12) The arithmetic mean of twenty-seven consecutive natural numbers is 2008. What is the smallest of these numbers?
(A) 1995,
(B) 1997
(C) 1999
(D) 2001,
(E) 2004.
12. The answer is $(\mathbf{A})$.

Let $n$ be the fourteenth of the twenty-seven numbers whose arithmetic mean is being calculated, i.e., $n$ is exactly in the middle of the sequence of consecutive numbers we are averaging; then the mean is given by

$$
\frac{(n-13)+(n-12)+\cdots+(n-1)+n+(n+1)+\cdots+(n+12)+(n+13)}{27}=\frac{27 n}{27}=n .
$$

Therefore, $n=2008$ and the smallest of the numbers being averaged is $n-13=1995$.