Given the partition function $p(n)$, Ramanujan found,
$$\begin{align} p(5k+4) & \equiv 0 \pmod 5 \\ p(7k+5) & \equiv 0 \pmod 7 \\ p(11k+6) & \equiv 0 \pmod {11} \end{align}$$
It was later proved in 2000 by Ken Ono that such congruences exist modulo every integer $m$ coprime to $6$. Some examples from the link above,
$$p(11^3 \cdot 13k + 237)\equiv 0 \pmod {13}$$
$$p(107^4\cdot 31k + 30064597)\equiv 0\pmod{31}$$
$$p(999959^4\cdot29k+ 28995221336976431135321047) \equiv 0 \pmod{29}$$
Q: Assume a Ramanujan congruence $$p(Ak+B) \equiv 0 \pmod m$$ with $A>11$. Must the factorization of $A$ involve an integer with a power $> 1$?
P.S. Is there a smaller example for $m=29$?