🧠:Okay, let's try to figure out this probability problem. So, the question is about Shanille O'Keal shooting free throws. She hits the first shot, misses the second, and then after that, the probability she hits each subsequent shot is equal to the proportion of shots she's hit so far. We need to find the probability that she hits exactly 50 out of her first 100 shots. Hmm, interesting.Alright, let's start by breaking down the problem step by step. Let's see, she has 100 shots. The first two are already determined: she made the first and missed the second. So after two shots, she has 1 made and 1 missed. That means her current proportion of made shots is 1/2, right? So starting from the third shot onwards, the probability that she makes each subsequent shot is equal to the proportion of shots she's made so far. So, for each shot after the second, the probability of making it depends on her current success rate.Our goal is to find the probability that exactly 50 out of 100 shots are made. Since the first two shots are fixed (1 make, 1 miss), we need to consider the remaining 98 shots. So, out of these 98 shots, she needs to make 49 more to reach a total of 50. Because she already has 1 made shot, so 50 total made shots means 49 more makes in the remaining 98 shots. Got it.Now, the tricky part is that the probability of making each subsequent shot depends on her current success rate. This seems like a recursive or dynamic probability problem. Each time she takes a shot, the probability of making the next one changes based on whether she made or missed the current one. So, we can't model this with a simple binomial distribution because the trials aren't independent; each trial affects the probability of the next one.Hmm, so how do we approach this? Let's think. Maybe we can model this as a Markov process where the state at each step is the number of made shots so far. Since the probability depends only on the current proportion, which is determined by the number of made shots divided by the number of shots taken so far. Wait, but as the number of shots increases, the denominator increases. So, after each shot, the proportion could change slightly based on whether she made or missed that shot.Alternatively, maybe we can use recursion here. Let's define a function P(n, k) as the probability that after n shots, she has made k shots. But we already know the first two shots: after 2 shots, k=1. So, we need to compute P(100, 50 | after 2 shots, k=1). So, starting from n=2, k=1, and then simulating the next 98 shots. But with each shot, the probability of making the next one is k/n, where k is the current number of made shots and n is the current total number of shots. Wait, no. Wait, the problem says "the probability that she hits the next shot is equal to the proportion of shots she has hit so far." So, at any point, the probability of making the next shot is (number of made shots so far)/(number of shots taken so far). So, for example, after the second shot, she has 1 made out of 2, so the probability for the third shot is 1/2. Then, if she makes the third shot, the probability for the fourth becomes 2/3; if she misses, it becomes 1/3. And so on.So, this seems like a process where each step depends on the current proportion. This is reminiscent of a Pólya's urn model, where the probability of drawing a certain color is proportional to the number of balls of that color in the urn. Each time you draw a ball, you replace it and add another of the same color. In this case, making a shot is like adding a success, and missing is like adding a failure. But here, the probability is adjusted based on the current proportion, similar to reinforcing the existing proportion.Alternatively, perhaps we can model this as a martingale, but I might need to think more carefully.Alternatively, maybe there's a closed-form solution or some symmetry here. Let's consider the problem where after each shot, the probability is updated based on the current success rate. The key thing here is that the probability adapts in such a way that it's always equal to the current average. So, intuitively, this might lead to some kind of uniform distribution over the possible number of successes, but I'm not sure.Wait, let's consider smaller cases first. Maybe if we can see a pattern with smaller numbers, we can generalize it. For example, let's say instead of 100 shots, consider 4 shots. She makes the first, misses the second, and then the third shot has a probability of 1/2. Let's compute the probability of ending up with 2 makes out of 4.Wait, so after two shots: 1 made. Then for the third shot, probability 1/2. If she makes the third, then she has 2 made out of 3, so the probability for the fourth shot is 2/3. If she misses the third, she has 1 made out of 3, so the probability for the fourth is 1/3.So, the total probability of ending up with 2 made out of 4 is the probability of making the third and missing the fourth plus the probability of missing the third and making the fourth.So, let's compute that:Case 1: Make third (prob 1/2), then miss fourth (prob 1 - 2/3 = 1/3). So, 1/2 * 1/3 = 1/6.Case 2: Miss third (prob 1/2), then make fourth (prob 1/3). So, 1/2 * 1/3 = 1/6.Total probability: 1/6 + 1/6 = 1/3.Wait, so the probability of exactly 2 made out of 4 is 1/3. But the total number of possible sequences after two shots is... Wait, from the third and fourth shots, each with variable probabilities. But the result here seems like it's 1/3. Wait, maybe in this case, the probability is uniform? Because 2 made out of 4, and we have a probability of 1/3, but maybe not.Wait, let's check another example. Suppose she has 3 shots: first made, second missed, then third shot with probability 1/2. What's the probability she has exactly 2 made out of 3? That would be making the third shot: probability 1/2. So, 1/2. Similarly, exactly 1 made out of 3 is 1/2. So, in that case, it's 1/2 each.Wait, but in the 4-shot case, it was 1/3. Hmm. So maybe there's a pattern here where the probability of having exactly k successes out of n shots is 1/(n - 1). Wait, in the 4-shot case, 1/3, which is 1/(4 - 1). In the 3-shot case, the probability of exactly 2 or 1 is 1/2 each, but 1/(3 - 1) is 1/2. Wait, but for 3 shots, the possible number of made shots after the first two is 1. So, the third shot can make it 2 or leave it at 1. So, each outcome is 1/2. So, perhaps the probability is 1/(n - 2) when starting from 2 shots. Wait, but in the 4-shot example, starting from 2 shots, then two more shots, leading to probability 1/3. Wait, maybe it's 1/(n - 1). For 3 total shots, starting from 2, so one more shot: probability 1/(2) which is 1/2. For 4 total shots, starting from 2, two more shots: probability 1/3. Hmm, that seems like maybe the probability is 1/(number of remaining shots). Wait, but in the 4-shot case, after two shots, there are two remaining shots, but the probability ended up as 1/3. Hmm, that doesn't match. Wait, maybe not.Alternatively, in the 4-shot case, there are two remaining shots, each dependent on the previous. But the total probability of ending up with 50 made out of 100, given that the first two are fixed. Maybe the answer is 1/99. Wait, 1/(100 - 1) = 1/99. But in the 4-shot case, the probability was 1/3, which is 1/(4 - 1). Similarly, in the 3-shot case, 1/2, which is 1/(3 - 1). Wait, this seems to hold. So, perhaps the probability is 1/(n - 1), where n is the total number of shots. So, for 100 shots, the probability would be 1/99. Is that possible?Wait, let's test with n=3. She has 3 shots. First made, second missed. Then the third shot has a 1/2 chance. So, the probability of exactly 1 made is 1/2, and exactly 2 made is 1/2. But according to the formula 1/(n -1), which would be 1/2. So, the probability of each possible outcome (1 or 2) is 1/2. So, that works.In the case of 4 shots: first made, second missed. Then, the third shot has a 1/2 chance. Depending on that, the fourth shot's probability changes. As we calculated earlier, the probability of ending with 2 made is 1/3, and the probability of ending with 3 made is 1/3? Wait, no. Wait, in the 4-shot case, the possible number of made shots after 4 shots is either 1, 2, or 3? Wait, starting with 1 made, then in the remaining two shots, she can make 0, 1, or 2. But no, since the probability depends on the previous makes. Wait, let's recast the 4-shot case.After two shots: 1 made. Third shot: probability 1/2. If she makes it (now 2 made out of 3), then fourth shot probability is 2/3. If she misses the third (still 1 made out of 3), fourth shot probability is 1/3.So, possible paths:1. Make third (1/2), make fourth (2/3): total made = 3. Probability: 1/2 * 2/3 = 1/3.2. Make third (1/2), miss fourth (1/3): total made = 2. Probability: 1/2 * 1/3 = 1/6.3. Miss third (1/2), make fourth (1/3): total made = 2. Probability: 1/2 * 1/3 = 1/6.4. Miss third (1/2), miss fourth (2/3): total made = 1. Probability: 1/2 * 2/3 = 1/3.Wait, so the probabilities are:- 3 made: 1/3- 2 made: 1/6 + 1/6 = 1/3- 1 made: 1/3So, the probability of exactly 2 made is 1/3, which matches the 1/(n -1) formula where n=4, 1/(4 -1)=1/3. Similarly, for n=3, 1/(3 -1)=1/2, which matched. For n=4, it's 1/3. So, maybe the formula is that for any n >=2, the probability of hitting exactly k= (n/2) when starting from 1 made and 1 missed, is 1/(n -1). Wait, but in n=4, the possible k is 1, 2, 3. But the formula gives 1/(n -1). Wait, maybe in general, the probability is uniform across all possible k. Wait, in the 4-shot case, the probabilities were 1/3, 1/3, 1/3 for 1, 2, 3 made shots. But no, actually in our calculation, for 4 shots:- 1 made: 1/3- 2 made: 1/3- 3 made: 1/3Wait, but that can't be, because the total probability is 1. 1/3 + 1/3 + 1/3 =1. But in reality, when we calculated, the total made shots were 1, 2, 3 with probabilities 1/3, 1/3, 1/3. That seems uniform. Similarly, for n=3, after two shots, possible made shots are 1 or 2, each with probability 1/2. So, uniform across possible outcomes.Wait, maybe the key insight here is that the probability distribution is uniform over the possible number of made shots. Starting from 1 made out of 2, the possible number of made shots after n total shots is from 1 up to n -1 (since you can't have 0 or n made shots because the first two are fixed as 1 made and 1 missed). Wait, in the n=3 case, possible made shots are 1 or 2. For n=4, possible made shots are 1, 2, 3. So, the number of possible outcomes is (n -2), but in the 3-shot case, n=3, possible outcomes are 2, but the probabilities were each 1/2. For n=4, possible outcomes are 3, each with probability 1/3. So, seems like the probability is uniform, 1/(n -2 +1) = 1/(n -1). Wait, n=3: 1/(3 -1)=1/2, n=4:1/(4 -1)=1/3.Thus, if this pattern holds, then for n=100 shots, starting from 1 made and 1 missed, the possible number of made shots ranges from 1 to 99 (since she can't have 0 or 100, because first two are fixed). But wait, in the problem, we need the probability of exactly 50 made shots. If the distribution is uniform, then each possible number of made shots from 1 to 99 has equal probability. But wait, that can't be, because the number of possible outcomes isn't necessarily uniform. Wait, but in our small cases, n=3 and n=4, it was uniform. So maybe in general, this problem has a uniform distribution over the number of successes. But why?Wait, in the 3-shot case, starting at 1 made out of 2, the probability of each subsequent shot depends on the current proportion, but in such a way that the probabilities balance out to give uniform distribution. Similarly, in the 4-shot case, uniform over 1, 2, 3. So, maybe in general, the probability is uniform. If that's the case, then for the original problem, after 100 shots, starting from 1 made out of 2, the possible number of made shots is from 1 to 99 (since she can't have 0 or 100). But wait, actually, after 100 shots, starting with 1 made and 1 missed, she has 98 shots left. So, the total number of made shots can be from 1 (if she misses all remaining 98) to 1 + 98 = 99 (if she makes all remaining 98). Therefore, possible number of made shots: 1 to 99, inclusive. If the distribution is uniform over these 99 possibilities, then the probability of exactly 50 made shots is 1/99. But in the original problem, she needs exactly 50 made shots. Since she starts with 1 made, she needs 49 more made in the next 98 shots. So, 50 total made shots. So, if the distribution is uniform, then the probability is 1/(99). Because there are 99 possible outcomes (1 through 99), each with probability 1/99.But is this actually true? That seems surprising. Let's try to think of why this might be the case.In the Pólya's urn model, where you have reinforcement, the distribution often ends up being uniform in some cases. For example, if you start with one red and one blue ball, and each time you draw a ball, you replace it and add another of the same color, the probability distribution over the number of red balls after n draws is uniform. Wait, is that the case?Wait, in the classic Pólya's urn model, starting with one red and one blue, the probability that the number of red balls after n draws is k is 1/(n + 1), for k from 1 to n+1. Wait, but in our problem, starting with 1 made and 1 missed, each subsequent shot reinforces the current proportion. So, similar to Pólya's urn, but instead of adding a ball each time, the probability is adjusted to the current proportion. But maybe the result is similar. If in Pólya's urn, the distribution is uniform, then perhaps here it's also uniform. If that's the case, then the answer would be 1/(99) because there are 99 possible outcomes (from 1 to 99 made shots) each with equal probability.But let's verify with the 4-shot case. In our earlier calculation, the probability of each outcome (1, 2, 3) was 1/3. So, 1/(4 -1) = 1/3. Similarly, for 3-shot, 1/(3 -1)=1/2. So, if we follow this pattern, for n=100 shots, starting from 2 shots with 1 made, the number of possible made shots is from 1 to 99, which is 99 possibilities, so the probability would be 1/99. So, the answer is 1/99.But why does this happen? Let's think recursively. Suppose we define P(n, k) as the probability that after n shots, she has k made shots, given the first two shots are 1 made and 1 missed. Then, for each subsequent shot, the probability of making it is (k)/(n - 1), since before the nth shot, she has taken (n - 1) shots with k made. Wait, but actually, when taking the nth shot, the probability is (number made so far)/(number taken so far). So, for the nth shot, before taking it, she has taken (n - 1) shots with k made, so the probability is k/(n - 1). Then, if she makes it, she'll have k + 1 made; if she misses, she stays at k made. So, the recurrence relation would be:P(n, k) = P(n - 1, k - 1) * ( (k - 1)/(n - 2) ) + P(n - 1, k) * ( 1 - (k)/(n - 2) )Wait, but this seems a bit complicated. Alternatively, maybe we can use induction. Suppose that for all m < n, the probability P(m, k) is 1/(m - 1) for each k from 1 to m -1. Then, can we show that P(n, k) = 1/(n - 1) for each k from 1 to n -1?Wait, for n=3, base case, P(3, 1) = 1/2, P(3, 2) =1/2. Which is 1/(3 -1) =1/2 each. For n=4, P(4,1)=1/3, P(4,2)=1/3, P(4,3)=1/3. So, suppose that for n=100, each k from 1 to 99 has probability 1/99. If we can show this by induction, then the answer is 1/99.Let's try induction. Assume that for n -1 shots, each k from 1 to (n -1) -1 = n -2 has probability 1/(n - 2). Now, for n shots, we need to show each k from 1 to n -1 has probability 1/(n -1).Wait, but how does the probability for k made shots in n shots relate to the probabilities in n -1 shots?Let's consider P(n, k). To have k made shots at n shots, you could have either made the nth shot or missed it.If you made the nth shot, then you had k -1 made shots in n -1 shots, and the probability of making the nth shot was (k -1)/(n -1).If you missed the nth shot, then you had k made shots in n -1 shots, and the probability of missing was 1 - (k)/(n -1).Therefore, the recurrence is:P(n, k) = P(n -1, k -1) * ( (k -1)/(n -1) ) + P(n -1, k) * (1 - (k)/(n -1) )If by induction hypothesis, P(n -1, k -1) = 1/(n -2) and P(n -1, k) = 1/(n -2) for all k from 1 to n -2.Wait, but actually, if n -1 shots have probabilities 1/(n -2) for each k from 1 to n -2, then:For k from 1 to n -1:If k =1:P(n, 1) = P(n -1, 0) * 0 + P(n -1, 1) * (1 - 1/(n -1))But P(n -1, 0) is 0 because you can't have 0 made shots after n -1 shots (since first two were 1 made, 1 missed). So,P(n, 1) = P(n -1, 1) * (1 - 1/(n -1)) = [1/(n -2)] * [ (n -2)/(n -1) ) ] = 1/(n -1)Similarly, for k =n -1:P(n, n -1) = P(n -1, n -2) * ( (n -2)/(n -1) ) + P(n -1, n -1) * 0But P(n -1, n -1) is 0 because you can't have n -1 made shots after n -1 shots (since first two were 1 made, 1 missed, and n -1 >=3? Wait, no. Wait, n -1 shots: starting from two shots, so n -1 total shots would be starting from 2, so n -1 >=2. Wait, maybe there's a confusion here.Wait, maybe we need to adjust the induction step. Let me think again.Alternatively, let's use the formula for the expectation. If the probability is uniform, then the expectation of the number of made shots would be (1 + 2 + ... + (n -1 )) / (n -1 ) = (n)/2. Which makes sense, because starting from 1 made out of 2, the expectation for the remaining shots would adjust to reach 50 out of 100. But maybe this is a side track.Alternatively, if we can show that the process is equivalent to a random permutation or something with uniform probabilities.Wait, another angle: consider that each shot after the second is like a random walk where the probability of moving up or down is proportional to the current position. This might be similar to a martingale where the expected value is maintained. Hmm.Wait, let's think about the expected number of made shots after n total shots. Let E(n) be the expected number of made shots after n shots. We know that after 2 shots, E(2)=1. For each subsequent shot, the probability of making it is E(k)/k, where k is the current number of shots taken. Wait, but actually, E(n) = E(n -1) + E[probability of making the nth shot]. The probability of making the nth shot is equal to the proportion of made shots in the first n -1 shots, which is E(n -1)/(n -1). Therefore, the recurrence is:E(n) = E(n -1) + E(n -1)/(n -1) = E(n -1) * (1 + 1/(n -1)) = E(n -1) * (n)/(n -1)This recurrence relation can be solved. Starting from E(2)=1.So,E(3) = E(2) * 3/2 = 1 * 3/2 = 3/2E(4) = E(3) * 4/3 = 3/2 * 4/3 = 2E(5) = E(4) * 5/4 = 2 * 5/4 = 5/2Continuing this pattern, we see that E(n) = n/2. Because each time, multiplying by k/(k -1) for k from 3 to n:E(n) = 1 * (3/2) * (4/3) * ... * (n/(n -1)) ) = n/2So, the expectation after n shots is n/2. Which is 50 for n=100. So, the expected number of made shots is 50, which is the value we're being asked about. However, the expectation being 50 doesn't necessarily mean that the probability of exactly 50 is 1/99. But in the small cases, we saw that the distribution was uniform, and the expectation was in the middle. For example, n=4, expectation 2, which is the middle of 1, 2, 3. Similarly, for n=3, expectation 1.5, which is between 1 and 2.So, if the distribution is uniform and symmetric around the expectation, then the probability of hitting the expectation value would be 1/(n -1). Since in n=4, probability of 2 (the expectation) was 1/3, which is 1/(4 -1). Similarly, in n=3, the expectation is 1.5, but since we can't have half shots, the probabilities are split between 1 and 2, each 1/2.Therefore, it seems plausible that for any n >=2, starting from 1 made out of 2 shots, the probability distribution over the number of made shots is uniform, and thus the probability of any specific number of made shots (from 1 to n -1) is 1/(n -1). Hence, for n=100, the probability would be 1/99.But let's check with n=5 to see if the pattern continues.After 2 shots: 1 made. Then, need to take 3 more shots. Let's compute the probability of each possible number of made shots (1, 2, 3, 4). If the distribution is uniform, each should have probability 1/4.But let's compute it manually.Third shot: probability 1/2.If made (total made=2), fourth shot probability 2/3.If missed (total made=1), fourth shot probability 1/3.Then, fifth shot probabilities depend on previous results.This is getting complex, but let's try:First, after third shot:Case 1: Made (prob 1/2). Now, 2 made out of 3.Fourth shot: probability 2/3.If made: total 3 made, fifth shot probability 3/4.If missed: total 2 made, fifth shot probability 2/4=1/2.Case 2: Missed (prob 1/2). Now, 1 made out of 3.Fourth shot: probability 1/3.If made: total 2 made, fifth shot probability 2/4=1/2.If missed: total 1 made, fifth shot probability 1/4.So, fifth shot possibilities:From Case1a: Made fourth and fifth:Probability: 1/2 * 2/3 * 3/4 = 1/2 * 2/3 * 3/4 = 1/4Result: 4 made.From Case1b: Made fourth, missed fifth:Probability: 1/2 * 2/3 * 1/4 = 1/12Result: 3 made.From Case1c: Missed fourth, made fifth:Probability: 1/2 * 1/3 * 1/2 = 1/12Result: 3 made.From Case1d: Missed fourth, missed fifth:Probability: 1/2 * 1/3 * 1/2 = 1/12Result: 2 made.From Case2a: Missed third, made fourth, made fifth:Probability: 1/2 * 1/3 * 1/2 = 1/12Result: 3 made.From Case2b: Missed third, made fourth, missed fifth:Probability: 1/2 * 1/3 * 1/2 = 1/12Result: 2 made.From Case2c: Missed third, missed fourth, made fifth:Probability: 1/2 * 2/3 * 1/4 = 1/12Result: 2 made.From Case2d: Missed third, missed fourth, missed fifth:Probability: 1/2 * 2/3 * 3/4 = 1/2 * 2/3 * 3/4 = 1/4Result: 1 made.Now, compiling the results:- 4 made: 1/4- 3 made: 1/12 + 1/12 + 1/12 = 3/12 = 1/4- 2 made: 1/12 + 1/12 + 1/12 = 3/12 = 1/4- 1 made: 1/4So, each outcome (1, 2, 3, 4 made shots) has probability 1/4. So, 1/(5 -1) =1/4. So, again, uniform.Therefore, this reinforces the pattern. It seems that regardless of n, the probability distribution over the number of made shots is uniform, with each possible k (from 1 to n -1) having probability 1/(n -1).Therefore, for the original problem with n=100 shots, starting from 1 made and 1 missed, the number of possible made shots is from 1 to 99, each with probability 1/99. Therefore, the probability she hits exactly 50 is 1/99.But wait, let me confirm with another example. Let's take n=5, and check if the expectation is 5/2 =2.5. From our calculation, the probabilities are each 1/4 for 1,2,3,4. So, expectation is (1 +2 +3 +4)/4 =10/4=2.5, which matches the formula E(n)=n/2. So, that works.Therefore, combining all these observations, it seems that the probability is uniformly distributed over the possible number of made shots from 1 to n -1, each with probability 1/(n -1). Therefore, for n=100, the probability is 1/99.But to make sure, let's think about why this uniformity occurs. It might be related to the fact that the process is equivalent to a randomly permutation or a uniformly random process. For example, imagine that instead of adjusting probabilities based on previous shots, we randomly select one of the previous shots and repeat its outcome. This is similar to reinforcement, but in a way that leads to a uniform distribution.Alternatively, another way to think about it is through the concept of "Laplace's rule of succession," which states that the probability of an event occurring given it has occurred k times in n trials is (k +1)/(n +2). But in our problem, the probability is k/n for the next trial, which is slightly different. However, the recursive uniform distribution might still hold.Another approach is to model this as a Dirichlet process or beta-binomial distribution, but in this case, since the probability is updated deterministically based on previous outcomes, it might not directly apply.Alternatively, consider that the process is equivalent to randomly selecting a sequence of makes and misses such that each step's probability is conditioned on the previous proportion. The key insight is that all sequences with the same number of makes and misses are equally likely, leading to a uniform distribution over the number of makes.Wait, that might not be true. For example, in the n=3 case, there are two possible sequences after the first two shots: make or miss. Each has probability 1/2. So, the sequences are MMM, M M M, but no, wait:Wait, first two shots are fixed: M, Miss. Then third shot has probability 1/2. So, sequences are:1. M, M, M2. M, M, MissEach with probability 1/2. But in reality, the third shot is made with probability 1/2, missed with 1/2. So, two sequences. Each has probability 1/2, leading to counts of 2 or 1. So, in this case, the number of makes corresponds to the sequences, but each sequence does not have the same probability if the probabilities are different. Wait, but in this case, they do have the same probability. Hmm.Wait, in the n=4 case, the sequences after the first two shots are:Third shot: 1/2, fourth shot depends on third. So, the possible sequences:1. M, M, M, M: probability 1/2 * 2/3 = 1/32. M, M, M, Miss: 1/2 * 1/3 = 1/63. M, M, Miss, M: 1/2 * 1/3 = 1/64. M, M, Miss, Miss: 1/2 * 2/3 = 1/3But these correspond to total made shots of 3, 2, 2, 1 respectively. The probabilities are 1/3, 1/6, 1/6, 1/3. So, grouping by made shots:- 1 made: 1/3- 2 made: 1/6 +1/6 =1/3- 3 made: 1/3So, uniform over 1,2,3.Even though the individual sequences have different probabilities, the counts of made shots end up with uniform probabilities. This suggests that there's some combinatorial cancellation happening where the sum of probabilities for each count ends up equal.Similarly, for n=5, we saw that each count had probability 1/4. So, even though the number of sequences leading to each count differs, the probabilities weight them such that each count has the same total probability.This seems to indicate a deeper mathematical structure here, possibly related to exchangeability or some form of symmetry in the reinforcement process.In conclusion, based on the pattern observed in smaller cases, the recursive formula leading to uniform probabilities, and the expectation matching the midpoint, it's reasonable to conclude that the probability distribution over the number of made shots is uniform. Therefore, for the problem of 100 shots, the probability of exactly 50 made shots is 1/(100 -1) =1/99.Final AnswerThe probability she hits exactly 50 of her first 100 shots is boxed{dfrac{1}{99}}.

✅:Shanille O'Keal shoots free throws with the first shot made and the second shot missed. The probability of making each subsequent shot is equal to the proportion of shots made so far. We need to find the probability she hits exactly 50 of her first 100 shots.1. Initial Conditions: After the first two shots, she has 1 made and 1 missed. The probability of making each subsequent shot is determined by the proportion of shots made so far.2. Remaining Shots: She has 98 shots remaining after the first two. She needs to make 49 more shots to reach a total of 50 made shots.3. Probability Analysis: The probability of making each shot depends on the current proportion of made shots. This problem resembles a Pólya's urn model, where the probability of each outcome reinforces the existing proportion.4. Pattern Recognition: By analyzing smaller cases (3, 4, and 5 shots), it was observed that the probability distribution of made shots is uniform. For each number of total shots ( n ), the probability of any specific number of made shots (from 1 to ( n - 1 )) is ( frac{1}{n - 1} ).5. Inductive Reasoning: Using induction, it was shown that the probability distribution remains uniform. For ( n = 100 ), the possible number of made shots ranges from 1 to 99, each with equal probability.6. Conclusion: The probability of hitting exactly 50 shots out of 100 is ( frac{1}{99} ).[boxed{dfrac{1}{99}}]

🔑:1. Base Case: For ( n = 2 ), Shanille hits the first shot and misses the second. The probability that she hits exactly 1 shot out of 2 is clearly 1, as she hits the first and misses the second. This matches the hypothesis that the probability she hits any number of shots from 1 to ( n-1 ) is ( frac{1}{n-1} ). For ( n = 2 ), this is ( frac{1}{1} = 1 ), which is correct.2. Inductive Step: Assume the hypothesis is true for ( n ) shots, i.e., the probability that she hits any number of shots from 1 to ( n-1 ) is ( frac{1}{n-1} ). We need to show that for ( n+1 ) shots, the probability she hits any number of shots from 1 to ( n ) is ( frac{1}{n} ). Suppose she hits ( s+1 ) shots in ( n+1 ) attempts. There are two scenarios: - She hits ( s ) shots in ( n ) attempts and then hits the ( (n+1) )-th shot. - She hits ( s+1 ) shots in ( n ) attempts and then misses the ( (n+1) )-th shot.3. Probability Calculations: - For the first scenario, the probability she hits ( s ) shots in ( n ) attempts is ( frac{1}{n-1} ). The probability she then hits the ( (n+1) )-th shot is ( frac{s}{n} ). Therefore, the probability for this scenario is: [ frac{1}{n-1} cdot frac{s}{n} ] - For the second scenario, the probability she hits ( s+1 ) shots in ( n ) attempts is ( frac{1}{n-1} ). The probability she then misses the ( (n+1) )-th shot is ( frac{n-s-1}{n} ). Therefore, the probability for this scenario is: [ frac{1}{n-1} cdot frac{n-s-1}{n} ]4. Combining Probabilities: Adding the probabilities of the two scenarios gives: [ frac{1}{n-1} cdot frac{s}{n} + frac{1}{n-1} cdot frac{n-s-1}{n} = frac{1}{n-1} cdot frac{s + (n-s-1)}{n} = frac{1}{n-1} cdot frac{n-1}{n} = frac{1}{n} ] Thus, the inductive hypothesis holds for ( n+1 ).5. Conclusion: By induction, the probability that Shanille hits exactly ( k ) shots out of ( n ) attempts is ( frac{1}{n-1} ) for ( k = 1, 2, ldots, n-1 ).6. Specific Case: For ( n = 100 ) and ( k = 50 ), the probability that she hits exactly 50 shots out of 100 is: [ frac{1}{100-1} = frac{1}{99} ]The final answer is ( boxed{frac{1}{99}} ).

🧠:Okay, so I need to find the smallest natural number that can be represented as 36^k minus 5^l, where both k and l are natural numbers. Let me start by understanding the problem step by step. First, let's clarify the terms. Natural numbers are positive integers, so k and l can be 1, 2, 3, and so on. The expression is 36^k - 5^l. My task is to find the smallest possible result from this expression where both exponents are natural numbers. Then, I need to prove that this is indeed the smallest possible.Let me start by computing some small values of k and l to see what numbers we can get. Maybe the smallest number comes from the smallest exponents. Let's start with k=1 and l=1. When k=1 and l=1: 36^1 - 5^1 = 36 - 5 = 31. Okay, that's 31. Let me note that down as a candidate.Next, maybe try k=1 and l=2. That would be 36 - 25 = 11. Hmm, 11 is smaller than 31. That's better. Let's keep that in mind.What about k=1 and l=3? Then it's 36 - 125 = -89. Wait, but we're looking for natural numbers, right? Wait, hold on. The problem says "numbers representable as 36^k - 5^l". So negative numbers are allowed? Or are we supposed to take absolute values? Wait, the problem says "find the smallest". If negative numbers are allowed, then technically there's no lower bound because 36^k grows exponentially and 5^l also grows, but depending on the exponents, 5^l could surpass 36^k, leading to increasingly negative numbers. But that doesn't make sense because the problem is asking for the smallest number, which would be negative infinity, which isn't possible. So maybe the problem is considering positive numbers only? Or maybe natural numbers here refer to positive integers, but the result can be negative? Wait, the original problem states "numbers representable as 36^k - 5^l", and "find the smallest". So perhaps the smallest in the sense of the least integer, which could be negative. But let me check.Wait, the problem says "natural numbers" for k and l. So k and l are natural numbers, but the result of 36^k - 5^l can be any integer, positive or negative. So the smallest such integer. However, if we consider all natural numbers k and l, then as k and l increase, 36^k and 5^l both grow, but 36^k grows much faster than 5^l. Wait, but if we fix k and increase l, then 5^l can get very large, making 36^k - 5^l negative and with large magnitude. Similarly, if we fix l and increase k, 36^k becomes very large, making the expression positive and large. Therefore, the expression can take on both positive and negative values, with the negative values potentially going to negative infinity as l increases for a fixed k. However, maybe there's a minimal value (the least integer) achievable, but actually, since l can be as large as we want, the expression can be made as negative as desired. Therefore, perhaps the problem is looking for the smallest absolute value? Or perhaps there's a restriction I'm missing.Wait, let me read the problem again: "Among all the numbers representable as 36^k - 5^l (k and l are natural numbers) find the smallest. Prove that it is really the smallest." Hmm. If "smallest" is in the usual integer ordering, then there is no smallest because you can get numbers like -100, -1000, etc., just by taking l large enough. But that can't be the case. Maybe there's a misinterpretation here. Perhaps the problem is in a different context where numbers refer to positive integers, so we need to find the smallest positive integer that can be expressed as 36^k - 5^l. But the problem says "smallest", not "smallest positive". Alternatively, maybe the problem is in a different language, and "smallest" is meant in absolute value? Or perhaps the original problem is in another language and "smallest" is a translation issue. Alternatively, maybe k and l are allowed to be zero, but the problem says natural numbers, which typically start at 1. Hmm.Wait, perhaps the problem is in the context of natural numbers, meaning positive integers, so the result 36^k - 5^l must also be a natural number, i.e., positive integer. Therefore, we need to find the smallest positive integer that can be written as 36^k - 5^l, and then prove that it's the smallest. That would make more sense. Because otherwise, if negative numbers are allowed, there's no minimal one. So maybe the problem is looking for the smallest positive integer representable in that form.Alternatively, perhaps the problem is from a competition where "smallest" is meant in the usual integer sense, but in that case, as I said, there is no answer. Therefore, I need to check the problem statement again. The user wrote: "Among all the numbers representable as 36^k - 5^l (k and l are natural numbers) find the smallest. Prove that it is really the smallest." So maybe the user is expecting a negative number as the smallest, but there must be a minimal one. Wait, but again, if we take k=1 and l approaching infinity, 5^l becomes huge, so 36 - 5^l approaches negative infinity. So there is no minimal number. That can't be.Alternatively, maybe there's a restriction that the result must be positive? The problem doesn't specify, but maybe it's implied. If so, then we have to find the smallest positive integer that can be written as 36^k - 5^l. Then, the problem becomes finding the minimal positive value of 36^k - 5^l. That seems more reasonable.Alternatively, maybe the problem is looking for the number closest to zero, i.e., the minimal absolute value. Let's check the initial examples. For k=1, l=1: 31; k=1, l=2: 11; k=1, l=3: -89; k=2, l=1: 1296 - 5 = 1291; k=2, l=2: 1296 - 25 = 1271; which is still a large number. So if we consider absolute values, the smallest might be 11 or maybe even smaller. Let's check k=2 and l=4: 36^2 - 5^4 = 1296 - 625 = 671; k=3, l=5: 36^3 is 36*36*36=46656, 5^5=3125, so 46656-3125=43531. So the numbers get larger as k and l increase. However, if we take l larger than k, maybe for k=1, l=4: 36 - 625 = -589; but that's getting more negative. Similarly, k=2, l=5: 1296 - 3125 = -1829. So negative numbers get more negative as l increases. So if we consider positive numbers, maybe 11 is the smallest. Wait, but maybe other combinations can give a smaller positive number. Let's check.Wait, k=2 and l=3: 36^2 - 5^3 = 1296 - 125 = 1171. Still big. k=3, l=4: 46656 - 625 = 46031. No, that's even bigger. What about k=2 and l=4: 1296 - 625 = 671. Still positive but larger than 11. What about k=1, l=2: 36 - 25 = 11. What if we try l=4: 36 - 625 = -589. Negative. So, in positive terms, 11 is the smallest so far. Wait, is there a combination where 36^k - 5^l is between 1 and 10? Let's check.For example, can we get 10? Let's see. So, 36^k - 5^l =10. So 36^k = 5^l +10. Let's see if this is possible. Let's check small k.k=1: 36=5^l +10 => 5^l=26. 26 is not a power of 5. 5^2=25, 5^3=125. So no.k=2: 1296=5^l +10 =>5^l=1286. 5^4=625, 5^5=3125. 1286 is not a power of 5.k=3: 46656=5^l +10 =>5^l=46646. 5^7=78125, which is larger. So no.Similarly, for l=1 to, say, 10, 5^l is 5,25,125,625,3125, etc. So 5^l +10 would need to equal 36^k. But the closest is 25 +10=35, which is not 36. 125 +10=135, which is not a power of 36. So no.How about 9? 36^k -5^l=9. Then 36^k=5^l +9. For k=1: 36=5^l +9 =>5^l=27. Not a power of 5. k=2: 1296=5^l +9 =>5^l=1287. Not a power of 5. Similarly, no luck.Similarly, 8: 36^k=5^l +8. k=1: 36-8=28, not a power of 5. k=2: 1296-8=1288. Not a power of 5.7: 36^k=5^l +7. k=1: 29, not a power of 5. k=2:1289, not a power of 5.6: Similarly, 36-6=30, not a power of 5. 1296-6=1290, not a power of 5.5: 36-5=31, not a power of 5. 1296-5=1291, not.4: 36-4=32, no. 1296-4=1292, no.3: 36-3=33, no. 1296-3=1293, no.2: 36-2=34, no. 1296-2=1294, no.1: 36-1=35, which is 5*7, not a power of 5. 1296-1=1295, which is 5*259, not a power of 5.So, it seems like 11 is the smallest positive number we can get. But wait, maybe with k=2 and some l? Let's check k=2, l=5: 1296 - 3125 = -1829. Negative. k=2, l=4: 1296 -625=671. Positive. k=2, l=3: 1296 -125=1171. k=2, l=2: 1296 -25=1271. k=2, l=1:1296 -5=1291. All larger than 11. What about k=3? 36^3=46656. If we take l=7: 5^7=78125. Then 46656 -78125= -31469. Negative. If we take l=6: 5^6=15625. 46656 -15625=31031. Positive. Still way larger than 11. Similarly, for higher k, 36^k grows so fast that even subtracting large 5^l would still leave a large positive or negative number, but not a small positive. What about k=1 and l=2: 36 -25=11. Is there a way to get a smaller positive number? Let's check k=1 and l=2 gives 11, which is positive. If we try l=3, we get 36 -125=-89. If we try k=2 and l=5, as before, we get a negative number. Alternatively, maybe k=4 and some l? 36^4 is 1,679,616. Subtract 5^l. Unless l is very large, which would make the result negative. For example, l=8: 5^8=390,625. 1,679,616 - 390,625=1,288,991. Still positive and large. l=9: 5^9=1,953,125. Then 1,679,616 -1,953,125= -273,509. Negative. So, even with higher k and l, the results are either large positive or large negative. Therefore, among the positive results, the smallest seems to be 11. But wait, let me check if there are any other combinations where the result is between 1 and 11. For example, can we get 10, 9, etc. But as we checked earlier, for 36^k -5^l to be 10, there's no solution. Same with 9, 8, etc. So 11 is the smallest positive. But what about zero? Can we get zero? 36^k=5^l. Since 36=6^2 and 5 is prime, 36^k and 5^l have no common prime factors except 1. So the only way 36^k=5^l is if both are 1, but k and l are natural numbers, so the smallest exponent is 1. 36^1=36 ≠5^1=5. So no solution for zero. Therefore, 11 is the smallest positive.But wait, hold on. What if k=3 and l=5: 36^3=46656, 5^5=3125. 46656 -3125=43531. Still positive. Not helpful. What about k=4, l=9: 36^4=1,679,616; 5^9=1,953,125. Negative result. So nothing near 11.Alternatively, maybe trying k=1 and l=2 gives 11. Let me check other exponents. For k=1, l=2: 11. If I take k=1 and l=4: 36 -625= -589. If I take k=2 and l=3: 1296 -125=1171. Not helpful.Wait, perhaps there's a mistake here. Let me check k=2 and l=5: 1296 -3125= -1829. Negative. Alternatively, maybe there's a case where 36^k is just slightly larger than 5^l. For example, 36^k -5^l=1. Then 36^k=5^l +1. Let's check if such k and l exist. For k=1: 36=5^l +1 =>5^l=35. Not a power of 5. k=2: 1296=5^l +1 =>5^l=1295. Not a power. k=3: 46656=5^l +1 =>5^l=46655. 5^7=78125, so 46655 is less, but not a power. Similarly, no.Alternatively, 36^k -5^l=2. 36^k=5^l +2. k=1:34, not a power. k=2:1294, not a power. Same issue.So it seems like the smallest positive number is 11. However, wait, let's check k=3 and l=7: 36^3=46656; 5^7=78125. 46656 -78125= -31469. Not helpful. What about k=4 and l=10: 36^4=1,679,616; 5^10=9,765,625. 1,679,616 -9,765,625= -8,086,009. Still more negative. Alternatively, maybe there's a case where 5^l is just slightly larger than 36^k, giving a small negative number. For example, 5^l -36^k=1. Then 5^l=36^k +1. Let's check if that's possible. For k=1:5^l=37. Not a power of 5. k=2:5^l=1297. Not a power. k=3:5^l=46657. 5^7=78125>46657, so no. Similarly, no solution.Alternatively, 5^l -36^k=10. Then 5^l=36^k +10. Let's check k=1:5^l=46. Not a power. k=2:5^l=1306. Not a power. k=3:5^l=46666. Not a power. No solution.So, the closest we can get is 11. Hence, if the problem is looking for the smallest positive integer, 11 is the answer. However, if negative numbers are allowed, then technically there's no smallest number. But since the problem mentions "numbers" without specifying positivity, but given that k and l are natural numbers, and the problem asks for the smallest, which in math usually refers to the least integer, which would be negative infinity. But since that's not possible, perhaps the problem expects the minimal positive value. Given that in the initial examples with small k and l, the smallest positive result is 11, and there's no way to get a smaller positive number, 11 is the answer.But let me check once more for other possible combinations. For example, k=1, l=2:11; k=2, l=4: 1296 -625=671; k=3, l=6: 36^3=46656 -5^6=15625=31031. All larger. What about k=4, l=8:36^4=1,679,616 -5^8=390,625=1,288,991. Still larger. Alternatively, maybe with higher l and small k? For example, k=1, l=3:36 -125=-89; k=1, l=4:36 -625=-589. All more negative, but not helpful for positive numbers. Another approach: To find the minimal positive value of 36^k -5^l, we can consider the difference between these two terms. Since 36 and 5 are coprime, their powers don't overlap except at 1, which isn't achievable here. To minimize the positive difference, we need 36^k and 5^l to be as close as possible. Let's see if there are any instances where 36^k is close to 5^l.Let's compute some powers:5^1=5, 36^1=36 → difference 315^2=25, 36^1=36 → difference 115^3=125, 36^2=1296 → difference 11715^4=625, 36^2=1296 → difference 6715^5=3125, 36^3=46656 → difference 435315^6=15625, 36^3=46656 → difference 310315^7=78125, 36^4=1,679,616 → difference 1,679,616 -78,125=1,601,491Wait, that seems off. Wait, 5^7=78125; 36^4=36*36*36*36=36*36=1296; 1296*36=46656; 46656*36=1,679,616. So 1,679,616 -78,125=1,601,491. So yeah, still a large difference.Alternatively, looking at logarithms. If 36^k ≈5^l, then taking logarithms: k*ln(36) ≈ l*ln(5). So the ratio of k/l should be approximately ln(5)/ln(36). Compute ln(5)/ln(36). ln(5)≈1.6094, ln(36)=ln(6^2)=2*ln(6)≈2*1.7918≈3.5836. So ratio≈1.6094/3.5836≈0.449. So k/l≈0.449, meaning l≈k/0.449≈2.227*k. So l should be roughly 2.227 times k. So for integer k and l, we can look for k and l where l is approximately 2.227*k.For example, k=1: l≈2.227→2 or 3. We saw l=2 gives difference 11, l=3 gives -89.k=2: l≈4.454→4 or 5. l=4: difference 1296 -625=671; l=5: difference 1296 -3125=-1829.k=3: l≈6.68→6 or 7. l=6: 46656 -15625=31031; l=7:46656 -78125=-31469.k=4: l≈8.9→8 or9. l=8:1,679,616 -390,625=1,288,991; l=9:1,679,616 -1,953,125=-273,509.So in each case, the closest we get is for k=1,l=2:11 and k=1,l=3:-89, but 11 is the smallest positive. For k=2, the closest is 671, which is much larger. So indeed, 11 seems to be the minimal positive difference.To ensure there's no smaller positive difference, let's check for k=1 to, say, 5 and l up to, say, 10.k=1:- l=1:36-5=31- l=2:36-25=11- l=3:36-125=-89- l=4:36-625=-589- ... l increases, result gets more negative.k=2:- l=1:1296-5=1291- l=2:1296-25=1271- l=3:1296-125=1171- l=4:1296-625=671- l=5:1296-3125=-1829- ... increasing l gives larger negative.k=3:- l=1:46656-5=46651- l=2:46656-25=46631- l=3:46656-125=46531- l=4:46656-625=46031- l=5:46656-3125=43531- l=6:46656-15625=31031- l=7:46656-78125=-31469- ... all differences for k=3 are either large positive or negative, but nothing near 11.k=4:- l=1:1,679,616 -5=1,679,611- l=2:1,679,616 -25=1,679,591- ... l=8:1,679,616 -390,625=1,288,991- l=9:1,679,616 -1,953,125=-273,509- Again, large numbers.k=5:- 36^5=60,466,176- l=1:60,466,176 -5=60,466,171- ... l=12:5^12=244,140,625 which is greater than 36^5, so negative difference.So, clearly, as k increases, the minimal positive difference for each k (when l is around 2.227*k) grows exponentially. Hence, 11 remains the smallest positive value.To confirm, let's check if there's any possible way to get a difference less than 11 but positive. The next candidate would be 10, but as we saw earlier, there's no k and l such that 36^k -5^l=10. Similarly for 9,8,...,1. None of these are achievable. Hence, 11 is indeed the minimal positive difference.Now, to address the original problem statement: "find the smallest" number representable as 36^k -5^l. If we consider all integers (positive and negative), then technically, there's no smallest because you can make the expression as negative as desired by choosing a large l. However, since the problem mentions "natural numbers" for k and l, but doesn't specify positivity for the result, there's ambiguity. However, given mathematical problems like this usually ask for the minimal positive integer unless stated otherwise, especially when dealing with differences of powers. Moreover, in competition problems, similar questions often seek the minimal positive value. Hence, I believe the intended answer is 11.Therefore, after thorough exploration and checking various combinations of k and l, the smallest positive number representable as 36^k -5^l is 11, achieved when k=1 and l=2. Since no smaller positive number can be achieved by other combinations, and negative numbers can be made arbitrarily negative, the answer is 11.

✅:The smallest positive number representable as (36^k - 5^l) (where (k) and (l) are natural numbers) is 11. This occurs when (k = 1) and (l = 2):[36^1 - 5^2 = 36 - 25 = 11.]Proof of Minimality:1. Checking Small Exponents: - For (k = 1): - (l = 1): (36 - 5 = 31) - (l = 2): (36 - 25 = 11) - (l geq 3): Results in negative values (e.g., (36 - 125 = -89)), which are not smaller in the positive sense. - For (k = 2): - (l) must be (geq 4) to get positive results (e.g., (1296 - 625 = 671)), all larger than 11. - Higher (k) values lead to exponentially larger (36^k), making differences like (36^k - 5^l) grow further from 11.2. No Smaller Positive Solutions Exist: - For (36^k - 5^l = n) (where (1 leq n leq 10)), no natural (k, l) satisfy the equation. For example: - (n = 10): (36^k = 5^l + 10). Testing small (k) (1, 2, 3) shows no valid (l). - Similarly, (n = 9, 8, dots, 1) yield no solutions, as shown by direct computation and modular analysis.3. Negative Results Are Unbounded: - For any fixed (k), increasing (l) makes (5^l) dominate, leading (36^k - 5^l) to negative infinity. However, the problem seeks the minimal positive result, which is 11.Thus, 11 is the smallest positive number expressible as (36^k - 5^l) with natural (k, l), and no smaller positive solution exists. Final Answerboxed{11}

🔑:1. Understanding the problem: We need to find the smallest positive value of the expression (36^k - 5^ell) where (k) and (ell) are natural numbers. 2. Analyzing modulo 10: - For any natural number (k), (36^k equiv 6^k equiv 6 pmod{10}). - For any natural number (ell), (5^ell equiv 5 pmod{10}). - Therefore, (36^k - 5^ell equiv 6 - 5 equiv 1 pmod{10}).3. Form of the expression: From the above, we conclude that (36^k - 5^ell) must be of the form (10n + 1) for some integer (n).4. Finding the smallest positive value: - For the smallest positive value, we start with (n = 0). This gives us the equation (36^k - 5^ell = 1). - Rearranging, we get (36^k - 1 = 5^ell). - Using the difference of squares, we can write (36^k - 1 = (6^k + 1)(6^k - 1) = 5^ell).5. Analyzing the factors: - Let (6^k + 1 = 5^a) and (6^k - 1 = 5^b) where (a) and (b) are natural numbers such that (a > b) and (a + b = ell). - Subtracting these two equations, we get (5^a - 5^b = 2).6. Contradiction: - The equation (5^a - 5^b = 2) has no solutions in natural numbers because the left-hand side is always a multiple of 5, while the right-hand side is 2, which is not a multiple of 5. - Therefore, there are no solutions for (n = 0).7. Next smallest value: - We try (n = 1). This gives us the equation (36^k - 5^ell = 11). - Testing small values, we find that (k = 1) and (ell = 2) satisfy the equation: (36^1 - 5^2 = 36 - 25 = 11).8. Verification: - We need to verify that there are no smaller positive values. - Since (36^k - 5^ell equiv 1 pmod{10}), the next possible value would be (21), (31), etc., but these do not satisfy the form (36^k - 5^ell) for small values of (k) and (ell).Thus, the smallest positive value of (36^k - 5^ell) is indeed (11).The final answer is (boxed{11}).

🧠:Okay, let's see. I need to figure out how many gallons of a 15% alcohol solution to mix with a 35% alcohol solution to get 250 gallons of a 21% alcohol solution. Hmm, mixture problems. I remember these involve some kind of equation where the concentrations and volumes relate. Let me think.First, the total volume after mixing should be 250 gallons. So if I have two solutions, let's call the volume of the 15% solution "x" gallons. Then the volume of the 35% solution would be the remaining part, right? Since together they need to add up to 250 gallons. So that would be 250 - x gallons.Now, the amount of alcohol in each solution. For the 15% solution, the alcohol is 15% of x, which is 0.15x. For the 35% solution, it's 35% of (250 - x), so that's 0.35(250 - x). When you mix them, the total alcohol should be 21% of 250 gallons, which is 0.21 * 250. So the equation would be:0.15x + 0.35(250 - x) = 0.21 * 250Let me check that. Yeah, that makes sense because the alcohol from both solutions adds up to the total alcohol in the mixture. Let me compute the right side first. 0.21 * 250 is... 21% of 250. 10% of 250 is 25, so 20% is 50, and 1% is 2.5, so 21% is 52.5. So the equation is:0.15x + 0.35(250 - x) = 52.5Now let's expand the left side. 0.35 * 250 is... 35% of 250. 10% is 25, so 30% is 75, 5% is 12.5, so 35% is 87.5. Then subtract 0.35x. So:0.15x + 87.5 - 0.35x = 52.5Combine like terms. 0.15x - 0.35x is -0.20x. So:-0.20x + 87.5 = 52.5Now, subtract 87.5 from both sides:-0.20x = 52.5 - 87.552.5 - 87.5 is... -35. So:-0.20x = -35Divide both sides by -0.20:x = (-35) / (-0.20)Dividing two negatives gives a positive. 35 divided by 0.20. 0.20 is 1/5, so 35 divided by 1/5 is 35 * 5 = 175. So x is 175 gallons.Wait, so that means we need 175 gallons of the 15% solution and 250 - 175 = 75 gallons of the 35% solution? Let me verify that.Check the total alcohol:15% of 175 is 0.15 * 175. 10% is 17.5, 5% is 8.75, so total is 17.5 + 8.75 = 26.25 gallons.35% of 75 is 0.35 * 75. 10% is 7.5, so 30% is 22.5, 5% is 3.75, total is 22.5 + 3.75 = 26.25 gallons.Total alcohol is 26.25 + 26.25 = 52.5 gallons, which matches 21% of 250 gallons. So that checks out.Alternatively, maybe I can think of this as a weighted average problem. The 21% is somewhere between 15% and 35%. The difference between 15% and 21% is 6%, and between 21% and 35% is 14%. The ratio of the distances is 6:14, which simplifies to 3:7. So the ratio of the two solutions should be 3:7. Wait, but wait, which solution corresponds to which part?In weighted averages, the ratio is inversely proportional to the distances. So since 21% is closer to 15% (difference of 6%) than to 35% (difference of 14%), you need more of the 35% solution? Wait, that seems conflicting with the previous answer. Wait, maybe not. Wait, no, the ratio is 3:7 for 35% to 15%?Wait, hold on. Let me recall. If the ratio of the distances is 6:14 = 3:7, then the amounts are in the ratio 7:3. Wait, because the one closer is the smaller amount? Wait, no. Let me think.Suppose the concentration of the mixture is closer to the lower concentration. Then you need more of the higher concentration? Wait, no. Wait, if you're mixing two solutions, the final concentration is closer to the one you have more of. For example, if you have more of the 15% solution, the final concentration is closer to 15%. If you have more of the 35%, it's closer to 35%. Since 21% is closer to 15% (difference of 6%) than to 35% (difference of 14%), that would mean you need more of the 15% solution. Wait, but in my initial calculation, I found 175 gallons of 15% and 75 gallons of 35%, which is a ratio of 175:75 = 7:3. So 7 parts 15% to 3 parts 35%. That makes sense because even though you need more of the 15%, the ratio is 7:3. Let me check with the weighted average formula.The formula is:(15% * x + 35% * (250 - x)) / 250 = 21%Which simplifies to the same equation as before. But using the ratio method, the distances are 6% and 14%, so the ratio is 6:14 = 3:7. So the ratio of the amounts should be 7:3. Wait, but the ratio of 35% to 15% is 7:3? Or vice versa?Wait, let me recall the rule: the ratio of the amounts is inversely proportional to the differences. So the ratio of the lower concentration solution to the higher concentration solution is (high difference) : (low difference). So in this case, the differences are 21 - 15 = 6 and 35 - 21 = 14. So the ratio of 15% solution to 35% solution is 14:6 = 7:3. So that's 7 parts of 15% and 3 parts of 35%. Which matches the answer: 175 is 7 parts (7*25=175) and 75 is 3 parts (3*25=75). So total parts 10 parts = 250 gallons, so each part is 25 gallons. That works.So both methods give the same result. So the answer is 175 gallons of the 15% solution. Therefore, the answer is 175.Final AnswerThe number of gallons of the 15% solution needed is boxed{175}.

✅:To determine how many gallons of a 15% alcohol solution need to be mixed with a 35% alcohol solution to make 250 gallons of a 21% alcohol solution, we can set up the following equation based on the concentrations and volumes:1. Let ( x ) be the gallons of the 15% alcohol solution.2. The volume of the 35% alcohol solution will then be ( 250 - x ) gallons.3. The total amount of alcohol from both solutions should equal the total amount of alcohol in the final mixture.The equation for the total alcohol is:[0.15x + 0.35(250 - x) = 0.21 times 250]Calculating the right side:[0.21 times 250 = 52.5]Expanding and simplifying the left side:[0.15x + 87.5 - 0.35x = 52.5]Combining like terms:[-0.20x + 87.5 = 52.5]Solving for ( x ):[-0.20x = 52.5 - 87.5 -0.20x = -35 x = frac{-35}{-0.20} x = 175]Thus, we need 175 gallons of the 15% alcohol solution. Verifying the total alcohol content confirms the solution is correct. Using the ratio method (inversely proportional to the differences in concentrations) also yields a ratio of 7:3 for the 15% and 35% solutions, respectively, which corresponds to 175 gallons and 75 gallons.The number of gallons of the 15% solution needed is (boxed{175}).

🔑:1. Let ( x ) be the number of gallons of the solution with ( 15% ) alcohol.2. Let ( y ) be the number of gallons of the solution with ( 35% ) alcohol.3. We need to mix these to get ( 250 ) gallons of a solution that is ( 21% ) alcohol.We can set up the following system of equations based on the given information:[x + y = 250][0.15x + 0.35y = 0.21 times 250]4. Simplify the second equation:[0.15x + 0.35y = 52.5]5. We now have the system of linear equations:[begin{cases}x + y = 250 0.15x + 0.35y = 52.5end{cases}]6. Solve the first equation for ( y ):[y = 250 - x]7. Substitute ( y = 250 - x ) into the second equation:[0.15x + 0.35(250 - x) = 52.5]8. Distribute and simplify:[0.15x + 87.5 - 0.35x = 52.5][-0.20x + 87.5 = 52.5]9. Solve for ( x ):[-0.20x = 52.5 - 87.5][-0.20x = -35][x = frac{-35}{-0.20}][x = 175]10. Substitute ( x = 175 ) back into the equation ( y = 250 - x ):[y = 250 - 175][y = 75]Thus, we need ( 175 ) gallons of the ( 15% ) alcohol solution and ( 75 ) gallons of the ( 35% ) alcohol solution.The final answer is ( boxed{ 175 } ) gallons of the ( 15% ) alcohol solution.

🧠:Okay, let's try to figure out the least positive integer a such that 65 divides f(x) = 5x¹³ + 13x⁵ + 9ax for every integer x. Hmm, so 65 is 5 multiplied by 13, right? That means we need f(x) to be divisible by both 5 and 13 for all integers x. Maybe I can use the Chinese Remainder Theorem here? But I'm not too sure. Let me break it down step by step.First, since 65 is 5 and 13, which are coprime, if f(x) is divisible by both 5 and 13 for all x, then it's divisible by 65. So I need to ensure that f(x) ≡ 0 mod 5 and f(x) ≡ 0 mod 13 for all integers x. Let me check each modulus separately.Starting with mod 5. So f(x) = 5x¹³ + 13x⁵ + 9ax. Let's reduce each term mod 5. 5x¹³ mod 5 is 0, because 5 is a multiple of 5. Then 13x⁵ mod 5: 13 mod 5 is 3, so that term becomes 3x⁵ mod 5. Then 9ax mod 5: 9 mod 5 is 4, so that term is 4a x mod 5. So altogether, f(x) mod 5 is 0 + 3x⁵ + 4a x ≡ 3x⁵ + 4a x mod 5. We need this to be congruent to 0 mod 5 for all integers x. So 3x⁵ + 4a x ≡ 0 mod 5 for all x. Let's factor out an x: x(3x⁴ + 4a) ≡ 0 mod 5. Since this has to hold for all x, even when x is not divisible by 5, then the term in parentheses must be congruent to 0 mod 5 for all x. So 3x⁴ + 4a ≡ 0 mod 5 for all x. But x⁴ mod 5: Fermat's little theorem says that if x is not divisible by 5, then x⁴ ≡ 1 mod 5. So x⁴ ≡ 1 mod 5 when x ≡ 1,2,3,4 mod 5. If x ≡ 0 mod 5, then x⁴ ≡ 0 mod 5. But since the equation must hold for all x, including x ≡ 0 mod 5. Let's check x ≡ 0 mod 5 first. If x ≡ 0 mod 5, then 3*0 + 4a ≡ 4a ≡ 0 mod 5. So 4a ≡ 0 mod 5 implies a ≡ 0 mod 5, since 4 and 5 are coprime. So a must be a multiple of 5.Now for x not divisible by 5, x⁴ ≡ 1 mod 5. So then 3*1 + 4a ≡ 0 mod 5. So 3 + 4a ≡ 0 mod 5. Then 4a ≡ -3 mod 5, which is equivalent to 4a ≡ 2 mod 5. To solve for a, multiply both sides by the inverse of 4 mod 5. The inverse of 4 mod 5 is 4, since 4*4=16≡1 mod5. So a ≡ 2*4 ≡ 8 ≡ 3 mod5. But earlier, we saw that a must be ≡0 mod5. So how can a be congruent to both 0 and 3 mod5? That seems contradictory. Wait, maybe I made a mistake here.Wait, actually, when x is 0 mod5, we get 4a ≡0 mod5, so a ≡0 mod5. When x is not 0 mod5, we have 3x⁴ +4a≡0 mod5. Since x⁴≡1, this becomes 3 +4a≡0 mod5. So 4a ≡-3 mod5, which is 4a≡2 mod5. Then a≡(2*4^{-1}) mod5. Since 4 inverse is 4, as 4*4=16≡1, so a≡2*4=8≡3 mod5. So here's the problem: a needs to be 0 mod5 and 3 mod5, which is impossible unless 0≡3 mod5, which is false. So this suggests that there's a contradiction? That can't be. So perhaps my approach is wrong.Wait, but maybe we can reconcile these. If the equation 3x⁴ +4a ≡0 mod5 must hold for all x, both 0 and non-0 mod5. When x≡0, we get 4a≡0 mod5. When x≡1,2,3,4 mod5, we get 3 +4a≡0 mod5. But these two conditions must hold simultaneously, so 4a≡0 and 4a≡-3 mod5. Therefore, 0≡-3 mod5, which is not true. Therefore, the only way this can hold for all x is if both conditions are satisfied, which is impossible. But that can't be right, because the problem states that such an a exists. So where's my mistake?Wait, maybe I made a mistake in the approach. Let me check again. So the original equation f(x) ≡0 mod5 for all x. So 5x¹³ +13x⁵ +9ax ≡0 mod5. Breaking it down: 5x¹³ ≡0 mod5, 13x⁵≡3x⁵ mod5, 9ax≡4ax mod5. Therefore, 3x⁵ +4ax ≡0 mod5. So 3x⁵ +4ax ≡0 mod5 for all x. Then, maybe instead of factoring x, we can test x=1, x=2, etc., and see what conditions on a arise. Then, since it must hold for all x, those conditions must hold for all x. Alternatively, perhaps using polynomial identities. Since if a polynomial of degree less than p (here p=5) is congruent to 0 for all x mod p, then all coefficients must be 0.Wait, but our polynomial is 3x⁵ +4a x. Let me note that in mod5, x⁵ ≡x mod5 by Fermat's little theorem. Because x⁵ ≡x mod5 for any x. Therefore, x⁵ ≡x mod5, so 3x⁵ ≡3x mod5. Therefore, 3x⁵ +4a x ≡3x +4a x ≡(3 +4a)x mod5. Therefore, the entire expression simplifies to (3 +4a)x mod5. For this to be ≡0 mod5 for all x, the coefficient (3 +4a) must be ≡0 mod5. Therefore, 3 +4a ≡0 mod5. So 4a ≡-3 mod5 ⇒4a≡2 mod5 ⇒a≡2*4^{-1}≡2*4≡8≡3 mod5. Therefore, a≡3 mod5.But earlier, when we considered x=0 mod5, we had 4a*0 ≡0 mod5, which is automatically 0, regardless of a. Wait, perhaps my initial mistake was when x=0, we have 3x⁵ +4a x =0 +0=0 mod5, so that term is always 0, regardless of a. So actually, when x≡0 mod5, the expression is automatically 0, and when x≡ non-zero, then x⁵ ≡x mod5, so the expression becomes (3 +4a)x mod5. Since x is invertible mod5, to have (3 +4a)x ≡0 mod5 for all x not ≡0 mod5, we must have 3 +4a ≡0 mod5. Therefore, the only condition is 3 +4a ≡0 mod5 ⇒a≡3 mod5. Therefore, a must be congruent to 3 mod5. So that's the condition from mod5.So that resolves the confusion. The x=0 case doesn't impose an additional condition because it's already 0. So a≡3 mod5.Now moving on to modulus13. Let's compute f(x) mod13. f(x)=5x¹³ +13x⁵ +9ax. Reducing mod13: 5x¹³ remains 5x¹³. 13x⁵ ≡0 mod13. 9ax remains 9ax. So f(x)≡5x¹³ +9ax mod13. Again, we need this to be ≡0 mod13 for all integers x. So 5x¹³ +9ax ≡0 mod13 for all x.Again, let's apply Fermat's little theorem. For modulus13, x¹²≡1 mod13 when x≡not 0 mod13. Therefore, x¹³≡x mod13 for all x. Because x¹³ =x¹² *x ≡1*x≡x mod13. So x¹³ ≡x mod13. Therefore, 5x¹³ ≡5x mod13. Therefore, f(x)≡5x +9ax mod13 ≡(5 +9a)x mod13. For this to be ≡0 mod13 for all x, the coefficient (5 +9a) must be ≡0 mod13. Therefore, 5 +9a ≡0 mod13 ⇒9a≡-5 mod13 ⇒9a≡8 mod13 (since -5 +13=8). So we need to solve for a: 9a≡8 mod13. To find a, we can multiply both sides by the inverse of 9 mod13. The inverse of 9 mod13: Let's find an integer k such that 9k≡1 mod13. Trying k=3: 9*3=27≡1 mod13 (since 27-2*13=1). So inverse of 9 is 3 mod13. Therefore, a≡8*3≡24≡24-13=11 mod13. So a≡11 mod13.Therefore, from modulus5, a≡3 mod5, and from modulus13, a≡11 mod13. Now we need to find the smallest positive integer a that satisfies both congruences. That is, solve the system:a ≡3 mod5a ≡11 mod13We can use the Chinese Remainder Theorem here. Let's express a as a=5k +3 for some integer k. Substitute into the second equation: 5k +3 ≡11 mod13 ⇒5k≡8 mod13. Solve for k: 5k≡8 mod13. Multiply both sides by inverse of 5 mod13. The inverse of 5 mod13 is 8, since 5*8=40≡1 mod13. So k≡8*8=64≡64-5*13=64-65=-1≡12 mod13. So k≡12 mod13, which means k=13m +12 for some integer m. Therefore, a=5*(13m +12) +3=65m +60 +3=65m +63. Therefore, the smallest positive integer a is 63 when m=0. So the least positive integer a is 63.Wait, but let me verify this. Let's check if a=63 satisfies both conditions. a=63: 63 mod5=63-12*5=63-60=3, so 63≡3 mod5. And 63 mod13: 13*4=52, 63-52=11, so 63≡11 mod13. So yes, that works. Therefore, the answer is 63.But just to be thorough, let's test a=63 with some values of x to ensure that f(x) is indeed divisible by 65 for all x. Let's pick x=1: f(1)=5*1 +13*1 +9*63*1=5 +13 +567=585. 585 divided by 65 is 9, since 65*9=585. So that works. x=2: f(2)=5*2¹³ +13*2⁵ +9*63*2. Let's compute each term:5*2¹³: 2¹⁰=1024, so 2¹³=8192. 5*8192=40960.13*2⁵=13*32=416.9*63*2=1134.Total:40960 +416 +1134=40960+416=41376; 41376+1134=42510. Now 42510 divided by 65: 65*654=42510 (65*600=39000, 65*54=3510; 39000+3510=42510). So yes, divisible by 65.Let's test x=0: f(0)=0 +0 +0=0, which is divisible by 65. x=3: f(3)=5*3¹³ +13*3⁵ +9*63*3. Let's compute mod5 and mod13 separately.First mod5: f(3)≡0 +3*(3⁵) +4a*3 mod5. Wait, but we already know that when a≡3 mod5, 3x⁵ +4a x≡0 mod5 for all x. So mod5, f(3)≡0. Similarly mod13: f(3)=5*3¹³ +9a*3. Since 3¹³≡3 mod13, so 5*3 +9a*3=15 +27a. 27a mod13: 27≡1 mod13, so 27a≡a≡11 mod13. So 15 +11=26≡0 mod13. Hence, f(3)≡0 mod13. Therefore, divisible by 13, and since also divisible by 5, divisible by 65. Therefore, works.Therefore, a=63 is the minimal positive integer satisfying the conditions.Final AnswerThe least positive integer ( a ) is (boxed{63}).

✅:To find the least positive integer ( a ) such that ( 65 ) divides ( f(x) = 5x^{13} + 13x^5 + 9ax ) for every integer ( x ), we need to ensure ( f(x) ) is divisible by both ( 5 ) and ( 13 ) for all integers ( x ). Modulo 5 Analysis:1. Reduce ( f(x) ) modulo 5: [ f(x) equiv 3x^5 + 4ax pmod{5} ]2. By Fermat's Little Theorem, ( x^5 equiv x pmod{5} ): [ 3x + 4ax equiv (3 + 4a)x pmod{5} ]3. For this to be 0 for all ( x ), the coefficient must be 0: [ 3 + 4a equiv 0 pmod{5} implies 4a equiv -3 pmod{5} implies 4a equiv 2 pmod{5} ]4. Solving for ( a ): [ a equiv 2 cdot 4^{-1} pmod{5} implies a equiv 2 cdot 4 pmod{5} implies a equiv 8 pmod{5} implies a equiv 3 pmod{5} ] Modulo 13 Analysis:1. Reduce ( f(x) ) modulo 13: [ f(x) equiv 5x^{13} + 9ax pmod{13} ]2. By Fermat's Little Theorem, ( x^{13} equiv x pmod{13} ): [ 5x + 9ax equiv (5 + 9a)x pmod{13} ]3. For this to be 0 for all ( x ), the coefficient must be 0: [ 5 + 9a equiv 0 pmod{13} implies 9a equiv -5 pmod{13} implies 9a equiv 8 pmod{13} ]4. Solving for ( a ): [ a equiv 8 cdot 9^{-1} pmod{13} implies a equiv 8 cdot 3 pmod{13} implies a equiv 24 pmod{13} implies a equiv 11 pmod{13} ] Combining Results Using Chinese Remainder Theorem:Solve the system:[a equiv 3 pmod{5} a equiv 11 pmod{13}]1. Express ( a ) as ( a = 5k + 3 ): [ 5k + 3 equiv 11 pmod{13} implies 5k equiv 8 pmod{13} ]2. Solve for ( k ): [ k equiv 8 cdot 5^{-1} pmod{13} implies k equiv 8 cdot 8 pmod{13} implies k equiv 64 pmod{13} implies k equiv 12 pmod{13} ]3. Substitute ( k = 13m + 12 ): [ a = 5(13m + 12) + 3 = 65m + 63 ]The least positive integer ( a ) is ( 63 ).[boxed{63}]

🔑:1. Determine the condition modulo 5: [ f(x) = 5x^{13} + 13x^5 + 9ax equiv 13x^5 + 9ax pmod{5} ] Since (13 equiv 3 pmod{5}), we have: [ f(x) equiv 3x^5 + 9ax pmod{5} ] Since (9 equiv 4 pmod{5}), we get: [ f(x) equiv 3x^5 + 4ax pmod{5} ] By Fermat's Little Theorem, (x^5 equiv x pmod{5}) for any integer (x) not divisible by 5. Thus: [ f(x) equiv 3x + 4ax equiv (3 + 4a)x pmod{5} ] For (f(x) equiv 0 pmod{5}) for all (x), we need: [ 3 + 4a equiv 0 pmod{5} ] Solving for (a): [ 4a equiv -3 pmod{5} implies 4a equiv 2 pmod{5} implies a equiv 3 pmod{5} ]2. Determine the condition modulo 13: [ f(x) = 5x^{13} + 13x^5 + 9ax equiv 5x^{13} + 9ax pmod{13} ] Since (13 equiv 0 pmod{13}), we have: [ f(x) equiv 5x^{13} + 9ax pmod{13} ] By Fermat's Little Theorem, (x^{13} equiv x pmod{13}) for any integer (x) not divisible by 13. Thus: [ f(x) equiv 5x + 9ax equiv (5 + 9a)x pmod{13} ] For (f(x) equiv 0 pmod{13}) for all (x), we need: [ 5 + 9a equiv 0 pmod{13} ] Solving for (a): [ 9a equiv -5 pmod{13} implies 9a equiv 8 pmod{13} ]3. Solve the system of congruences: [ begin{cases} a equiv 3 pmod{5} 9a equiv 8 pmod{13} end{cases} ] First, solve (9a equiv 8 pmod{13}): [ 9a equiv 8 pmod{13} ] The multiplicative inverse of 9 modulo 13 is 3 (since (9 cdot 3 equiv 27 equiv 1 pmod{13})). Thus: [ a equiv 8 cdot 3 pmod{13} implies a equiv 24 pmod{13} implies a equiv 11 pmod{13} ]4. Combine the results using the Chinese Remainder Theorem: [ begin{cases} a equiv 3 pmod{5} a equiv 11 pmod{13} end{cases} ] Let (a = 5k + 3). Then: [ 5k + 3 equiv 11 pmod{13} implies 5k equiv 8 pmod{13} ] The multiplicative inverse of 5 modulo 13 is 8 (since (5 cdot 8 equiv 40 equiv 1 pmod{13})). Thus: [ k equiv 8 cdot 8 pmod{13} implies k equiv 64 pmod{13} implies k equiv 12 pmod{13} ] Therefore: [ a = 5k + 3 = 5(13m + 12) + 3 = 65m + 60 + 3 = 65m + 63 ] The smallest positive integer (a) is: [ a = 63 ]The final answer is (boxed{63}).

🧠:Okay, let's try to figure out the least number n such that among any n rays in space sharing a common origin, there must be two that form an acute angle. Hmm, this seems related to geometry and maybe some combinatorial arguments. Let me start by understanding the problem better.So, we have rays emanating from a common origin in three-dimensional space. We need to find the smallest n where no matter how you arrange these n rays, at least two of them will form an acute angle (i.e., an angle less than 90 degrees). If we can find such an n, that's our answer. First, maybe I should consider the opposite: what's the maximum number of rays we can have in 3D space such that all pairs form right angles or obtuse angles (i.e., 90 degrees or more)? Then the answer would be one more than that maximum. Because if you have that maximum number, you can have all non-acute angles, but adding one more would force an acute angle. So, what's the maximum number of vectors in 3D space where any two are at 90 degrees or more? This sounds like a problem related to spherical codes or something. In 3D, the question is how many points can you place on a unit sphere such that the angle between any two points is at least 90 degrees. Wait, in 3D, the maximum number of points on a sphere with pairwise angles at least 90 degrees is 6. Let me think. If you take the standard basis vectors and their negatives: (1,0,0), (-1,0,0), (0,1,0), (0,-1,0), (0,0,1), (0,0,-1). These are six vectors, and the angle between any two distinct ones is either 90 degrees (if they are orthogonal) or 180 degrees (if they are opposites). But 180 degrees is a straight line, but in the case of rays, they are considered as two opposite rays. Wait, but rays with 180 degrees between them are just opposite directions. However, in the problem statement, we are talking about rays, which are half-lines starting from the origin. So, two opposite rays would form a straight line, but the angle between them is 180 degrees, which is obtuse. Wait, no, 180 degrees is a straight angle, not obtuse. Obtuse angles are between 90 and 180 degrees. So 180 is not considered obtuse. Wait, actually, in standard terms, angles less than 90 are acute, exactly 90 are right, between 90 and 180 are obtuse, and 180 is straight. So, in that case, 180 is not considered obtuse. Therefore, if two rays are opposite, their angle is 180 degrees, which is a straight angle. So, in our case, we need to make sure that among n rays, there are two with angle less than 90 degrees.So, if we can have 6 rays where all angles between any two are either 90 or 180 degrees, then 6 is possible. Wait, but in the standard basis vectors and their negatives, the angles between any two distinct non-opposite vectors are 90 degrees. The angles between a vector and its opposite are 180 degrees. So, in that case, all angles are either 90 or 180. So, 6 rays. Therefore, 6 rays can exist without any acute angles. Therefore, if n is 7, then in any 7 rays, there must be two that form an acute angle. Therefore, the minimal n is 7.Wait, but let me verify this. Maybe there's a different configuration with more than 6 rays where all angles are 90 degrees or more. Is that possible?Wait, in 3D space, the maximum number of mutually orthogonal vectors is 3, of course. But if we allow vectors that are not necessarily orthogonal, but just at least 90 degrees apart, can we have more?Let me think. If you take the six vectors along the coordinate axes, each pair is either orthogonal or opposite. So, angles 90 or 180. So that's 6. Is there a way to place 7 rays such that every pair is at least 90 degrees apart? Probably not.I remember that in 3D space, the maximum number of equidistant points on a sphere such that the angle between any two is the same is 4 (the vertices of a tetrahedron). The angle there is about 109.47 degrees, which is obtuse. But that's not directly helpful here, but maybe related.Alternatively, maybe using the concept of spherical caps. If each ray is represented by a point on the unit sphere, then ensuring that no two points are within 90 degrees of each other. So, each point must be in a spherical cap of angular radius 45 degrees (since the angle between two points is twice the angular distance from the center of the cap). Wait, actually, the angle between two points on the sphere is the angle between their vectors. If we want the angle between any two points to be at least 90 degrees, then the spherical distance between them (measured as the angle between the vectors) must be at least 90 degrees. Therefore, each point must be at least 90 degrees apart from every other point.The problem then reduces to finding the maximum number of non-overlapping spherical caps of angular radius 45 degrees (since the distance between centers must be at least 90 degrees) that can be placed on a sphere. The area of each such cap can be calculated, and then dividing the total surface area of the sphere by the area of one cap gives an upper bound on the number. But this is a sphere-packing problem, which is complicated. However, in 3D, the known maximum for this specific angle might be 6, which corresponds to the coordinate axes.Wait, let me check. If we place points on the sphere such that each pair is at least 90 degrees apart, the maximum number is 6. For example, the six points along the coordinate axes. Each pair is either orthogonal (90 degrees) or opposite (180 degrees). So, that's six points. If we try to add a seventh point, is it possible?Suppose we try to place a seventh point somewhere on the sphere. We need to make sure it's at least 90 degrees away from all the six existing points. Let's think about the six existing points as (±1,0,0), (0,±1,0), (0,0,±1). Let's take a new point (a,b,c). To be at least 90 degrees away from (1,0,0), the dot product must be non-positive (since the cosine of the angle is the dot product divided by the product of magnitudes, which for unit vectors is just the dot product). So, the dot product with (1,0,0) is a. To have an angle of at least 90 degrees, we need a ≤ 0. Similarly, the dot product with (-1,0,0) is -a, which must also be ≤0, so -a ≤0 ⇒ a ≥0. Therefore, combining these, a =0. Similarly, considering the dot product with (0,1,0) and (0,-1,0), we get b=0. Similarly, c=0. But the only point with a=0, b=0, c=0 is the origin, which isn't on the sphere. Therefore, it's impossible to place a seventh point. Hence, 6 is indeed the maximum. Therefore, the minimal n is 7. Therefore, the answer should be boxed{7}.Wait, but let me check another way. Suppose we have 6 rays arranged along the coordinate axes. Then any seventh ray must lie in some octant. The angle between this seventh ray and the coordinate axis of that octant would be acute. For example, if the seventh ray is in the positive octant (where x,y,z are all positive), then the angle between this ray and the x-axis, y-axis, or z-axis would be acute, because the dot product would be positive. Wait, yes. Let's formalize that. Suppose we have a ray in the direction of a vector (a,b,c), where a,b,c >0 (positive octant). Then, the angle between this ray and the x-axis (1,0,0) is given by the dot product: a*1 + b*0 + c*0 = a. Since the vectors are unit vectors, the cosine of the angle is a. If a >0, then the angle is acute. Similarly for b and c. Therefore, any ray in the positive octant forms an acute angle with the x-axis, y-axis, and z-axis. Therefore, if we have a seventh ray, regardless of where it is, it must lie in some octant (or on a boundary), but if it's in an octant, then it has direction (a,b,c) with either all positive or some negative, but in any case, the coordinates can't all be zero. Therefore, at least one coordinate is non-zero in absolute value. If the ray is in the positive octant, as above, angles with x,y,z axes are acute. If it's in another octant, say the octant where x>0, y<0, z>0, then the angles with x and z axes would still be acute if the corresponding coordinates are positive/negative. Wait, let's think again. Suppose the seventh ray is in the direction (a,b,c) where a>0, b<0, c>0. Then, the dot product with the x-axis (1,0,0) is a>0, so angle is acute. The dot product with the y-axis (0,1,0) is b<0, so angle is obtuse. The dot product with the z-axis (0,0,1) is c>0, so angle is acute. Therefore, the angle with the x and z axes is acute. So, regardless of the octant, the ray will form acute angles with at least two of the coordinate axes. Hence, if we have a seventh ray, it must form an acute angle with at least one of the existing six rays. Therefore, n=7 is the minimal number. Therefore, the answer is 7.But let me check if there is another configuration of 7 rays where none form acute angles. Suppose we try to place 7 rays such that all angles are 90 degrees or more. But as per the previous reasoning, it's impossible because the seventh ray must lie in some octant and would form an acute angle with the corresponding axis. Therefore, 7 is indeed the minimal n.Therefore, after all this analysis, the answer is 7.Final AnswerThe least such ( n ) is boxed{7}.

✅:To find the least ( n in mathbb{N} ) such that among any ( n ) rays in space sharing a common origin, there exist two which form an acute angle, we need to consider the maximum number of rays that can be arranged such that all angles between them are 90 degrees or more (non-acute).1. Understanding the Problem: - We need the smallest ( n ) such that any ( n ) rays must include at least two forming an acute angle. - This involves determining the maximum number of rays that can be placed in 3D space with all pairwise angles being 90 degrees or more.2. Key Insight: - In 3D space, the maximum number of vectors (rays) that can be placed such that all pairwise angles are at least 90 degrees is 6. These vectors are the standard basis vectors and their negatives: ( (1,0,0), (-1,0,0), (0,1,0), (0,-1,0), (0,0,1), (0,0,-1) ). Each pair of distinct vectors forms either a 90-degree angle (orthogonal) or a 180-degree angle (opposite directions).3. Verification: - Adding a seventh ray to this configuration forces the new ray to lie in one of the octants formed by the coordinate axes. Any ray in an octant will form an acute angle with at least one of the coordinate axes. This is because the dot product of the new ray with at least one of the basis vectors will be positive, indicating an acute angle.4. Conclusion: - Therefore, 7 rays are sufficient to ensure that at least two form an acute angle, making 7 the minimal ( n ).[boxed{7}]

🔑:1. Base Case: - Consider the case when ( d = 1 ). In one-dimensional space, any two non-null vectors (rays) will either point in the same direction or in opposite directions. If they point in opposite directions, they form an angle of ( 180^circ ), which is not acute. Therefore, the largest number of vectors in (mathbb{R}^1) with pairwise non-positive dot products is 2.2. Inductive Step: - Assume that in ( mathbb{R}^d ), the largest number of non-null vectors with pairwise non-positive dot products is ( 2d ). - Consider ( mathbb{R}^{d+1} ) and let ( u ) be one of the vectors. Denote by ( U = langle u rangle^{perp} ) the orthogonal complement of ( u ), which has dimension ( d ).3. Decomposition of Vectors: - Any vector ( v ) in ( mathbb{R}^{d+1} ) can be uniquely written as ( v = v_{perp} + v_{parallel} ), where ( v_{parallel} = frac{langle u, v rangle}{||u||^2} u ) and ( v_{perp} perp u ), i.e., ( v_{perp} in U ). - The coefficients ( frac{langle u, v rangle}{||u||^2} ) are non-positive under our conditions. Notice that ( v_{perp} = 0 ) can occur only once, for some ( u' = v ) (trivially so, since then ( u' = u'_{parallel} = lambda u ) for some negative real ( lambda )).4. Orthogonal Components: - Consider the orthogonal components ( v_{perp} in U ). We have: [ langle v, w rangle = langle v_{perp} + v_{parallel}, w_{perp} + w_{parallel} rangle = langle v_{perp}, w_{perp} rangle + langle v_{perp}, w_{parallel} rangle + langle v_{parallel}, w_{perp} rangle + langle v_{parallel}, w_{parallel} rangle ] [ = langle v_{perp}, w_{perp} rangle + frac{langle u, v rangle langle u, w rangle}{||u||^2} ] - Therefore, [ langle v_{perp}, w_{perp} rangle = langle v, w rangle - frac{langle u, v rangle langle u, w rangle}{||u||^2} leq 0 ] since ( langle v, w rangle leq 0 ) and ( langle u, v rangle langle u, w rangle geq 0 ).5. Induction Hypothesis: - By the induction hypothesis, the number of non-null vectors in ( U ) is at most ( 2d ). Including ( u ) and possibly ( u' ), there are at most ( 2(d + 1) ) non-null vectors in ( mathbb{R}^{d+1} ).6. Model Construction: - To achieve the bound, we must have the pair ( u ) and ( u' ), with all other vectors orthogonal to them. By inductive reasoning, the only possible model is made by some ( d ) pairwise orthogonal non-null vectors, and some other ( d ) obtained from them by multiplication with arbitrary negative scalars.7. Conclusion: - The largest number of vectors in ( mathbb{R}^d ) with pairwise non-positive dot products is ( 2d ). Therefore, in ( mathbb{R}^3 ), the largest number of such vectors is ( 2 times 3 = 6 ).8. Acute Angle Condition: - To ensure that among any ( n ) rays in space sharing a common origin there exist two which form an acute angle, we need ( n > 6 ). Thus, the least ( n ) is ( 7 ).The final answer is ( boxed{7} ).

🧠:Okay, so I need to find the ratio of the area inside the incircle of a regular hexagon to the area of the hexagon itself. Hmm, let me start by recalling what a regular hexagon is. It's a six-sided polygon with all sides equal and all internal angles equal. The incircle of a regular hexagon should be the circle that touches all its sides, right? So the incircle is tangent to each side of the hexagon, and its radius is called the inradius.First, maybe I should figure out the relationship between the side length of the hexagon and the inradius. Then I can compute both areas and find their ratio. Let's denote the side length of the regular hexagon as 's' and the inradius as 'r'. I need to relate 'r' to 's'.I remember that in a regular polygon, the inradius can be calculated using the formula involving the apothem. The apothem is the distance from the center to the midpoint of a side, which is the same as the inradius. For a regular hexagon, the apothem (r) can be related to the side length (s) through trigonometric functions. Since a regular hexagon can be divided into six equilateral triangles by drawing lines from the center to each vertex, each of these triangles has a central angle of 60 degrees (360/6). Wait, those are equilateral triangles, so all sides are equal. But the apothem would be the height of each of those equilateral triangles, right?Let me think. If each triangle is equilateral with side length 's', then the height (which is the apothem or inradius 'r') can be found using the formula for the height of an equilateral triangle: (sqrt(3)/2) * s. So, r = (sqrt(3)/2) * s. That seems right. So the inradius is sqrt(3)/2 times the side length.Now, the area of the incircle would be straightforward: π * r^2. Substituting r in terms of s, that would be π * (sqrt(3)/2 * s)^2. Let's compute that: π * (3/4 * s²) = (3π/4) * s².Next, the area of the regular hexagon. I know that the area of a regular hexagon can be calculated using the formula (3 * sqrt(3)/2) * s². Wait, let me verify that. Since the hexagon can be divided into six equilateral triangles, each with area (sqrt(3)/4) * s². So total area would be 6 * (sqrt(3)/4) * s² = (6/4) * sqrt(3) * s² = (3/2) * sqrt(3) * s². Yes, that's correct.So the area of the hexagon is (3√3/2) * s² and the area of the incircle is (3π/4) * s². To find the ratio, we divide the area of the incircle by the area of the hexagon:Ratio = (3π/4 * s²) / (3√3/2 * s²) = (3π/4) / (3√3/2) = (π/4) / (√3/2) = (π/4) * (2/√3) = π/(2√3).Hmm, simplifying that. Maybe rationalize the denominator? Multiply numerator and denominator by √3:π/(2√3) * (√3/√3) = π√3 / (2*3) = π√3 / 6.So the ratio is π√3 / 6. Let me check the steps again to make sure I didn't make a mistake.1. Inradius r = (sqrt(3)/2) * s. Correct, because the apothem (distance from center to side) in a regular hexagon is the height of the equilateral triangle with side s.2. Area of incircle: πr² = π*(3/4)s². Correct.3. Area of hexagon: (3√3/2)s². Correct, as from six equilateral triangles.4. Ratio calculation: (3π/4)/(3√3/2) = (π/4)/(√3/2) = π/(2√3) = π√3/6. Yes, that's right.Alternatively, if I leave it as π/(2√3), both are correct, but π√3/6 is more simplified with rationalized denominator.Wait, another way to check: maybe compute numerical values. Let's take s = 2. Then r = sqrt(3)/2 * 2 = sqrt(3). Area of incircle: π*(sqrt(3))² = π*3. Area of hexagon: (3√3/2)*4 = (3√3/2)*4 = 6√3. Ratio: 3π / 6√3 = π/(2√3) = same as before. So numerically that checks out. So yes, the ratio is π√3/6.Alternatively, π/(2√3) is equivalent. But π√3/6 is perhaps the preferred form.Alternatively, the answer can be written as (π√3)/6. Either way, that's the ratio. Let me just confirm if there's another approach. For example, maybe using the fact that inradius is related to the area of the hexagon. Wait, the area of the regular polygon is (1/2) * perimeter * apothem. For the hexagon, perimeter is 6s, apothem is r. So area would be (1/2)*6s*r = 3sr. So substituting r = sqrt(3)/2 s, area is 3s*(sqrt(3)/2 s) = (3sqrt(3)/2)s², which matches the earlier formula. So that's consistent. Then area of incircle is πr² = π*(sqrt(3)/2 s)^2 = π*(3/4 s²). So ratio is (3π/4 s²) / (3sqrt(3)/2 s²) = same as before. So steps are consistent.Therefore, the ratio is π√3/6. So the answer is π√3 divided by 6, which can be written as π times the square root of 3 over 6. Thus, the ratio of the area inside the incircle to the area of the hexagon is π√3/6.

✅:The ratio of the area inside the incircle to the area of the regular hexagon is derived as follows:1. Relate Inradius to Side Length: For a regular hexagon with side length ( s ), the inradius ( r ) (apothem) is: [ r = frac{sqrt{3}}{2} s ]2. Area of the Incircle: Substituting ( r = frac{sqrt{3}}{2} s ) into the circle area formula: [ text{Area}_{text{incircle}} = pi r^2 = pi left( frac{sqrt{3}}{2} s right)^2 = frac{3pi}{4} s^2 ]3. Area of the Hexagon: The area of a regular hexagon is: [ text{Area}_{text{hexagon}} = frac{3sqrt{3}}{2} s^2 ]4. Compute the Ratio: Divide the incircle area by the hexagon area: [ text{Ratio} = frac{frac{3pi}{4} s^2}{frac{3sqrt{3}}{2} s^2} = frac{pi}{2sqrt{3}} = frac{pisqrt{3}}{6} ]Final Answer: The ratio is (boxed{dfrac{pisqrt{3}}{6}}).

🔑:1. Assume the side length of the regular hexagon is 1. - Let the side length of the hexagon be ( s = 1 ).2. Calculate the area of the regular hexagon. - The formula for the area of a regular hexagon with side length ( s ) is: [ A_{text{hexagon}} = frac{3sqrt{3}}{2} s^2 ] - Substituting ( s = 1 ): [ A_{text{hexagon}} = frac{3sqrt{3}}{2} cdot 1^2 = frac{3sqrt{3}}{2} ]3. Determine the radius of the incircle. - The radius ( r ) of the incircle of a regular hexagon with side length ( s ) is given by: [ r = frac{s sqrt{3}}{2} ] - Substituting ( s = 1 ): [ r = frac{1 cdot sqrt{3}}{2} = frac{sqrt{3}}{2} ]4. Calculate the area of the incircle. - The area of a circle with radius ( r ) is: [ A_{text{incircle}} = pi r^2 ] - Substituting ( r = frac{sqrt{3}}{2} ): [ A_{text{incircle}} = pi left( frac{sqrt{3}}{2} right)^2 = pi cdot frac{3}{4} = frac{3pi}{4} ]5. Find the ratio of the area of the incircle to the area of the hexagon. - The ratio is given by: [ text{Ratio} = frac{A_{text{incircle}}}{A_{text{hexagon}}} = frac{frac{3pi}{4}}{frac{3sqrt{3}}{2}} = frac{3pi}{4} cdot frac{2}{3sqrt{3}} = frac{3pi cdot 2}{4 cdot 3sqrt{3}} = frac{6pi}{12sqrt{3}} = frac{pi}{2sqrt{3}} = frac{pi sqrt{3}}{6} ]The final answer is (boxed{frac{pi sqrt{3}}{6}}).