💡Okay, so I have this problem where there are six people racing: Alice, Bob, Carol, Dan, Eve, and Frank. I need to figure out how many different outcomes there can be for the first, second, third, and fourth places, and there are no ties. Hmm, okay, so each position from first to fourth has to be occupied by a different person.Let me start by thinking about the first place. There are six people, so any one of them could win first place. That means there are six possible choices for who comes in first.Now, once someone has won first place, there are five people left for second place. So, for each of the six possible first-place finishers, there are five possible second-place finishers. So, that would be 6 times 5, which is 30 possible combinations for first and second places.Moving on to third place, after first and second places have been determined, there are four people remaining. So, for each of the 30 combinations of first and second places, there are four possible third-place finishers. That would be 30 times 4, which equals 120 possible combinations for first, second, and third places.Finally, for fourth place, there are three people left. So, for each of the 120 combinations of first, second, and third places, there are three possible fourth-place finishers. That would be 120 times 3, which equals 360 possible outcomes for the first four places.Wait, let me make sure I didn't make a mistake here. So, it's like a permutation problem where order matters, and we're selecting four positions out of six without any repetition. The formula for permutations is P(n, k) = n! / (n - k)!, where n is the total number of items, and k is the number of positions to fill.In this case, n is 6 and k is 4. So, P(6, 4) = 6! / (6 - 4)! = 6! / 2! = (6 × 5 × 4 × 3 × 2 × 1) / (2 × 1) = 720 / 2 = 360. Yeah, that matches what I got earlier.So, it seems like the total number of different outcomes for the first four places is 360.

💡First, determine the total number of people staying in the hotel. Since there are 10 rooms and each room holds a family of 3, the total number of people is 10 multiplied by 3, which equals 30.Next, calculate the number of towels each person receives by dividing the total number of towels by the total number of people. With 60 towels distributed among 30 people, each person receives 60 divided by 30, which equals 2 towels.

💡First, I need to simplify the square root of 27. To do this, I'll start by finding the prime factors of 27.27 can be factored into 3 multiplied by 9, and 9 can be further factored into 3 multiplied by 3. So, 27 is equal to 3 times 3 times 3, which is 3 cubed.Next, I'll express the square root of 27 using these prime factors:√27 = √(3³)I know that the square root of a number raised to a power can be simplified by dividing the exponent by 2. In this case, 3³ can be written as 3² multiplied by 3.So, √(3³) becomes √(3² × 3).Using the property of square roots that √(a × b) = √a × √b, I can separate the square root into two parts:√(3²) × √3The square root of 3 squared is simply 3, so this simplifies to:3 × √3Therefore, the simplified form of √27 is 3√3.

💡First, we identify the probabilities of persons A and B deciphering the code, which are ( frac{1}{2} ) and ( frac{1}{3} ) respectively.Next, we consider the opposite event where neither person can decipher the code. The probability that person A cannot decipher the code is ( 1 - frac{1}{2} = frac{1}{2} ), and the probability that person B cannot decipher the code is ( 1 - frac{1}{3} = frac{2}{3} ).Since the events are independent, we multiply these probabilities to find the probability that neither can decipher the code:[frac{1}{2} times frac{2}{3} = frac{1}{3}]Finally, we subtract this probability from 1 to find the probability that at least one person can decipher the code:[1 - frac{1}{3} = frac{2}{3}]

💡Okay, so I'm trying to figure out this problem about the sages and the hats. There are 1000 sages lined up, and each has a hat numbered from 1 to 1001, but one number is missing. They can see the hats in front of them, and they have to announce a number from 1 to 1001 without repeating any number. The goal is to maximize the number of sages who correctly state their own hat number. Part a) asks if they can guarantee more than 500 correct answers. Hmm, okay. I remember hearing about similar problems where people use parity or some kind of modular arithmetic to communicate information. Maybe the last sage can use some kind of code to tell the others about the sum or something.Let me think. If the last sage can see all the hats in front of him, he could calculate the sum of all those numbers modulo 1001. Then, he can announce a number that encodes this sum. The next sage can then subtract the sum of the hats he sees from the total sum modulo 1001 to figure out his own hat number. If they all do this, each sage can figure out their own number based on the previous announcements.Wait, but since one number is missing, the total sum isn't fixed. So the last sage would have to account for the missing number somehow. Maybe he can use the fact that the sum of numbers from 1 to 1001 is known, so he can subtract the sum he sees from that total to figure out the missing number. But he can't announce the missing number because he doesn't know it yet. Hmm, maybe he can use the modulo operation to encode the missing number.If the total sum of all hats is S, and the sum of the hats the last sage sees is S', then the missing number is S - S'. But since S is fixed, the last sage can compute S - S' modulo 1001 and announce that number. Then, the next sage can use this information to figure out his own number by subtracting the sum of the hats in front of him from the announced number.Wait, but if the last sage announces a number, the next sage can use that to find his own number. Then, each subsequent sage can do the same by keeping track of the numbers that have been announced so far. So, in theory, every sage except maybe the last one could figure out their own number. But since the last sage is just announcing a number that encodes the missing number, he might not be able to figure out his own number. So, maybe only the first 999 sages can figure out their numbers, and the last one has a 1/1001 chance of guessing correctly.But the question is asking if they can guarantee more than 500 correct answers. If 999 can figure it out, then yes, they can guarantee more than 500. So, part a) seems possible.Now, part b) asks if they can guarantee at least 999 correct answers. From what I just thought, if the last sage can't figure out his own number, but everyone else can, then they can guarantee 999 correct answers. So, part b) should also be possible.But wait, I need to make sure. Let me think again. The last sage announces a number that encodes the missing number. The next sage can figure out his own number by subtracting the sum he sees from the total sum minus the missing number. But does this work? Let me do a small example.Suppose there are 3 sages and hats numbered 1 to 4, missing one number. The total sum is 1+2+3+4=10. The last sage sees the two hats in front of him, say 1 and 2. The sum he sees is 3. The missing number is 10 - 3 - his own hat. But he doesn't know his own hat. Wait, maybe he can announce the sum modulo 4. So, 10 - 3 = 7, 7 modulo 4 is 3. So he announces 3. The next sage knows the total sum modulo 4 is 3. He sees the hat in front of him, say 1. So, his own hat should be 3 - 1 = 2. Then the last sage can figure out his own hat is 10 - 3 - 2 = 5, but 5 is not in the range. Wait, maybe I messed up.Alternatively, maybe the last sage announces the missing number. If he can calculate the missing number, he can announce it, and then everyone else can use that information. But how? If he sees the sum of the hats in front of him, and he knows the total sum, he can figure out the missing number. Then he can announce the missing number. Then, the next sage knows the missing number, and he can subtract the sum of the hats in front of him from the total sum minus the missing number to find his own hat.Wait, let me try again. Total sum is 10. Last sage sees sum S'. Missing number is M. So, S' + his own hat + M = 10. So, his own hat = 10 - S' - M. But he doesn't know M. So, he can't directly compute his own hat. But if he can announce M, then the next sage can use that. But how does he announce M? He can't see M, because M is the missing number. So, he can't directly announce M.Hmm, maybe the last sage can use the fact that the sum of all hats is known, and announce a number that encodes the missing number. For example, he can announce the sum of all hats modulo 1001, which would be equal to the missing number modulo 1001. Then, the next sage can use that information to figure out his own hat.Wait, let me think. The total sum of all hats is fixed, say T. The last sage sees sum S'. Then, T = S' + his own hat + M. So, his own hat = T - S' - M. But he doesn't know M. However, if he can encode M in his announcement, then the next sage can use that. So, he can announce a number that is equal to M. But how? Because he doesn't know M.Wait, maybe he can announce the sum of all hats modulo 1001, which is equal to M modulo 1001. Since M is between 1 and 1001, this would uniquely determine M. So, he can announce (T - S') modulo 1001, which is equal to M modulo 1001. Then, the next sage knows M, and can compute his own hat as T - S' - M.Wait, but T is known, right? Because it's the sum from 1 to 1001, which is a fixed number. So, T = (1001)(1002)/2. So, the last sage can compute T - S', which is equal to his own hat + M. Then, he can announce (T - S') modulo 1001, which is equal to (his own hat + M) modulo 1001. But since his own hat and M are both between 1 and 1001, their sum is between 2 and 2002. So, modulo 1001, it would be between 1 and 1001. So, he can announce that number.Then, the next sage knows that the number announced is equal to (his own hat + M) modulo 1001. But he can see all the hats in front of him, so he can compute the sum of those hats, say S''. Then, he knows that T = S'' + his own hat + M + the hats behind him. Wait, no, because he can only see the hats in front of him. So, he knows S'', and he knows that T = S'' + his own hat + M + the hats behind him. But he doesn't know the hats behind him. Hmm, this seems complicated.Wait, maybe I need to think differently. If the last sage announces M, the missing number, then the next sage can figure out his own hat. But how does the last sage announce M? He can't see M, because M is the missing number. So, he can't directly announce M. But he can announce a number that encodes M. For example, he can announce the sum of all hats modulo 1001, which would be equal to M modulo 1001. Since M is between 1 and 1001, this uniquely determines M.So, the last sage computes T - S', which is equal to his own hat + M. Then, he announces (T - S') modulo 1001, which is equal to (his own hat + M) modulo 1001. Since his own hat and M are both between 1 and 1001, their sum modulo 1001 is equal to M if his own hat is 0 modulo 1001, but his own hat is at least 1. So, maybe this doesn't work.Wait, maybe I need to adjust the approach. Instead of the last sage announcing the sum modulo 1001, maybe he can announce a number that encodes the missing number in a way that the next sage can use. For example, he can announce the missing number directly, but how? He can't see it.Alternatively, maybe the sages can agree on a system where the last sage's announcement encodes the missing number, and then each subsequent sage can use that information to deduce their own hat. For example, they can use the concept of parity or some kind of error-correcting code.Wait, another idea. Since there are 1001 numbers and 1000 hats, the missing number is known to be one of the numbers from 1 to 1001. The last sage can see 999 hats, so he can compute the sum of those 999 hats. The total sum of all 1001 numbers is fixed, so the missing number plus the last sage's hat plus the sum of the 999 hats he sees equals the total sum. Therefore, the last sage can compute the missing number plus his own hat as total sum minus the sum he sees. Then, he can announce a number that encodes this sum, perhaps modulo 1001.Then, the next sage can use this information. He knows the total sum, and he can see the sum of the hats in front of him. He also knows the announced number from the last sage, which is the sum of the missing number and the last sage's hat modulo 1001. So, he can compute his own hat as the announced number minus the sum he sees modulo 1001.Wait, let me formalize this. Let T be the total sum of numbers from 1 to 1001. Let S be the sum of the hats the last sage sees. Then, T = S + a_1000 + M, where a_1000 is the last sage's hat and M is the missing number. So, a_1000 + M = T - S. The last sage can announce (a_1000 + M) modulo 1001. Let's call this number C.Then, the next sage, sage 999, sees the sum S' of the hats in front of him. He knows that T = S' + a_999 + a_1000 + M. But he also knows that a_1000 + M = C. So, he can write T = S' + a_999 + C. Therefore, a_999 = T - S' - C. Since T and S' are known, and C was announced by sage 1000, sage 999 can compute his own hat.Similarly, sage 998 can use the same logic. He knows that T = S'' + a_998 + a_999 + a_1000 + M. But he knows a_999 and a_1000 + M = C, so he can compute a_998 = T - S'' - a_999 - C.Wait, but this seems like it could work. Each sage can use the previously announced numbers and the sum of the hats in front of them to compute their own hat. The only problem is that the last sage, sage 1000, announced C = (a_1000 + M) modulo 1001. But a_1000 and M are both between 1 and 1001, so their sum is between 2 and 2002. Therefore, modulo 1001, C can be between 1 and 1001. So, C uniquely determines a_1000 + M modulo 1001, but not necessarily the exact value. However, since a_1000 and M are both positive integers, and their sum is T - S, which is known, sage 1000 can compute a_1000 + M exactly, and then announce C = (a_1000 + M) modulo 1001.But then, sage 999 can compute a_999 = T - S' - C. Since T and S' are known, and C is known, a_999 can be computed exactly. Similarly, sage 998 can compute a_998 = T - S'' - a_999 - C, and so on.Wait, but T is a fixed number, right? It's the sum from 1 to 1001, which is (1001)(1002)/2. So, yes, T is known to all sages. Therefore, this method should work. Each sage can compute their own hat number based on the total sum, the sum of the hats in front of them, and the previously announced numbers.So, in this case, the last sage (sage 1000) announces C = (a_1000 + M) modulo 1001. Then, sage 999 can compute his own hat as a_999 = T - S' - C. Sage 998 can compute a_998 = T - S'' - a_999 - C, and so on. This way, every sage except possibly the last one can figure out their own hat number. The last sage, however, only announces C, which is (a_1000 + M) modulo 1001, but he doesn't know his own hat number or the missing number M. So, he can't determine his own hat number for sure. Therefore, he might guess, but he has a 1/1001 chance of being correct.But the question is about guaranteeing a certain number of correct answers. Since the last sage can't guarantee his own hat number, but everyone else can, then we can guarantee that at least 999 sages will correctly state their hat numbers. Therefore, for part b), the answer is yes.Wait, but part a) asks if they can guarantee more than 500. If they can guarantee 999, then certainly more than 500. So, both a) and b) are possible.But let me double-check. Is there any flaw in this reasoning? Suppose the last sage announces C = (a_1000 + M) modulo 1001. Then, sage 999 can compute a_999 = T - S' - C. But T is known, S' is the sum of the hats in front of sage 999, which he can see, and C is known from the last sage's announcement. So, yes, he can compute a_999 exactly. Similarly, sage 998 can compute a_998 = T - S'' - a_999 - C, and so on.Wait, but what if C is not unique? For example, if a_1000 + M is greater than 1001, then C would be a_1000 + M - 1001. But since a_1000 and M are both between 1 and 1001, their sum is between 2 and 2002. So, C can be between 1 and 1001. But since T - S is equal to a_1000 + M, which is between 2 and 2002, and T is fixed, the last sage can compute a_1000 + M exactly, and then announce C = (a_1000 + M) modulo 1001. Therefore, C uniquely determines a_1000 + M modulo 1001, but not the exact value. However, since a_1000 and M are both positive integers, and their sum is T - S, which is known, sage 1000 can compute a_1000 + M exactly, and then announce C = (a_1000 + M) modulo 1001.But then, sage 999 can compute a_999 = T - S' - C. Since T and S' are known, and C is known, a_999 can be computed exactly. Similarly, sage 998 can compute a_998 = T - S'' - a_999 - C, and so on. This way, every sage except the last one can figure out their own hat number. The last sage, however, only announces C, which is (a_1000 + M) modulo 1001, but he doesn't know his own hat number or the missing number M. So, he can't determine his own hat number for sure. Therefore, he might guess, but he has a 1/1001 chance of being correct.But the question is about guaranteeing a certain number of correct answers. Since the last sage can't guarantee his own hat number, but everyone else can, then we can guarantee that at least 999 sages will correctly state their hat numbers. Therefore, for part b), the answer is yes.Wait, but part a) asks if they can guarantee more than 500. If they can guarantee 999, then certainly more than 500. So, both a) and b) are possible.I think this makes sense. The key idea is that the last sage encodes the missing number plus his own hat number modulo 1001, and then each subsequent sage can use that information along with the sum of the hats in front of them to deduce their own hat number. This way, all sages except the last one can correctly state their hat numbers, guaranteeing at least 999 correct answers. Therefore, both a) and b) are possible.

💡First, I need to determine the total number of votes cast in the election. Let's denote the total number of votes as ( V ).The candidate received 35% of the total votes, which can be expressed as ( 0.35V ).The rival won by 2400 votes more than the candidate. Therefore, the rival received ( 0.35V + 2400 ) votes.Since the sum of the votes received by both the candidate and the rival equals the total number of votes cast, I can set up the equation:[0.35V + (0.35V + 2400) = V]Combining like terms:[0.7V + 2400 = V]To isolate ( V ), I'll subtract ( 0.7V ) from both sides of the equation:[2400 = V - 0.7V][2400 = 0.3V]Finally, solving for ( V ) by dividing both sides by 0.3:[V = frac{2400}{0.3} = 8000]Therefore, the total number of votes cast is 8000.