4. For which natural numbers $n$ does the equality $\operatorname{GCD}(6, n)+\operatorname{GCD}(8,2 n)=10$ hold?
4. Solution. Note that the possible values of $GCD(6, n)$ are $1,2,3,6$, and the possible values of $GCD(8, n)$ are $1,2,4,8$. Their sum, equal to 10, is possible in two cases.

Case 1. $\left\{\begin{array}{l}G C D(8,2 n)=8 \\ G C D(6, n)=2\end{array}\right.$. From the first equation, we get $n=4 k$, and from the second $n=2(3 m+1)$ or $n=2(3 m+2)$. Both conditions are satisfied by numbers of the form $n=12 t+4, \quad n=12 t+8, \quad t \in Z$.

Case 2. $\left\{\begin{array}{l}G C D(8,2 n)=4 \\ G C D(6, n)=6\end{array}\right.$. From the first equation, we get $n=2(2 k+1)$, and from the second $-n=6 m$. Both conditions are satisfied by numbers of the form $n=12 t+6, \quad t \in Z$.

Thus, the equality $GCD(6, n)+GCD(8,2 n)=10$ holds for $n=12 t+4, n=12 t+6, n=12 t+8, t \geq 0, t \in Z$.