Problem (Algebra):
Let $T=9$. The sequence $a_{1}, a_{2}, a_{3}, \ldots$ is an arithmetic progression, $d$ is the common difference, $a_{T}=10$, and $a_{K}=2010$, where $K>T$. If $d$ is an integer, compute the value of $K$ such that $|K-d|$ is minimal.

Solution:
Note that $a_{T}=a_{1}+(T-1) d$ and $a_{K}=a_{1}+(K-1) d$, hence $a_{K}-a_{T}=(K-T) d=2010-10=$ 2000. Thus $K=\frac{2000}{d}+T$, and to minimize $\left|T+\frac{2000}{d}-d\right|$, choose a positive integer $d$ such that $\frac{2000}{d}$ is also an integer and $\frac{2000}{d}-d$ is as close as possible to $-T$. Note that $T>0$, so $\frac{2000}{d}-d$ should be negative, i.e., $d^{2}>2000$ or $d>44$. The value of $T$ determines how far apart $\frac{2000}{d}$ and $d$ need to be. For example, if $T$ is close to zero, then choose $d$ such that $\frac{2000}{d}$ and $d$ are close to each other. With $T=9$, take $d=50$ so that $\frac{2000}{d}=40$ and $|K-d|=|49-50|=1$. Thus $K=49$.

Final answer: 49