💡First, I need to determine the total time Trey has for completing all his chores, which is 2 hours. Converting this into minutes gives me 120 minutes.Next, I'll calculate the time spent on each category of chores. For cleaning the house, there are 7 tasks, each taking 10 minutes, totaling 70 minutes. Taking a shower involves 1 task, which takes 10 minutes.Adding the time spent on cleaning the house and taking a shower gives me 80 minutes.Subtracting this from the total available time leaves me with 40 minutes for making dinner.Since each task takes 10 minutes, dividing the remaining time by 10 minutes per task tells me that Trey has to do 4 things to make dinner.

💡Alright, so I have this problem here where I need to figure out the sequence ( {c_n} ) after removing certain numbers from the set of positive integers. Let me try to break it down step by step.First, the problem defines ( a_n ) as the integer closest to ( sqrt{n} ). So, for each positive integer ( n ), I need to find the closest integer to its square root. That makes sense. For example, if ( n = 1 ), then ( sqrt{1} = 1 ), so ( a_1 = 1 ). If ( n = 2 ), ( sqrt{2} ) is approximately 1.414, so the closest integer is 1, hence ( a_2 = 1 ). Similarly, for ( n = 3 ), ( sqrt{3} ) is about 1.732, so ( a_3 = 2 ). Okay, I think I get how ( a_n ) is determined.Next, ( b_n ) is defined as ( n + a_n ). So, for each ( n ), I add ( a_n ) to ( n ) to get ( b_n ). Let me compute a few examples to see the pattern.- For ( n = 1 ): ( a_1 = 1 ), so ( b_1 = 1 + 1 = 2 ).- For ( n = 2 ): ( a_2 = 1 ), so ( b_2 = 2 + 1 = 3 ).- For ( n = 3 ): ( a_3 = 2 ), so ( b_3 = 3 + 2 = 5 ).- For ( n = 4 ): ( a_4 = 2 ) (since ( sqrt{4} = 2 )), so ( b_4 = 4 + 2 = 6 ).- For ( n = 5 ): ( a_5 = 2 ) (since ( sqrt{5} approx 2.236 )), so ( b_5 = 5 + 2 = 7 ).- For ( n = 6 ): ( a_6 = 2 ) (since ( sqrt{6} approx 2.449 )), so ( b_6 = 6 + 2 = 8 ).- For ( n = 7 ): ( a_7 = 3 ) (since ( sqrt{7} approx 2.645 )), so ( b_7 = 7 + 3 = 10 ).- For ( n = 8 ): ( a_8 = 3 ) (since ( sqrt{8} approx 2.828 )), so ( b_8 = 8 + 3 = 11 ).- For ( n = 9 ): ( a_9 = 3 ) (since ( sqrt{9} = 3 )), so ( b_9 = 9 + 3 = 12 ).Hmm, so the sequence ( b_n ) starts as 2, 3, 5, 6, 7, 8, 10, 11, 12, and so on. I notice that some numbers are missing from the natural numbers. For example, 1 is missing, 4 is missing, 9 is missing, 16 is missing, etc. Wait a minute, those are perfect squares! So, it seems like ( b_n ) is constructed in such a way that all the perfect squares are excluded.Let me test this hypothesis. If ( n ) is a perfect square, say ( n = k^2 ), then ( a_n = k ) because ( sqrt{k^2} = k ). Therefore, ( b_n = k^2 + k ). So, ( b_n ) for perfect squares is ( k^2 + k ), which is not a perfect square. Hence, perfect squares are not in the ( b_n ) sequence, meaning they are the ones being removed from the set of positive integers.Therefore, when we remove all ( b_n ) from the positive integers, the remaining numbers are exactly the perfect squares. So, the sequence ( {c_n} ) is just the sequence of perfect squares. That is, ( c_n = n^2 ).Wait, let me make sure I didn't miss anything. If ( b_n ) removes numbers like 2, 3, 5, 6, 7, 8, 10, 11, 12, etc., then the remaining numbers are 1, 4, 9, 16, 25, and so on. Yes, those are the perfect squares. So, arranging them in ascending order, we get ( c_1 = 1 ), ( c_2 = 4 ), ( c_3 = 9 ), ( c_4 = 16 ), etc. So, indeed, ( c_n = n^2 ).But just to be thorough, let me check a few more terms. For ( n = 10 ): ( a_{10} = 3 ) (since ( sqrt{10} approx 3.162 )), so ( b_{10} = 10 + 3 = 13 ). So, 13 is removed. The next perfect square is 16, which should be ( c_4 ). Let me check ( n = 11 ): ( a_{11} = 3 ), ( b_{11} = 11 + 3 = 14 ). So, 14 is removed. Then ( n = 12 ): ( a_{12} = 3 ), ( b_{12} = 12 + 3 = 15 ). So, 15 is removed. Then ( n = 13 ): ( a_{13} = 4 ) (since ( sqrt{13} approx 3.606 )), so ( b_{13} = 13 + 4 = 17 ). So, 17 is removed. The next perfect square is 16, which is ( c_4 ), and then 25, which is ( c_5 ).Yes, this seems consistent. So, the sequence ( {c_n} ) is indeed the sequence of perfect squares. Therefore, ( c_n = n^2 ).I think I've got it. The key was realizing that ( b_n ) removes all numbers that are not perfect squares, leaving only the perfect squares behind. So, the remaining numbers are the squares, and arranging them in order gives ( c_n = n^2 ).**Final Answer**The expression for ( c_n ) in terms of ( n ) is boxed{n^2}.

💡Okay, so I have this problem about the spread of influenza in an area where people visit their friends every day. The rules are that healthy people who visit sick friends become ill the next day. Each person recovers after one day of illness and then has at least one day of immunity, during which they can't get infected again. I need to prove two things:1. If some people have immunity on the first day, the flu can spread indefinitely.2. If on any given day no one has immunity, the flu will eventually cease spreading.Alright, let's break this down. First, I need to understand the model. It seems like a simple SIR (Susceptible, Infected, Recovered) model but with some specific rules. People can be healthy (susceptible), infected, or immune (recovered). The key points are:- Each day, everyone visits their friends.- Healthy people who visit sick friends become ill the next day.- Infected people recover after one day and become immune for at least one day.So, the immune state lasts for at least one day, meaning that after recovery, they can't get sick again the next day. That seems important.For the first part, if some people have immunity on day one, can the flu spread forever? Intuitively, if there's always someone immune, maybe the flu can keep cycling through the population because the immune people can't get sick again, but maybe they can help sustain the spread somehow. Hmm.Let me think about a simple example. Suppose there are three people: A, B, and C. On day one, A is immune, B is infected, and C is healthy. So, on day two, B recovers and becomes immune, and C, who visited B, becomes infected. On day three, C recovers and becomes immune, and A, who was immune on day one, is now healthy again because immunity only lasts one day. So, A is healthy on day three, and if C was infected on day two, then on day three, C recovers, and A, being healthy, could get infected again if someone is sick.Wait, but on day three, C recovers, so who is infected? Maybe I need to track this more carefully.Let me make a table:- Day 1: A (immune), B (infected), C (healthy)- Day 2: A (healthy), B (immune), C (infected)- Day 3: A (infected), B (healthy), C (immune)- Day 4: A (immune), B (infected), C (healthy)Oh, so it cycles every three days. So, the flu keeps going around the three people indefinitely because each time someone recovers, they become immune for one day, and then become susceptible again. So, with immunity on day one, the cycle can continue forever.That seems to make sense. So, if there's at least one immune person on day one, the flu can keep spreading in a cycle. Therefore, part 1 is proven by this example.Now, for part 2: If on any given day no one has immunity, the flu will eventually cease spreading. So, if there are no immune people on a particular day, does that mean the flu can't continue?Wait, the wording is a bit confusing. It says, "If on any given day no one has immunity, the flu will eventually cease spreading." So, does it mean that if there's a day where no one is immune, then eventually the flu stops? Or does it mean that if on every day, there's no one immune, then the flu stops? I think it's the first interpretation: if there exists a day where no one is immune, then the flu will eventually stop.But in the first part, we saw that if there's always someone immune, the flu can continue. So, if there's a day with no immunity, maybe the flu can't sustain itself.Let me think about this. Suppose on day k, no one is immune. That means everyone is either susceptible or infected. Since there's no one immune, everyone who was infected on day k-1 has either recovered and is now susceptible again or is still infected.Wait, but each person recovers after one day of illness and has at least one day of immunity. So, if someone was infected on day k-1, they recover on day k and become immune on day k. So, on day k, they are immune, not susceptible.But the problem says, "If on any given day no one has immunity," which would mean that on day k, no one is immune. But according to the rules, if someone was infected on day k-1, they should be immune on day k. So, the only way for no one to be immune on day k is if no one was infected on day k-1. Because if someone was infected on day k-1, they would be immune on day k.Therefore, if on day k, no one is immune, that means on day k-1, no one was infected. So, if no one was infected on day k-1, then on day k, no one is infected either, because you can't get infected if no one is sick.Wait, but that seems a bit circular. Let me try to formalize this.Let S(t) be the set of susceptible people on day t, I(t) be the set of infected people on day t, and R(t) be the set of immune people on day t.Given the rules:- Each day, everyone visits their friends.- Healthy (susceptible) people who visit sick friends become ill the next day.- Infected people recover after one day and become immune for at least one day.So, the transitions are:- S(t) ∩ (friends of I(t)) become I(t+1)- I(t) becomes R(t+1)- R(t) becomes S(t+1) (since immunity lasts at least one day, so after being immune, they become susceptible again)Therefore, the state transitions can be written as:I(t+1) = S(t) ∩ (friends of I(t))R(t+1) = I(t)S(t+1) = (S(t) I(t+1)) ∪ R(t)Now, if on some day k, R(k) = ∅, meaning no one is immune on day k. Then, according to the transition, R(k) = I(k-1). So, R(k) = ∅ implies I(k-1) = ∅.Therefore, if no one is immune on day k, then no one was infected on day k-1. If no one was infected on day k-1, then on day k, no one can get infected because there are no infected people to infect others. Therefore, I(k) = ∅.Similarly, if I(k) = ∅, then R(k+1) = I(k) = ∅, and S(k+1) = S(k) ∪ R(k) = S(k) ∪ ∅ = S(k). So, the susceptible population remains the same, and no new infections occur.Therefore, once there is a day with no immunity, the flu cannot spread anymore because there were no infections the previous day, leading to no new infections the next day, and so on. Hence, the flu ceases to spread.Wait, but what if the susceptible population is still there? If no one is infected, but there are susceptible people, can the flu start again? But according to the problem, the flu is spreading from infected people. If there are no infected people, the flu can't spread. So, once there's a day with no immunity, meaning no one was infected the previous day, the flu stops.Therefore, part 2 is proven because if there's a day with no immunity, the flu can't continue.But wait, in the first part, we had a cycle where immunity was always present, so the flu could continue. In the second part, if there's a day with no immunity, the flu stops. So, these two results are consistent.I think I've got it. So, summarizing:1. If some people have immunity on day one, the flu can cycle through the population indefinitely because the immune people become susceptible again after their immunity period, allowing the flu to keep spreading.2. If on any day there's no immunity, that means no one was infected the previous day, so the flu can't spread anymore, leading to it ceasing.Yeah, that makes sense. I think I've worked through it correctly.

💡To determine the range of values for ( x ) such that ( sqrt{x - 3} ) is meaningful within the real number range, we need to ensure that the expression inside the square root is non-negative. This is because the square root of a negative number is not defined in the set of real numbers.1. Start with the expression inside the square root: [ x - 3 ]2. Set up the inequality to ensure the expression is non-negative: [ x - 3 geq 0 ]3. Solve the inequality for ( x ) by adding 3 to both sides: [ x - 3 + 3 geq 0 + 3 ] [ x geq 3 ]Therefore, the range of values for ( x ) is ( x geq 3 ).

💡Okay, so I have this system of inequalities to solve for x, and I need to find the range of values for m such that the system has a solution. The system is:2x - 6 + m < 0 4x - m > 0Alright, let me try to figure this out step by step. I'm a bit nervous because inequalities can sometimes be tricky, especially when variables are involved on both sides. But I think I can handle this by isolating x in each inequality and then seeing what conditions on m make the system solvable.Starting with the first inequality: 2x - 6 + m < 0. I want to solve for x, so I'll move the constants to the other side. Let me rewrite it:2x - 6 + m < 0 => 2x < 6 - mNow, to get x by itself, I'll divide both sides by 2:x < (6 - m)/2Hmm, that simplifies to x < 3 - (m/2). Let me write that down as inequality (1):(1) x < 3 - (m/2)Okay, moving on to the second inequality: 4x - m > 0. Again, I'll solve for x. So, I'll add m to both sides:4x - m > 0 => 4x > mNow, divide both sides by 4:x > m/4So, that gives me x > m/4. Let me note that as inequality (2):(2) x > m/4Now, I have two inequalities:(1) x < 3 - (m/2) (2) x > m/4For the system to have a solution, there must be some x that satisfies both inequalities simultaneously. That means the lower bound from inequality (2) must be less than the upper bound from inequality (1). In other words, m/4 must be less than 3 - (m/2). If m/4 is not less than 3 - (m/2), then there would be no x that satisfies both inequalities, and the system would have no solution.So, setting up the inequality:m/4 < 3 - (m/2)Now, I need to solve this inequality for m. Let me do that step by step. First, I'll get rid of the fractions by finding a common denominator. The denominators are 4 and 2, so the least common denominator is 4. I'll multiply every term by 4 to eliminate the fractions:4*(m/4) < 4*3 - 4*(m/2)Simplifying each term:m < 12 - 2mOkay, so now I have:m < 12 - 2mI want to get all the m terms on one side. So, I'll add 2m to both sides:m + 2m < 12Which simplifies to:3m < 12Now, divide both sides by 3:m < 4So, this tells me that m must be less than 4 for the system to have a solution. Let me just double-check my steps to make sure I didn't make a mistake.Starting with the two inequalities:2x - 6 + m < 0 4x - m > 0Solving the first for x gives x < 3 - (m/2). Solving the second for x gives x > m/4. Then, setting m/4 < 3 - (m/2) to ensure there's an overlap in the solutions. Multiplying by 4 gives m < 12 - 2m, leading to 3m < 12, so m < 4.That seems correct. So, the range of values for m is all real numbers less than 4.Looking back at the options provided:A: m ≤ 4 B: m < 4 C: m ≥ 4 D: m > 4The correct answer should be B: m < 4.Wait, but just to be thorough, what happens if m equals 4? Let me test m = 4 in the original inequalities.First inequality: 2x - 6 + 4 < 0 => 2x - 2 < 0 => 2x < 2 => x < 1Second inequality: 4x - 4 > 0 => 4x > 4 => x > 1So, if m = 4, the first inequality gives x < 1 and the second gives x > 1. There's no x that satisfies both x < 1 and x > 1 simultaneously. Therefore, when m = 4, the system has no solution. So, m must be strictly less than 4, not less than or equal to.Therefore, the correct answer is indeed B: m < 4.I think that's solid. I don't see any mistakes in my reasoning, and testing m = 4 confirmed that it doesn't work, so m has to be strictly less than 4.**Final Answer**boxed{B}

💡Okay, so I have this problem about essential subsets and deviations modulo a prime number p of the form 12n + 11. Let me try to unpack this step by step.First, the set M is defined as {1, 2, ..., p-1}, and an essential subset S is a subset of M with (p-1)/2 elements. The product of all elements in S is denoted as Π_s, and the product of all the other elements in M (which would be the complement of S) is denoted as Π_s_bar. The deviation Δ_s is then defined as Π_s - Π_s_bar. The goal is to find the least possible remainder of this deviation modulo p.Alright, so I need to find the minimal value of Δ_s mod p, where Δ_s = Π_s - Π_s_bar, and S is an essential subset of size (p-1)/2.Let me recall Wilson's theorem, which says that for a prime p, (p-1)! ≡ -1 mod p. That might come in handy here because the product of all elements in M is (p-1)!.Given that S is a subset of size (p-1)/2, the complement of S, let's call it S', will also be of size (p-1)/2. So, both Π_s and Π_s_bar are products of (p-1)/2 elements each.From Wilson's theorem, we have that Π_s * Π_s_bar ≡ (p-1)! ≡ -1 mod p. So, Π_s * Π_s_bar ≡ -1 mod p.Let me denote Π_s as 'a'. Then Π_s_bar would be the product of the remaining elements, which is equal to -1/a mod p because a * Π_s_bar ≡ -1 mod p. So, Π_s_bar ≡ -a^{-1} mod p.Therefore, the deviation Δ_s = a - Π_s_bar ≡ a - (-a^{-1}) ≡ a + a^{-1} mod p.So, now the problem reduces to finding the minimal value of a + a^{-1} mod p, where 'a' is the product of (p-1)/2 elements from M.Hmm, interesting. So, I need to find the minimal value of a + a^{-1} mod p.Let me think about possible values of a + a^{-1} mod p.First, note that a + a^{-1} can be rewritten as (a^2 + 1)/a. So, if a^2 + 1 ≡ 0 mod p, then a + a^{-1} ≡ 0 mod p. But does this happen?Well, a^2 ≡ -1 mod p. For this to be possible, -1 must be a quadratic residue modulo p. However, for primes p ≡ 3 mod 4, -1 is not a quadratic residue. Since p is of the form 12n + 11, which is 3 mod 4 (because 12n is 0 mod 4, and 11 is 3 mod 4), so p ≡ 3 mod 4. Therefore, -1 is not a quadratic residue modulo p. So, a^2 ≡ -1 mod p has no solution. Therefore, a + a^{-1} ≡ 0 mod p is impossible.Next, let's consider if a + a^{-1} ≡ 1 mod p. That would mean a^2 - a + 1 ≡ 0 mod p. Let's see if this equation has solutions.The discriminant of this quadratic equation is (-1)^2 - 4*1*1 = 1 - 4 = -3. So, for this equation to have solutions, -3 must be a quadratic residue modulo p.But p ≡ 11 mod 12, which is also 11 mod 12. Let me recall that -3 is a quadratic residue modulo p if and only if p ≡ 1 mod 3. Since p ≡ 11 mod 12, which is 11 mod 3 (because 12 ≡ 0 mod 3, so 12n + 11 ≡ 0 + 2 ≡ 2 mod 3). So, p ≡ 2 mod 3. Therefore, -3 is not a quadratic residue modulo p. Hence, the equation a^2 - a + 1 ≡ 0 mod p has no solutions. Therefore, a + a^{-1} ≡ 1 mod p is also impossible.So, moving on, let's check if a + a^{-1} ≡ 2 mod p is possible. That would mean a^2 - 2a + 1 ≡ 0 mod p, which factors as (a - 1)^2 ≡ 0 mod p. So, a ≡ 1 mod p.Is it possible for a product of (p-1)/2 elements to be congruent to 1 mod p? Well, certainly, because 1 is an element in M, and if we take S to be a subset that includes 1 and other elements whose product is 1. But wait, the product of all elements in S is 1 mod p.But wait, is 1 the only element that can give a product of 1? No, because for example, if we take pairs of elements that are inverses of each other, their product would be 1. So, if we can pair up elements in M such that each pair multiplies to 1, then the product of all elements in S would be 1.But M is the set {1, 2, ..., p-1}, and every element except 1 and -1 has a unique inverse different from itself. Since p ≡ 3 mod 4, -1 is not a quadratic residue, but that doesn't directly affect the inverses. Wait, actually, for p ≡ 3 mod 4, -1 is not a square, but inverses still exist for all elements except 0, which isn't in M.So, in M, every element except 1 and -1 has a distinct inverse. Since p is odd, p-1 is even, so (p-1)/2 is an integer. Therefore, we can pair up the elements into (p-1)/2 pairs, each multiplying to 1. Therefore, if we take S to be one such pair, the product of S would be 1.Wait, but S is supposed to have (p-1)/2 elements, so if we take one element from each pair, then the product would be 1. Alternatively, if we take all the elements in one pair, but that would be two elements, which is not necessarily (p-1)/2.Wait, no, actually, if we have (p-1)/2 pairs, each multiplying to 1, then taking one element from each pair would give us a subset S with (p-1)/2 elements, and the product of S would be 1, because each element is paired with its inverse, and we're taking one from each pair.Wait, but actually, if we take one element from each pair, the product would be the product of all the elements in S, which would be the product of one element from each inverse pair. But since each pair multiplies to 1, the product of all elements in S would be the product of one element from each pair, which is not necessarily 1. Hmm, maybe I need to think differently.Alternatively, perhaps if we take S to be the set of quadratic residues modulo p. Since p ≡ 3 mod 4, the number of quadratic residues is (p-1)/2, which matches the size of S. The product of all quadratic residues modulo p is known to be 1 if p ≡ 3 mod 4. Wait, is that true?Let me recall that the product of all quadratic residues modulo p is equal to 1 if p ≡ 3 mod 4. Yes, that's a result from number theory. So, if S is the set of quadratic residues, then Π_s ≡ 1 mod p. Therefore, Π_s_bar ≡ -1/a ≡ -1 mod p, since a ≡ 1 mod p.Therefore, Δ_s = Π_s - Π_s_bar ≡ 1 - (-1) ≡ 2 mod p.So, in this case, the deviation is 2 mod p. Is this the minimal possible deviation?Well, earlier, we saw that deviations of 0 and 1 are impossible. So, the next possible minimal deviation is 2. Therefore, the least possible remainder is 2.But let me double-check if there could be a subset S where the deviation is less than 2. Since we're working modulo p, the possible remainders are from 0 to p-1. But since 0 and 1 are impossible, the next smallest is 2.Therefore, the least possible remainder of the deviation modulo p is 2.