answer:We are given the system of equations: 1. ( x + y^2 + z^3 = 3 ) 2. ( y + z^2 + x^3 = 3 ) 3. ( z + x^2 + y^3 = 3 ) We need to find solutions for ( x, y, ) and ( z ) in positive real numbers. Step-by-Step Solution: 1. **Assumption and Comparison:** Assume ( x ge 1 ) and ( y ge 1 ). Notice the inequality: [ x + y^2 le x^2 + y^3 qquad text{(since } x, y ge 1 text{ and equality holds when } x = y = 1 text{)} ] 2. **Substitution and Simplification:** From the first equation, we have: [ z^3 = 3 - (x + y^2) ] Given ( x ge 1 ) and ( y ge 1 ), let's substitute to check the inequality: [ z^3 = 3 - (x + y^2) le 3 - (x^2 + y^3) ] 3. **Comparing and Deducing Values:** [ z^3 le 3 - (x^2 + y^3) = z implies z^3 le z ] Since ( z ) is a positive real number, ( z ge 1 ). We now have ( x, y, z ge 1 ). [ z le 1 qquad text{(since } z^3 le z text{ implies } 0 le z le 1 text{)} ] 4. **Conclusion of Comparison:** Combining the above results, ( x = y = z = 1 ) must hold true. If any one of ( x, y, ) or ( z ) are greater than 1, the inequalities lead to a contradiction, as shown. Similarly, if ( x, y, le 1 ) are assumed, then all must equal 1 due to the structure of the equations. 5. **Consistency Check:** Check ( x = y = z = 1 ): [ x + y^2 + z^3 = 1 + 1^2 + 1^3 = 1 + 1 + 1 = 3 ] [ y + z^2 + x^3 = 1 + 1^2 + 1^3 = 1 + 1 + 1 = 3 ] [ z + x^2 + y^3 = 1 + 1^2 + 1^3 = 1 + 1 + 1 = 3 ] The equations hold true when ( x = y = z = 1 ). # Conclusion: Thus, the only solution in positive real numbers to the system of equations is: [ boxed{x = y = z = 1} ]

answer:To find the gain in the transaction per year, we need to calculate the amount the person owes after 2 years at 4% p.a. compound interest compounded quarterly, and the amount he earns from lending it at 6% p.a. compound interest compounded semi-annually. Then we'll subtract the owed amount from the earned amount and divide by 2 to find the annual gain. First, let's calculate the amount owed after 2 years at 4% p.a. compounded quarterly. The formula for compound interest is: A = P(1 + r/n)^(nt) Where: A = the future value of the investment/loan, including interest P = the principal investment amount (the initial deposit or loan amount) r = the annual interest rate (decimal) n = the number of times that interest is compounded per year t = the number of years the money is invested or borrowed for For the borrowed amount: P = Rs. 8000 r = 4% p.a. = 0.04 per annum n = 4 (since the interest is compounded quarterly) t = 2 years A = 8000(1 + 0.04/4)^(4*2) A = 8000(1 + 0.01)^(8) A = 8000(1.01)^8 A = 8000 * 1.082856 A ≈ Rs. 8662.85 Now, let's calculate the amount earned after 2 years at 6% p.a. compounded semi-annually. For the lent amount: P = Rs. 8000 r = 6% p.a. = 0.06 per annum n = 2 (since the interest is compounded semi-annually) t = 2 years A = 8000(1 + 0.06/2)^(2*2) A = 8000(1 + 0.03)^(4) A = 8000(1.03)^4 A = 8000 * 1.125509 A ≈ Rs. 9004.07 Now, let's find the gain over the 2 years: Gain = Amount earned - Amount owed Gain = Rs. 9004.07 - Rs. 8662.85 Gain ≈ Rs. 341.22 To find the gain per year, we divide the total gain by 2: Gain per year = Rs. 341.22 / 2 Gain per year ≈ Rs. 170.61 So, the person's gain in the transaction per year is approximately Rs. boxed{170.61} .

answer:We are asked to show that for any sequence of positive reals, a_n, we have: [ limsup_{n to infty} n left(frac{a_{n+1} + 1}{a_n} - 1right) geq 1 ] Additionally, we need to find a sequence where the equality holds. Part 1: Showing the Inequality 1. **Assume for contradiction that limsup_{n to infty} n left(frac{a_{n+1} + 1}{a_n} - 1right) < 1.** This means there exists some constant c < 1 such that for all sufficiently large n, we have: [ n left(frac{a_{n+1} + 1}{a_n} - 1right) < c ] 2. **Rewriting the inequality:** For n geq N where N is sufficiently large, we can write: [ n left(frac{a_{n+1} + 1}{a_n} - 1right) < c ] Dividing both sides by n yields: [ frac{a_{n+1} + 1}{a_n} - 1 < frac{c}{n} ] Adding 1 to both sides, we get: [ frac{a_{n+1} + 1}{a_n} < 1 + frac{c}{n} ] 3. **Isolating a_{n+1}:** [ a_{n+1} + 1 < a_n left(1 + frac{c}{n}right) ] [ a_{n+1} < a_n left(1 + frac{c}{n}right) - 1 ] 4. **Applying this recursively:** By continuing this reasoning, for all k geq N: [ a_{N+k} < a_N prod_{j=0}^{k-1} left(1 + frac{c}{N+j}right) - k ] 5. **Estimate the product prod_{j=0}^{k-1} left(1 + frac{c}{N+j}right):** [ prod_{j=0}^{k-1} left(1 + frac{c}{N+j}right) approx expleft(sum_{j=0}^{k-1} frac{c}{N+j}right) ] Since sum_{j=0}^{k-1} frac{1}{N+j} is approximately logfrac{N+k}{N}: [ sum_{j=0}^{k-1} frac{c}{N+j} approx c log frac{N+k}{N} ] [ prod_{j=0}^{k-1} left(1 + frac{c}{N+j}right) approx expleft(c log frac{N+k}{N}right) = left(frac{N+k}{N}right)^c ] 6. **Combining the results:** [ a_{N+k} < a_N left(frac{N+k}{N}right)^c - k ] As k to infty, the term left(frac{N+k}{N}right)^c grows less than (1+k/N)^c, but the subtracted k term increases, potentially making a_{N+k} negative, which is impossible since a_n are positive reals. 7. **Conclusion: The assumption limsup_{n to infty} n left(frac{a_{n+1} + 1}{a_n} - 1right) < 1 leads to a contradiction. Therefore,** [ boxed{limsup_{n to infty} n left(frac{a_{n+1} + 1}{a_n} - 1right) geq 1} ] Part 2: Finding a Sequence Where Equality Holds 1. **Consider the sequence a_n = n log n.** 2. **Calculate the term n left(frac{a_{n+1} + 1}{a_n} - 1right) for a_n = n log n:** [ a_{n+1} = (n+1) log(n+1) ] [ frac{a_{n+1} + 1}{a_n} = frac{(n+1) log(n+1) + 1}{n log n} ] 3. **Simplifying:** [ n left( frac{(n+1) log(n+1) + 1}{n log n} - 1 right) ] [ = n left( frac{(n+1) log(n+1)}{n log n} + frac{1}{n log n} - 1 right) ] [ = n left( frac{n+1}{n} frac{log(n+1)}{log n} + frac{1}{n log n} - 1 right) ] [ = n left( frac{n+1}{n} frac{log(n+1)}{log n} - 1 + frac{1}{n log n} right) ] 4. **As n to infty:** We know log(n+1) approx log n + frac{1}{n} for large n, thus: [ frac{log(n+1)}{log n} approx 1 + frac{1}{n log n} ] [ frac{n+1}{n} = 1 + frac{1}{n} ] Substituting these approximations: [ n left( left(1 + frac{1}{n}right) left(1 + frac{1}{n log n}right) - 1 + frac{1}{n log n} right) ] Ignoring higher-order small terms: [ approx n left( 1 + frac{1}{n} + frac{1}{n log n} - 1 + frac{1}{n log n} right) ] [ = n left( frac{1}{n} + frac{2}{n log n} right) ] [ = 1 + frac{2}{log n} ] 5. **Taking n to infty:** We see as n to infty, frac{2}{log n} to 0, leading us to: [ lim_{n to infty} n left(frac{a_{n+1} + 1}{a_n} - 1right) = 1 ] 6. **Conclusion: This verifies that with a_n = n log n, equality holds.** [ boxed{limsup_{n to infty} n left(frac{a_{n+1} + 1}{a_n} - 1right) = 1} ]

answer:1. **Define the problem:** - We have small square rods that can be arranged in three different formations. - Formation 1: 1 times k (one row) - Formation 2: 2 times m (two rows) - Formation 3: n times n (square) - All formations should strictly use the same number of rods with no remainder. - We need to find the smallest number greater than 1000 that satisfies these conditions. 2. **Formulate the mathematical constraints:** - We need a number that satisfies: [ 3k+1 = 5m+2 = 2n(n+1) ] - This implies: - ( text{Number} equiv 1 pmod{3} ) - ( text{Number} equiv 2 pmod{5} ) - ( text{Number} = 2n(n+1) ) 3. **Analyze the properties:** - ( 2n(n+1) ) shows this number must be even, since it’s twice the product of two consecutive integers. - Must be a multiple of 4. 4. **Test possibilities greater than 1000:** - We need to find the smallest 2n(n+1) greater than 1000 that matches the given modulo constraints. - We’ll check numbers in sequence: **Step 1:** - Start trying values for 2n(n+1). **Step 2:** - Begin with n = 22: [ text{Number} = 2 times 22 times 23 = 2 times 506 = 1012 ] **Check the values:** - Check modulo 3: [ 1012 div 3 implies 1012 equiv 1 pmod{3} ] As 1012 mod 3 = 1, it matches the requirement (3k+1). - Check modulo 5: [ 1012 div 5 implies 1012 equiv 2 pmod{5} ] As 1012 mod 5 = 2, it matches the requirement (5m+2). 5. **Conclusion:** Since 1012 meets all conditions: [ 1012 = 2(22)(23) ] And satisfies: [ 1012 equiv 1 pmod{3} ] [ 1012 equiv 2 pmod{5} ] The smallest number greater than 1000 that can be arranged in all three specified formations is: [ boxed{1012} ]