<s>[INST] Find all primes $p \geq 5$ such that $6^p \cdot (p - 4)! + 10^{3p}$ is divisible by $p$ 
Find all primes $p \geq 5$ such that $6^p \cdot (p - 4)! + 10^{3p}$ is divisible by $p$

I've tried this : First check $(p - 4)!:$
\begin{align*}
(p - 1)! &\equiv -1 \text{(mod p)}\tag{by Wilson's Theorem} \\
(p - 1)(p - 2)(p - 3)(p - 4)! &\equiv -1 \text{(mod p)} \\
(-1)(-2)(-3)(p - 4)! &\equiv -1 \text{(mod p)} \\
6(p - 4)! &\equiv 1
\end{align*}
By Fermat's Little Theorem, $6^{p - 1} \equiv 1$ (mod p), since $5 \nmid 6$ and the next primes are all greater than 6, so no prime $p \geq 5$ can divide 6. Then 
\begin{align*}
6^{p - 1}6(p - 4)! &\equiv 1 \text{(mod p)} \\
6^p(p - 4)! &\equiv 1 \text{(mod p)}
\end{align*}
Now how would I apply Fermat's Little Theorem on $10^{3p}$? I've tried writing it as $(10^p)^3$, but since $5\mid 10$, it doesnt work. Would I discard the case when $p = 5$ to be able to use the theorem?
 [/INST] HINT.-After you have $6^p(p-4)!\equiv 1\pmod p$ you need
$$1+(10^p)^3\equiv1+10^3=1001=7\cdot11\cdot13\equiv0$$ then you have three solutions $7,11$ and $13$.
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