Ms. Math's kindergarten class has $16$ registered students. The classroom has a very large number, $N$, of play blocks which satisfies the conditions:
(a) 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
(b) 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 some multiple of $\text{lcm}(14, 15, 16)= 2^{4}\cdot 3\cdot 5\cdot 7$ ; this $lcm$ is hereby denoted $k$ and $N = qk$.
$1$, $2$, $3$, $4$, $5$, $6$, $7$, $8$, $10$, and $12$ all divide $k$, so $x, y, z = 9, 11, 13$
We have the following three modulo equations:
$nk\equiv 3\pmod{9}$
$nk\equiv 3\pmod{11}$
$nk\equiv 3\pmod{13}$
To solve the equations, you can notice the answer must be of the form $9\cdot 11\cdot 13\cdot m + 3$ where $m$ is an integer. 
This must be divisible by $lcm$ $(14, 15, 16)$, which is $560\cdot 3$. 
Therefore, $\frac{9\cdot 11\cdot 13\cdot m + 3}{560\cdot 3} = q$, which is an integer. Factor out $3$ and divide to get $\frac{429m+1}{560} = q$. 
Therefore, $429m+1=560q$. We can use [Bezout's Identity](https://artofproblemsolving.com/wiki/index.php/Bezout%27s_Lemma) or a [Euclidean algorithm](https://artofproblemsolving.com/wiki/index.php/Euclidean_algorithm) bash to solve for the least of $m$ and $q$. 
We find that the least $m$ is $171$ and the least $q$ is $131$. 
Since we want to factor $1680\cdot q$, don't multiply: we already know that the prime factors of $1680$ are $2$, $3$, $5$, and $7$, and since $131$ is prime, we have $2 + 3 + 5 + 7 + 131 = \boxed{148}$.