1. (2 points) Boy Vasya tried to recall the distributive law of multiplication and wrote the formula: $a+(b \times c)=(a+b) \times(a+c)$. Then he substituted three non-zero numbers into this formula and found that it resulted in a true equality. Find the sum of these numbers.
Answer: 1

Solution: Expand the brackets on the right side: $a+b c=a^{2}+a c+a b+b c$, from which we get $a(a+b+c-1)=0$. Since $a \neq 0$, we obtain $a+b+c=1$.