💡First, I need to determine how much work Jose and Raju can complete individually in one day. Jose completes the work in 10 days, so his work rate is 1/10 of the work per day. Raju completes the work in 40 days, so his work rate is 1/40 of the work per day.Next, I'll add their individual work rates to find their combined work rate when working together. To add the fractions 1/10 and 1/40, I need a common denominator, which is 40.Adding the fractions:1/10 + 1/40 = 4/40 + 1/40 = 5/40Simplifying the combined work rate:5/40 = 1/8This means that together, Jose and Raju can complete 1/8 of the work per day. To find out how many days it will take them to complete the entire work, I take the reciprocal of the combined work rate:Number of days = 1 / (1/8) = 8Therefore, it will take Jose and Raju 8 days to complete the work together.

💡Okay, so I have this problem with two propositions, p and q, and I need to figure out if the compound proposition p ∨ q is true or false. Let me try to break this down step by step.First, let's look at proposition p: "There exists n ∈ ℝ, such that for all m ∈ ℝ, m · n = m." Hmm, this sounds like it's talking about some special number n that, when you multiply it by any real number m, you just get m back. That seems familiar. Isn't that the multiplicative identity? Yeah, I think the multiplicative identity in real numbers is 1 because 1 times any number is that number itself. So, if n is 1, then for any m, m · 1 = m. That makes sense. So, proposition p is saying that there exists such an n, which is 1, and since 1 is a real number, p is true.Now, moving on to proposition q: "For all n ∈ ℝ, there exists m ∈ ℝ, such that m² < n." Okay, this is a bit trickier. It's saying that for every real number n, I can find some real number m where m squared is less than n. Let me think about this. If n is a positive number, then m can be, say, half of the square root of n. So, m = (√n)/2. Then, m squared would be (n)/4, which is definitely less than n if n is positive. What if n is zero or negative? If n is zero, then m squared has to be less than zero. But wait, m squared is always non-negative because squaring any real number gives a non-negative result. So, if n is zero or negative, there's no real number m such that m squared is less than n. That means proposition q isn't true for all real numbers n. Specifically, it fails when n is zero or negative.So, putting it all together: proposition p is true because there exists n = 1 that satisfies the condition, and proposition q is false because it doesn't hold for n ≤ 0. Now, the compound proposition is p ∨ q, which means "p or q." In logic, "or" is true if at least one of the propositions is true. Since p is true, even though q is false, the whole statement p ∨ q is true.Wait, but I want to make sure I didn't miss anything with proposition q. Maybe there's a way to interpret it differently? Let me double-check. For all n ∈ ℝ, there exists m ∈ ℝ such that m² < n. If n is negative, m² is always non-negative, so m² can't be less than a negative number. If n is zero, m² can't be less than zero. So, yes, q is definitely false for n ≤ 0. Therefore, my initial conclusion seems correct.So, in summary, p is true, q is false, and since p ∨ q only requires one of them to be true, the compound proposition is true.

💡Okay, so I need to prove that for P = 23, there exists an arrangement satisfying the given conditions. The problem involves splitting piles of stones in such a way that any two piles on the table at the same time must be "similar." I'm not entirely sure what "similar" means here, but I think it might refer to having the same number of stones or some proportional relationship. First, I should probably understand the rules better. We start with a pile of n stones. In one move, we can split any pile into two smaller piles. The key condition is that at any time, all piles on the table must be "similar." If "similar" means they have the same number of stones, then we're looking to split the pile into smaller piles of equal size. But that might not always be possible, especially if the number of stones isn't a power of two or something. Wait, maybe "similar" doesn't mean exactly the same size but some proportional relationship. Maybe it means that the sizes are in a specific ratio or follow a certain pattern. Since the problem mentions arranging the stones, it might be about partitioning them into groups where each group follows a particular rule or sequence.Looking back at the problem statement, it says that in one move, you can split any existing pile into two smaller ones. So, starting from one pile, you can split it into two, then split those into more, and so on. The condition is that at any time, the sizes of any two piles must be "similar." I think "similar" might mean that the sizes are multiples of each other or follow a specific sequence. Maybe it's related to the Fibonacci sequence or some other recursive sequence. The hint mentioned using tables with rows formed by multiplying the first row by a number, choosing 1 and a multiplier of 3. That makes me think of geometric sequences or something like that.The formula given is a_{i,j} = (-1)^{j+1} j F_i, where F_i is defined recursively by F_0 = 0, F_1 = 1, F_i = 3F_{i-1} - F_{i-2}. So, this is a linear recurrence relation similar to the Fibonacci sequence but with a different multiplier. Let me compute the first few terms of F_i to see the pattern:F_0 = 0F_1 = 1F_2 = 3*F_1 - F_0 = 3*1 - 0 = 3F_3 = 3*F_2 - F_1 = 3*3 - 1 = 9 - 1 = 8F_4 = 3*F_3 - F_2 = 3*8 - 3 = 24 - 3 = 21F_5 = 3*F_4 - F_3 = 3*21 - 8 = 63 - 8 = 55F_6 = 3*F_5 - F_4 = 3*55 - 21 = 165 - 21 = 144F_7 = 3*F_6 - F_5 = 3*144 - 55 = 432 - 55 = 377F_8 = 3*F_7 - F_6 = 3*377 - 144 = 1131 - 144 = 987F_9 = 3*F_8 - F_7 = 3*987 - 377 = 2961 - 377 = 2584F_10 = 3*F_9 - F_8 = 3*2584 - 987 = 7752 - 987 = 6765F_11 = 3*F_10 - F_9 = 3*6765 - 2584 = 20295 - 2584 = 17711F_12 = 3*F_11 - F_10 = 3*17711 - 6765 = 53133 - 6765 = 46368Hmm, these numbers are growing quite rapidly. I notice that they resemble Fibonacci numbers but scaled up. In fact, F_i seems to be related to the Fibonacci sequence with even indices. Let me check:The Fibonacci sequence is F_0=0, F_1=1, F_2=1, F_3=2, F_4=3, F_5=5, F_6=8, F_7=13, F_8=21, F_9=34, F_10=55, F_11=89, F_12=144, etc.Comparing with the sequence we have:Our F_0=0, F_1=1, F_2=3, F_3=8, F_4=21, F_5=55, F_6=144, F_7=377, F_8=987, F_9=2584, F_10=6765, F_11=17711, F_12=46368.Yes, these are Fibonacci numbers with even indices multiplied by some factor. Specifically, F_i in our sequence corresponds to the (2i)th Fibonacci number. For example, F_2=3 is F_4 in the standard Fibonacci sequence, F_3=8 is F_6, and so on.So, the sequence defined by F_i = 3F_{i-1} - F_{i-2} is generating Fibonacci numbers with even indices. That's interesting. Now, how does this relate to the problem?The problem mentions that the sizes of any two piles on the table at the same time must be "similar." If "similar" means that their sizes are in a specific ratio, perhaps related to the Fibonacci sequence, then this sequence might help in constructing such piles.The formula a_{i,j} = (-1)^{j+1} j F_i suggests that each element in the table is a product of (-1)^{j+1}, j, and F_i. The (-1)^{j+1} term alternates the sign depending on the column, and j is the column index. So, for each row i, the entries alternate in sign and increase linearly with the column index.The problem also mentions verifying the sum condition:a_{i, j} ≡ a_{i-1, j} + a_{i+1, j} + a_{i, j-1} + a_{i, j+1} (mod P)Substituting the formula into this condition, we get:(-1)^{j+1} j F_i ≡ (-1)^{j+1} j F_{i-1} + (-1)^{j+1} j F_{i+1} + (-1)^j (j-1) F_i + (-1)^{j+2} (j+1) F_i (mod P)Simplifying each term:First term: (-1)^{j+1} j F_iSecond term: (-1)^{j+1} j F_{i-1}Third term: (-1)^{j+1} j F_{i+1}Fourth term: (-1)^j (j-1) F_i = (-1)^{j+1} (j-1) F_iFifth term: (-1)^{j+2} (j+1) F_i = (-1)^{j} (j+1) F_i = (-1)^{j+1} (j+1) F_iSo, combining the fourth and fifth terms:(-1)^{j+1} (j-1) F_i + (-1)^{j+1} (j+1) F_i = (-1)^{j+1} [ (j-1) + (j+1) ] F_i = (-1)^{j+1} (2j) F_iNow, combining all terms:Left-hand side: (-1)^{j+1} j F_iRight-hand side: (-1)^{j+1} j F_{i-1} + (-1)^{j+1} j F_{i+1} + (-1)^{j+1} (2j) F_iFactor out (-1)^{j+1} j:Right-hand side = (-1)^{j+1} j (F_{i-1} + F_{i+1} + 2F_i)But from the recurrence relation, we have F_i = 3F_{i-1} - F_{i-2}, which can be rearranged to F_{i+1} = 3F_i - F_{i-1}So, F_{i-1} + F_{i+1} = F_{i-1} + (3F_i - F_{i-1}) = 3F_iTherefore, the right-hand side becomes:(-1)^{j+1} j (3F_i + 2F_i) = (-1)^{j+1} j (5F_i)Wait, that doesn't seem to match the left-hand side. Did I make a mistake?Let me double-check:From the recurrence, F_{i+1} = 3F_i - F_{i-1}So, F_{i-1} + F_{i+1} = F_{i-1} + 3F_i - F_{i-1} = 3F_iTherefore, F_{i-1} + F_{i+1} + 2F_i = 3F_i + 2F_i = 5F_iSo, the right-hand side is (-1)^{j+1} j * 5F_iBut the left-hand side is (-1)^{j+1} j F_iSo, we have:(-1)^{j+1} j F_i ≡ (-1)^{j+1} j * 5F_i (mod P)Which simplifies to:F_i ≡ 5F_i (mod P)Subtracting F_i from both sides:0 ≡ 4F_i (mod P)So, 4F_i ≡ 0 (mod P)Given that P = 23, which is a prime number, and 4 and 23 are coprime, this implies that F_i ≡ 0 (mod 23)Therefore, for the sum condition to hold, we need F_i ≡ 0 (mod 23) for all i.But wait, looking back at the sequence we computed earlier, F_12 = 46368. Let's compute 46368 mod 23.First, divide 46368 by 23:23 * 2000 = 4600046368 - 46000 = 368Now, 368 divided by 23:23 * 16 = 368So, 46368 = 23 * (2000 + 16) = 23 * 2016Therefore, 46368 ≡ 0 mod 23So, F_12 ≡ 0 mod 23Similarly, let's check F_6 = 144144 divided by 23:23 * 6 = 138144 - 138 = 6So, 144 ≡ 6 mod 23Not zero.But according to our earlier condition, we need F_i ≡ 0 mod 23 for the sum condition to hold. However, F_6 ≡ 6 mod 23, which is not zero.This suggests that my earlier reasoning might be flawed. Maybe the condition is not that F_i ≡ 0 mod 23 for all i, but only for a specific i.Looking back, the problem mentions that P = 23, and we need to show that there exists an arrangement. The hint suggests that a sufficient condition is for F_i ≡ 0 when i = (P + 1)/2, which for P = 23 would be i = 12.Indeed, F_12 = 46368 ≡ 0 mod 23, as we saw earlier.So, perhaps the condition is that F_{(P+1)/2} ≡ 0 mod P. For P = 23, this is F_12 ≡ 0 mod 23, which holds.Therefore, if we construct the table using the formula a_{i,j} = (-1)^{j+1} j F_i, and ensure that F_{(P+1)/2} ≡ 0 mod P, then the sum condition holds, and thus the arrangement is valid.So, to summarize, by defining the sequence F_i with the recurrence F_i = 3F_{i-1} - F_{i-2}, starting with F_0 = 0 and F_1 = 1, and noting that F_{(P+1)/2} ≡ 0 mod P for P = 23, we can construct a table where the sum condition is satisfied. This implies that there exists an arrangement of piles where all piles are "similar" as per the problem's conditions.Therefore, for P = 23, such an arrangement exists.

💡First, determine how many chocolate bars Ron purchased by dividing the total amount spent by the cost per chocolate bar: 15 ÷ 1.50 = 10 chocolate bars.Next, calculate the total number of s'mores by multiplying the number of chocolate bars by the number of s'mores each bar can make: 10 chocolate bars × 3 s'mores per bar = 30 s'mores.Finally, divide the total number of s'mores by the number of scouts to find out how many s'mores each scout gets: 30 s'mores ÷ 15 scouts = 2 s'mores per scout.

💡Alright, so I have this problem about analogical reasoning between real numbers and complex numbers. There are four propositions, and I need to figure out which ones are correct. Let me go through each one step by step.First, let me understand what analogical reasoning means in this context. It seems like it's about taking a property that's true for real numbers and seeing if it's also true for complex numbers. So, for each proposition, I need to check if the statement holds in both R (real numbers) and C (complex numbers).Let's start with proposition ①: "If a, b ∈ R, then a - b = 0 ⇒ a = b." This is obviously true for real numbers because if the difference between two numbers is zero, they must be equal. Now, is this also true for complex numbers? Well, complex numbers have a real part and an imaginary part. If a - b = 0 in complex numbers, that means both the real parts and the imaginary parts of a and b are equal. So yes, a and b must be the same complex number. Therefore, ① seems correct.Moving on to proposition ②: "If a, b ∈ R, then ab = 0 ⇒ a = 0 or b = 0." This is the zero product property for real numbers. It states that if the product of two numbers is zero, at least one of them must be zero. Does this hold for complex numbers? Well, complex numbers can be multiplied similarly, and the zero product property still applies because the product of two complex numbers is zero only if at least one of them is zero. So, ② also seems correct.Now, proposition ③: "If a, b ∈ R, then a - b > 0 ⇒ a > b." This is about the ordering of real numbers. If the difference between two real numbers is positive, then the first number is greater than the second. But in complex numbers, can we say something similar? Hmm, complex numbers don't have a natural ordering like real numbers do. You can't compare two complex numbers in the same way because they have both real and imaginary parts. So, even if a - b is a positive real number, it doesn't necessarily mean that a is greater than b in the complex plane. For example, if a = 1 + i and b = i, then a - b = 1, which is positive, but a and b are complex numbers, and we can't say a > b. Therefore, ③ is incorrect.Finally, proposition ④: "If a, b ∈ R, then a² + b² ≥ 0." For real numbers, squares are always non-negative, so the sum of two squares is also non-negative. But for complex numbers, squaring can lead to negative or even complex results. Let me test this with an example. If a = i and b = i, then a² = (i)² = -1 and b² = -1, so a² + b² = -2, which is less than zero. This contradicts the statement that a² + b² ≥ 0. Therefore, ④ is incorrect.So, summarizing my thoughts:- ① is correct because equality holds in both real and complex numbers.- ② is correct because the zero product property holds in both real and complex numbers.- ③ is incorrect because complex numbers can't be ordered in the same way as real numbers.- ④ is incorrect because the sum of squares can be negative in complex numbers.Therefore, the correct analogical conclusions are ① and ②.**Final Answer**The correct analogical conclusion numbers are boxed{①②}.

💡First, I need to find the average age of the cricket team. Let's denote this average age as ( A ). Since there are 11 team members, the total age of the team is ( 11A ).The captain is 28 years old, and the wicket keeper is 3 years older, making the wicket keeper 31 years old. Together, their combined age is ( 28 + 31 = 59 ) years.If we exclude the captain and the wicket keeper, there are ( 11 - 2 = 9 ) remaining players. The average age of these remaining players is one year less than the average age of the whole team, which is ( A - 1 ). Therefore, the total age of the remaining players is ( 9 times (A - 1) ).The total age of the team can also be expressed as the sum of the ages of the captain, wicket keeper, and the remaining players:[11A = 59 + 9(A - 1)]Expanding and simplifying the equation:[11A = 59 + 9A - 9][11A - 9A = 59 - 9][2A = 50][A = 25]Therefore, the average age of the cricket team is 25 years.