17 If $2^{n+2} \cdot 3^{n}+5 n-a$ can be divisible by 25, then the smallest positive value of $a$ is
17 4. $\quad 4 \times 6^{n}+5 n-a=4\left(5^{n}+C_{n}^{1} \cdot 5^{n-1}+\cdots+C_{n}^{n-1} \times 5+1\right)+5 n-a$, so $25 \mid 4-a, a=4$