## Task $5 / 89$

Determine all pairs $(n ; k)$ with $n ; k \in N$, that satisfy the equation $n!-56 k+10^{n}=0$!
Due to $56 k \equiv 0(\bmod 7)$, it must be that $10^{n} \equiv -n!(\bmod 7)$. Since $10^{n} \neq 0(\bmod 7)$, it must also be that $n! \neq 0(\bmod 7)$, so $n < 7$.

Checking the numbers $n=0 ; 1 ; 2 ; 3 ; 4 ; 5 ; 6$ with respect to the congruence $10^{n} \equiv -n!(\bmod 7)$, one finds that only $n_{1}=4$ and $n_{2}=6$ satisfy it.

From the given equation, the corresponding values for $k$ are obtained. Thus, there are exactly two pairs $(n ; k):\left(n_{1} ; k 1\right)=(4 ; 179),\left(n_{2} ; k_{2}\right)=(6 ; 17870)$.