Problem (Number Theory):
Let $T=\frac{8}{5}$. Let $z_{1}=15+5 i$ and $z_{2}=1+K i$. Compute the smallest positive integral value of $K$ such that $\left|z_{1}-z_{2}\right| \geq 15 T$.

Solution:
Note that $z_{1}-z_{2}=14+(5-K) i$, hence $\left|z_{1}-z_{2}\right|=\sqrt{14^{2}+(5-K)^{2}}$. With $T=8 / 5,15 T=24$, hence $14^{2}+(5-K)^{2} \geq 24^{2}$. Thus $|5-K| \geq \sqrt{24^{2}-14^{2}}=\sqrt{380}$. Because $K$ is a positive integer, it follows that $K-5 \geq 20$, hence the desired value of $K$ is $\mathbf{2 5}$.

Final answer: 25