Ms. Math's kindergarten class has $16$ registered students. The classroom has a very large number, $N$, of play blocks which satisfies the conditions:
- If $16$, $15$, or $14$ students are present in the class, then in each case all the blocks can be distributed in equal numbers to each student, and
- There are three integers $0 < x < y < z < 14$ such that when $x$, $y$, or $z$ students are present and the blocks are distributed in equal numbers to each student, there are exactly three blocks left over.
Find the sum of the distinct prime divisors of the least possible value of $N$ satisfying the above conditions.
$N$ must be a common multiple of $14$, $15$, and $16$. Let $k$ be their least common multiple, i.e. $k = 2^{4} \cdot 3 \cdot 5 \cdot 7$. Then, $N = nk$ where $n$ is a positive integer.
Because $1$, $2$, $3$, $4$, $5$, $6$, $7$, $8$, $10$, and $12$ all divide $k$, so $x, y, z = 9, 11, 13$. Then we have the following three modulo equations: $$\left\{\begin{array} {rl} nk&\equiv 3 \pmod{9}\\ nk&\equiv 3 \pmod{11}\\ nk&\equiv 3 \pmod{13} \end{array}\right.$$
This means that $nk$ must be three more than a common multiple of $9$, $11$, and $13$, or $nk=9\times 11\times 13 \times p + 3$ where $p$ is an integer. Meanwhile, it must be divisible by $k$. Therefore there exists an integer $q$ such that $$\begin{array}{rcl} 9\times 11\times 13\times p + 3 &=& (2^4\times 3\times 5\times 7 \times 1)\times q \\ 429\times p + 1 &=& 560\times q \end{array}$$
This is a classic linear indeterminate equation. The least positive integer solution is $(p, q)=(171, 131)$. Setting $p=171$ into $(9\times 11\times 13\times p + 3)$ and factorizing find the distinct prime divisors are $2$, $3$, $5$, $7$, and $131$. Their sum is $\boxed{148}$.