💡Alright, so I have this problem where I need to find the number of 100-digit positive integers where each digit is one of 2, 3, 4, 5, 6, or 7, and the entire number is divisible by (2^{100}). Hmm, that seems pretty complex, but let me try to break it down step by step.First, I know that for a number to be divisible by (2^n), its last (n) digits must be divisible by (2^n). But in this case, the entire number is 100 digits long, and we need it to be divisible by (2^{100}). So, does that mean all 100 digits somehow contribute to this divisibility? I'm not entirely sure, but maybe I can think about it recursively or use some combinatorial approach.Let me consider smaller cases first to get a sense of the pattern. Let's say for (n = 1), how many 1-digit numbers composed of digits 2, 3, 4, 5, 6, or 7 are divisible by (2^1 = 2)? Well, the even digits in that set are 2, 4, and 6. So, there are 3 such numbers. That seems straightforward.Now, what about (n = 2)? We need 2-digit numbers where each digit is from 2, 3, 4, 5, 6, or 7, and the number is divisible by (2^2 = 4). For a number to be divisible by 4, its last two digits must form a number divisible by 4. But since each digit is from 2, 3, 4, 5, 6, or 7, let's list all possible 2-digit combinations and check which are divisible by 4.Wait, that might take too long. Maybe there's a smarter way. I remember that for divisibility by 4, only the last two digits matter. So, if I fix the first digit, the second digit needs to make the entire number divisible by 4. But since the first digit can be any of 2, 3, 4, 5, 6, or 7, and the second digit also can be any of those, but the combination needs to be divisible by 4.Alternatively, maybe I can think in terms of modular arithmetic. If I have a number (N = 10a + b), where (a) is the first digit and (b) is the second digit, then (N equiv 10a + b mod 4). Since (10 equiv 2 mod 4), this simplifies to (N equiv 2a + b mod 4). So, for (N) to be divisible by 4, (2a + b) must be congruent to 0 modulo 4.Given that (a) and (b) are digits from 2, 3, 4, 5, 6, or 7, let's see what possible values (2a + b) can take modulo 4.First, let's list the possible values of (a) and (b) modulo 4:- 2 mod 4 = 2- 3 mod 4 = 3- 4 mod 4 = 0- 5 mod 4 = 1- 6 mod 4 = 2- 7 mod 4 = 3So, possible residues for (a) and (b) are 0, 1, 2, or 3.Now, (2a) modulo 4:- If (a equiv 0 mod 4), then (2a equiv 0 mod 4)- If (a equiv 1 mod 4), then (2a equiv 2 mod 4)- If (a equiv 2 mod 4), then (2a equiv 0 mod 4)- If (a equiv 3 mod 4), then (2a equiv 2 mod 4)So, (2a) can only be 0 or 2 modulo 4, depending on whether (a) is even or odd. Wait, actually, looking back, (a) can be 2,3,4,5,6,7. So, (a) is even if it's 2,4,6 and odd if it's 3,5,7.So, if (a) is even (2,4,6), (2a equiv 0 mod 4). If (a) is odd (3,5,7), (2a equiv 2 mod 4).Therefore, for (N = 10a + b) to be divisible by 4, we have:- If (a) is even, then (2a + b equiv 0 + b equiv 0 mod 4). So, (b) must be 0 mod 4. Looking back, the digits that are 0 mod 4 are 4. So, (b) must be 4.- If (a) is odd, then (2a + b equiv 2 + b equiv 0 mod 4). So, (b equiv 2 mod 4). The digits that are 2 mod 4 are 2 and 6. So, (b) can be 2 or 6.Therefore, for each even (a) (which are 2,4,6), there's only 1 choice for (b) (which is 4). For each odd (a) (which are 3,5,7), there are 2 choices for (b) (which are 2 and 6).So, total number of 2-digit numbers is:- Number of even (a): 3 (2,4,6), each with 1 choice for (b): 3 * 1 = 3- Number of odd (a): 3 (3,5,7), each with 2 choices for (b): 3 * 2 = 6Total: 3 + 6 = 9So, (f_2 = 9). Hmm, interesting. So, for (n=1), (f_1=3), and for (n=2), (f_2=9). That seems like (f_n = 3^n). Wait, 3^1=3, 3^2=9, so maybe (f_n = 3^n). Let me test for (n=3).For (n=3), we need 3-digit numbers where each digit is 2,3,4,5,6,7, and the number is divisible by (2^3=8). For divisibility by 8, the last three digits must be divisible by 8. But since the entire number is 3 digits, the whole number must be divisible by 8.Again, let's use modular arithmetic. Let (N = 100a + 10b + c), where (a, b, c in {2,3,4,5,6,7}). Then, (N equiv 100a + 10b + c mod 8).Since (100 mod 8 = 4), (10 mod 8 = 2), and (1 mod 8 = 1), so:(N equiv 4a + 2b + c mod 8)We need (4a + 2b + c equiv 0 mod 8).This seems more complicated. Maybe I can think recursively. If I already know (f_2 = 9), how can I build (f_3) from (f_2)?Wait, perhaps there's a pattern here. For each additional digit, we have some number of choices that maintain the divisibility condition. If for (n=1), (f_1=3), for (n=2), (f_2=9=3^2), maybe for (n=3), (f_3=27=3^3), and so on, leading to (f_{100}=3^{100}).But I should verify this for (n=3) to be sure.Let me try to see how (f_3) relates to (f_2). Suppose I have a 2-digit number (y) that is divisible by (4), and I want to prepend a digit (x) to make a 3-digit number (100x + y). For (100x + y) to be divisible by (8), we need (100x + y equiv 0 mod 8).Since (100 equiv 4 mod 8), this becomes (4x + y equiv 0 mod 8).But (y) is already divisible by (4), so (y = 4k) for some integer (k). Therefore, (4x + 4k equiv 0 mod 8) simplifies to (x + k equiv 0 mod 2). So, (x + k) must be even.Now, (x) is a digit from 2,3,4,5,6,7. The parity of (x) depends on whether it's even or odd. Let's see:- Even digits: 2,4,6 (even)- Odd digits: 3,5,7 (odd)So, if (x) is even, then (k) must be even for (x + k) to be even. If (x) is odd, then (k) must be odd.But (y = 4k) is a 2-digit number. Since (y) is divisible by 4, (k) can be any integer such that (4k) is a 2-digit number with digits from 2,3,4,5,6,7.Wait, but (k) is determined by (y). For each (y in F_2), which has (f_2 = 9) elements, (k = y / 4). So, (k) can be either even or odd depending on (y).But how many (y in F_2) have (k) even and how many have (k) odd?Let me list all possible 2-digit numbers (y) with digits from 2,3,4,5,6,7 and divisible by 4.From earlier, for each even (a) (2,4,6), (b=4), so the numbers are 24, 44, 64.For each odd (a) (3,5,7), (b=2) or (6), so the numbers are 32, 36, 52, 56, 72, 76.So, the 9 numbers are: 24, 44, 64, 32, 36, 52, 56, 72, 76.Now, let's compute (k = y / 4) for each:- 24 / 4 = 6 (even)- 44 / 4 = 11 (odd)- 64 / 4 = 16 (even)- 32 / 4 = 8 (even)- 36 / 4 = 9 (odd)- 52 / 4 = 13 (odd)- 56 / 4 = 14 (even)- 72 / 4 = 18 (even)- 76 / 4 = 19 (odd)So, among the 9 numbers, (k) is even for 24, 64, 32, 56, 72, which is 5 numbers, and odd for 44, 36, 52, 76, which is 4 numbers.Wait, that's 5 even (k) and 4 odd (k). Hmm, not exactly equal, but close.So, for each (y in F_2), depending on whether (k) is even or odd, we can choose (x) such that (x + k) is even.If (k) is even, then (x) must be even (since even + even = even). There are 3 choices for even (x): 2,4,6.If (k) is odd, then (x) must be odd (since odd + odd = even). There are 3 choices for odd (x): 3,5,7.So, for each (y in F_2), regardless of whether (k) is even or odd, there are 3 choices for (x) such that (100x + y) is divisible by 8.Therefore, (f_3 = 3 * f_2 = 3 * 9 = 27).So, indeed, (f_3 = 3^3 = 27). This seems to confirm the pattern that (f_n = 3^n).Wait, but in the case of (n=2), we had 9 numbers, which is 3^2, and for (n=3), we have 27, which is 3^3. So, it seems that each time we add a digit, we multiply the number of valid numbers by 3.Is this a coincidence, or is there a general pattern here?Let me try to generalize. Suppose we have an (n)-digit number (y) that is divisible by (2^n). We want to prepend a digit (x) to make an ((n+1))-digit number (10^n x + y) that is divisible by (2^{n+1}).So, (10^n x + y equiv 0 mod 2^{n+1}).Since (y) is divisible by (2^n), we can write (y = 2^n k) for some integer (k). Then, the equation becomes:(10^n x + 2^n k equiv 0 mod 2^{n+1}).Dividing both sides by (2^n), we get:(5^n x + k equiv 0 mod 2).So, (5^n x + k) must be even.Now, (5^n) is always odd because 5 is odd and any power of an odd number is odd. Therefore, (5^n x) has the same parity as (x). So, the equation simplifies to:(x + k equiv 0 mod 2).This means that (x) and (k) must have the same parity. So, if (k) is even, (x) must be even; if (k) is odd, (x) must be odd.Now, in our set of digits, we have 3 even digits (2,4,6) and 3 odd digits (3,5,7). So, regardless of whether (k) is even or odd, there are exactly 3 choices for (x) that satisfy the parity condition.Therefore, for each (y in F_n), there are 3 choices for (x) such that (10^n x + y in F_{n+1}). Hence, (f_{n+1} = 3 f_n).Given that (f_1 = 3), we can use induction to show that (f_n = 3^n) for all (n geq 1).Base case: (n=1), (f_1=3=3^1).Inductive step: Assume (f_k = 3^k) for some (k geq 1). Then, (f_{k+1} = 3 f_k = 3 * 3^k = 3^{k+1}).Therefore, by induction, (f_n = 3^n) for all (n geq 1).So, for (n=100), the number of 100-digit numbers with digits from 2,3,4,5,6,7 that are divisible by (2^{100}) is (f_{100} = 3^{100}).Wait, but let me double-check this reasoning. Is there any step where I might have made a mistake?In the inductive step, I assumed that for each (y in F_n), there are exactly 3 choices for (x) such that (10^n x + y) is divisible by (2^{n+1}). This relies on the fact that (x) must have the same parity as (k), where (y = 2^n k). Since we have 3 even and 3 odd digits, regardless of (k)'s parity, there are 3 choices for (x). That seems correct.Also, when I tested for (n=2) and (n=3), the pattern held, so it's consistent.Therefore, I think the conclusion is correct: the number of such 100-digit numbers is (3^{100}).

💡Alright, so I have this geometry problem here about triangle ABC. It says that the sides a, b, and c satisfy the equation a² + b² - c² = ab. I need to find the measure of angle C. Hmm, okay, let me think about how to approach this.First off, I remember that in triangles, when dealing with the relationships between sides and angles, the Law of Cosines is really useful. The Law of Cosines relates the lengths of the sides of a triangle to the cosine of one of its angles. The formula is c² = a² + b² - 2ab cos(C), where C is the angle opposite side c. Wait, the problem gives me an equation that looks similar to the Law of Cosines. Let me write it down again: a² + b² - c² = ab. If I compare this to the Law of Cosines formula, which is c² = a² + b² - 2ab cos(C), I can see that if I rearrange the Law of Cosines, it becomes a² + b² - c² = 2ab cos(C). So, in the problem, a² + b² - c² is equal to ab, and according to the Law of Cosines, it should also be equal to 2ab cos(C). That means I can set these two expressions equal to each other: ab = 2ab cos(C). Hmm, okay, so ab equals 2ab cos(C). I can simplify this equation by dividing both sides by ab, assuming that a and b are not zero, which makes sense because they are sides of a triangle and must have positive lengths. So, dividing both sides by ab gives me 1 = 2 cos(C). Now, solving for cos(C), I can divide both sides by 2: cos(C) = 1/2. Alright, so the cosine of angle C is 1/2. I need to find the measure of angle C. I remember that cos(60°) is 1/2, and cos(π/3 radians) is also 1/2. Since in a triangle, all angles must be between 0 and 180 degrees (or 0 and π radians), and angle C is one of those angles, it must be 60 degrees or π/3 radians. Let me double-check my steps to make sure I didn't make any mistakes. I started with the given equation a² + b² - c² = ab, recognized it as similar to the Law of Cosines, set them equal, simplified, and solved for cos(C). Everything seems to check out. Just to be thorough, I can think about what kind of triangle this might be. If angle C is 60 degrees, then triangle ABC could be an equilateral triangle if all sides are equal, but since the equation doesn't specify that a, b, and c are equal, it's just a general triangle with one angle being 60 degrees. I also recall that in a triangle, the larger the angle, the longer the side opposite it. Since angle C is 60 degrees, which is a relatively small angle, the side opposite it, which is c, should be shorter than the other sides if the other angles are larger. But without more information about the sides, I can't say much more about the specific lengths. So, to sum it up, by using the Law of Cosines and the given equation, I was able to determine that angle C has a cosine of 1/2, which corresponds to an angle of 60 degrees or π/3 radians. I think that's solid. I don't see any errors in my reasoning, and each step logically follows from the previous one. It's always good to verify, but I feel confident about this answer.

💡First, calculate the total number of apples delivered by multiplying the number of apples per crate by the number of crates.Next, subtract the number of rotten apples from the total to find out how many apples are left.Finally, divide the remaining apples by the number of apples that fit in each box to determine how many boxes are needed.

💡First, determine the total number of car washes Jake performs over 20 weeks, which is 20 washes.Next, calculate how many bottles of car wash soap Jake needs by dividing the total number of washes by the number of washes per bottle:20 washes ÷ 4 washes per bottle = 5 bottles.Finally, find the cost per bottle by dividing the total amount spent by the number of bottles:20 ÷ 5 bottles = 4 per bottle.

💡Okay, so I have this function f(theta) which is a ratio of some sine and cosine terms. It looks a bit complicated, but maybe I can simplify it using trigonometric identities. Let me write it down again to see it clearly:f(theta) = [sin(theta - 5π) * cos(-π/2 - theta) * cos(8π - theta)] / [sin(theta - 3π/2) * sin(-theta - 4π)]Hmm, okay. I remember that sine and cosine have periodicity and some angle addition formulas. Let me try to simplify each term step by step.First, looking at the numerator:1. sin(theta - 5π): I know that sine has a period of 2π, so sin(theta - 5π) is the same as sin(theta - π - 4π), which simplifies to sin(theta - π) because 4π is two full periods. And sin(theta - π) is equal to -sin(theta) because sin(theta - π) = -sin(theta). So that term simplifies to -sin(theta).2. cos(-π/2 - theta): Cosine is an even function, so cos(-x) = cos(x). Therefore, cos(-π/2 - theta) = cos(π/2 + theta). Now, cos(π/2 + theta) is equal to -sin(theta) because cosine of (π/2 + theta) is -sin(theta). So that term becomes -sin(theta).3. cos(8π - theta): Cosine is also periodic with period 2π, so cos(8π - theta) is the same as cos(-theta) because 8π is four full periods. Cos(-theta) is equal to cos(theta) because cosine is even. So that term is cos(theta).So putting the numerator together: (-sin(theta)) * (-sin(theta)) * cos(theta) = sin^2(theta) * cos(theta).Now, the denominator:1. sin(theta - 3π/2): Let's see, sin(theta - 3π/2). I can write this as sin(theta - π - π/2). Using the identity sin(theta - π) = -sin(theta), but wait, actually, sin(theta - 3π/2) can be rewritten using the sine periodicity. Alternatively, I can use the identity sin(theta - 3π/2) = sin(theta - π - π/2) = sin(theta - π - π/2). Hmm, maybe it's easier to think of it as sin(theta - 3π/2) = sin(theta + π/2) because subtracting 3π/2 is the same as adding π/2 (since 3π/2 = π + π/2). Wait, no, actually, sin(theta - 3π/2) = sin(theta + π/2) because subtracting 3π/2 is the same as adding π/2 (since 3π/2 = π + π/2, so subtracting that is like adding π/2). But wait, sin(theta + π/2) is equal to cos(theta). So sin(theta - 3π/2) = cos(theta).Wait, let me double-check that. Let's use the identity sin(theta - 3π/2) = sin(theta - π - π/2). Using the sine subtraction formula: sin(a - b) = sin(a)cos(b) - cos(a)sin(b). So sin(theta - π - π/2) = sin(theta - π)cos(π/2) - cos(theta - π)sin(π/2). Now, cos(π/2) is 0 and sin(π/2) is 1. So this simplifies to -cos(theta - π). And cos(theta - π) is equal to -cos(theta) because cosine is even and cos(theta - π) = cos(π - theta) = -cos(theta). So putting it together: -(-cos(theta)) = cos(theta). So yes, sin(theta - 3π/2) = cos(theta).2. sin(-theta - 4π): Let's simplify this. First, sin is an odd function, so sin(-x) = -sin(x). So sin(-theta - 4π) = -sin(theta + 4π). Now, sine has a period of 2π, so sin(theta + 4π) = sin(theta). Therefore, sin(-theta - 4π) = -sin(theta).So the denominator becomes cos(theta) * (-sin(theta)) = -sin(theta)cos(theta).Putting it all together, the function f(theta) is:Numerator: sin^2(theta) * cos(theta)Denominator: -sin(theta)cos(theta)So f(theta) = [sin^2(theta) * cos(theta)] / [-sin(theta)cos(theta)].Simplify this: The sin(theta) in the numerator and denominator cancels out, as does the cos(theta). So we have sin(theta) / (-1) = -sin(theta).So f(theta) simplifies to -sin(theta).Now, for part (2), we need to find f(4π/3). So we substitute theta = 4π/3 into f(theta):f(4π/3) = -sin(4π/3).Now, sin(4π/3) is sin(π + π/3) = -sin(π/3) because sine is negative in the third quadrant. And sin(π/3) is √3/2. So sin(4π/3) = -√3/2. Therefore, f(4π/3) = -(-√3/2) = √3/2.Wait, hold on, let me double-check that step. f(theta) is -sin(theta), so f(4π/3) = -sin(4π/3). sin(4π/3) is indeed -√3/2, so -sin(4π/3) = -(-√3/2) = √3/2. Yes, that's correct.So the final answers are:(1) f(theta) = -sin(theta)(2) f(4π/3) = √3/2

💡First, I need to identify the terms in each polynomial that will contribute to the ( x^2 ) term when multiplied together.Looking at the first polynomial ( 2x^3 + 5x^2 - 3x ), the relevant terms are ( 5x^2 ) and ( -3x ).In the second polynomial ( 3x^2 - 4x - 5 ), the relevant terms are ( -5 ) and ( 4x ).Next, I will multiply these corresponding terms:1. ( 5x^2 times (-5) = -25x^2 )2. ( -3x times 4x = -12x^2 )Finally, I will add these results to find the total coefficient of the ( x^2 ) term:[-25x^2 - 12x^2 = -37x^2]Therefore, the coefficient of the ( x^2 ) term is (-37).