2012 AMC 10A Problem 10

1. Call the smallest angle $a$ and the step between neighbouring angles $d$. Both are whole numbers. The twelve angles are $a,\ a + d,\ a + 2d,\ \dots,\ a + 11d$.
2. Together they fill the whole circle, $360^\circ$. Add them up: there are twelve copies of $a$, plus $d$ taken $0 + 1 + 2 + \cdots + 11 = 66$ times. So
$$12a + 66d = 360.$$
Divide everything by $6$: $2a + 11d = 60$.
3. We want $a$ as small as possible, so we want $d$ as large as possible, because a bigger $11d$ leaves less for $2a$.
4. Now see which values of $d$ are allowed:
- $2a$ and $60$ are both even, so $11d$ must be even. That means $d$ is even.
- $a$ must be at least $1$, so $11d$ can be at most $58$. That means $d$ is at most $5$.
5. The largest even number that is at most $5$ is $4$. Then $2a = 60 - 44 = 16$, so $a = 8$.
6. Check: the angles are $8, 12, 16, \dots, 52$. There are twelve of them; the first and last add to $60$, and there are six such pairs, so the total is $6 \times 60 = 360$. It works.

The answer is $\boxed{\textbf{(C)}\ 8}$.

When a list of equally spaced numbers has a fixed total, the first number and the step are tied together by one equation. Making the step bigger forces the first number down. Whole-number rules then decide which steps are allowed; here, looking at even and odd shows that the step must be even.

Test the choices from the smallest. From $2a + 11d = 60$ we get $11d = 60 - 2a$. For $a = 5$: $11d = 50$, not a multiple of $11$. For $a = 6$: $11d = 48$, no. For $a = 8$: $11d = 44$, so $d = 4$, which works. The smallest choice that works is $8$.