# Problem 6. (4 points)

Solve the equation $a b c d e f=a+b+c+d+e+f$ in natural numbers.
Answer: $(1,1,1,1,2,6)$ in any order

Solution:

Let the largest number be $a$. The total sum of the numbers does not exceed $6a$.

Suppose there are fewer than three ones in our set. Then the total product is at least $2 \cdot 2 \cdot 2 \cdot a = 8a > 6a$. This leads to a contradiction.

Therefore, there must be at least three ones in the set, and the sum is already strictly less than $6a$.

Thus, we need to consider the cases: $(1,1,1,1,2, a); (1,1,1,1,3, a); (1,1,1,2,2, a)$; $(1,1,1,1,4, a); (1,1,1,1,5, a)$.

To do this, we need to solve the equations $2a = a + 6; 3a = a + 7; 4a = a + 7; 4a = a + 8$ and $5a = a + 9$. It is easy to verify that only the first equation has a solution when $a = 6$.