answer:We use the Euclidean Algorithm. begin{align*} text{gcd},(3a^3+a^2+4a+57,a) &=text{gcd},(3a^3+a^2+4a+57-(3a^2+a+4)a,a) &=text{gcd},(57,a). end{align*}Since 57 is a divisor of 456, and a is a multiple of 456, the greatest common divisor is boxed{57}.

answer:Writing out the recursive statement for a_n, a_{n-1}, dots, a_{10} and summing them gives[a_n+dots+a_{10}=100(a_{n-1}+dots+a_{10})+n+dots+10]Which simplifies to[a_n=99(a_{n-1}+dots+a_{10})+frac{1}{2}(n+10)(n-9)]Therefore, a_n is divisible by 99 if and only if frac{1}{2}(n+10)(n-9) is divisible by 99, so (n+10)(n-9) needs to be divisible by 9 and 11. Assume that n+10 is a multiple of 11. Writing out a few terms, n=12, 23, 34, 45, we see that n=45 is the smallest n that works in this case. Next, assume that n-9 is a multiple of 11. Writing out a few terms, n=20, 31, 42, 53, we see that n=53 is the smallest n that works in this case. The smallest n is boxed{45}. Note that we can also construct the solution using CRT by assuming either 11 divides n+10 and 9 divides n-9, or 9 divides n+10 and 11 divides n-9, and taking the smaller solution.

answer:Note firstly that f(i) must be an integer, so this means that i must be a perfect square in order for sqrt{i} to be an integer. Out of the perfect squares, we claim that i must be the square of some prime p. For if sqrt{i} is composite, then it can be written as the product of two integers a and b and we find f(i) ge 1 + sqrt{i} + i + a + b > 1 + sqrt{i} + i. Moreover, if sqrt{i} is prime, then the only factors of i are 1, sqrt{i}, and i, so f(i) = 1 + sqrt{i} + i as desired. It follows that we only need to calculate the number of primes less than sqrt{2010}. Since sqrt{2010} < 45, the desired set of primes is {2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31, 37, 41, 43}. The set has boxed{14} elements.

answer:Let our answer be n. Write n = 42a + b, where a, b are positive integers and 0 leq b < 42. Then note that b, b + 42, ... , b + 42(a-1) are all primes. If b is 0mod{5}, then b = 5 because 5 is the only prime divisible by 5. We get n = 215 as our largest possibility in this case. If b is 1mod{5}, then b + 2 times 42 is divisible by 5 and thus a leq 2. Thus, n leq 3 times 42 = 126 < 215. If b is 2mod{5}, then b + 4 times 42 is divisible by 5 and thus a leq 4. Thus, n leq 5 times 42 = 210 < 215. If b is 3mod{5}, then b + 1 times 42 is divisible by 5 and thus a = 1. Thus, n leq 2 times 42 = 84 < 215. If b is 4mod{5}, then b + 3 times 42 is divisible by 5 and thus a leq 3. Thus, n leq 4 times 42 = 168 < 215. Our answer is boxed{215}.