1. The members of an infinite geometric sequence are natural numbers, at least two of which are not divisible by 4. Write the general term of this sequence if you know that one of the terms is equal to 2004.
IV/1. The sequence is of the form $a_{n}=a \cdot r^{n}, n \geq 0$. First, we need to prove that all terms of the sequence are integers. Since $r=\frac{a_{n}}{a_{n-1}}$, $r$ is a rational number. Let's write $r=\frac{\alpha}{\beta}, \beta>0$, where the numbers $\alpha$ and $\beta$ are coprime. We need to prove that $b=1$. Suppose there exists a prime number $p$ that divides $b$. Then the number $a_{n}=a \frac{\alpha^{n}}{\beta^{n}}$ will not be an integer for sufficiently large $n$.

If a term of the sequence is divisible by 4, then all subsequent terms are also divisible by 4, so $a_{0}$ and $a_{1}$ are not divisible by 4. Similarly, $r \neq 1$, otherwise all terms would be equal to 2004. Suppose the $k$-th term is equal to 2004. Then $a_{k}=a \cdot r^{k}=2004=2 \cdot 2 \cdot 3 \cdot 167$. Since 2004 is an even number, $k \geq 2$. If $k>2$, it would follow that $r=1$, which is not possible. Therefore, $k=2$ and $r>1$. The only possibility is $r=2$, $a=3 \cdot 167=501$. The desired sequence is $a_{n}=501 \cdot 2^{n}$.