of the expression:[log_{10}(3^{15} times 5^{10}) = log_{10}(3^{15}) + log_{10}(5^{10})]Using the logarithm power rule, this becomes:[15 cdot log_{10}(3) + 10 cdot log_{10}(5)]I know that (log_{10}(3) approx 0.4771) and (log_{10}(5) approx 0.6990). Plugging these values in:[15 times 0.4771 + 10 times 0.6990 = 7.1565 + 6.9900 = 14.1465]The number of digits is the integer part of this value plus one:[lfloor 14.1465 rfloor + 1 = 14 + 1 = 15]Therefore, the expression (3^{15} times 5^{10}) has 15 digits.