💡Okay, so I have this problem here: "How many more digits does the base-4 representation of 1234 base 10 have than the base-9 representation of 1234 base 10?" Hmm, I need to figure out the number of digits in both base-4 and base-9 for the number 1234, and then find the difference between those two numbers of digits.Alright, let's start by recalling how to convert a base-10 number to another base. I think the general method involves dividing the number by the base and keeping track of the remainders. But since I'm only interested in the number of digits, maybe I don't need to fully convert the number. Instead, I can find the highest power of the base that fits into the number, and that will give me the number of digits.Let me verify that thought. For example, in base-10, the number of digits in a number is determined by the highest power of 10 that is less than or equal to the number. So, for 1234, the highest power of 10 less than or equal to 1234 is 10^3 = 1000, which means it has 4 digits. That makes sense. So, applying the same logic, for base-4 and base-9, I need to find the highest power of 4 and 9 that is less than or equal to 1234, respectively.Alright, let's tackle base-4 first. I need to find the largest exponent n such that 4^n ≤ 1234. Let me compute the powers of 4:4^1 = 44^2 = 164^3 = 644^4 = 2564^5 = 10244^6 = 4096Wait, 4^6 is 4096, which is greater than 1234. So, the largest power of 4 less than or equal to 1234 is 4^5 = 1024. That means the number of digits in base-4 is 5 + 1 = 6 digits. Okay, so base-4 representation of 1234 has 6 digits.Now, moving on to base-9. Similarly, I need to find the largest exponent m such that 9^m ≤ 1234. Let me compute the powers of 9:9^1 = 99^2 = 819^3 = 7299^4 = 6561Wait, 9^4 is 6561, which is way larger than 1234. So, the largest power of 9 less than or equal to 1234 is 9^3 = 729. Therefore, the number of digits in base-9 is 3 + 1 = 4 digits. So, base-9 representation of 1234 has 4 digits.Now, to find how many more digits the base-4 representation has compared to base-9, I subtract the number of digits in base-9 from the number of digits in base-4: 6 - 4 = 2.Wait, let me double-check my calculations to make sure I didn't make a mistake. For base-4, 4^5 is 1024, which is less than 1234, and 4^6 is 4096, which is more than 1234. So, yes, 5 + 1 = 6 digits. For base-9, 9^3 is 729, less than 1234, and 9^4 is 6561, which is way more. So, 3 + 1 = 4 digits. The difference is 6 - 4 = 2.Hmm, seems straightforward. But just to be thorough, maybe I should actually convert 1234 into base-4 and base-9 to confirm the number of digits.Starting with base-4. To convert 1234 to base-4, I can use the division method:1234 ÷ 4 = 308 with a remainder of 2308 ÷ 4 = 77 with a remainder of 077 ÷ 4 = 19 with a remainder of 119 ÷ 4 = 4 with a remainder of 34 ÷ 4 = 1 with a remainder of 01 ÷ 4 = 0 with a remainder of 1So, writing the remainders from last to first, 1234 in base-4 is 1 0 3 1 0 2, which is 103102_4. That's 6 digits, which matches my earlier calculation.Now, converting 1234 to base-9:1234 ÷ 9 = 137 with a remainder of 1137 ÷ 9 = 15 with a remainder of 215 ÷ 9 = 1 with a remainder of 61 ÷ 9 = 0 with a remainder of 1So, writing the remainders from last to first, 1234 in base-9 is 1 6 2 1, which is 1621_9. That's 4 digits, which also matches my earlier calculation.Therefore, the base-4 representation has 6 digits, and the base-9 representation has 4 digits. The difference is 2 digits.I think that's solid. I don't see any mistakes in my reasoning or calculations. So, the answer should be 2 more digits in base-4 compared to base-9.

💡First, identify the number of fruit salads at Alaya's restaurant, which is 200.Next, calculate the number of fruit salads at Angel's restaurant by doubling Alaya's amount: 2 * 200 = 400.Then, determine the number of fruit salads at Betty's restaurant by tripling Angel's amount: 3 * 400 = 1200.Finally, add the number of fruit salads from all three restaurants to find the total: 200 + 400 + 1200 = 1800.

💡First, I need to understand what the "opposite" of a number means. The opposite of a number is its additive inverse, which means it is the number that, when added to the original number, results in zero.Given the number (frac{1}{6}), I need to find its additive inverse. To do this, I simply change the sign of the number. Since (frac{1}{6}) is positive, its opposite will be negative.Therefore, the opposite of (frac{1}{6}) is (-frac{1}{6}).Now, I will review the answer choices:- A: (frac{1}{6})- B: (-6)- C: (6)- D: (-frac{1}{6})The correct answer is D: (-frac{1}{6}).

💡Okay, so I have this equation: x³ - 12x² + 48x + 28 = 0. I need to find the product of its roots. Hmm, I remember something about Vieta's formulas from algebra class. Let me try to recall how that works.Vieta's formulas relate the coefficients of a polynomial to sums and products of its roots. For a cubic equation like ax³ + bx² + cx + d = 0, there are three roots, say r1, r2, and r3. I think the product of the roots is related to the constant term and the leading coefficient. Wait, is it -d/a or d/a? I'm a bit confused. Let me think. For a quadratic equation, like ax² + bx + c = 0, the product of the roots is c/a. So for a cubic equation, maybe it's similar but with a sign change? Let me check the formula again. For a cubic equation ax³ + bx² + cx + d = 0, the product of the roots r1 * r2 * r3 is equal to -d/a. Yeah, I think that's right because in the quadratic case, it's c/a, and for cubic, it's -d/a. So in my equation, the coefficients are a = 1, b = -12, c = 48, and d = 28. Plugging into the formula, the product should be -d/a, which is -28/1, so -28. Wait, let me make sure I didn't mix up anything. The equation is x³ - 12x² + 48x + 28 = 0, so a is positive 1, d is positive 28. So -d/a is -28. That seems right. I don't think I need to find the actual roots because Vieta's formula gives me the product directly. So, yeah, the product of the roots is -28.

💡Okay, so I have this problem about triangle ABC. The sides opposite angles A, B, and C are a, b, c respectively. There are two vectors given: vector m is (cos A, cos B) and vector n is (a, 2c - b). It's also given that vector m is parallel to vector n. Part I asks me to find the measure of angle A. Hmm, okay. Since the vectors are parallel, that means one is a scalar multiple of the other. So, if vector m is parallel to vector n, there must be some scalar k such that m = k * n. That would mean each component of m is k times the corresponding component of n. So, cos A = k * a and cos B = k * (2c - b). Wait, so if I write that out, it's:cos A = k * a cos B = k * (2c - b)Since both equalities involve k, I can set up a ratio. If I divide the first equation by the second, I get:cos A / cos B = (k * a) / (k * (2c - b)) The k's cancel out, so cos A / cos B = a / (2c - b)So, that gives me an equation: (2c - b) * cos A = a * cos B.Hmm, okay. Now, I remember the Law of Sines, which 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 here.Let me express a, b, and c in terms of the sines of their opposite angles. So, a = 2R sin A, b = 2R sin B, c = 2R sin C.Substituting these into the equation (2c - b) * cos A = a * cos B:(2*(2R sin C) - 2R sin B) * cos A = (2R sin A) * cos B Simplify that:(4R sin C - 2R sin B) * cos A = 2R sin A cos B I can factor out 2R on both sides:2R (2 sin C - sin B) * cos A = 2R sin A cos B Divide both sides by 2R:(2 sin C - sin B) * cos A = sin A cos BOkay, so now I have:(2 sin C - sin B) * cos A = sin A cos BHmm, let's see. In triangle ABC, the sum of angles is π, so A + B + C = π. Therefore, C = π - A - B.So, sin C = sin(π - A - B) = sin(A + B). Because sin(π - x) = sin x.So, sin C = sin(A + B). Let me substitute that into the equation:(2 sin(A + B) - sin B) * cos A = sin A cos BLet me expand sin(A + B):sin(A + B) = sin A cos B + cos A sin BSo, substituting that in:[2(sin A cos B + cos A sin B) - sin B] * cos A = sin A cos BLet me distribute the 2:[2 sin A cos B + 2 cos A sin B - sin B] * cos A = sin A cos BFactor sin B in the last two terms:[2 sin A cos B + sin B (2 cos A - 1)] * cos A = sin A cos BNow, let's distribute cos A:2 sin A cos B cos A + sin B (2 cos A - 1) cos A = sin A cos BHmm, this is getting a bit complicated. Let me see if I can rearrange terms.Bring everything to one side:2 sin A cos B cos A + sin B (2 cos A - 1) cos A - sin A cos B = 0Factor sin A cos B from the first and last terms:sin A cos B (2 cos A - 1) + sin B (2 cos A - 1) cos A = 0Wait, both terms have a factor of (2 cos A - 1). Let me factor that out:(2 cos A - 1)(sin A cos B + sin B cos A) = 0Oh, interesting! So, either 2 cos A - 1 = 0 or sin A cos B + sin B cos A = 0.But sin A cos B + sin B cos A is equal to sin(A + B). Because sin(x + y) = sin x cos y + cos x sin y.So, sin(A + B) = 0 or 2 cos A - 1 = 0.But in a triangle, angles A and B are between 0 and π, so A + B is between 0 and π. Therefore, sin(A + B) can't be zero because that would imply A + B = 0 or π, which isn't possible in a triangle. So, sin(A + B) ≠ 0, which means the other factor must be zero:2 cos A - 1 = 0 So, 2 cos A = 1 cos A = 1/2Therefore, angle A is arccos(1/2). The arccos of 1/2 is π/3 or 60 degrees.Okay, so that's part I. Angle A is π/3 radians or 60 degrees.Now, moving on to part II. It says if a = 2√5, find the maximum area of triangle ABC.Alright, so a is given as 2√5. We need to find the maximum area. I remember that the area of a triangle can be given by (1/2)ab sin C, or in this case, maybe using sides and angles.But since we know angle A is π/3, maybe we can express the area in terms of sides b and c.Wait, let's recall the formula for area: (1/2) * b * c * sin A. Since A is π/3, sin A is √3/2. So, area = (1/2) * b * c * (√3/2) = (√3/4) * b * c.So, to maximize the area, we need to maximize the product b * c.But we have constraints because it's a triangle with side a = 2√5 opposite angle A = π/3.From the Law of Cosines, we can relate sides a, b, c. The Law of Cosines says:a² = b² + c² - 2bc cos AWe know a = 2√5, so a² = (2√5)² = 4 * 5 = 20And cos A = cos(π/3) = 1/2So, plugging into the formula:20 = b² + c² - 2bc*(1/2) Simplify:20 = b² + c² - bcSo, we have the equation: b² + c² - bc = 20We need to maximize bc given this constraint.Hmm, okay. So, we have an equation involving b², c², and bc, and we need to maximize bc.I think this is a problem that can be approached using the method of Lagrange multipliers, but maybe there's a simpler way.Alternatively, we can consider that for fixed a and angle A, the area is maximized when the triangle is such that sides b and c are arranged to make the area as large as possible.Wait, another thought: perhaps using the AM-GM inequality or some other inequality to find the maximum of bc.But let's see.We have b² + c² - bc = 20We can write this as b² + c² = bc + 20We need to maximize bc.Let me denote x = bc.So, b² + c² = x + 20But I also know that for any real numbers b and c, b² + c² ≥ 2bc, by the AM-GM inequality.So, x + 20 = b² + c² ≥ 2xSo, x + 20 ≥ 2x Subtract x from both sides:20 ≥ xSo, x ≤ 20Therefore, the maximum value of bc is 20.Wait, so the maximum product bc is 20.Therefore, the maximum area is (√3/4) * 20 = (√3/4)*20 = 5√3.But wait, is this achievable? Because in the AM-GM inequality, equality holds when b = c.So, if b = c, then we can check if that satisfies the equation.If b = c, then the equation becomes:b² + b² - b*b = 20 2b² - b² = 20 b² = 20 b = √20 = 2√5So, b = c = 2√5. Then, a = 2√5 as given.So, in this case, the triangle would have sides a = 2√5, b = 2√5, c = 2√5. Wait, that's an equilateral triangle? Because all sides are equal.But angle A is π/3, which is 60 degrees, which is consistent with an equilateral triangle.But wait, in an equilateral triangle, all angles are 60 degrees, so that's consistent.Therefore, the maximum area is indeed 5√3.Wait, but let me double-check.If b = c = 2√5, then the area is (√3/4) * (2√5)^2 = (√3/4) * 20 = 5√3. Yep, that's correct.But hold on, is that the only case when bc is maximized? Because sometimes, depending on the constraints, the maximum might not be achieved at equality in AM-GM.But in this case, since we have the equation b² + c² - bc = 20, and when b = c, it satisfies the equation, so it's achievable.Therefore, the maximum area is 5√3.So, summarizing:I. Angle A is π/3 radians or 60 degrees.II. The maximum area is 5√3.**Final Answer**(I) The measure of angle ( A ) is (boxed{dfrac{pi}{3}}).(II) The maximum area of (triangle ABC) is (boxed{5sqrt{3}}).

💡Okay, so I have this problem where a sequence is defined by a₁ = 1, a₂ = 2, and then a recursive formula a_{n+2} = 2a_{n+1} - p a_n for all n ≥ 1, where p is a prime number. I need to find all primes p such that there exists some m where a_m = -3.Hmm, let me start by understanding the sequence better. The first two terms are given, and each subsequent term is a linear combination of the two previous terms. This looks like a linear recurrence relation. Maybe I can solve it using characteristic equations?The characteristic equation for the recurrence relation a_{n+2} = 2a_{n+1} - p a_n would be r² - 2r + p = 0. Let me find the roots of this equation.Using the quadratic formula, the roots are [2 ± sqrt(4 - 4p)] / 2 = 1 ± sqrt(1 - p). So, the roots are 1 + sqrt(1 - p) and 1 - sqrt(1 - p). Depending on whether the discriminant is positive, zero, or negative, the roots will be real and distinct, repeated, or complex.But since p is a prime number, and primes are positive integers greater than 1, 1 - p will be negative for p ≥ 2. So, the roots are complex numbers. That means the solution to the recurrence relation will involve sines and cosines or can be expressed in terms of complex exponentials.Alternatively, maybe I can write the solution in terms of real numbers using the amplitude and phase form. But perhaps that's getting too ahead of myself.Wait, maybe instead of solving the recurrence relation directly, I can compute the first few terms of the sequence for different primes p and see if -3 appears.Let me try p = 2 first. Then the recurrence is a_{n+2} = 2a_{n+1} - 2a_n.Compute the terms:a₁ = 1a₂ = 2a₃ = 2*2 - 2*1 = 4 - 2 = 2a₄ = 2*2 - 2*2 = 4 - 4 = 0a₅ = 2*0 - 2*2 = 0 - 4 = -4a₆ = 2*(-4) - 2*0 = -8 - 0 = -8a₇ = 2*(-8) - 2*(-4) = -16 + 8 = -8a₈ = 2*(-8) - 2*(-8) = -16 + 16 = 0a₉ = 2*0 - 2*(-8) = 0 + 16 = 16a₁₀ = 2*16 - 2*0 = 32 - 0 = 32Hmm, I don't see -3 appearing here. Maybe p=2 isn't the right prime.Let's try p=3. Then the recurrence is a_{n+2} = 2a_{n+1} - 3a_n.Compute the terms:a₁ = 1a₂ = 2a₃ = 2*2 - 3*1 = 4 - 3 = 1a₄ = 2*1 - 3*2 = 2 - 6 = -4a₅ = 2*(-4) - 3*1 = -8 - 3 = -11a₆ = 2*(-11) - 3*(-4) = -22 + 12 = -10a₇ = 2*(-10) - 3*(-11) = -20 + 33 = 13a₈ = 2*13 - 3*(-10) = 26 + 30 = 56a₉ = 2*56 - 3*13 = 112 - 39 = 73a₁₀ = 2*73 - 3*56 = 146 - 168 = -22Still, I don't see -3 here. Maybe p=3 isn't the right prime either.Let's try p=5. Then the recurrence is a_{n+2} = 2a_{n+1} - 5a_n.Compute the terms:a₁ = 1a₂ = 2a₃ = 2*2 - 5*1 = 4 - 5 = -1a₄ = 2*(-1) - 5*2 = -2 - 10 = -12a₅ = 2*(-12) - 5*(-1) = -24 + 5 = -19a₆ = 2*(-19) - 5*(-12) = -38 + 60 = 22a₇ = 2*22 - 5*(-19) = 44 + 95 = 139a₈ = 2*139 - 5*22 = 278 - 110 = 168a₉ = 2*168 - 5*139 = 336 - 695 = -359a₁₀ = 2*(-359) - 5*168 = -718 - 840 = -1558Hmm, -3 doesn't show up here either. Maybe p=5 isn't the right prime.Wait, let's try p=7. Then the recurrence is a_{n+2} = 2a_{n+1} - 7a_n.Compute the terms:a₁ = 1a₂ = 2a₃ = 2*2 - 7*1 = 4 - 7 = -3Oh! There it is. a₃ = -3. So, for p=7, we have a term equal to -3 at m=3.Let me check p=11 just to be thorough. Then the recurrence is a_{n+2} = 2a_{n+1} - 11a_n.Compute the terms:a₁ = 1a₂ = 2a₃ = 2*2 - 11*1 = 4 - 11 = -7a₄ = 2*(-7) - 11*2 = -14 - 22 = -36a₅ = 2*(-36) - 11*(-7) = -72 + 77 = 5a₆ = 2*5 - 11*(-36) = 10 + 396 = 406a₇ = 2*406 - 11*5 = 812 - 55 = 757a₈ = 2*757 - 11*406 = 1514 - 4466 = -2952a₉ = 2*(-2952) - 11*757 = -5904 - 8327 = -14231a₁₀ = 2*(-14231) - 11*(-2952) = -28462 + 32472 = 4010No -3 here either. So, p=11 doesn't work.Wait, so p=7 works because a₃ = -3. Let me see if there are other primes where this could happen.Looking back at p=2,3,5,7,11, only p=7 gives a term equal to -3. Maybe I should check p=13.For p=13, recurrence is a_{n+2}=2a_{n+1} -13a_n.Compute terms:a₁=1a₂=2a₃=2*2 -13*1=4-13=-9a₄=2*(-9)-13*2=-18-26=-44a₅=2*(-44)-13*(-9)=-88+117=29a₆=2*29 -13*(-44)=58 + 572=630a₇=2*630 -13*29=1260 - 377=883a₈=2*883 -13*630=1766 - 8190=-6424a₉=2*(-6424)-13*883=-12848 -11479=-24327a₁₀=2*(-24327)-13*(-6424)=-48654 +83512=34858No -3 here either. So p=13 doesn't work.Wait, maybe I should think about this differently. Instead of computing term by term for each prime, perhaps there's a pattern or a property that can help me find p.Given the recurrence relation, the characteristic equation is r² - 2r + p = 0, with roots 1 ± sqrt(1 - p). Since p is prime, 1 - p is negative, so the roots are complex: 1 ± i*sqrt(p - 1).Therefore, the general solution is a_n = α*(1 + i*sqrt(p - 1))^n + β*(1 - i*sqrt(p - 1))^n.But since the coefficients are real, we can express this in terms of real numbers using Euler's formula. So, the solution can be written as a_n = C*λ^n * cos(nθ) + D*λ^n * sin(nθ), where λ is the modulus of the complex roots and θ is the argument.Wait, but maybe that's complicating things. Alternatively, since the roots are complex, the solution will involve sines and cosines with a damping factor. However, since the modulus of the roots is sqrt(1 + (sqrt(p - 1))^2) = sqrt(p), which is greater than 1 for p > 1, the terms will grow in magnitude unless the coefficients C and D are zero, which they aren't because a₁ and a₂ are given.Hmm, but in our case, the sequence can have negative terms, so maybe the oscillatory nature can lead to negative values. But I need to find when a specific term is -3.Alternatively, perhaps I can use modular arithmetic to find constraints on p. Let me consider the sequence modulo 3 because -3 is a multiple of 3.If I can find p such that a_m ≡ 0 mod 3 and a_m = -3, then p must satisfy certain conditions.Let me compute the sequence modulo 3 for different p.Case 1: p ≡ 0 mod 3. Since p is prime, p=3.Compute the sequence modulo 3:a₁=1, a₂=2a₃=2*2 -3*1=4-3=1 mod3a₄=2*1 -3*2=2-6=2 mod3a₅=2*2 -3*1=4-3=1 mod3a₆=2*1 -3*2=2-6=2 mod3So, the sequence modulo3 cycles between 1 and 2. Therefore, a_m can never be 0 mod3, so a_m can never be -3.Case2: p ≡1 mod3. So p=3k+1.Compute the sequence modulo3:a₁=1, a₂=2a₃=2*2 - (3k+1)*1=4 -3k -1=3 -3k≡0 mod3a₄=2*0 - (3k+1)*2=0 -6k -2≡-2≡1 mod3a₅=2*1 - (3k+1)*0=2 -0=2 mod3a₆=2*2 - (3k+1)*1=4 -3k -1=3 -3k≡0 mod3So, the sequence modulo3 cycles as 1,2,0,1,2,0,...Therefore, every third term is 0 mod3. So, if a_m = -3, then m must be a multiple of3.But let's see if a_m can be exactly -3.Looking at p=7, which is 1 mod3 (7=3*2+1). We saw earlier that a₃=-3.Similarly, for p=13=3*4+1, a₃=-9, which is -3 mod3 but not exactly -3.Wait, but for p=7, a₃=-3 exactly. So, maybe p=7 is the only prime where a_m=-3.Wait, let's check p=7 again.a₁=1, a₂=2a₃=2*2 -7*1=4-7=-3Yes, exactly -3.For p=13, a₃=2*2 -13*1=4-13=-9, which is -3 mod3 but not exactly -3.So, p=7 works because a₃=-3.Wait, is there another prime p ≡1 mod3 where a₃=-3?Let me see: a₃=2*2 -p*1=4 -p.Set 4 -p = -3 => p=7.So, p=7 is the only prime where a₃=-3.Therefore, p=7 is the only prime for which there exists m such that a_m=-3.I think that's the answer.**Final Answer**The only prime ( p ) is boxed{7}.