answer:To calculate the amount of water formed, we need to consider the stoichiometry of the reactions. For the main reaction, the stoichiometry is 2 moles of HCl reacting with 1 mole of Ca(OH)2 to produce 2 moles of H2O. Since we have 4 moles of HCl and 2 moles of Ca(OH)2, the reaction will consume all of the Ca(OH)2 and 4 moles of HCl, producing 4 moles of H2O. For the side reaction, the stoichiometry is 1 mole of HCl reacting with 1 mole of NaOH to produce 1 mole of H2O. However, we do not have the initial amount of NaOH given, so we cannot calculate the amount of water produced from this reaction without additional information. Assuming no NaOH is present or the amount of NaOH is not enough to react with all the HCl, the total amount of water produced from the main reaction is 4 moles. To calculate the final temperature and pressure of the resulting mixture, we would need to know the heat of reaction and the specific heat capacities of the reactants and products, as well as the initial temperature and pressure conditions. Since we are not given this information, we cannot calculate the final temperature and pressure accurately. However, we can make some general statements. The reaction between HCl and boxed{Ca(OH)2} is exothermic, so the temperature of the resulting mixture will likely increase. The pressure may also change due to the consumption of gaseous HCl and the formation of liquid water, which has a much lower volume than the gas. Without the specific values for the heat of reaction and the heat capacities, we cannot provide a numerical answer for the final temperature and pressure.

answer:1. **Boundary Lattice Points Configuration**: For a square to have exactly three lattice points on its boundary, it must be positioned such that three of its vertices are lattice points, with the fourth vertex not on a lattice point. 2. **Determining the Size of the Square**: Without loss of generality, let the square be positioned such that its vertices are (0,0), (a,0), (0,a), and (a,a), where a must be an integer. To ensure exactly three lattice points on the boundary, one of the sides must miss the lattice points except at the ends. For simplicity, assume the square is rotated, so a side is diagonal across lattice units. 3. **Diagonal as a Lattice Line**: If the diagonal is from (0,0) to (a,a), the diagonal's length d = asqrt{2}. This diagonal must pass through exactly one additional lattice point besides the endpoints. The smallest integer a for which this occurs is when a=2. This diagonal will pass through (1,1). 4. **Calculate the Area**: The area of the square is a^2. So for a=2, the area is 2^2 = 4. 5. **Conclusion**: Thus, the area of the smallest square that contains exactly three lattice points on its boundary is 4. The final answer is boxed{B}

answer:1. **Recursive Definition and Setup**: Let N be the initial number of coins. For each pirate k, from k=0 to k=14, the k^text{th} pirate will take frac{k+1}{15} of the remaining coins. Consequently, the recursive formula for the coins remaining after each pirate is x_k = frac{15-k-1}{15} x_{k-1} for k geq 0. 2. **Expression for the 15^text{th} Pirate**: Pirate 15 takes whatever is left after the 14th pirate. We need to find x_{14}: [ x_{14} = left(frac{14}{15}right)left(frac{13}{15}right)... left(frac{1}{15}right)N ] Simplifying, we find: [ x_{14} = frac{14!}{15^{14}} N ] 3. **Determining the Smallest N**: To ensure x_{14} is a whole number, N must be a multiple of the denominator of frac{14!}{15^{14}} when reduced to lowest terms. Given: [ 15^{14} = (3 cdot 5)^{14} = 3^{14} cdot 5^{14} ] [ 14! = 14 cdot 13 cdot 12 cdot ... cdot 2 cdot 1 ] Factoring out powers of 3 and 5 from 14!, and finding the remaining factors, we obtain: [ frac{14!}{3^{4} cdot 5^{2}} = 7 cdot 13 cdot 11 cdot 9 cdot 8 cdot 4 cdot 2 = 208080 ] Thus, the smallest N is 208080 for each pirate, including the 15th, to get whole coins. The 15^text{th} pirate receives: [ 208080 ] The final answer is boxed{208080}

answer:1. **Define the sequence and initial conditions:** The sequence ((a_n)_{n in mathbb{N}}) is defined by: [ a_1 = 8, quad a_2 = 18, quad a_{n+2} = a_{n+1}a_n ] 2. **Express (a_n) in terms of prime factors:** Notice that (a_n) can be expressed in terms of its prime factors: [ a_n = 2^{b_n} cdot 3^{c_n} ] where (b_n) and (c_n) are sequences of exponents of 2 and 3, respectively. 3. **Determine the initial values for (b_n) and (c_n):** From the initial conditions: [ a_1 = 8 = 2^3 cdot 3^0 implies b_1 = 3, quad c_1 = 0 ] [ a_2 = 18 = 2^1 cdot 3^2 implies b_2 = 1, quad c_2 = 2 ] 4. **Find the recurrence relations for (b_n) and (c_n):** Using the recurrence relation (a_{n+2} = a_{n+1}a_n), we get: [ a_{n+2} = 2^{b_{n+2}} cdot 3^{c_{n+2}} = (2^{b_{n+1}} cdot 3^{c_{n+1}}) cdot (2^{b_n} cdot 3^{c_n}) = 2^{b_{n+1} + b_n} cdot 3^{c_{n+1} + c_n} ] Therefore, the recurrence relations for (b_n) and (c_n) are: [ b_{n+2} = b_{n+1} + b_n ] [ c_{n+2} = c_{n+1} + c_n ] 5. **Analyze the sequence (c_n):** The sequence (c_n) follows the Fibonacci sequence but shifted: [ c_n = 2F_{n-1} ] where (F_n) is the Fibonacci sequence. Since (F_n) is defined as: [ F_0 = 0, quad F_1 = 1, quad F_{n+2} = F_{n+1} + F_n ] it follows that (c_n) is always even because it is twice a Fibonacci number. 6. **Analyze the sequence (b_n):** The sequence (b_n) can be expressed as: [ b_n = F_n + 2F_{n-2} ] where (F_{-1} = 1) and (F_0 = 0). We need to determine when (b_n) is even. It is known that (F_n) is even if and only if (n) is a multiple of 3. 7. **Determine when (a_n) is a perfect square:** For (a_n) to be a perfect square, both (b_n) and (c_n) must be even. Since (c_n) is always even, we only need to check when (b_n) is even. From the analysis, (b_n) is even if and only if (n) is a multiple of 3. Conclusion: [ a_n text{ is a perfect square if and only if } 3 mid n. ] The final answer is ( boxed{ a_n text{ is a perfect square if and only if } 3 mid n. } )