Problem 115

Let $a_1$ be the first term of the arithmetic progression and $a_{12}$ be the last term of the arithmetic progression. From the formula of the sum of an arithmetic progression (or arithmetic series), we have $12*\frac{a_1+a_{12}}{2}=360$, which leads us to $a_1 + a_{12} = 60$. $a_{12}$, the largest term of the progression, can also be expressed as $a_1+11d$, where $d$ is the common difference. Since each angle measure must be an integer, $d$ must also be an integer. We can isolate $d$ by subtracting $a_1$ from $a_{12}$ like so: $a_{12}-a_1=a_1+11d-a_1=11d$. Since $d$ is an integer, the difference between the first and last terms, $11d$, must be divisible by $11.$ Since the total difference must be less than $60$, we can start checking multiples of $11$ less than $60$ for the total difference between $a_1$ and $a_{12}$. We start with the largest multiple, because the maximum difference will result in the minimum value of the first term. If the difference is $55$, $a_1=\frac{60-55}{2}=2.5$, which is not an integer, nor is it one of the five options given. If the difference is $44$, $a_1=\frac{60-44}{2}$, or $\boxed{\textbf{(C)}\ 8}$
-Solution by Rhiju