💡To determine the probability of rolling a sum of 17 with three standard 6-faced dice, we need to identify all possible combinations of three numbers (each between 1 and 6) that add up to 17.1. **Identify Possible Combinations**: - The highest possible number on a die is 6. To reach a sum of 17, we need at least two dice showing 6, as 6 + 6 + 5 = 17. - The valid combinations are: - (6, 6, 5) - (6, 5, 6) - (5, 6, 6)2. **Calculate the Number of Favorable Outcomes**: - There are 3 distinct arrangements of the numbers 6, 6, and 5.3. **Determine the Total Number of Possible Outcomes**: - Each die has 6 faces, so the total number of possible outcomes when rolling three dice is (6 times 6 times 6 = 216).4. **Compute the Probability**: - The probability is the number of favorable outcomes divided by the total number of possible outcomes: [ text{Probability} = frac{3}{216} = frac{1}{72} ]

💡First, I need to determine the total amount of water the sprinkler system uses each day. The system runs twice a day: once in the morning and once in the evening.In the morning, it uses 4 liters of water, and in the evening, it uses 6 liters of water. To find the total daily water usage, I add these two amounts together.Next, I need to find out how many days it will take for the sprinkler system to use a total of 50 liters of water. I can do this by dividing the total amount of water needed (50 liters) by the amount of water used each day (10 liters).This calculation will give me the number of days required for the sprinkler system to use 50 liters of water.

💡First, let's determine the cost price of the house and the store.For the house:- The selling price is 9,000, which is a 10% loss.- Let ( h ) be the cost price of the house.- A 10% loss means the selling price is 90% of the cost price.- So, ( 0.9h = 9,000 ).- Solving for ( h ): ( h = frac{9,000}{0.9} = 10,000 ).For the store:- The selling price is 13,000, which is a 30% gain.- Let ( s ) be the cost price of the store.- A 30% gain means the selling price is 130% of the cost price.- So, ( 1.3s = 13,000 ).- Solving for ( s ): ( s = frac{13,000}{1.3} = 10,000 ).Now, let's calculate the total cost price and total selling price.- Total cost price: ( 10,000 + 10,000 = 20,000 ).- Total selling price: ( 9,000 + 13,000 = 22,000 ).The difference between the total selling price and total cost price is:- ( 22,000 - 20,000 = 2,000 ).Since the total selling price is higher than the total cost price, there is a gain of 2,000.

💡First, I need to determine how many green hats were purchased out of the total 85 hats. I know that blue hats cost 6 each and green hats cost 7 each, with the total cost being 548.I'll start by defining variables for the number of blue hats (B) and green hats (G). According to the problem, the sum of blue and green hats is 85, which gives me the first equation: B + G = 85.Next, I'll use the total cost to form a second equation. The cost of blue hats is 6B and the cost of green hats is 7G, so the total cost equation is 6B + 7G = 548.To solve these equations, I'll express B in terms of G from the first equation: B = 85 - G. Then, I'll substitute this expression into the second equation: 6(85 - G) + 7G = 548.Expanding and simplifying this equation, I'll solve for G to find the number of green hats.

💡First, I need to determine how long Jane takes to complete the task alone. Let's denote this time as ( J ) days.John can complete the task in 20 days, so his work rate is ( frac{1}{20} ) of the task per day. Jane's work rate is ( frac{1}{J} ) of the task per day.When they work together, their combined work rate is ( frac{1}{20} + frac{1}{J} ) of the task per day.Jane was indisposed 4 days before the work got over, so she worked together with John for ( 10 - 4 = 6 ) days.In those 6 days, they completed ( 6 times left(frac{1}{20} + frac{1}{J}right) ) of the task.For the remaining 4 days, only John worked, completing ( 4 times frac{1}{20} ) of the task.The total work done is the sum of the work done by both when they worked together and the work done by John alone, which equals 1 (the whole task).So, the equation is:[6 times left(frac{1}{20} + frac{1}{J}right) + 4 times frac{1}{20} = 1]Simplifying the equation:[frac{6}{20} + frac{6}{J} + frac{4}{20} = 1][frac{10}{20} + frac{6}{J} = 1][frac{1}{2} + frac{6}{J} = 1][frac{6}{J} = frac{1}{2}][J = 6 times 2 = 12]Therefore, Jane can complete the task alone in 12 days.

💡Okay, so I have this problem where six friends each visited the swimming pool exactly 75 times over 100 days, and each day they could visit at most once. I need to find the maximum and minimum possible values of ( n ), which is the number of days where at least five of them visited the pool.First, let me try to understand the problem better. Each friend goes to the pool 75 times in 100 days, so each friend misses the pool 25 times. Since there are six friends, the total number of visits is ( 6 times 75 = 450 ) visits over 100 days.Now, I need to figure out how these visits can be distributed across the 100 days to maximize and minimize the number of days where at least five friends are present.Starting with the maximum value of ( n ). To maximize ( n ), I want as many days as possible where at least five friends are visiting. The more days with five or six friends, the higher ( n ) will be. But I have to make sure that the total number of visits doesn't exceed 450.If I have ( n ) days where five friends visit, each of those days contributes 5 visits. So, the total visits from these days would be ( 5n ). The remaining ( 100 - n ) days can have at most four friends visiting, contributing ( 4(100 - n) ) visits. So, the total visits would be ( 5n + 4(100 - n) ). This total must be at least 450 because that's the total number of visits we have.So, setting up the inequality:[ 5n + 4(100 - n) geq 450 ]Simplifying:[ 5n + 400 - 4n geq 450 ][ n + 400 geq 450 ][ n geq 50 ]Wait, that gives me a lower bound, but I was trying to find the maximum. Hmm, maybe I need to approach it differently. If I want to maximize ( n ), I should consider that on each of these ( n ) days, as many friends as possible are visiting. So, if on each of these days, six friends visit, that would contribute more visits, but since each friend can only visit 75 times, I have to make sure that no friend exceeds their limit.Alternatively, maybe I should think about the maximum number of days where at least five friends are present. Each day with five friends contributes 5 visits, and each day with six friends contributes 6 visits. To maximize ( n ), I should minimize the number of days where fewer than five friends are present, which would be days with four or fewer.But perhaps another approach is to consider the total number of visits. If I have ( n ) days with at least five friends, the minimum number of visits on those days is ( 5n ). The remaining ( 100 - n ) days can have at most 4 visits each, contributing ( 4(100 - n) ) visits. So, the total visits would be at least ( 5n + 4(100 - n) ).But since the total visits are exactly 450, we have:[ 5n + 4(100 - n) leq 450 ]Wait, no, that's not right. The total visits must be exactly 450, so:[ 5n + 4(100 - n) leq 450 ]But actually, the total visits can't exceed 450, so:[ 5n + 4(100 - n) leq 450 ]Simplifying:[ 5n + 400 - 4n leq 450 ][ n + 400 leq 450 ][ n leq 50 ]Wait, that's conflicting with my earlier thought. I think I need to clarify.If I want to maximize ( n ), the number of days with at least five friends, I should consider that on those days, the minimum number of visits is 5, but to maximize ( n ), I should minimize the number of visits on the other days. So, the other days can have at most 4 visits each. Therefore, the total visits would be ( 5n + 4(100 - n) ). This must be equal to 450.So:[ 5n + 4(100 - n) = 450 ][ 5n + 400 - 4n = 450 ][ n + 400 = 450 ][ n = 50 ]But wait, that gives me ( n = 50 ), but I think I can have more days with five friends if I adjust the visits. Maybe I can have some days with six friends, which would allow me to have more days with five friends.Let me think again. Suppose on ( n ) days, five friends visit, contributing ( 5n ) visits. On the remaining ( 100 - n ) days, let's say ( k ) days have six friends, contributing ( 6k ) visits, and the rest have four friends, contributing ( 4(100 - n - k) ) visits. The total visits would be ( 5n + 6k + 4(100 - n - k) = 450 ).Simplifying:[ 5n + 6k + 400 - 4n - 4k = 450 ][ n + 2k + 400 = 450 ][ n + 2k = 50 ]So, ( n + 2k = 50 ). Since ( k ) can be at most ( 100 - n ), but also, each friend can only visit 75 times. So, the total number of visits from days with six friends is ( 6k ), and from days with five friends is ( 5n ). Each friend can't exceed 75 visits.So, for each friend, the number of days they visit is 75. If a friend visits on ( k ) days with six friends and ( m ) days with five friends, then ( k + m leq 75 ). But since there are six friends, the total number of visits from six friends on ( k ) days is ( 6k ), and from five friends on ( n ) days is ( 5n ).But I'm getting a bit tangled here. Maybe another approach is to consider the maximum number of days with at least five friends. Since each friend can only miss 25 days, the maximum number of days where a friend is present is 75.If I want to maximize the number of days where at least five friends are present, I need to arrange the visits such that as many days as possible have five or six friends.Let me consider that on ( n ) days, five friends are present, and on the remaining ( 100 - n ) days, four or fewer are present. The total visits would be ( 5n + 4(100 - n) ). This must be equal to 450.So:[ 5n + 400 - 4n = 450 ][ n + 400 = 450 ][ n = 50 ]But wait, that suggests that the maximum ( n ) is 50, but I think that's not correct because if I have some days with six friends, I can have more days with five friends.Let me try again. Suppose on ( n ) days, five friends visit, and on ( m ) days, six friends visit. Then, the total visits are ( 5n + 6m ). The remaining ( 100 - n - m ) days have at most four friends, contributing ( 4(100 - n - m) ) visits.So, total visits:[ 5n + 6m + 4(100 - n - m) = 450 ]Simplify:[ 5n + 6m + 400 - 4n - 4m = 450 ][ n + 2m + 400 = 450 ][ n + 2m = 50 ]So, ( n + 2m = 50 ). To maximize ( n ), I need to minimize ( m ). The minimum ( m ) can be is 0, which would give ( n = 50 ). But if I allow ( m ) to be positive, I can have ( n ) larger than 50.Wait, no. If ( m ) increases, ( n ) decreases because ( n = 50 - 2m ). So, to maximize ( n ), I need to minimize ( m ), which is 0, giving ( n = 50 ). But that contradicts my earlier thought that having some days with six friends could allow more days with five friends.Wait, maybe I'm misunderstanding. If I have some days with six friends, those days contribute more visits, which might allow me to have more days with five friends because the total visits are fixed.Let me think differently. Suppose I have ( n ) days with five friends and ( m ) days with six friends. Then, the total visits are ( 5n + 6m ). The remaining ( 100 - n - m ) days have at most four friends, contributing ( 4(100 - n - m) ) visits.So, total visits:[ 5n + 6m + 4(100 - n - m) = 450 ]Simplify:[ 5n + 6m + 400 - 4n - 4m = 450 ][ n + 2m + 400 = 450 ][ n + 2m = 50 ]So, ( n = 50 - 2m ). To maximize ( n ), I need to minimize ( m ). The minimum ( m ) can be is 0, so ( n = 50 ). But if I set ( m = 1 ), then ( n = 48 ), which is less. So, actually, the maximum ( n ) is 50 when ( m = 0 ).But wait, that doesn't make sense because if I have some days with six friends, those days contribute more visits, which would allow me to have more days with five friends because the total visits are fixed.Wait, no, because the total visits are fixed at 450. If I have more days with six friends, those days contribute more visits, which would mean that the remaining days can have fewer visits, potentially allowing more days with five friends.Wait, let me think again. If I have ( m ) days with six friends, those contribute ( 6m ) visits. Then, the remaining visits are ( 450 - 6m ). These remaining visits have to be covered by ( n ) days with five friends and ( 100 - n - m ) days with four or fewer friends.So, the remaining visits after six friends days are ( 450 - 6m ). These need to be covered by ( n ) days with five friends and ( 100 - n - m ) days with four friends.So:[ 5n + 4(100 - n - m) = 450 - 6m ]Simplify:[ 5n + 400 - 4n - 4m = 450 - 6m ][ n + 400 - 4m = 450 - 6m ][ n = 50 - 2m ]So, again, ( n = 50 - 2m ). To maximize ( n ), set ( m = 0 ), so ( n = 50 ). If ( m = 1 ), ( n = 48 ), etc. So, the maximum ( n ) is 50.But wait, that seems counterintuitive because if I have some days with six friends, I might be able to have more days with five friends. Maybe I'm missing something.Alternatively, perhaps I should consider the total number of days each friend can be present. Each friend can be present on 75 days. If I have ( n ) days with five friends, each of those days, a friend is either present or not. To maximize ( n ), I need to arrange the visits so that as many days as possible have five friends, without exceeding each friend's 75 visits.Let me think about it from the perspective of each friend. Each friend can be present on 75 days. If I have ( n ) days with five friends, each friend can be present on some of those days. The more days a friend is present on five-friend days, the fewer days they can be present on other days.But to maximize ( n ), I need to have as many five-friend days as possible, while ensuring that no friend exceeds their 75 visits.Suppose on ( n ) days, five friends are present. Each of these days, a different set of five friends is present. But since there are six friends, each friend would miss at least one of these days. So, each friend would be present on ( n - 1 ) days, because they miss one day out of the ( n ) days.Wait, that might not be the case. If the same five friends are present on all ( n ) days, then those five friends would each have ( n ) visits, and the sixth friend would have 0 visits on those days. But that's not possible because each friend must have 75 visits.So, we need to distribute the five-friend days among the six friends in such a way that no friend exceeds 75 visits.Let me denote the number of days each friend is present on five-friend days as ( a_1, a_2, a_3, a_4, a_5, a_6 ). Each ( a_i ) is the number of days friend ( i ) is present on five-friend days. Since each five-friend day has exactly five friends, the sum of all ( a_i ) is ( 5n ).Also, each friend can be present on at most 75 days, so ( a_i + b_i = 75 ), where ( b_i ) is the number of days friend ( i ) is present on non-five-friend days (i.e., days with four or fewer friends).But the total number of visits on non-five-friend days is ( 450 - 5n ). These visits are distributed across the ( 100 - n ) days, each contributing at most 4 visits.So, the total visits on non-five-friend days is ( 4(100 - n) ). Therefore:[ 4(100 - n) geq 450 - 5n ][ 400 - 4n geq 450 - 5n ][ -4n + 5n geq 450 - 400 ][ n geq 50 ]Wait, that's interesting. So, ( n geq 50 ). But earlier, I thought ( n leq 50 ). Now, this inequality suggests that ( n ) must be at least 50. But I'm trying to find the maximum ( n ).This is confusing. Let me try to clarify.If I have ( n ) days with five friends, contributing ( 5n ) visits, and the remaining ( 100 - n ) days with at most four friends, contributing ( 4(100 - n) ) visits, then the total visits are ( 5n + 4(100 - n) ). This must be equal to 450.So:[ 5n + 400 - 4n = 450 ][ n + 400 = 450 ][ n = 50 ]So, ( n = 50 ) is the exact value if all remaining days have exactly four friends. But if some days have fewer than four friends, then ( 5n + 4(100 - n) ) would be greater than 450, which is not possible because the total visits are exactly 450. Therefore, to have the total visits exactly 450, the remaining days must have exactly four friends each.Wait, but that would mean that the number of days with five friends is exactly 50, and the remaining 50 days have four friends each. But that would give a total of ( 5 times 50 + 4 times 50 = 250 + 200 = 450 ), which matches.But then, how can ( n ) be more than 50? Because if I have more than 50 days with five friends, say 51, then the remaining 49 days would have to contribute ( 450 - 5 times 51 = 450 - 255 = 195 ) visits. But 49 days with four friends would contribute ( 4 times 49 = 196 ) visits, which is more than 195. So, it's possible to have 51 days with five friends and 49 days with four friends, but the total visits would be ( 5 times 51 + 4 times 49 = 255 + 196 = 451 ), which is more than 450. Therefore, it's not possible.Alternatively, if I have 51 days with five friends, the remaining 49 days must contribute ( 450 - 255 = 195 ) visits. Since 49 days can contribute at most ( 4 times 49 = 196 ), but we need exactly 195, which is possible by having 48 days with four friends and 1 day with three friends. So, total visits would be ( 5 times 51 + 4 times 48 + 3 times 1 = 255 + 192 + 3 = 450 ).Wait, that works. So, ( n = 51 ) is possible. Similarly, ( n = 52 ) would require the remaining 48 days to contribute ( 450 - 260 = 190 ) visits. 48 days with four friends contribute ( 192 ), so we need to reduce by 2. That can be done by having two days with three friends instead of four. So, total visits: ( 5 times 52 + 4 times 46 + 3 times 2 = 260 + 184 + 6 = 450 ).Continuing this logic, the maximum ( n ) would be when the remaining days contribute the minimum number of visits, which is 0. But since each day must have at least one visitor, the minimum is 1 visit per day. Wait, no, the problem doesn't specify that someone has to visit every day. It just says each friend visited 75 times, no more than once a day.So, actually, some days could have zero visits. But the problem says "at least five of them visited the pool" on ( n ) days. So, the other days can have any number of visitors, including zero.Wait, but if we allow some days to have zero visitors, then the total visits can be distributed more flexibly. Let me reconsider.If I have ( n ) days with at least five friends, and the remaining ( 100 - n ) days can have any number of visitors, including zero. So, the total visits are ( 5n + V ), where ( V ) is the total visits on the remaining ( 100 - n ) days. Since ( V ) can be as low as 0, but we need ( 5n leq 450 ), so ( n leq 90 ).Wait, that makes more sense. Because if I have 90 days with five friends, that's ( 5 times 90 = 450 ) visits, and the remaining 10 days can have zero visits. So, that would satisfy the total visits.But wait, each friend must visit exactly 75 times. So, if I have 90 days with five friends, each friend must be present on 75 of those days. But there are six friends, so each day with five friends, one friend is missing. To ensure that each friend is missing exactly 25 days, we need to distribute the missing days evenly.So, over 90 days, each friend must miss exactly 25 days. Since there are six friends, each day one friend is missing, so over 90 days, each friend would be missing ( 90 / 6 = 15 ) days. But 15 is less than 25. Therefore, we need to have more days where a friend is missing.Wait, that's a problem. If I have 90 days with five friends, each friend is missing 15 days. But each friend needs to miss 25 days. So, we need an additional 10 days where each friend is missing 10 more days. But those days can't have five friends, because we've already used up all 90 days for five-friend days.So, those additional 10 days must have fewer than five friends. But if we have 10 days with zero friends, then those days contribute nothing to the total visits, but each friend would be missing those 10 days, adding to their total missed days.So, each friend would be missing 15 days from the five-friend days and 10 days from the zero-friend days, totaling 25 missed days, which is correct. So, this works.Therefore, the maximum ( n ) is 90.Wait, that makes sense now. So, to maximize ( n ), we have 90 days with five friends, each friend missing 15 of those days, and 10 days with zero friends, each friend missing those 10 days, totaling 25 missed days per friend.So, the maximum ( n ) is 90.Now, for the minimum ( n ). We need to find the smallest number of days where at least five friends visited the pool. To minimize ( n ), we need to maximize the number of days where fewer than five friends visited.So, we want as many days as possible with four or fewer friends. Let me denote ( n ) as the number of days with at least five friends, and ( 100 - n ) as the number of days with four or fewer friends.The total visits would be ( 5n + 4(100 - n) ). But this must be equal to 450.Wait, no, because the days with four or fewer friends can have varying numbers of visitors. To minimize ( n ), we need to maximize the number of days with four friends, because that would contribute the most visits without requiring ( n ) to be large.So, if we have ( 100 - n ) days with four friends, contributing ( 4(100 - n) ) visits, and ( n ) days with at least five friends, contributing at least ( 5n ) visits, the total visits would be at least ( 5n + 4(100 - n) ).But the total visits are exactly 450, so:[ 5n + 4(100 - n) leq 450 ][ 5n + 400 - 4n leq 450 ][ n + 400 leq 450 ][ n leq 50 ]Wait, that's the same as before, but now we're trying to find the minimum ( n ). So, this inequality tells us that ( n leq 50 ), but we need the minimum ( n ).Wait, perhaps I need to set up the inequality differently. To find the minimum ( n ), we need to ensure that the total visits are at least 450. So, if we have ( n ) days with five friends and ( 100 - n ) days with four friends, the total visits would be ( 5n + 4(100 - n) ). This must be at least 450.So:[ 5n + 400 - 4n geq 450 ][ n + 400 geq 450 ][ n geq 50 ]So, ( n geq 50 ). Therefore, the minimum ( n ) is 50.But wait, that can't be right because earlier, I thought the maximum ( n ) was 90, and now the minimum is 50. But the problem asks for both maximum and minimum, so maybe that's correct.But let me think again. If I try to minimize ( n ), I want as many days as possible with four friends. So, let me set ( n ) as small as possible, say ( n = 25 ). Then, the remaining 75 days would have four friends each, contributing ( 4 times 75 = 300 ) visits. The ( n = 25 ) days would contribute ( 5 times 25 = 125 ) visits. Total visits: ( 125 + 300 = 425 ), which is less than 450. So, that's not enough.Therefore, I need to increase ( n ) so that the total visits reach 450. Let me set up the equation:Total visits = ( 5n + 4(100 - n) = 450 )[ 5n + 400 - 4n = 450 ][ n + 400 = 450 ][ n = 50 ]So, ( n = 50 ) is the minimum number of days needed to reach the total visits of 450 if the remaining days have four friends each. But wait, can we have some days with more than four friends to allow ( n ) to be smaller?Wait, if we have some days with six friends, those days contribute more visits, which might allow us to have fewer days with five friends. Let me try that.Suppose we have ( n ) days with five friends and ( m ) days with six friends. The remaining ( 100 - n - m ) days have four friends each. The total visits would be ( 5n + 6m + 4(100 - n - m) = 450 ).Simplifying:[ 5n + 6m + 400 - 4n - 4m = 450 ][ n + 2m + 400 = 450 ][ n + 2m = 50 ]To minimize ( n ), we need to maximize ( m ). The maximum ( m ) can be is when ( n = 0 ), so ( 2m = 50 ), ( m = 25 ). So, ( n = 0 ) and ( m = 25 ). But is that possible?If ( n = 0 ), all days with at least five friends are actually days with six friends. So, 25 days with six friends, contributing ( 6 times 25 = 150 ) visits. The remaining 75 days have four friends each, contributing ( 4 times 75 = 300 ) visits. Total visits: ( 150 + 300 = 450 ). That works.But wait, each friend must visit exactly 75 times. On the 25 days with six friends, each friend is present on all those days, contributing 25 visits. Then, on the remaining 75 days with four friends, each friend must be present on ( 75 - 25 = 50 ) days. But there are only 75 days with four friends, and each day has four friends. So, the total number of visits on those 75 days is ( 4 times 75 = 300 ). These 300 visits need to be distributed among the six friends, each needing 50 visits.So, each friend needs 50 visits on the 75 days with four friends. That's possible because ( 6 times 50 = 300 ), which matches the total visits on those days.Therefore, it's possible to have ( n = 0 ), but wait, the problem defines ( n ) as the number of days where at least five friends visited. If all days with at least five friends are actually days with six friends, then ( n ) would be 25, not 0. Because ( n ) counts days with at least five friends, which includes days with six friends.Wait, that's a crucial point. If ( n ) is the number of days with at least five friends, then days with six friends are included in ( n ). So, in the case where we have 25 days with six friends and 75 days with four friends, ( n = 25 ), not 0.So, in that case, the minimum ( n ) is 25.Wait, let me verify. If I have 25 days with six friends, each friend is present on all 25 days, contributing 25 visits. Then, on the remaining 75 days, each friend needs to be present on 50 days. Since each of those 75 days has four friends, the total number of visits on those days is ( 4 times 75 = 300 ), which is exactly ( 6 times 50 = 300 ). So, it works.Therefore, the minimum ( n ) is 25.Wait, but earlier, when I tried ( n = 25 ), I thought the total visits would be 425, but that was when I assumed all remaining days had four friends. But in reality, if some days have six friends, the total visits can be adjusted.So, to summarize:- Maximum ( n ): 90 days with five friends, 10 days with zero friends. Each friend misses 15 days from the five-friend days and 10 days from the zero-friend days, totaling 25 missed days.- Minimum ( n ): 25 days with six friends, 75 days with four friends. Each friend is present on 25 days with six friends and 50 days with four friends, totaling 75 visits.Therefore, the maximum ( n ) is 90, and the minimum ( n ) is 25.