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For any natural numbers $m, n$, and $k$, such that $k$ divides both $m+2 n$ and $3 m$ $+4 n, k$ must be a common divisor of

A school has less than $5000$ students and if the students are divided equally into teams of either $9$ or $10$ or $12$ or $25$ each, exactly $4$ are always left out. However, if they are divided into teams of $11$ each, no one is left out. The maximum number of teams of $12$ each that can be formed out of the students in the school is

Let A be the largest positive integer that divides all the numbers of the form $3^k + 4^k + 5^k$, and B be the largest positive integer that divides all the numbers of the form $4^k + 3(4^k) + 4^{k + 2}$, where k is any positive integer. Then (A + B) equals