1. Let $a_{1}, a_{2}, \ldots, a_{n}$ be integers $(n>1)$ satisfying $a_{1}+a_{2}+\cdots+a_{n}=a_{1} a_{2} \cdots a_{n}=2005$. Find the smallest possible value of $n$.
(1 mark)
Let $a_{1}, a_{2}, \ldots, a_{n}$ be integers $(n>1)$, such that $a_{1}+a_{2}+\cdots+a_{n}=a_{1} a_{2} \cdots a_{n}=2005$. Find the smallest possible value of $n$.
1. 5
1. Since 2005 is odd, all $a_{i}$'s must be odd. Since the odd $a_{i}$'s add up to 2005, $n$ must be odd as well. Consider the case $n=3$ with $a_{1} \geq a_{2} \geq a_{3}$. Then $a_{1} \geq \frac{2005}{3}$ and this forces $a_{1}=2005$ by considering the factors of 2005. Then we must have $a_{2}+a_{3}=0$ and $a_{2} a_{3}=1$, which means $a_{2}^2+1=0$ and hence leads to no solution. Finally, we see that $n=5$ is possible since
$$
2005+1+(-1)+1+(-1)=2005 \times 1 \times(-1) \times 1 \times(-1)=2005 \text{. }
$$

Hence the answer is 5.