6. The number 25 is expressed as the sum of positive integers $x_{1}, x_{2}, \cdots, x_{k}$, where $k \leq 25$. What is the maximum value of the product of $x_{1}, x_{2}, x_{3}, \cdots$, and $x_{k}$ ?
6. Answer. 8748
Solution. Clearly, no $x_{i}$ should be 1 . If $x_{i} \geq 4$, then splitting it into two factors 2 and $x_{i}-2$ will give a product of $2 x_{i}-4$ which is at least as large as $x_{i}$. Further, $3 \times 3>2 \times 2 \times 2$, so any three factors of 2 should be replaced by two factors of 3 . Thus, split 25 into factors of 3 , retaining two 2 's, which means $25=7 \times 3+2 \times 2$. The maximum product is thus $3^{7} 2^{2}=8748$