[ equations in integers ]

A shepherd was tending a herd of 100 heads. For this, he was paid 200 r. For each bull, he was paid 20 r, for each cow - 10 r, and for each calf - 1 r.

How many bulls, how many cows, and how many calves are in the herd?
The number of calves is a multiple of 10.

## Solution

From the condition, it follows that the number of calves is a multiple of 10. Let the number of bulls be $a$, cows be $b$, and calves be $10 c$. Then $a+b+10 c=100$ and $20 a+10 b+10 c=200$, which simplifies to $2 a+b+c=20$.

From the first equation, it is clear that $a+b$ is a multiple of 10. From the second equation, $-a+b<20$. Therefore, $a+b=10, c=9, 2 a+b=11$. From this, we get $a=1, b=9, c=9$.

## Answer

1 bull, 9 cows, and 90 calves.