Question:
John counts up from 1 to 13, and then immediately counts down again to 1, and then back up to 13, and so on, alternately counting up and down: \begin{align*}
&(1, 2, 3,4,5,6,7,8,9,10,11,12,13,\\
&\qquad\qquad12,11,10,9,8,7,6,5,4,3,2,1,2,3,4,\ldots ).
\end{align*} What is the $5000^{\text{th}}$ integer in his list?

Answer:
We can treat this list as a sequence with a repetitive pattern. We see the sequence repeats itself every 24 elements (from 1 to 13 then back to 2). When 5000 is divided by 24, its remainder is 8. Therefore we see the $5000^{\text{th}}$ integer is the same as the $8^{\text{th}}$ integer, which is $\boxed{8}$.