💡Okay, so I have this problem where I need to find the maximum value of ( A + B ) given that ( A, B, C, D ) are distinct digits from 0 to 9. The fraction ( frac{A+B}{C+D} ) has to be an integer, and the total sum ( A + B + C + D ) must be exactly 17. Hmm, let's break this down step by step.First, my goal is to maximize ( A + B ). Since ( A ) and ( B ) are digits, the maximum they can be is 9 and 8, which would give ( A + B = 17 ). But wait, if ( A + B = 17 ), then ( C + D ) would have to be ( 17 - 17 = 0 ). But ( C ) and ( D ) are digits, and the smallest they can be is 0 and 1, which adds up to 1. So, ( C + D = 0 ) isn't possible because they have to be distinct. That means ( A + B = 17 ) is not feasible.Okay, let's try the next highest sum for ( A + B ). If I take ( A = 9 ) and ( B = 6 ), then ( A + B = 15 ). This would mean ( C + D = 17 - 15 = 2 ). Now, I need to find two distinct digits that add up to 2. The only possibilities are 0 and 2, or 1 and 1. But 1 and 1 aren't distinct, so ( C = 0 ) and ( D = 2 ) or vice versa. That works because all digits are distinct: 9, 6, 0, 2.Now, let's check the fraction ( frac{A+B}{C+D} = frac{15}{2} ). Wait, that's 7.5, which isn't an integer. So, this doesn't satisfy the condition that the fraction has to be an integer. Hmm, so ( A + B = 15 ) might not be the answer either.Let me try another combination. If I take ( A = 9 ) and ( B = 7 ), then ( A + B = 16 ). This would mean ( C + D = 17 - 16 = 1 ). The only way to get 1 with two distinct digits is 0 and 1. So, ( C = 0 ) and ( D = 1 ) or vice versa. That works because all digits are distinct: 9, 7, 0, 1.Now, let's check the fraction ( frac{A+B}{C+D} = frac{16}{1} = 16 ). That's an integer, which is perfect. So, this seems to satisfy all the conditions: ( A + B = 16 ), ( C + D = 1 ), all digits are distinct, and the total sum is 17.Wait, but let me double-check if there's a way to get a higher ( A + B ) that still works. If I try ( A = 8 ) and ( B = 7 ), then ( A + B = 15 ). Then ( C + D = 2 ), which we already saw doesn't work because ( frac{15}{2} ) isn't an integer. What if I take ( A = 9 ) and ( B = 5 ), then ( A + B = 14 ), and ( C + D = 3 ). The fraction would be ( frac{14}{3} ), which isn't an integer either.It seems like ( A + B = 16 ) is the highest possible sum that meets all the conditions. Let me just make sure there aren't any other combinations. If I take ( A = 9 ) and ( B = 8 ), ( A + B = 17 ), but as we saw earlier, ( C + D = 0 ) isn't possible. So, 16 is indeed the maximum.Therefore, the maximum value of ( A + B ) is 16.

💡Okay, so I have this problem: For how many positive integers ( n ) does the sum ( 1 + 2 + cdots + n ) evenly divide ( 10n )? Hmm, let me try to figure this out step by step.First, I remember that the sum of the first ( n ) positive integers is given by the formula ( frac{n(n+1)}{2} ). So, the sum ( 1 + 2 + cdots + n ) is ( frac{n(n+1)}{2} ). The problem says that this sum should evenly divide ( 10n ). That means when we divide ( 10n ) by ( frac{n(n+1)}{2} ), the result should be an integer.Let me write that down as an equation:[frac{10n}{frac{n(n+1)}{2}} = frac{20n}{n+1}]So, we need ( frac{20n}{n+1} ) to be an integer. Hmm, okay. So, for this fraction to be an integer, ( n+1 ) must divide ( 20n ). That is, ( n+1 ) is a divisor of ( 20n ).Now, I need to find all positive integers ( n ) such that ( n+1 ) divides ( 20n ). Let me think about how to approach this. Maybe I can consider the divisors of 20 and see how they relate to ( n+1 ).The divisors of 20 are: 1, 2, 4, 5, 10, 20. So, ( n+1 ) must be one of these divisors. That means ( n ) can be ( 0, 1, 3, 4, 9, 19 ). But wait, ( n ) has to be a positive integer, so ( n = 0 ) is out of the question. So, the possible values of ( n ) are 1, 3, 4, 9, 19.Let me check each of these to make sure:1. For ( n = 1 ): - Sum = ( 1 ) - ( 10n = 10 ) - ( 10 / 1 = 10 ), which is an integer. Good.2. For ( n = 3 ): - Sum = ( 1 + 2 + 3 = 6 ) - ( 10n = 30 ) - ( 30 / 6 = 5 ), which is an integer. Good.3. For ( n = 4 ): - Sum = ( 1 + 2 + 3 + 4 = 10 ) - ( 10n = 40 ) - ( 40 / 10 = 4 ), which is an integer. Good.4. For ( n = 9 ): - Sum = ( 1 + 2 + cdots + 9 = 45 ) - ( 10n = 90 ) - ( 90 / 45 = 2 ), which is an integer. Good.5. For ( n = 19 ): - Sum = ( 1 + 2 + cdots + 19 = 190 ) - ( 10n = 190 ) - ( 190 / 190 = 1 ), which is an integer. Good.So, all these values of ( n ) work. Therefore, there are 5 positive integers ( n ) that satisfy the condition.Wait, but let me think again. Is there a possibility that ( n+1 ) could be a divisor of 20n but not necessarily a divisor of 20? For example, if ( n+1 ) shares a common factor with ( n ), maybe there are more solutions?Let me consider that. So, ( n+1 ) divides ( 20n ). That means ( n+1 ) divides ( 20n ). Since ( n ) and ( n+1 ) are consecutive integers, they are coprime. That is, their greatest common divisor is 1. Therefore, ( n+1 ) must divide 20. So, actually, ( n+1 ) must be a divisor of 20, which brings us back to the original list of divisors.Therefore, the only possible values of ( n ) are 1, 3, 4, 9, 19, which are 5 in total.Hmm, that seems consistent. I don't think I missed any cases because of the coprimality of ( n ) and ( n+1 ). So, yeah, I think 5 is the correct answer.**Final Answer**The number of positive integers ( n ) is boxed{5}.

💡First, I need to find the vector (overrightarrow{k} + overrightarrow{i}). Given that (overrightarrow{k} = 2overrightarrow{i} - 4overrightarrow{j}), adding (overrightarrow{i}) to it results in:[overrightarrow{k} + overrightarrow{i} = 2overrightarrow{i} - 4overrightarrow{j} + overrightarrow{i} = 3overrightarrow{i} - 4overrightarrow{j}]Next, to find the magnitude of this resultant vector, I use the formula for the magnitude of a vector in terms of its components. Since (overrightarrow{i}) and (overrightarrow{j}) are perpendicular unit vectors, the magnitude is calculated as:[|overrightarrow{k} + overrightarrow{i}| = sqrt{(3)^2 + (-4)^2} = sqrt{9 + 16} = sqrt{25} = 5]Therefore, the magnitude of (overrightarrow{k} + overrightarrow{i}) is 5.

💡Alright, so I have this problem here about primes and subsets. Let me try to wrap my head around it. The problem says:Let ( P ) be the set of all primes, and let ( M ) be a non-empty subset of ( P ). Suppose that for any non-empty subset ( {p_{1}, p_{2}, ldots, p_{k}} ) of ( M ), all prime factors of ( p_{1} p_{2} ldots p_{k}+1 ) are also in ( M ). Prove that ( M = P ).Hmm, okay. So, ( M ) is a subset of primes, and whenever I take any non-empty subset of ( M ), multiply them together, add 1, and then factor that result, all the prime factors of that result must also be in ( M ). And I need to show that ( M ) must be all primes.Let me think about what this means. If ( M ) isn't all primes, then there's some prime ( p ) not in ( M ). Maybe I can use that to create a contradiction.So, suppose ( M ) is not equal to ( P ). Then there exists at least one prime ( p ) that's not in ( M ). Now, what can I do with this prime ( p )?Well, since ( M ) is non-empty, let's pick some prime ( q ) from ( M ). Then, consider the subset ( {q} ) of ( M ). According to the problem statement, the prime factors of ( q + 1 ) must also be in ( M ).Wait, so ( q + 1 ) is a number, and all its prime factors must be in ( M ). But ( q ) is a prime, so ( q + 1 ) is one more than a prime. For example, if ( q = 2 ), then ( q + 1 = 3 ), which is prime. If ( q = 3 ), ( q + 1 = 4 ), whose prime factor is 2. If ( q = 5 ), ( q + 1 = 6 ), whose prime factors are 2 and 3.So, in each case, the prime factors of ( q + 1 ) are in ( M ). That doesn't immediately help me, but maybe if I consider larger subsets.What if I take two primes from ( M ), say ( p_1 ) and ( p_2 ). Then, the prime factors of ( p_1 p_2 + 1 ) must be in ( M ). Hmm.I remember something about Euclid's proof of the infinitude of primes. He used the idea that if you take a finite set of primes, multiply them together, add 1, and then the resulting number must have a prime factor not in the original set. Maybe I can use a similar idea here.But in this problem, the prime factors of ( p_1 p_2 ldots p_k + 1 ) must be in ( M ). So, if ( M ) is not all primes, then there's some prime ( p ) not in ( M ). If I can create a situation where ( p ) divides ( p_1 p_2 ldots p_k + 1 ), then that would mean ( p ) is a prime factor, which should be in ( M ), but it's not. Contradiction.So, how can I ensure that ( p ) divides ( p_1 p_2 ldots p_k + 1 ) for some subset ( {p_1, p_2, ldots, p_k} ) of ( M )?Maybe I can use modular arithmetic. Let's think about it. Suppose ( p ) is a prime not in ( M ). Let me consider the numbers ( 1, p + 1, p^2 + 1, p^3 + 1, ldots ). These are numbers of the form ( p^k + 1 ).Wait, but I need to relate this to products of primes in ( M ). Maybe I can construct a product of primes in ( M ) such that when I add 1, it's divisible by ( p ).Alternatively, think about the multiplicative inverse. If I can find a product of primes in ( M ) that is congruent to -1 modulo ( p ), then adding 1 would make it divisible by ( p ).So, if I can find ( p_1 p_2 ldots p_k equiv -1 mod p ), then ( p_1 p_2 ldots p_k + 1 equiv 0 mod p ), meaning ( p ) divides that product plus one, so ( p ) must be in ( M ), which is a contradiction.But how do I ensure that such a product exists? Maybe using the Chinese Remainder Theorem or something related to residues.Wait, another idea: if ( M ) is missing some prime ( p ), then ( p ) doesn't divide any of the primes in ( M ). So, each prime in ( M ) is either congruent to 1 modulo ( p ) or some other residue.But since ( M ) is a set of primes, they can't all be congruent to 1 modulo ( p ), unless ( p = 2 ) and all primes in ( M ) are odd, but even then, 2 is a prime.Wait, maybe I need to construct a sequence where each term is built from the previous one by multiplying some primes in ( M ) and adding 1, ensuring that eventually, I get a multiple of ( p ).This seems a bit abstract. Let me try with an example. Suppose ( M ) is missing the prime 2. Then, all primes in ( M ) are odd. If I take any product of odd primes, it's odd, and adding 1 makes it even, so 2 divides that number. But 2 is not in ( M ), which would mean that 2 must be in ( M ), a contradiction. So, in this case, ( M ) must contain 2.Similarly, suppose ( M ) is missing 3. Then, take a prime in ( M ), say 2. Then, 2 + 1 = 3, which is not in ( M ), but according to the problem statement, 3 should be in ( M ). So, 3 must be in ( M ).Wait, that seems to work. Let me try another one. Suppose ( M ) is missing 5. Then, take 2 and 3 in ( M ). Their product is 6, plus 1 is 7, which is prime. So, 7 must be in ( M ). Then, take 2, 3, 7. Their product is 42, plus 1 is 43, which is prime. So, 43 must be in ( M ). Continuing this way, I might generate more primes that must be in ( M ). But does this process necessarily lead to 5 being included?Alternatively, maybe I can construct a product that is congruent to -1 modulo 5. For example, 2 * 3 = 6 ≡ 1 mod 5. 2 * 3 * 7 = 42 ≡ 2 mod 5. 2 * 3 * 7 * 43 = 1806 ≡ 1 mod 5. Hmm, not sure if this is getting me closer.Wait, maybe I need to use the fact that the multiplicative group modulo ( p ) is cyclic. So, if I can find a product of primes in ( M ) that generates the entire multiplicative group, then I can get an element congruent to -1 mod ( p ).But I'm not sure about the details here. Maybe another approach.Suppose ( M ) is missing some prime ( p ). Then, consider the set of all primes in ( M ). Since ( p ) is not in ( M ), none of the primes in ( M ) are congruent to 0 mod ( p ). So, each prime in ( M ) is congruent to some non-zero residue mod ( p ).Now, consider the multiplicative inverses of these residues. If I can find a combination of these residues whose product is congruent to -1 mod ( p ), then the product of the corresponding primes in ( M ) would be congruent to -1 mod ( p ), and adding 1 would make it divisible by ( p ), forcing ( p ) to be in ( M ).But how do I ensure that such a combination exists? Maybe using the pigeonhole principle or something related to the size of ( M ).Wait, another thought: if ( M ) is missing ( p ), then the primes in ( M ) are all coprime to ( p ). So, their residues mod ( p ) are all invertible. Therefore, the set of residues forms a multiplicative subgroup of the multiplicative group modulo ( p ).If this subgroup is the entire multiplicative group, then it contains -1, and hence, there exists a product of primes in ( M ) congruent to -1 mod ( p ). If it's a proper subgroup, then its order divides ( p - 1 ). But since ( M ) is infinite (as it's a subset of all primes), the number of residues would be infinite, which can't form a finite subgroup. Therefore, the subgroup must be the entire multiplicative group, implying that -1 is in the subgroup.Thus, there exists a product of primes in ( M ) congruent to -1 mod ( p ), so ( p ) divides that product plus 1, forcing ( p ) to be in ( M ), a contradiction.Therefore, ( M ) cannot miss any prime ( p ), so ( M = P ).Wait, does this make sense? Let me check.If ( M ) is missing ( p ), then the primes in ( M ) are all coprime to ( p ), so their residues mod ( p ) are invertible. The set of these residues forms a multiplicative subgroup of ( (mathbb{Z}/pmathbb{Z})^* ). Since ( M ) is infinite, the number of residues is infinite, but ( (mathbb{Z}/pmathbb{Z})^* ) is finite of order ( p - 1 ). Therefore, the subgroup generated by these residues must be the entire group because you can't have an infinite subgroup in a finite group. Hence, the subgroup is the entire multiplicative group, which includes -1. Therefore, there exists a product of primes in ( M ) congruent to -1 mod ( p ), so ( p ) divides that product plus 1, implying ( p ) is in ( M ), which is a contradiction. Therefore, ( M ) must contain all primes.Yeah, that seems to work. So, the key idea is that if ( M ) misses any prime ( p ), then the multiplicative subgroup generated by the residues of primes in ( M ) modulo ( p ) must be the entire multiplicative group, which includes -1, leading to a contradiction. Therefore, ( M ) cannot miss any prime, so ( M = P ).I think that's the gist of it. Maybe I should write it more formally, but the intuition is there.

💡First, calculate the sums in the numerators and denominators:- The sum of the numerators in the first fraction: (3^2 + 5^2 + 7^2 = 9 + 25 + 49 = 83).- The sum of the denominators in the first fraction: (2^2 + 4^2 + 6^2 = 4 + 16 + 36 = 56).- The sum of the numerators in the second fraction (same as previous denominator): (2^2 + 4^2 + 6^2 = 56).- The sum of the denominators in the second fraction (same as previous numerator): (3^2 + 5^2 + 7^2 = 83).Next, form the fractions and simplify each:- The first fraction becomes (frac{83}{56}).- The second fraction becomes (frac{56}{83}).Both fractions are already in their simplest forms since 83 and 56 are coprime.Subtract the second fraction from the first fraction:[frac{83}{56} - frac{56}{83}]Find a common denominator to perform the subtraction:- The common denominator of (frac{83}{56}) and (frac{56}{83}) is (56 times 83).- Convert (frac{83}{56}) to (frac{83 times 83}{56 times 83} = frac{6889}{4656}).- Convert (frac{56}{83}) to (frac{56 times 56}{56 times 83} = frac{3136}{4656}).Perform the subtraction:[frac{6889}{4656} - frac{3136}{4656} = frac{6889 - 3136}{4656} = frac{3753}{4656}]Thus, the final result of the expression is (frac{3753}{4656}).

💡Okay, so I need to figure out the value of sin 1200 degrees. Hmm, that's a pretty big angle. I remember that sine is a periodic function, which means it repeats its values after a certain interval. I think the period for sine is 360 degrees, right? So, if I can reduce 1200 degrees to an angle between 0 and 360 degrees, that should make it easier to find the sine value.Let me start by dividing 1200 by 360 to see how many full rotations that is. So, 1200 divided by 360... Let me do that calculation. 360 times 3 is 1080, and 1200 minus 1080 is 120. So, 1200 degrees is the same as 3 full rotations plus 120 degrees. That means sin 1200 degrees is the same as sin 120 degrees.Alright, now I need to find sin 120 degrees. I remember that 120 degrees is in the second quadrant of the unit circle, where sine is positive. Also, 120 degrees can be expressed as 180 degrees minus 60 degrees. So, sin 120 degrees is the same as sin (180 - 60) degrees.I recall that sin (180 - θ) is equal to sin θ. So, sin 120 degrees is equal to sin 60 degrees. Now, sin 60 degrees is a standard value that I should remember. It's √3/2. Therefore, sin 120 degrees is also √3/2.Wait, let me double-check that. If I think about the unit circle, at 60 degrees, the coordinates are (1/2, √3/2). So, the sine value is the y-coordinate, which is √3/2. Since 120 degrees is in the second quadrant and sine is positive there, it makes sense that sin 120 degrees is also √3/2.So, putting it all together, sin 1200 degrees is equal to sin 120 degrees, which is √3/2. Looking back at the options provided:A: 1/2B: √3/2C: √2/2D: -√2/2The correct answer should be option B: √3/2.I think that makes sense. I don't see any mistakes in my reasoning. I reduced the angle by subtracting full rotations, identified the reference angle, and used the properties of the unit circle to find the sine value. Everything checks out.