answer：Okay, so I have this function g(x) defined piecewise. It's one thing when x is less than 15, and another when x is greater than or equal to 15. I need to find the inverse of g at 8 and at 52, and then add those two inverses together. Hmm, okay, let's break this down step by step.First, let me write down the function again to make sure I have it right:[g(x) = left{begin{array}{cl}x + 4 & text{if } x < 15, 3x - 5 & text{if } x ge 15.end{array}right.]So, for any x less than 15, g(x) is just x plus 4. And for x greater than or equal to 15, g(x) is 3x minus 5. Got it.Now, I need to find g inverse at 8 and at 52. That means I need to find the x values such that g(x) equals 8 and 52, respectively. Then, add those x values together.Let me start with g inverse of 8. So, I need to solve g(x) = 8.Since g is piecewise, I have to check both pieces to see which one gives me a valid solution.First, let's check the first piece: x + 4 = 8.Solving for x: x = 8 - 4 = 4.Now, I need to check if this x satisfies the condition for that piece. The first piece is for x < 15. Is 4 less than 15? Yes, it is. So, x = 4 is a valid solution.Now, let's check the second piece: 3x - 5 = 8.Solving for x: 3x = 8 + 5 = 13, so x = 13/3 ≈ 4.333...Wait, that's approximately 4.333, which is still less than 15. But the second piece is defined for x ≥ 15. So, x = 13/3 is not in the domain of the second piece. Therefore, this solution isn't valid for the second piece.So, the only valid solution is x = 4. Therefore, g inverse of 8 is 4.Alright, that seems straightforward. Now, moving on to g inverse of 52. So, I need to solve g(x) = 52.Again, I'll check both pieces.First, the first piece: x + 4 = 52.Solving for x: x = 52 - 4 = 48.Now, check if this x satisfies the condition for the first piece. The first piece is for x < 15. Is 48 less than 15? No, it's way bigger. So, this solution isn't valid for the first piece.Now, let's check the second piece: 3x - 5 = 52.Solving for x: 3x = 52 + 5 = 57, so x = 57/3 = 19.Now, check if this x satisfies the condition for the second piece. The second piece is for x ≥ 15. Is 19 greater than or equal to 15? Yes, it is. So, x = 19 is a valid solution.Therefore, g inverse of 52 is 19.Now, I need to add these two inverses together: g inverse of 8 is 4, and g inverse of 52 is 19. So, 4 + 19 equals 23.Wait, let me just double-check my calculations to make sure I didn't make any mistakes.For g inverse of 8:- First piece: x + 4 = 8 ⇒ x = 4. 4 < 15, so valid.- Second piece: 3x - 5 = 8 ⇒ x = 13/3 ≈ 4.333. But 4.333 < 15, so invalid for the second piece. So, only x = 4 is valid.For g inverse of 52:- First piece: x + 4 = 52 ⇒ x = 48. 48 is not less than 15, so invalid.- Second piece: 3x - 5 = 52 ⇒ x = 19. 19 ≥ 15, so valid.Adding them together: 4 + 19 = 23.Hmm, seems correct. I don't think I made any mistakes here. The key was to check both pieces for each value and ensure that the solution falls within the correct domain for that piece.Just to be thorough, let me verify by plugging the values back into the original function.For x = 4: Since 4 < 15, we use the first piece: g(4) = 4 + 4 = 8. Correct.For x = 19: Since 19 ≥ 15, we use the second piece: g(19) = 3*19 - 5 = 57 - 5 = 52. Correct.So, both inverses are correct, and their sum is indeed 23.I think that's it. I can't see any errors in my reasoning or calculations. It makes sense that the inverse function would require checking both pieces and ensuring the solution fits the domain of that piece. It's important not to assume which piece the solution comes from without checking both.Another thing to note is that the function g(x) is piecewise linear, so it's invertible on each piece as long as the pieces are strictly increasing or decreasing. In this case, both pieces are linear with positive slopes, so they are invertible on their respective domains.Also, it's worth mentioning that the function g(x) is continuous at x = 15? Let me check that quickly.For x approaching 15 from the left: g(15-) = 15 + 4 = 19.For x approaching 15 from the right: g(15+) = 3*15 - 5 = 45 - 5 = 40.Wait, that's a big jump. So, at x = 15, the function jumps from 19 to 40. That means there's a discontinuity at x = 15. So, the function isn't continuous there. But since we're dealing with inverses, as long as each piece is invertible and the ranges don't overlap, the inverse is well-defined.Looking at the ranges:- For x < 15, g(x) = x + 4. So, as x approaches 15 from the left, g(x) approaches 19. So, the range for the first piece is (-∞, 19).- For x ≥ 15, g(x) = 3x - 5. Starting at x = 15, g(15) = 40, and as x increases, g(x) increases without bound. So, the range for the second piece is [40, ∞).Therefore, the overall range of g(x) is (-∞, 19) union [40, ∞). So, there's a gap between 19 and 40 where g(x) doesn't take any values. That means, for values between 19 and 40, there's no x such that g(x) equals those values. But in our problem, we're dealing with 8 and 52, which are outside that gap. 8 is less than 19, and 52 is greater than 40, so both are in the range of g(x). Therefore, their inverses exist.If we had a value between 19 and 40, say 25, we wouldn't be able to find a corresponding x because g(x) doesn't take that value. But in this case, both 8 and 52 are fine.So, just to recap:- g inverse of 8 is 4 because g(4) = 8, and 4 is in the domain of the first piece.- g inverse of 52 is 19 because g(19) = 52, and 19 is in the domain of the second piece.Adding them together gives 23.I think that's solid. I don't see any issues with this approach. It's crucial when dealing with piecewise functions to check each piece and ensure the solution lies within the correct domain. Otherwise, you might end up with an incorrect inverse.Another point to consider is whether the function is one-to-one on each piece. Since both pieces are linear with positive slopes, they are strictly increasing, which means they are one-to-one on their respective domains. Therefore, each piece has an inverse, and the overall function is invertible on its entire domain, considering the pieces don't overlap in their ranges.It's also interesting to note that because of the jump discontinuity at x = 15, the function isn't continuous, but it's still invertible in parts. The key is that each piece is invertible on its own, and their ranges don't interfere with each other.So, in summary, the process was:1. For each target value (8 and 52), set up the equation g(x) = target.2. Solve for x in both pieces of the function.3. Check if the solution x falls within the domain specified for each piece.4. If it does, that's the valid inverse; if not, discard that solution.5. Once both inverses are found, add them together.Following these steps carefully ensures that you don't make mistakes, especially when dealing with piecewise functions where the domain restrictions are crucial.I think I've covered all the bases here. I feel confident that my answer is correct, and I've double-checked each step to make sure there are no errors. It's a good reminder of how important it is to verify solutions, especially when working with piecewise functions and inverses.**Final Answer**The value of (g^{-1}(8) + g^{-1}(52)) is boxed{23}).

answer：Alright, so I've got this problem here where I have a bunch of positive integers a, b, c, ..., k. Each one divides twice the next one minus one. So, a divides 2b - 1, b divides 2c - 1, and so on, until k divides 2a - 1. The goal is to show that all of these numbers must be 1.Hmm, okay. Let me start by understanding what it means for one number to divide another. If a divides 2b - 1, that means 2b - 1 is a multiple of a. So, 2b - 1 = a * m for some integer m. Similarly, b divides 2c - 1, so 2c - 1 = b * n for some integer n, and so on.Since all these numbers are positive integers, they must be at least 1. So, the smallest possible value for each is 1. Let me see what happens if I assume that one of them is greater than 1. Maybe that will lead to a contradiction.Let's suppose, for example, that a is greater than 1. Then, since a divides 2b - 1, 2b - 1 must be at least a. So, 2b - 1 ≥ a. But since a is greater than 1, 2b - 1 must be at least 2. Therefore, 2b - 1 ≥ 2 implies that b ≥ (2 + 1)/2 = 1.5. But b is an integer, so b must be at least 2.Okay, so if a is greater than 1, then b must be at least 2. Now, let's look at the next condition: b divides 2c - 1. If b is at least 2, then 2c - 1 must be at least 2b - 1. Wait, no, that's not necessarily true. It just needs to be a multiple of b. So, 2c - 1 could be b, 2b, 3b, etc.But let's think about the size of these numbers. If a is greater than 1, then b is at least 2. Then, 2c - 1 must be a multiple of b, which is at least 2. So, 2c - 1 must be at least 2. Therefore, c must be at least (2 + 1)/2 = 1.5, so c is at least 2.Continuing this way, each subsequent number must be at least 2. But then, when we get back to k dividing 2a - 1, since k is at least 2, 2a - 1 must be at least 2k - 1. But since k is part of the cycle, this might create a loop where each number is forcing the next to be larger, but since we have a finite number of variables, this could lead to a contradiction.Wait, maybe I need to think about this differently. Let's consider the properties of numbers that divide 2^n - 1. These are numbers that are coprime to 2, meaning they are odd. So, all these a, b, c, ..., k must be odd numbers because if any of them were even, 2b - 1 would be odd, and an even number can't divide an odd number.So, all the numbers are odd. Now, let's think about the multiplicative order of 2 modulo these numbers. For a number m, the multiplicative order of 2 modulo m is the smallest positive integer d such that 2^d ≡ 1 mod m. If m divides 2^n - 1, then the order of 2 modulo m must divide n.So, in our case, since a divides 2b - 1, the order of 2 modulo a must divide b. Similarly, the order of 2 modulo b must divide c, and so on, until the order of 2 modulo k must divide a.This seems like a cyclic dependency where each order divides the next number in the sequence. If I can show that all these orders are 1, then 2^1 ≡ 1 mod m implies that m divides 1, so m must be 1.But how do I show that the orders are 1? Well, if the order of 2 modulo m is 1, then 2 ≡ 1 mod m, which implies m divides 1, so m = 1.Alternatively, if the order is greater than 1, then m must be greater than 1. But if all these orders are greater than 1, then we have a cycle where each number is forcing the next to have an order that divides it, but since they are all greater than 1, this might not be possible.Wait, maybe I can use induction or some kind of minimality argument. Suppose that all the numbers are greater than 1. Then, each number has a prime divisor. Let's take the smallest prime divisor among all these numbers. Let's say p is the smallest prime that divides any of a, b, c, ..., k.Since p is the smallest prime, it must divide one of the numbers, say a. Then, since a divides 2b - 1, we have 2b ≡ 1 mod a. Therefore, 2b ≡ 1 mod p. This implies that 2 has an inverse modulo p, which it does since p is odd (as we established earlier).So, b ≡ 2^{-1} mod p. But since p divides a, and a divides 2b - 1, we have that 2b ≡ 1 mod p. Therefore, b ≡ (1/2) mod p. But b is an integer, so b must be congruent to some integer modulo p.But since p is the smallest prime in the sequence, and b is part of the sequence, b must be at least p. Wait, no, b could be smaller than p. Hmm, maybe I need to think about the order of 2 modulo p.The order of 2 modulo p must divide b, as we saw earlier. But since p is the smallest prime, the order of 2 modulo p must be less than p. But the order must also divide b. If b is less than p, then the order must divide b, which is less than p, but p is prime, so the order must be 1 or something else.Wait, if the order is 1, then 2 ≡ 1 mod p, which implies p divides 1, which is impossible. So, the order must be greater than 1. But since p is the smallest prime, and the order divides b, which is less than p, this is a contradiction.Therefore, our assumption that there is a prime divisor p greater than 1 must be wrong. Hence, all the numbers must be 1.Wait, let me make sure I followed that correctly. If p is the smallest prime divisor, and it divides a, then the order of 2 modulo p divides b. But since p is the smallest prime, b must be less than p or something? Not necessarily, but if b is greater than or equal to p, then the order could still divide b.Hmm, maybe I need to think about the multiplicative order more carefully. If p divides a, then 2b ≡ 1 mod p, so 2b ≡ 1 mod p implies that 2 has an inverse modulo p, which is b. So, the order of 2 modulo p divides b.But since p is the smallest prime, and b is part of the sequence, b could be equal to p or greater. If b is equal to p, then the order of 2 modulo p divides p. But p is prime, so the order must be 1 or p. If the order is 1, then 2 ≡ 1 mod p, which is impossible. So, the order must be p, which means that 2^p ≡ 1 mod p.But by Fermat's little theorem, 2^{p-1} ≡ 1 mod p. So, the order of 2 modulo p must divide both p and p-1. The only common divisor is 1, which is a contradiction because we already saw that the order can't be 1. Therefore, our assumption that p divides a leads to a contradiction, so p cannot divide any of the numbers, meaning all numbers must be 1.Yeah, that seems to make sense. So, by assuming that there is a prime divisor, we end up with a contradiction, which means that all the numbers must be 1.

answer：First, I need to establish the relationships between the amounts that P, Q, and R have. I know that the total amount among the three friends is Rs. 4,000. R has two-thirds of the total amount that P and Q have together. Additionally, the ratio of P's amount to Q's amount is 3:2.I'll start by expressing P's amount in terms of Q's amount using the given ratio. If the ratio of P to Q is 3:2, then P = (3/2) * Q.Next, I'll use the information about R's amount. Since R has two-thirds of the total amount that P and Q have together, I can write R = (2/3) * (P + Q). Substituting P = (3/2) * Q into this equation, I get R in terms of Q.Now, I'll substitute both P and R in terms of Q into the total amount equation (P + Q + R = 4,000) to solve for Q. After finding Q, I can easily find P and R.Once I have the amounts for P, Q, and R, I'll calculate 10% of P's amount to determine the charity donation. Finally, I'll subtract this donation from P's amount to find the remaining amount with P and confirm that R's amount remains unchanged.

answer：To determine the greatest integer less than (-frac{19}{3}), I first convert the fraction to a mixed number.[-frac{19}{3} = -6frac{1}{3}]This shows that (-frac{19}{3}) is between (-7) and (-6). Since (-6frac{1}{3}) is closer to (-6) than to (-7), the greatest integer less than (-6frac{1}{3}) is (-7).

answer：First, I need to determine the total number of sheep in Gabriel's herd. I know that half of the sheep have lice, and 94 sheep have lice in total. So, if half of the sheep have lice, the total number of sheep must be twice the number of sheep with lice.Next, I'll calculate the total number of sheep:Total sheep = 2 × 94 = 188Now, I need to find out how many sheep have only lice. I know that 84 sheep have both fleas and lice. Therefore, the number of sheep with only lice is the total number of sheep with lice minus those with both pests:Sheep with only lice = 94 - 84 = 10Finally, to find the number of sheep with only fleas, I'll subtract the number of sheep with only lice and those with both pests from the total number of sheep:Sheep with only fleas = Total sheep - Sheep with only lice - Sheep with both pestsSheep with only fleas = 188 - 10 - 84 = 94Therefore, there are 94 sheep with only fleas.

answer：Okay, so I have this problem here where I need to find the largest product by multiplying two numbers from the set {-6, -4, -2, 1, 5}. Hmm, let me think about how to approach this.First, I remember that when multiplying two numbers, the product can be positive or negative depending on the signs of the numbers. If I multiply two negative numbers, the result is positive. If I multiply two positive numbers, the result is also positive. But if I multiply a negative and a positive number, the result is negative. Since I'm looking for the largest product, I probably want to focus on the positive products because they can be larger than negative ones.So, let me list out all possible products of two numbers from this set and see which one is the largest. But wait, that might take a while. Maybe there's a smarter way to do this without checking every single pair.I recall that to get the largest positive product, I can either multiply the two largest positive numbers or the two smallest negative numbers. That makes sense because multiplying two large positive numbers gives a big positive result, and multiplying two small negative numbers (which are actually large in magnitude) also gives a big positive result.Looking at the set, the positive numbers are 1 and 5. So, multiplying those gives 1 * 5 = 5. That's a positive product, but maybe not the largest.Now, looking at the negative numbers: -6, -4, and -2. The two smallest (most negative) numbers are -6 and -4. If I multiply those, I get (-6) * (-4) = 24. That's a positive product and seems larger than 5.Let me check the other possible products to make sure I haven't missed anything. If I multiply -6 and -2, that's (-6) * (-2) = 12. That's still positive but smaller than 24. Similarly, multiplying -4 and -2 gives (-4) * (-2) = 8, which is also positive but smaller than 24.What about multiplying a negative and a positive number? For example, (-6) * 5 = -30, which is negative. Similarly, (-4) * 5 = -20, (-2) * 5 = -10, (-6) * 1 = -6, (-4) * 1 = -4, and (-2) * 1 = -2. All of these are negative, so they can't be the largest product.So, after considering all possibilities, the largest product comes from multiplying the two smallest negative numbers, -6 and -4, which gives 24. That seems to be the maximum product possible from this set.Wait, just to double-check, is there any other pair that could give a larger product? Let's see:- 5 * 1 = 5- 5 * (-2) = -10- 5 * (-4) = -20- 5 * (-6) = -30- 1 * (-2) = -2- 1 * (-4) = -4- 1 * (-6) = -6- (-2) * (-4) = 8- (-2) * (-6) = 12- (-4) * (-6) = 24Yep, 24 is indeed the largest product here. So, I think I'm confident that 24 is the correct answer.**Final Answer**boxed{A}