💡Okay, so I have this problem about signatures of positive integers. The signature is defined as the sum of the exponents in the prime factorization of a number. For example, if a number n has a prime factorization of ( p_1^{alpha_1} p_2^{alpha_2} cdots p_k^{alpha_k} ), then its signature is ( alpha_1 + alpha_2 + cdots + alpha_k ). The question is asking whether there exist 2020 consecutive positive integers such that exactly 1812 of them have signatures strictly smaller than 11. Hmm, that's a mouthful. Let me try to break it down.First, I need to understand what the signature represents. It's essentially the total number of prime factors, counting multiplicities. So, for example, the number 12 has a prime factorization of ( 2^2 times 3^1 ), so its signature is ( 2 + 1 = 3 ). Similarly, a prime number like 7 would have a signature of 1 because it's just ( 7^1 ).So, the problem is about finding a sequence of 2020 consecutive numbers where 1812 of them have a signature less than 11. That means in this block of 2020 numbers, 1812 have a relatively small number of prime factors (counting multiplicities), and the remaining 208 have 11 or more prime factors.I think the key here is to use some number theory concepts, maybe the Chinese Remainder Theorem or something related to prime distributions. Let me think about how signatures behave in consecutive numbers.If I consider a range of numbers, their signatures can vary quite a bit. For example, numbers like primes have a signature of 1, while highly composite numbers can have much higher signatures. But in general, most numbers aren't going to have extremely high signatures. So, having 208 numbers in 2020 with signatures of 11 or more might be challenging, but maybe possible.Wait, the problem is asking if such a sequence exists, not necessarily to construct it. So maybe I can use some sort of counting argument or density argument.Let me think about the distribution of signatures. The signature function is related to the concept of "smoothness" in numbers, where a number is called smooth if all its prime factors are below a certain bound. But here, we're dealing with the sum of exponents, not the size of the prime factors.I recall that the number of integers with a given signature can be estimated using the concept of multiplicative functions and Dirichlet generating functions. But I'm not sure if that's the right approach here.Alternatively, maybe I can use the Chinese Remainder Theorem to construct a sequence where many numbers have high signatures. For example, if I can make many of the numbers in the sequence divisible by high powers of primes, their signatures would be large.Let me try to formalize this idea. Suppose I want 208 numbers in my sequence to have signatures of at least 11. If I can find 208 distinct primes and set up congruences such that each of these 208 numbers is divisible by a high power of one of these primes, then their signatures would be at least 11.But wait, how do I ensure that the other 1812 numbers have signatures less than 11? I need to make sure that these numbers are not divisible by high powers of primes, or that their prime factorizations don't accumulate too many exponents.This seems tricky. Maybe I can use the Chinese Remainder Theorem to construct a number n such that n, n+1, ..., n+2019 are each divisible by a distinct prime raised to the 11th power. Then, each of these numbers would have a signature of at least 11. But that would mean all 2020 numbers have signatures of at least 11, which is the opposite of what we want.Wait, but the problem is asking for exactly 1812 numbers with signatures less than 11. So, actually, we need 1812 numbers with low signatures and 208 numbers with high signatures.Maybe I can use the Chinese Remainder Theorem to construct a sequence where 208 specific numbers in the sequence have high signatures, while the rest are arranged to have low signatures. For example, set up the sequence so that n + k is divisible by a high power of a prime for k in some subset of size 208, and the rest are arranged to be primes or have low exponents.But how do I ensure that the rest of the numbers have low signatures? If I don't control their prime factors, they might still have high signatures by chance. Maybe I need to make sure that the other numbers are primes or products of a small number of primes with low exponents.Alternatively, perhaps I can use the concept of "adjoining" primes in such a way that only specific numbers in the sequence have high signatures. For example, if I can make n + k divisible by a high power of a prime for k in a specific set, and the other numbers are constructed to be primes or have minimal exponents.But this seems complicated. Maybe there's a simpler way. Let me think about the function f(n) which counts the number of integers in [n, n+2019] with signature less than 11. We need to show that f(n) can take the value 1812 for some n.We know that for small n, say n=1, f(n)=2020 because all numbers up to 2020 have signatures less than 11. As n increases, the numbers in the interval [n, n+2019] can have higher signatures, so f(n) can decrease.Eventually, for very large n, we might have f(n)=0, meaning all numbers in the interval have signatures of at least 11. But does f(n) decrease smoothly from 2020 to 0, or does it jump in some places?If f(n) can only decrease by at most 1 each time n increases by 1, then by the intermediate value theorem, it must take every integer value between 2020 and 0, including 1812. But is this the case?Wait, when n increases by 1, the interval [n, n+2019] shifts by 1. So, the number leaving the interval is n, and the number entering is n+2020. The change in f(n) depends on whether n had a signature less than 11 and whether n+2020 has a signature less than 11.Therefore, f(n+1) = f(n) - (1 if σ(n) < 11 else 0) + (1 if σ(n+2020) < 11 else 0).So, the change in f(n) is either -1, 0, or +1. It can decrease by 1 if n had σ(n) < 11 and n+2020 has σ(n+2020) ≥ 11. It can stay the same if both n and n+2020 have σ < 11 or both have σ ≥ 11. It can increase by 1 if n had σ(n) ≥ 11 and n+2020 has σ(n+2020) < 11.Therefore, f(n) is a function that changes by at most 1 when n increases by 1. So, it's a function that can decrease, stay the same, or increase by 1 each time.We know that f(1)=2020, and as n increases, f(n) can decrease. But is there a point where f(n)=0? If yes, then since f(n) changes by at most 1 each time, it must pass through every integer value between 2020 and 0, including 1812.So, the key is to show that f(n) can reach 0. If we can find an n such that all numbers in [n, n+2019] have signatures of at least 11, then f(n)=0, and by the intermediate value theorem, there must be some n where f(n)=1812.How can we ensure that all numbers in [n, n+2019] have signatures of at least 11? One way is to use the Chinese Remainder Theorem to construct such an n. For each number in the interval, we can set it to be divisible by a distinct prime raised to the 11th power. This way, each number in the interval has at least one prime factor with exponent 11, making their signature at least 11.Let me elaborate. Suppose we choose 2020 distinct primes p_1, p_2, ..., p_{2020}. Then, we can set up the following system of congruences:n ≡ 0 mod p_1^{11}n + 1 ≡ 0 mod p_2^{11}n + 2 ≡ 0 mod p_3^{11}...n + 2019 ≡ 0 mod p_{2020}^{11}By the Chinese Remainder Theorem, since all the moduli p_i^{11} are pairwise coprime (because the p_i are distinct primes), there exists a solution n to this system. This n will satisfy that n + k is divisible by p_{k+1}^{11} for each k from 0 to 2019. Therefore, each number in [n, n+2019] is divisible by a distinct prime raised to the 11th power, ensuring that their signatures are at least 11.Thus, f(n)=0 for such an n. Since f(n) starts at 2020 and can decrease to 0, and changes by at most 1 each time, it must take every integer value in between, including 1812. Therefore, there exists an n such that f(n)=1812.Wait, but does this construction actually work? I mean, the Chinese Remainder Theorem guarantees a solution, but does it necessarily make all the numbers in the interval have signatures exactly 11, or could they have higher signatures?Well, in this construction, each number n + k is divisible by p_{k+1}^{11}, so their signatures are at least 11. However, they might have higher signatures if they have other prime factors with exponents. For example, n + k could be divisible by p_{k+1}^{11} and another prime, which would add to the signature.But in our case, we only care that the signature is at least 11, not exactly 11. So, even if some numbers have higher signatures, it doesn't affect our count. We just need at least 208 numbers in the interval to have signatures ≥11, which is satisfied by this construction.Therefore, the argument holds. Since f(n) can reach 0, and it changes by at most 1 each time, it must pass through 1812. Hence, such a sequence of 2020 consecutive integers exists.I think this makes sense. The key steps are:1. Define the function f(n) counting the number of integers in [n, n+2019] with signature <11.2. Show that f(n) starts at 2020 and can decrease to 0.3. Use the Chinese Remainder Theorem to construct an n where f(n)=0.4. Since f(n) changes by at most 1 each time, it must take every integer value between 2020 and 0, including 1812.Therefore, the answer should be yes, such a sequence exists.

💡Okay, so I have this triangle ABC, and I know the sides opposite angles A, B, and C are a, b, and c respectively. The given information is that b is 2√3, c is 2, and there's this equation: b sin C - 2c sin B cos A = 0. I need to find the area of the triangle first, and then work on this function f(x).Starting with part (1), finding the area. I remember that the area of a triangle can be found using the formula (1/2)ab sin C, but in this case, I might need to use some trigonometric identities or laws like the Law of Sines or Cosines. Let me see.Given the equation: b sin C - 2c sin B cos A = 0. Let me write that down:b sin C - 2c sin B cos A = 0.I know that in a triangle, the Law of Sines says that a/sin A = b/sin B = c/sin C = 2R, where R is the radius of the circumscribed circle. Maybe I can use that to express sin C and sin B in terms of sides.So, from Law of Sines:sin B = b/(2R) and sin C = c/(2R).But I'm not sure if that's the best approach. Alternatively, maybe I can express sin C in terms of sin B and cos A.Wait, another idea: maybe I can use the Law of Cosines on angle A. The Law of Cosines says that a² = b² + c² - 2bc cos A. But I don't know a yet, so maybe that's not directly helpful.Alternatively, let's see if I can manipulate the given equation.Given: b sin C - 2c sin B cos A = 0.Let me rearrange it:b sin C = 2c sin B cos A.Hmm, can I express sin C in terms of sin B? Or maybe express sin C using the Law of Sines.From Law of Sines, sin C / c = sin B / b, so sin C = (c / b) sin B.So substituting into the equation:b * (c / b) sin B = 2c sin B cos A.Simplify:c sin B = 2c sin B cos A.Assuming sin B ≠ 0 (since B is an angle in a triangle, it can't be 0 or π), we can divide both sides by sin B:c = 2c cos A.Divide both sides by c (c ≠ 0):1 = 2 cos A.So cos A = 1/2.Therefore, angle A is π/3 or 60 degrees.Okay, that's a key piece of information. So angle A is 60 degrees.Now, to find the area. The area S of triangle ABC can be found using the formula:S = (1/2) * b * c * sin A.We know b = 2√3, c = 2, and sin A = sin(60°) = √3/2.So plugging in:S = (1/2) * 2√3 * 2 * (√3/2).Let me compute that step by step.First, 1/2 * 2√3 = √3.Then, √3 * 2 = 2√3.Then, 2√3 * (√3/2) = (2√3 * √3)/2 = (2*3)/2 = 6/2 = 3.So the area S is 3.Alright, that seems straightforward. So part (1) is done, the area is 3.Now, moving on to part (2). The function f(x) is given as:f(x) = 4 cos x (sin x cos A + cos x sin A).And x is in the interval [0, π/3]. I need to find the maximum value of f(x) and the corresponding x.First, let's note that angle A is π/3, as found earlier. So cos A = 1/2 and sin A = √3/2.So let's substitute cos A and sin A into f(x):f(x) = 4 cos x (sin x * (1/2) + cos x * (√3/2)).Simplify inside the parentheses:sin x * (1/2) + cos x * (√3/2) = (1/2) sin x + (√3/2) cos x.So f(x) = 4 cos x [ (1/2) sin x + (√3/2) cos x ].Let me factor out the 1/2:= 4 cos x * [ (1/2)(sin x + √3 cos x) ]= 4 * (1/2) cos x (sin x + √3 cos x)= 2 cos x (sin x + √3 cos x).So f(x) simplifies to 2 cos x (sin x + √3 cos x).Let me expand this:= 2 cos x sin x + 2√3 cos² x.I can recognize that 2 cos x sin x is sin(2x), and cos² x can be expressed using the double-angle identity.So let's rewrite:= sin(2x) + 2√3 cos² x.But cos² x = (1 + cos 2x)/2, so:= sin(2x) + 2√3 * (1 + cos 2x)/2= sin(2x) + √3 (1 + cos 2x)= sin(2x) + √3 + √3 cos 2x.So f(x) = sin(2x) + √3 cos 2x + √3.Hmm, this looks like a sinusoidal function. Maybe I can write it in the form R sin(2x + φ) + √3, where R is the amplitude and φ is the phase shift.Let me recall that a sin θ + b cos θ = R sin(θ + φ), where R = √(a² + b²) and tan φ = b/a.In this case, a = 1 (coefficient of sin 2x) and b = √3 (coefficient of cos 2x).So R = √(1² + (√3)²) = √(1 + 3) = √4 = 2.And tan φ = b/a = √3 / 1 = √3, so φ = π/3.Therefore, sin(2x) + √3 cos 2x = 2 sin(2x + π/3).So f(x) = 2 sin(2x + π/3) + √3.Now, to find the maximum value of f(x), we know that the sine function has a maximum value of 1. So the maximum of 2 sin(2x + π/3) is 2*1 = 2. Therefore, the maximum of f(x) is 2 + √3.Now, we need to find the value of x in [0, π/3] where this maximum occurs.The maximum of sin(θ) is 1 when θ = π/2 + 2π k, where k is an integer.So, set 2x + π/3 = π/2.Solving for x:2x = π/2 - π/3 = (3π/6 - 2π/6) = π/6.So x = π/12.We need to check if x = π/12 is within the interval [0, π/3]. Since π/12 ≈ 0.2618 and π/3 ≈ 1.0472, yes, it is within the interval.Therefore, the maximum value of f(x) is 2 + √3, and it occurs at x = π/12.Wait, let me double-check my steps to make sure I didn't make a mistake.Starting from f(x) = 4 cos x (sin x cos A + cos x sin A). Substituted cos A = 1/2 and sin A = √3/2 correctly.Simplified to 2 cos x (sin x + √3 cos x), then expanded to 2 cos x sin x + 2√3 cos² x, which is sin(2x) + √3 (1 + cos 2x). Then, combined to sin(2x) + √3 cos 2x + √3.Expressed as 2 sin(2x + π/3) + √3. That seems correct because sin(2x) + √3 cos(2x) is 2 sin(2x + π/3). Then, the maximum is 2 + √3 at x = π/12.Yes, that seems correct.So, summarizing:1) The area of triangle ABC is 3.2) The maximum value of f(x) is 2 + √3, occurring at x = π/12.**Final Answer**(1) The area of the triangle is boxed{3}.(2) The maximum value of the function is boxed{2 + sqrt{3}}, and it occurs at ( x = boxed{dfrac{pi}{12}} ).

💡First, I need to determine what 25% of 88 is. To do this, I'll calculate 25% of 88 by multiplying 88 by 0.25.Next, I'll add the result from the first step to the original number, 88, to find the value of x that is 25% greater than 88.Finally, I'll present the final value of x in a clear and concise manner.

💡Okay, so I have this problem here: If b over a is 3 and c over b is 2, what is the ratio of a plus b to b plus c? The options are A) 3/8, B) 2/5, C) 4/9, D) 2/3, and E) 3/4. Hmm, let me try to figure this out step by step.First, let's write down what we know. We have two ratios: b/a = 3 and c/b = 2. I need to find the ratio (a + b)/(b + c). Maybe I can express a and c in terms of b to make things easier.Starting with b/a = 3. If I rearrange this, I can solve for a. So, if b/a = 3, then a = b/3. Got that. So, a is one-third of b.Next, c/b = 2. Similarly, solving for c gives me c = 2b. So, c is twice as big as b. Okay, so now I have a in terms of b and c in terms of b.Now, I need to find (a + b)/(b + c). Let's substitute the expressions we found for a and c. So, a is b/3, and c is 2b. Plugging those in:(a + b) becomes (b/3 + b), and (b + c) becomes (b + 2b). Let me compute each part separately.For the numerator, b/3 + b. To add these, they need a common denominator. So, b is the same as 3b/3. Therefore, b/3 + 3b/3 = (1b + 3b)/3 = 4b/3.For the denominator, b + 2b is straightforward. That's just 3b.So now, the ratio becomes (4b/3) divided by (3b). Dividing fractions is the same as multiplying by the reciprocal, so (4b/3) * (1/(3b)). Let's compute that.Multiplying the numerators: 4b * 1 = 4b.Multiplying the denominators: 3 * 3b = 9b.So, the ratio is 4b/9b. The b's cancel out, leaving 4/9.Wait, so the ratio of a + b to b + c is 4/9. Looking back at the options, that's option C.Let me double-check my steps to make sure I didn't make a mistake. Starting from b/a = 3, so a = b/3. Then c/b = 2, so c = 2b. Substituting into (a + b)/(b + c) gives (b/3 + b)/(b + 2b). Simplifying numerator: b/3 + 3b/3 = 4b/3. Denominator: 3b. So, 4b/3 divided by 3b is 4/9. Yep, that seems right.I think I got it. The answer should be 4/9, which is option C.

💡First, I need to determine how many hours Mike watches TV each day. Let's denote the number of hours he watches TV every day as ( T ).On the days he plays video games, he plays for half as long as he watches TV, which means he plays for ( frac{T}{2} ) hours.Mike plays video games 3 days a week, so the total time he spends playing video games in a week is:[3 times frac{T}{2} = frac{3T}{2} text{ hours}]Since he watches TV every day, the total time he spends watching TV in a week is:[7 times T = 7T text{ hours}]The total time spent on both activities in a week is given as 34 hours. Therefore, we can set up the equation:[7T + frac{3T}{2} = 34]To solve for ( T ), I'll first find a common denominator for the terms on the left side of the equation. The common denominator for 7 and ( frac{3}{2} ) is 2. Multiplying the first term by ( frac{2}{2} ) gives:[frac{14T}{2} + frac{3T}{2} = 34]Combining the terms on the left side:[frac{17T}{2} = 34]Next, I'll multiply both sides by 2 to eliminate the denominator:[17T = 68]Finally, I'll divide both sides by 17 to solve for ( T ):[T = frac{68}{17} = 4]Therefore, Mike watches TV for 4 hours every day.

💡Okay, so I have this quadratic equation: ( x^2 + 2kx + 3k^2 = 0 ). I need to find the real values of ( k ) for which this equation has real and distinct roots. Hmm, I remember that for a quadratic equation ( ax^2 + bx + c = 0 ), the discriminant ( Delta = b^2 - 4ac ) determines the nature of the roots. If ( Delta > 0 ), the equation has two distinct real roots. If ( Delta = 0 ), it has exactly one real root (a repeated root), and if ( Delta < 0 ), it has no real roots, meaning the roots are complex.So, applying this to my equation, let's identify the coefficients first. Here, ( a = 1 ), ( b = 2k ), and ( c = 3k^2 ). Now, let's compute the discriminant:( Delta = (2k)^2 - 4 times 1 times 3k^2 )Calculating that, ( (2k)^2 = 4k^2 ) and ( 4 times 1 times 3k^2 = 12k^2 ). So,( Delta = 4k^2 - 12k^2 = -8k^2 )Wait a minute, that's negative because ( -8k^2 ) is always negative or zero, right? Since ( k^2 ) is always non-negative, multiplying it by -8 makes it non-positive. So, ( Delta = -8k^2 ) is less than or equal to zero for all real ( k ).But we need ( Delta > 0 ) for real and distinct roots. Since ( -8k^2 ) can never be positive, there are no real values of ( k ) that satisfy this condition. That means the equation doesn't have real and distinct roots for any real ( k ).Let me double-check my calculations to be sure I didn't make a mistake. Coefficients: ( a = 1 ), ( b = 2k ), ( c = 3k^2 ). Discriminant: ( (2k)^2 - 4(1)(3k^2) = 4k^2 - 12k^2 = -8k^2 ). Yep, that's correct. So, the discriminant is negative for all real ( k ), meaning the equation has no real roots, let alone distinct ones.Therefore, the answer must be that there are no real values of ( k ) for which the equation has real and distinct roots. That corresponds to option E.