A number theory problem by Swadesh Rath

The function $f(n)$ can be expressed as;

$f(n) = n! + (n+1)! + (n+ 2)! = n! +(n+1)n! + (n+2)(n+1)n! ;$

or $[1 + (n+1) + (n+2)(n+1)]n! = (n^2 + 4n + 4)n! = (n+2)^2 \cdot n!$

If we are interested in $n \in {0, 1,2 ,3,..., 16}$ such that $49 | f(n)$ , then we obtain:

$\frac{f(n)}{49} = \frac{(n+2)^2 \cdot n!}{7^2} = (\frac{n+2}{7})^2 \cdot n!$

which is an integer iff $n = 5, 12, 14, 15, 16$ . The final sum is just $5 + 12 + 14 + 15 + 16 = \boxed{62}.$