# Problem 6. (3 points)

Solve the equation $a b c d e f g=a+b+c+d+e+f+g$ in natural numbers.
Answer: $(1,1,1,1,1,2,7)$ and $(1,1,1,1,1,3,4)$. In any order.

Solution:

Let the largest number be $a$. The total sum of the numbers does not exceed $7a$.

Suppose there are fewer than four ones in our set. Then the total product is at least $2 \cdot 2 \cdot 2 \cdot a = 8a > 7a$. This leads to a contradiction.

Therefore, there must be at least four ones in the set, and the sum is already strictly less than $7a$.

Thus, we need to consider the cases: $(1,1,1,1,1,2, a) ; (1,1,1,1,1,3, a) ; (1,1,1,1,2,2, a)$; $(1,1,1,1,1,4, a) ; (1,1,1,1,1,5, a) ; (1,1,1,1,2,3, a) ; (1,1,1,1,1,6, a)$.

To do this, we need to solve the equations $2a = a + 7 ; 3a = a + 8 ; 4a = a + 8 ; 4a = a + 9$; $5a = a + 10 ; 6a = a + 9$ and $6a = a + 11$. It is easy to verify that only the first equation has a solution at $a = 7$, the second at $a = 4$, and the fourth at $a = 3$, which gives us two sets: $(1,1,1,1,1,2,7)$ and $(1,1,1,1,1,3,4)$.