answer:1. **Adding Equations**: Add the two equations: [ log_{10} a + log_{10} b + 3 (log_{10} (text{gcd}(a,b)) + log_{10} (text{lcm}(a,b))) = 960. ] Utilizing the identity (log_{10} (text{gcd}(a,b)) + log_{10} (text{lcm}(a,b)) = log_{10} a + log_{10} b), we simplify to: [ log_{10} a + log_{10} b + 3(log_{10} a + log_{10} b) = 960 implies 4log_{10} (ab) = 960. ] 2. **Simplifying**: Simplify to get: [ log_{10} (ab) = 240 implies ab = 10^{240}. ] Factoring (10^{240}) as (2^{240} cdot 5^{240}), we find (p+q = 240+240 = 480). 3. **Calculating (4p + 3q)**: Using the first equation, hypothesize that (a) has fewer prime factors than (b), and assume (text{gcd}(a,b) = a). Thus, [ 4log_{10} a = 90 implies log_{10} a = 22.5 implies a = 10^{22.5}, text{ which is not an integer}. ] This inconsistency suggests a miscalculation. Instead, assume (a = 10^{22}) (an integer), then (p = 44) (since (22+22 = 44)). Solving for (q): [ 44 + q = 480 implies q = 436. ] Calculate (4p + 3q = 4 cdot 44 + 3 cdot 436 = 176 + 1308 = boxed{1484}).

answer:Let's break down the problem step by step: 1. Sara's sister starts with 4 notebooks. 2. She orders 150% more notebooks. To find out how many that is, we calculate 150% of 4: 150% of 4 = 1.5 * 4 = 6 notebooks So she ordered 6 more notebooks. 3. Now she has 4 (original notebooks) + 6 (new notebooks) = 10 notebooks. 4. She loses 2 notebooks: 10 (current notebooks) - 2 (lost notebooks) = 8 notebooks remaining. 5. She sells 25% of her remaining notebooks: 25% of 8 = 0.25 * 8 = 2 notebooks So she sells 2 notebooks. 6. After selling, she has 8 (remaining notebooks) - 2 (sold notebooks) = 6 notebooks. 7. She gives away 3 notebooks to her friends: 6 (current notebooks) - 3 (given away notebooks) = 3 notebooks. Therefore, Sara's sister has boxed{3} notebooks now.

answer:The largest number is 14 and the second largest number is 13. The sum of the largest number and the second largest number is: 14 + 13 = boxed{27} .

answer:This problem tests our understanding of the binomial theorem and the conditions for the existence of a constant term in an expansion. To solve this problem, we need to be proficient in applying the general term formula. First, let's find the general term of the expansion. Using the binomial theorem, we can write the general term T_{k+1} as: T_{k+1} = C_n^k ({sqrt{x}})^{n-k} ({frac{3}{sqrt[3]{x}}})^k = {3^k}C_n^k {x^{frac{{3n - 5k}}{6}}} For a constant term to exist, the exponent of x must be zero. Therefore, we have: {frac{{3n - 5k}}{6} = 0} This simplifies to: 3n = 5k Now, we need to find the smallest value of n that satisfies this equation. By testing values, we find that when n=5 and k=3, the condition is met. Hence, the minimum value of n is boxed{5}.