answer：Okay, so I have this vector problem here, and I need to find the cosine of the angle θ between the vectors (a - b) and a. Let me see what information I have and how I can use it.First, the problem states that the magnitude of vector a is 1, so |a| = 1. The magnitude of vector b is 2, so |b| = 2. There's also this dot product given: (a + b) · b = 3. Hmm, okay, so I can use this to find some relationship between vectors a and b.Let me write down what I know:1. |a| = 12. |b| = 23. (a + b) · b = 34. The angle between (a - b) and a is θ, and I need to find cosθ.Alright, starting with the third point: (a + b) · b = 3. I can expand this dot product. Remember, the dot product is distributive, so:(a + b) · b = a · b + b · bSo that's a · b + |b|² because the dot product of a vector with itself is the square of its magnitude. We know |b| is 2, so |b|² is 4. Therefore:a · b + 4 = 3Subtracting 4 from both sides gives:a · b = 3 - 4 = -1Okay, so the dot product of a and b is -1. That's useful information.Now, I need to find the angle θ between (a - b) and a. To find the cosine of this angle, I can use the dot product formula:cosθ = [(a - b) · a] / (|a - b| |a|)So, I need to compute two things: the dot product of (a - b) and a, and the magnitudes of (a - b) and a.First, let's compute the dot product (a - b) · a. Again, using the distributive property:(a - b) · a = a · a - b · aWe already know that a · a is |a|², which is 1² = 1. And we found earlier that a · b = -1, so b · a is also -1 because the dot product is commutative. Therefore:(a - b) · a = 1 - (-1) = 1 + 1 = 2Great, so the numerator of our cosine formula is 2.Next, we need the magnitudes of (a - b) and a. We already know |a| is 1, so that's straightforward. Now, let's find |a - b|.The magnitude of a vector difference can be found using the formula:|a - b| = sqrt(|a|² + |b|² - 2(a · b))Plugging in the values we have:|a - b| = sqrt(1² + 2² - 2*(-1)) = sqrt(1 + 4 + 2) = sqrt(7)So, |a - b| is sqrt(7).Now, putting it all together into the cosine formula:cosθ = [(a - b) · a] / (|a - b| |a|) = 2 / (sqrt(7) * 1) = 2 / sqrt(7)But, usually, we rationalize the denominator, so multiplying numerator and denominator by sqrt(7):cosθ = (2 * sqrt(7)) / (sqrt(7) * sqrt(7)) = (2 sqrt(7)) / 7So, cosθ is 2 sqrt(7) over 7.Wait, let me double-check my steps to make sure I didn't make a mistake. Starting from the beginning:1. (a + b) · b = 32. Expanded to a · b + |b|² = 33. Substituted |b|² = 4, so a · b = -14. Then, (a - b) · a = a · a - a · b = 1 - (-1) = 25. Calculated |a - b| using the formula, got sqrt(7)6. Plugged into cosθ formula, got 2 / sqrt(7) which simplifies to 2 sqrt(7) / 7Everything seems to check out. I don't see any errors in my calculations or reasoning.So, the final answer should be 2 sqrt(7) divided by 7.**Final Answer**The value of costheta is boxed{dfrac{2sqrt{7}}{7}}.

answer：Okay, so I need to figure out the largest integer ( n ) such that ( 7^{2048} - 1 ) is divisible by ( 2^n ). Hmm, that means I need to find the highest power of 2 that divides ( 7^{2048} - 1 ). First, I remember that when dealing with expressions like ( a^k - 1 ), it's often useful to factor them. Maybe I can use the difference of powers formula. Let me recall that ( a^k - 1 ) can be factored as ( (a - 1)(a^{k-1} + a^{k-2} + dots + a + 1) ). But since 2048 is a power of 2, maybe there's a better way to factor it.Wait, 2048 is ( 2^{11} ), right? So, ( 7^{2048} - 1 ) can be written as ( (7^{1024} - 1)(7^{1024} + 1) ). Then, each of those terms can be factored further. Let me try to factor it completely.Starting with ( 7^{2048} - 1 ), I can factor it as:[7^{2048} - 1 = (7^{1024} - 1)(7^{1024} + 1)]Then, ( 7^{1024} - 1 ) can be factored as:[7^{1024} - 1 = (7^{512} - 1)(7^{512} + 1)]Continuing this process, each time factoring the difference of squares, I get:[7^{2048} - 1 = (7 - 1)(7 + 1)(7^2 + 1)(7^4 + 1)dots(7^{1024} + 1)]So, that's a total of 12 factors: ( (7 - 1) ) is the first, ( (7 + 1) ) is the second, and then each subsequent term is ( 7^{2^k} + 1 ) for ( k ) from 1 to 10.Now, I need to find how many factors of 2 are in each of these terms. Let's start with the first two factors:1. ( 7 - 1 = 6 ). That's ( 2 times 3 ), so it has one factor of 2.2. ( 7 + 1 = 8 ). That's ( 2^3 ), so it has three factors of 2.Now, for the remaining terms ( 7^{2^k} + 1 ) where ( k ) ranges from 1 to 10. Let me see if I can find a pattern here. Maybe I can compute each term modulo 4 or 8 to see how many factors of 2 they contribute.Starting with modulo 4:- ( 7 equiv -1 mod 4 )- So, ( 7^{2^k} equiv (-1)^{2^k} mod 4 )- Since ( 2^k ) is even for ( k geq 1 ), ( (-1)^{2^k} = 1 mod 4 )- Therefore, ( 7^{2^k} + 1 equiv 1 + 1 = 2 mod 4 )This means each of these terms is congruent to 2 modulo 4, which implies each has exactly one factor of 2. So, each ( 7^{2^k} + 1 ) contributes one more factor of 2.Since there are 10 such terms (from ( k = 1 ) to ( k = 10 )), each contributing one factor of 2, that's 10 additional factors.Adding up all the factors of 2:- From ( 7 - 1 ): 1- From ( 7 + 1 ): 3- From each ( 7^{2^k} + 1 ): 10Total factors of 2: ( 1 + 3 + 10 = 14 )Wait, let me double-check that. The first two factors give 1 and 3, which is 4. Then, the remaining 10 factors each give 1, so 10 more. So, 4 + 10 is 14. Yeah, that seems right.But just to be thorough, maybe I should check one of the higher terms modulo a higher power of 2, like 8 or 16, to see if any of them contribute more than one factor of 2.Let's take ( 7^2 + 1 = 49 + 1 = 50 ). 50 is ( 2 times 25 ), so only one factor of 2.Next, ( 7^4 + 1 = 2401 + 1 = 2402 ). 2402 divided by 2 is 1201, which is odd. So, again, only one factor of 2.Similarly, ( 7^8 + 1 ) would be a larger number, but modulo 8:- ( 7 equiv -1 mod 8 )- ( 7^8 equiv (-1)^8 = 1 mod 8 )- So, ( 7^8 + 1 equiv 1 + 1 = 2 mod 8 )- Which means it's divisible by 2 but not by 4, so again only one factor of 2.So, it seems consistent that each of these terms contributes exactly one factor of 2. Therefore, my initial calculation of 14 factors of 2 should be correct.Just to recap:- ( 7 - 1 = 6 ): 1 factor- ( 7 + 1 = 8 ): 3 factors- ( 7^2 + 1 = 50 ): 1 factor- ( 7^4 + 1 = 2402 ): 1 factor- ( 7^8 + 1 ): 1 factor- ...- Up to ( 7^{1024} + 1 ): 1 factorTotal: 1 + 3 + 10*1 = 14I think that's solid. I don't see any mistakes in my reasoning. Each step checks out, and the modular arithmetic confirms that the higher terms don't contribute more than one factor of 2. So, the largest integer ( n ) such that ( 7^{2048} - 1 ) is divisible by ( 2^n ) is 14.

answer：Okay, so I have this problem: I need to find the coefficient of the term containing ( x^5 y^3 ) in the expansion of ( (2x^2 - 3x - y)^6 ). Hmm, that seems a bit complicated, but I think I can break it down step by step.First, I remember that when expanding expressions raised to a power, the binomial theorem is useful. But wait, this isn't just a binomial; it's a trinomial because there are three terms: ( 2x^2 ), ( -3x ), and ( -y ). So, maybe I need to use the multinomial theorem here. I'm a bit fuzzy on the details, so let me recall.The multinomial theorem generalizes the binomial theorem for expressions with more than two terms. It states that:[(a_1 + a_2 + dots + a_k)^n = sum_{n_1 + n_2 + dots + n_k = n} frac{n!}{n_1! n_2! dots n_k!} a_1^{n_1} a_2^{n_2} dots a_k^{n_k}]In this case, ( k = 3 ) because we have three terms, and ( n = 6 ). So, each term in the expansion will be of the form:[frac{6!}{n_1! n_2! n_3!} (2x^2)^{n_1} (-3x)^{n_2} (-y)^{n_3}]where ( n_1 + n_2 + n_3 = 6 ).I need the term that contains ( x^5 y^3 ). Let's figure out what exponents ( n_1 ), ( n_2 ), and ( n_3 ) must satisfy to give me ( x^5 y^3 ).Looking at each part:- ( (2x^2)^{n_1} ) contributes ( x^{2n_1} )- ( (-3x)^{n_2} ) contributes ( x^{n_2} )- ( (-y)^{n_3} ) contributes ( y^{n_3} )So, combining the exponents of ( x ):[2n_1 + n_2 = 5]And for ( y ):[n_3 = 3]Since ( n_1 + n_2 + n_3 = 6 ), and ( n_3 = 3 ), that simplifies to:[n_1 + n_2 = 3]So now I have two equations:1. ( 2n_1 + n_2 = 5 )2. ( n_1 + n_2 = 3 )Let me solve these equations. Subtracting the second equation from the first:[(2n_1 + n_2) - (n_1 + n_2) = 5 - 3 n_1 = 2]Then, substituting back into the second equation:[2 + n_2 = 3 n_2 = 1]So, ( n_1 = 2 ), ( n_2 = 1 ), and ( n_3 = 3 ).Now, let's compute the coefficient for this term. The general term is:[frac{6!}{n_1! n_2! n_3!} (2x^2)^{n_1} (-3x)^{n_2} (-y)^{n_3}]Plugging in the values:[frac{6!}{2! 1! 3!} (2x^2)^2 (-3x)^1 (-y)^3]First, compute the multinomial coefficient:[frac{720}{2 times 1 times 6} = frac{720}{12} = 60]Now, compute each part:- ( (2x^2)^2 = 4x^4 )- ( (-3x)^1 = -3x )- ( (-y)^3 = -y^3 )Multiply them together:[4x^4 times (-3x) times (-y^3) = 4 times (-3) times (-1) times x^{4+1} y^3 = 12x^5 y^3]Now, multiply by the multinomial coefficient:[60 times 12x^5 y^3 = 720x^5 y^3]So, the coefficient is 720.Wait, let me double-check my calculations to make sure I didn't make a mistake.Multinomial coefficient:- ( 6! = 720 )- ( 2! = 2 ), ( 1! = 1 ), ( 3! = 6 )- So, ( 720 / (2 times 1 times 6) = 720 / 12 = 60 ). That seems correct.Calculating each part:- ( (2x^2)^2 = 4x^4 )- ( (-3x)^1 = -3x )- ( (-y)^3 = -y^3 )Multiplying these:- ( 4 times (-3) = -12 )- ( (-12) times (-1) = 12 )- ( x^4 times x = x^5 )- ( y^3 )So, 12x^5 y^3. Then, 60 times 12 is indeed 720.Hmm, seems solid. But just to be thorough, let me consider if there's another way to approach this problem, maybe using binomial theorem in a different grouping.Alternatively, I could think of the expression as ( (2x^2 - 3x - y)^6 ) and group terms differently. Maybe group ( 2x^2 - 3x ) as one term and ( -y ) as another. So, it becomes ( (A - y)^6 ) where ( A = 2x^2 - 3x ).Using the binomial theorem, the expansion would be:[sum_{k=0}^{6} binom{6}{k} A^{6 - k} (-y)^k]We need the term where ( y^3 ), so ( k = 3 ). Thus, the term is:[binom{6}{3} A^{3} (-y)^3]Compute ( binom{6}{3} = 20 ), and ( (-y)^3 = -y^3 ), so the term is:[20 times A^3 times (-y^3) = -20 A^3 y^3]Now, ( A = 2x^2 - 3x ), so ( A^3 = (2x^2 - 3x)^3 ). Let's expand this.Using the binomial theorem again:[(2x^2 - 3x)^3 = sum_{m=0}^{3} binom{3}{m} (2x^2)^{3 - m} (-3x)^m]We need the term in ( A^3 ) that gives ( x^5 ) because when multiplied by ( y^3 ), it will give ( x^5 y^3 ).So, in ( (2x^2 - 3x)^3 ), the general term is:[binom{3}{m} (2x^2)^{3 - m} (-3x)^m = binom{3}{m} 2^{3 - m} (-3)^m x^{2(3 - m) + m} = binom{3}{m} 2^{3 - m} (-3)^m x^{6 - m}]We need the exponent of ( x ) to be 5, so:[6 - m = 5 implies m = 1]So, the term is:[binom{3}{1} 2^{2} (-3)^1 x^5 = 3 times 4 times (-3) x^5 = 3 times 4 times (-3) x^5 = -36 x^5]Therefore, ( A^3 ) has a term ( -36 x^5 ). Plugging back into the earlier expression:[-20 times (-36 x^5) y^3 = 720 x^5 y^3]So, the coefficient is indeed 720. That matches my previous result. I think this confirms that 720 is the correct coefficient.Just to ensure I haven't missed any other terms that could contribute to ( x^5 y^3 ), I should consider if there are other combinations of ( n_1, n_2, n_3 ) that could result in ( x^5 y^3 ). From earlier, I found ( n_1 = 2 ), ( n_2 = 1 ), ( n_3 = 3 ), and that was the only solution to the system of equations. So, there are no other terms contributing to ( x^5 y^3 ). Therefore, 720 is the only coefficient for that term.I think I've covered all bases here. I used two different approaches—multinomial theorem and binomial theorem with grouping—and both led me to the same answer. I also double-checked my calculations step by step to make sure there were no arithmetic errors. It seems solid.**Final Answer**The coefficient is boxed{720}.

answer：First, I need to simplify the given expression by combining like terms. I'll start by distributing the negative sign to the terms inside the second parenthesis.Next, I'll combine the ( x^4 ) terms, the ( x^2 ) terms, and the constant terms separately to simplify the expression further.After simplifying, I'll look for the greatest common factor among the terms to factor it out. This will help in reducing the expression to its simplest form.Finally, I'll ensure that the remaining polynomial inside the parentheses cannot be factored any further, completing the factoring process.

answer：First, I need to find the prime factorization of 360. I know that 360 is divisible by 2, 3, and 5. Dividing 360 by these primes, I get:[ 360 = 2^3 cdot 3^2 cdot 5^1 ]Next, to determine the number of perfect square factors, I need to consider how exponents in the prime factorization contribute to a perfect square. A perfect square has even exponents for all its prime factors.For the prime factor 2, which has an exponent of 3, the possible even exponents are 0 and 2.For the prime factor 3, with an exponent of 2, the possible even exponents are 0 and 2.For the prime factor 5, with an exponent of 1, the only possible even exponent is 0.Now, I calculate the number of combinations of these exponents:- For 2: 2 choices (0 or 2)- For 3: 2 choices (0 or 2)- For 5: 1 choice (0)Multiplying these together gives:[ 2 times 2 times 1 = 4 ]Therefore, there are 4 perfect square factors of 360.

answer：Okay, so I have this problem where there are 5 representatives from each of 5 different companies at a conference. During a networking event, each person shakes hands with every other attendee except their own company's representatives. I need to figure out how many handshakes occur in total.Alright, let's start by understanding the setup. There are 5 companies, and each company has 5 representatives. So, in total, there are 5 times 5, which is 25 people at the conference. That part seems straightforward.Now, each person is supposed to shake hands with every other attendee except the ones from their own company. So, for any given person, they won't shake hands with the 4 other people from their company. That means each person will shake hands with 25 total people minus themselves and the 4 from their company. So, 25 minus 1 is 24, and then minus 4 is 20. So, each person shakes hands with 20 others.If I think about this, if each of the 25 people shakes hands with 20 others, then the total number of handshakes would be 25 times 20. Let me calculate that: 25 times 20 is 500. But wait, this counts each handshake twice because when person A shakes hands with person B, it's one handshake, but it's counted once for A and once for B. So, to get the actual number of unique handshakes, I need to divide this number by 2.So, 500 divided by 2 is 250. That seems like the total number of unique handshakes. But let me double-check to make sure I didn't make a mistake.Another way to think about it is using combinations. The total number of possible handshakes if everyone shook hands with everyone else would be the combination of 25 people taken 2 at a time. The formula for combinations is n(n-1)/2, where n is the total number of people. So, 25 times 24 divided by 2 is 300. But this includes handshakes between people from the same company, which we need to exclude.How many handshakes are there within each company? For one company, there are 5 representatives, and the number of handshakes among them is 5 choose 2, which is 10. Since there are 5 companies, the total number of intra-company handshakes is 5 times 10, which is 50.So, if I subtract the intra-company handshakes from the total possible handshakes, I get 300 minus 50, which is 250. That matches the previous result. So, both methods give me the same answer, which is a good sign.Wait, let me make sure I didn't confuse intra-company with inter-company. The problem says each person doesn't shake hands with their own company's representatives, so we're excluding intra-company handshakes. So, subtracting those from the total gives me the correct number of inter-company handshakes.Another way to visualize this is to imagine each person only shaking hands with people from other companies. Since there are 5 companies, and each company has 5 people, for any given person, they can shake hands with 5 times 4, which is 20 people (since they don't shake hands with their own company). So, 20 handshakes per person, 25 people, but again, this counts each handshake twice, so dividing by 2 gives 250.I think I've approached this problem from a couple of different angles, and each time I arrive at the same answer: 250 handshakes. I feel pretty confident that this is the correct solution.Just to recap:1. Total number of people: 25.2. Total possible handshakes if everyone shook hands: 300.3. Subtract the intra-company handshakes: 50.4. Result: 250.Alternatively:1. Each person shakes hands with 20 others.2. Total handshakes counted twice: 500.3. Divide by 2 to get unique handshakes: 250.Both methods lead to the same conclusion, so I think 250 is the right answer.