[Algebra]
Problem: The arithmetic sequences $a_{1}, a_{2}, a_{3}, \ldots, a_{20}$ and $b_{1}, b_{2}, b_{3}, \ldots, b_{20}$ consist of 40 distinct positive integers, and $a_{20}+b_{14}=1000$. Compute the least possible value for $b_{20}+a_{14}$.

Solution: Write $a_{n}=a_{1}+r(n-1)$ and $b_{n}=b_{1}+s(n-1)$. Then $a_{20}+b_{14}=a_{1}+b_{1}+19 r+13 s$, while $b_{20}+a_{14}=a_{1}+b_{1}+13 r+19 s=a_{20}+b_{14}+6(s-r)$. Because both sequences consist only of integers, $r$ and $s$ must be integers, so $b_{20}+a_{14} \equiv a_{20}+b_{14} \bmod 6$. Thus the least possible value of $b_{20}+a_{14}$ is 4 . If $b_{20}=3$ and $a_{14}=1$, then $\left\{a_{n}\right\}$ must be a decreasing sequence (else $a_{13}$ would not be positive) and $a_{20} \leq-5$, which is impossible. The case $b_{20}=a_{14}=2$ violates the requirement that the terms be distinct, and by reasoning analogous to the first case, $b_{20}=1, a_{14}=3$ is also impossible. Hence the sum $b_{20}+a_{14}$ is at least 10 . To show that 10 is attainable, make $\left\{a_{n}\right\}$ decreasing and $b_{20}$ as small as possible: set $b_{20}=1, a_{14}=9$, and $a_{n}=23-n$. Then $a_{20}=3$, yielding $b_{14}=997$. Hence $s=\frac{997-1}{14-20}=\frac{996}{-6}=-166$ and



$b_{1}=997-(13)(-166)=3155$, yielding $b_{n}=3155-166(n-1)$. Because $b_{20}=1 \leq a_{20}$ and $b_{19}=167 \geq a_{1}$, the sequences $\left\{b_{n}\right\}$ and $\left\{a_{n}\right\}$ are distinct for $1 \leq n \leq 20$, completing the proof. Hence the minimum possible value of $b_{20}+a_{14}$ is 10.

Final answer: 10