[Number Theory]
Problem: Let $T=98721$, and let $K$ be the sum of the digits of $T$. Let $A_{n}$ be the number of ways to tile a $1 \times n$ rectangle using $1 \times 3$ and $1 \times 1$ tiles that do not overlap. Tiles of both types need not be used; for example, $A_{3}=2$ because a $1 \times 3$ rectangle can be tiled with three $1 \times 1$ tiles or one $1 \times 3$ tile. Compute the smallest value of $n$ such that $A_{n} \geq K$.

Solution: Consider the rightmost tile of the rectangle. If it's a $1 \times 1$ tile, then there are $A_{n-1}$ ways to tile the remaining $1 \times(n-1)$ rectangle, and if it's a $1 \times 3$ tile, then there are $A_{n-3}$ ways to tile the remaining $1 \times(n-3)$ rectangle. Hence $A_{n}=A_{n-1}+A_{n-3}$ for $n>3$, and $A_{1}=A_{2}=1, A_{3}=2$. Continuing the sequence gives the following values:

| $n$ | 1 | 2 | 3 | 4 | 5 | 6 | 7 | 8 | 9 | 10 |
| :---: | :---: | :---: | :---: | :---: | :---: | :---: | :---: | :---: | :---: | :---: |
| $A_{n}$ | 1 | 1 | 2 | 3 | 4 | 6 | 9 | 13 | 19 | 28 |

With $T=98721, K=27$, hence the answer is 10 .

Final answer: 10