answer:To prove the claim, we will proceed by contradiction. Let's assume the opposite of what we want to prove, namely, that n+1 is a composite number. 1. Assume ( n+1 ) is a composite number. - By definition, a composite number has at least one prime factor other than 1 and itself. 2. Let ( p ) be a prime factor of ( n+1 ). This means ( p leq n ) because if ( p ) were greater than ( n ), then ( p ) cannot be a factor of a number that is at most ( n+1 ) and thus violating the property of ( p ) being a factor of ( n+1 ). 3. Since ( p leq n ), ( p ) must be a divisor of ( n! ), because ( n! ) is the product of all positive integers up to ( n ). Hence, we have: [ n! equiv 0 pmod{p} ] 4. Now, consider ( n! + 1 ). For ( n! + 1 ) to be divisible by ( n+1 ), it should also be divisible by ( p ). However, since ( n! equiv 0 pmod{p} ), it follows that: [ n! + 1 equiv 1 pmod{p} ] This means ( n! + 1 equiv 1 pmod{p} ), so ( n! + 1 ) cannot be divisible by ( p ). 5. This leads to a contradiction because if ( p ) is a divisor of ( n+1 ) but ( p ) does not divide ( n! + 1 ), then ( n! + 1 ) cannot be divisible by ( n+1 ). 6. Given this contradiction, our initial assumption that ( n+1 ) is a composite number must be false. Therefore, ( n+1 ) must be a prime number. Conclusion: [ boxed{n+1 text{ is a prime number.}} ]

answer:To solve this problem, we should first simplify the function using trigonometric identities and then analyze the monotonicity of the resulting function, which will be a transformed sine function. We can express f(x) as a linear combination of sine and cosine functions: f(x) = sin x - sqrt{3}cos x. Now, we can use the sine sum identity sin(A+B) = sin A cos B + cos A sin B to rewrite f(x) as a shifted sine function: f(x) = 2left( frac{1}{2}sin x - frac{sqrt{3}}{2}cos x right) = 2sinleft(x - frac{pi}{3}right). Since x in [-pi, 0], the argument of the sine function x - frac{pi}{3} will be in the range [-frac{4pi}{3}, -frac{pi}{3}]. The sine function is increasing in the interval [-frac{pi}{2}, frac{pi}{2}]. To find when 2sinleft(x - frac{pi}{3}right) is increasing within the given domain [-pi, 0], we need to locate the part of [-frac{4pi}{3}, -frac{pi}{3}] that lies within [-frac{pi}{2}, frac{pi}{2}]. We find that -frac{pi}{2} leq x - frac{pi}{3} leq frac{pi}{2} corresponds to -frac{pi}{6} leq x leq frac{2pi}{3}. This interval, however, must be restricted to the original domain [-pi, 0], leading to the intersection [-frac{pi}{6}, 0] as the domain where f(x) is increasing. Thus, the monotonic increasing interval of f(x) is boxed{[-frac{pi}{6}, 0]}, which corresponds to option C.

answer:Assume n^2 + 5n + 13 = (n + k)^2, thus (n + k)^2 = n^2 + 2nk + k^2, so 2nk + k^2 = 5n + 13, hence n = frac{k^2 - 13}{5 - 2k}, If n is a natural number, it should be no less than 0 (as seen from the equation, it cannot be 0), therefore, ① k^2 > 13 and 5 > 2k (does not exist); ② k^2 < 13 and 5 < 2k, the only possibility is k = 3, At this point, n = 4. Hence, the answer is: boxed{4}.

answer:First, let's calculate the discount for each lens and the final price Mark would pay for each. For the 300 lens with a 20% discount: Discount = 20% of 300 = 0.20 * 300 = 60 Final price for 300 lens = 300 - 60 = 240 For the 220 lens with a 10% discount: Discount = 10% of 220 = 0.10 * 220 = 22 Final price for 220 lens = 220 - 22 = 198 Now, let's calculate how much money Mark saves by buying the cheaper lens after applying the discounts: Savings = Final price for 300 lens - Final price for 220 lens Savings = 240 - 198 = 42 Mark saves boxed{42} by buying the cheaper lens after applying the discounts.