2003 AMC 10B Problem 3

1. The first $8$ odd counting numbers are $1, 3, 5, 7, 9, 11, 13, 15$. Pair the first with the last, the second with the second-to-last, and so on: $1+15$, $3+13$, $5+11$, $7+9$. That is $4$ pairs of $16$, so the sum is $64$.
(A handy fact: the first $n$ odd numbers always add to $n \times n$. Here $8 \times 8 = 64$.)
2. The five even integers add to $4$ less than that: $64 - 4 = 60$.
3. Five numbers that are equally spaced are balanced around the middle one, so the middle one equals the average: $60 \div 5 = 12$.
4. Step down by $2$ twice from the middle: the five numbers are $8, 10, 12, 14, 16$. The smallest is $8$.
5. Check: $8 + 10 + 12 + 14 + 16 = 60$.

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

Two facts do all the work. The first $n$ odd numbers add to $n^2$, and in any list of equally spaced numbers with an odd count, the middle number equals the average. Using the middle number saves you from writing an equation with five unknown-looking terms.

Call the smallest even number $n$. The five numbers are $n, n+2, n+4, n+6, n+8$, which add to $5n + 20$. Setting $5n + 20 = 60$ gives $5n = 40$, so $n = 8$.

The first $8$ odd counting numbers are $1, 3, 5, \dots, 15$. Their sum is $8^2 = 64$ (or, pairing first with last, $8 \cdot \frac{1 + 15}{2} = 64$).

So the five consecutive even integers add up to $64 - 4 = 60$. Five equally spaced numbers have average equal to the middle one, so the middle even integer is $\frac{60}{5} = 12$. The five integers are $8, 10, 12, 14, 16$, and the smallest is $8$.

Check: $8 + 10 + 12 + 14 + 16 = 60$.

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