💡Okay, so I have this problem about cities connected by one-way flights, and I need to prove that the cities can be divided into two groups, A and B, such that no city in B can fly to any city in A. Hmm, interesting. Let me try to wrap my head around this.First, let me restate the problem to make sure I understand it. There are several cities, and between some of them, there are one-way flight routes. It's given that starting from any city, you can't reach every other city through transfers. So, no matter which city you start at, there's always at least one city you can't get to. That means the graph of cities and flights isn't strongly connected. Got it.Now, I need to show that we can split all these cities into two groups, A and B, such that if you're in group B, you can't fly to any city in group A. So, from B, you can't reach A, but maybe from A you can reach B? Or maybe not? The problem doesn't specify, just that from B, you can't reach A.Let me think about graph theory concepts here. Since the graph isn't strongly connected, it must have multiple strongly connected components. A strongly connected component is a maximal subset of cities where every city is reachable from every other city in the subset. So, if the whole graph isn't strongly connected, there must be at least two such components.Maybe I can use this idea of strongly connected components to divide the cities into groups A and B. If I can find two such components where one can't reach the other, that might work. But how?Wait, the problem says that starting from any city, you can't reach every city. So, for every city, there's at least one city it can't reach. That suggests that there's no universal reachability from any single city. So, perhaps the graph has a certain structure that allows us to partition it into two sets with the desired property.Let me consider selecting a particular city, say city X, and then look at all the cities that X can reach. Let's call this set B. Then, the remaining cities would be set A. If I can show that no city in B can reach any city in A, that would satisfy the condition.But how do I know that such a partition exists? Maybe I need to use some kind of induction or maybe properties of directed graphs. Alternatively, perhaps I can think about the condensation of the graph, which is the graph formed by collapsing each strongly connected component into a single node. In the condensation, the graph is a directed acyclic graph (DAG).In a DAG, there's always at least one source node (a node with no incoming edges) and at least one sink node (a node with no outgoing edges). Maybe I can use this property. If I take all the sink components and put them in group A, and the rest in group B, then perhaps group B can't reach group A.Wait, but the problem says that starting from any city, you can't reach every city. So, in terms of the condensation, there must be more than one component, and none of them can reach all the others. So, if I take group A as one of the sink components and group B as the rest, then group B can't reach group A because group A is a sink.But I'm not sure if that's the right approach. Maybe I need to think differently.Another idea: since the graph isn't strongly connected, there exists at least two strongly connected components. Let me pick one component and call it A, and the rest B. If I can show that no city in B can reach A, then I'm done.But how do I ensure that? Maybe by considering the reachability from B to A. If there's no edge from B to A, then obviously, no one in B can reach A. But if there are edges, then perhaps A is a sink component.Wait, in the condensation graph, if A is a sink component, then there are no edges from other components to A. So, if I take A as a sink component and B as the rest, then indeed, no city in B can reach A.But is A necessarily a sink component? Or can it be any component?Hmm, maybe I need to formalize this. Let me try to outline the steps:1. Consider the graph of cities and one-way flights.2. Since the graph isn't strongly connected, it has multiple strongly connected components.3. The condensation of the graph is a DAG.4. In a DAG, there exists at least one sink component (a component with no outgoing edges).5. Let A be this sink component, and B be the union of all other components.6. Since A is a sink, there are no edges from B to A, meaning no city in B can reach any city in A.Wait, that seems to work! So, by choosing A as a sink component in the condensation, and B as the rest, we ensure that no city in B can reach A. Therefore, the partition exists.But let me double-check. Suppose A is a sink component. Then, by definition, there are no edges from any other component to A. So, if you're in B, which is all the other components, you can't reach A. That makes sense.But what if there are multiple sink components? Then, I could choose any one of them as A, and the rest as B. But the problem doesn't specify anything about the sizes of A and B, just that they exist. So, as long as I can find at least one sink component, I can make that A and the rest B.Yes, that seems to cover it. So, the key idea is to use the fact that in a directed graph that isn't strongly connected, the condensation has at least one sink component, and we can use that to partition the graph into A and B with the desired property.I think that makes sense. Let me try to write this up more formally.

💡First, I need to determine the total number of persimmons in the five baskets that each contain 5 persimmons. To do this, I'll multiply the number of baskets by the number of persimmons in each basket.Next, I'll add the number of persimmons in the sixth basket, which contains 8 persimmons, to the total from the first five baskets. This will give me the overall total number of persimmons in all six baskets.By performing these calculations, I can find the final answer.

💡First, I need to determine which of the given fractions is the arithmetic mean of the other two. The fractions provided are (frac{8}{12}), (frac{5}{6}), and (frac{9}{12}).To make it easier to compare them, I'll convert all fractions to have a common denominator. The denominators are 12 and 6, so 12 is a suitable common denominator.1. (frac{8}{12}) is already in the desired form.2. (frac{5}{6}) can be converted to (frac{10}{12}) by multiplying both the numerator and the denominator by 2.3. (frac{9}{12}) is already in the desired form.Now, I have the fractions (frac{8}{12}), (frac{10}{12}), and (frac{9}{12}). Arranging them in order, they are (8), (10), and (9) when considering the numerators with the common denominator of 12.The arithmetic mean of two numbers is calculated by adding them together and dividing by 2. I'll check each fraction to see if it is the mean of the other two.1. Checking if (frac{8}{12}) is the mean of (frac{10}{12}) and (frac{9}{12}): [ frac{frac{10}{12} + frac{9}{12}}{2} = frac{19}{24} neq frac{8}{12} ] This is not equal to (frac{8}{12}).2. Checking if (frac{10}{12}) is the mean of (frac{8}{12}) and (frac{9}{12}): [ frac{frac{8}{12} + frac{9}{12}}{2} = frac{17}{24} neq frac{10}{12} ] This is not equal to (frac{10}{12}).3. Checking if (frac{9}{12}) is the mean of (frac{8}{12}) and (frac{10}{12}): [ frac{frac{8}{12} + frac{10}{12}}{2} = frac{18}{24} = frac{3}{4} = frac{9}{12} ] This is equal to (frac{9}{12}).Therefore, (frac{9}{12}) is the arithmetic mean of (frac{8}{12}) and (frac{10}{12}).

💡To determine ( S_6 ) for the arithmetic sequence, we start with the given information: ( a_1 = frac{1}{2} ) and ( S_4 = 20 ).1. **Use the sum formula for the first ( n ) terms of an arithmetic sequence:** [ S_n = frac{n}{2} left(2a_1 + (n-1)dright) ] 2. **Plug in the known values for ( S_4 ):** [ 20 = frac{4}{2} left(2 times frac{1}{2} + 3dright) ] 3. **Simplify the equation:** [ 20 = 2 left(1 + 3dright) ] [ 20 = 2 + 6d ] 4. **Solve for ( d ):** [ 6d = 18 ] [ d = 3 ] 5. **Calculate ( S_6 ) using the common difference ( d = 3 ):** [ S_6 = frac{6}{2} left(2 times frac{1}{2} + 5 times 3right) ] [ S_6 = 3 left(1 + 15right) ] [ S_6 = 3 times 16 = 48 ]

💡Alright, so I have this problem here: If 525 divided by 2027 equals approximately 0.258973, what is the 157th digit to the right of the decimal point in the decimal equivalent of the fraction? Hmm, okay, let's see how to approach this.First, I know that when you divide two numbers, sometimes the decimal repeats, and sometimes it doesn't. If it repeats, there's a pattern that cycles over and over. If it doesn't repeat, then it's a non-repeating decimal, which means it goes on forever without any repeating pattern. So, to find the 157th digit, I need to figure out if the decimal repeats and, if it does, what the repeating cycle is.Given that 525 divided by 2027 is approximately 0.258973, I can assume that the decimal might start repeating after some digits. But I don't know the full repeating cycle yet. Maybe it's just repeating the digits 258973 over and over, or maybe it's a longer cycle. I need to find out.One way to find out if the decimal repeats is to perform the division manually and see if a pattern emerges. But dividing 525 by 2027 manually would be time-consuming, especially since I need to go up to the 157th digit. That's a lot! Maybe there's a smarter way to figure this out without doing all that work.I remember that for fractions, if the denominator has prime factors other than 2 and 5, the decimal will eventually repeat. In this case, 2027 is the denominator. Let me check if 2027 can be factored into primes. Hmm, 2027 divided by 7 is approximately 289.571, which isn't an integer. Divided by 11? 2027 divided by 11 is about 184.27, still not an integer. How about 13? 2027 divided by 13 is roughly 155.923, not an integer either. Maybe 17? 2027 divided by 17 is about 119.235, still not an integer. Hmm, this is taking too long. Maybe 2027 is a prime number? I'm not sure, but for the sake of this problem, let's assume that 2027 is a prime number, which would mean that the decimal repeats with a certain period.If 2027 is prime, then the length of the repeating cycle would be related to 2027. Specifically, the period of the repeating decimal is the smallest positive integer n such that 10^n ≡ 1 mod 2027. That means 10 raised to the power of n leaves a remainder of 1 when divided by 2027. Finding this n would give me the length of the repeating cycle.But calculating 10^n mod 2027 for n up to 2026 (since the maximum period for a prime denominator p is p-1) is a bit too much for me to handle manually. Maybe there's a pattern or a shortcut I can use. Alternatively, I could use the fact that the decimal repeats every certain number of digits and use that to find the 157th digit.Given that the decimal starts as 0.258973, maybe the repeating cycle is 258973. Let's check how many digits that is. 2,5,8,9,7,3 – that's 6 digits. If the cycle is 6 digits long, then every 6 digits, the pattern repeats. So, to find the 157th digit, I can divide 157 by 6 and find the remainder. The remainder will tell me which digit in the cycle corresponds to the 157th position.Let's do that. 157 divided by 6 is 26 with a remainder of 1. So, 6 times 26 is 156, and 157 minus 156 is 1. That means the 157th digit corresponds to the first digit in the repeating cycle. Looking back at the decimal, the first digit after the decimal is 2. So, the 157th digit should be 2.Wait a minute, but I assumed that the repeating cycle is 258973. What if the cycle is longer? Maybe the decimal doesn't start repeating immediately after those 6 digits. How can I be sure? Well, without performing the full division, it's hard to know for certain. But given the information provided, 0.258973, it's reasonable to assume that this is the start of the repeating cycle.Alternatively, if the cycle is longer, say, 12 digits, then I would need to adjust my calculation accordingly. But since I don't have more information, I'll stick with the 6-digit cycle assumption.So, to recap, if the decimal repeats every 6 digits, then the 157th digit corresponds to the first digit in the cycle, which is 2. Therefore, the 157th digit to the right of the decimal point is 2.But just to be thorough, let's consider if the cycle is longer. Suppose the cycle is 12 digits. Then, 157 divided by 12 is 13 with a remainder of 1. So, again, the first digit in the cycle would be 2. If the cycle is 18 digits, 157 divided by 18 is 8 with a remainder of 13. Then, the 13th digit in the cycle would be... Wait, I don't know the cycle beyond the first 6 digits. So, without knowing the full cycle, I can't be certain.Given that, I think the safest assumption is that the cycle is 6 digits long, as provided by the initial decimal expansion. Therefore, the 157th digit is 2.Another way to think about it is to list out the decimal places:1: 22: 53: 84: 95: 76: 37: 28: 59: 810: 911: 712: 3And so on. So, every 6 digits, it repeats. Therefore, the 157th digit would be the same as the 1st digit in the cycle, which is 2.I think that's a reasonable conclusion based on the information given.

💡First, I need to determine the cash realized on selling the stock before deducting the brokerage. Let's denote this amount as ( C ).The brokerage rate is ( frac{1}{4} % ) of ( C ), which can be expressed as ( frac{1}{4} times frac{1}{100} times C = frac{C}{400} ).The net amount received after deducting the brokerage is Rs. 108, so I can set up the equation:[C - frac{C}{400} = 108]To solve for ( C ), I'll first find a common denominator and combine the terms:[frac{400C - C}{400} = 108][frac{399C}{400} = 108]Next, I'll multiply both sides by 400 to isolate ( 399C ):[399C = 108 times 400][399C = 43200]Finally, I'll divide both sides by 399 to solve for ( C ):[C = frac{43200}{399}][C = 108.27]Therefore, the cash realized on selling the stock before deducting the brokerage is Rs. (boxed{108.27}).