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

Question

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?

Answer 1

As suggested by barry you need to find $p$ such that $p|1000+6.(p-4)!\implies p|1000 (p-1)(p-2)(p-3)+6(p-1)!\implies p| 1000(p-1)(p-2)(p-3)-6\implies p|6006$

can you do the rest? or should i show you further?

Answer 2

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 $13$.