Problem (Algebra):
Let $T=5$. Suppose that $a_{1}=1$, and that for all positive integers $n, a_{n+1}=$ $\left\lceil\sqrt{a_{n}^{2}+34}\right\rceil$. Compute the least value of $n$ such that $a_{n}>100 T$.

Solution:
Start by computing the first few terms of the sequence: $a_{1}=1, a_{2}=\lceil\sqrt{35}\rceil=6, a_{3}=$ $\lceil\sqrt{70}\rceil=9$, and $a_{4}=\lceil\sqrt{115}\rceil=11$. Note that when $m \geq 17,(m+1)^{2}=m^{2}+2 m+1>$ $m^{2}+34$, so if $a_{n} \geq 17, a_{n+1}=\left[\sqrt{a_{n}^{2}+34}\right\rceil=a_{n}+1$. So it remains to continue the sequence until $a_{n} \geq 17: a_{5}=13, a_{6}=15, a_{7}=17$. Then for $n>7, a_{n}=17+(n-7)=n+10$, and $a_{n}>100 T \Rightarrow n>100 T-10$. With $T=5, n>490$, and the least value of $n$ is 491 .

Final answer: 491