# 8.2. (7 points)

Twelve people are carrying 12 loaves of bread. Each man carries 2 loaves, each woman carries half a loaf, and each child carries a quarter of a loaf. How many men, women, and children were there?
Answer: 5 men, one woman, and 6 children.

Solution: Let $x$ be the number of men, $y$ be the number of women, and $z$ be the number of children; $x, y, z$ are natural numbers.

Then $x+y+z=12$ and $2 x+\frac{y}{2}+\frac{z}{4}=12$. From the last equation, it follows that $8 x+2 y+z=48$.

Transform the last equation: $7 x+y+(x+y+z)=48$. Considering the first equality, we get: $7 x+y=36, x=\frac{36-y}{7}$.

Since $x, y$ are natural numbers, $y$ can take the values 1, 8, 15, 22, 29. The values of $y$, equal to 15, 22, and 29, do not meet the condition of the problem $(y<12)$.

When $y=8: x=4, z=0$ - this contradicts the condition of the problem.

Let $y=1$, then $x=5, z=6$.