### Problem 1
Let the number of men be \(x\), the number of women be \(y\), and the number of children be \(z\). Then we have the following system of equations:
\[
\begin{cases}
x + y + z = 36 \quad (1) \\
4x + 3y + \frac{z}{2} = 36 \quad (2)
\end{cases}
\]
Multiply equation \((2)\) by \(2\) to get rid of the fraction: \(8x + 6y + z = 72 \quad (3)\)
Subtract equation \((1)\) from equation \((3)\): \((8x + 6y + z) - (x + y + z) = 72 - 36\)
\(7x + 5y = 36\)
Since \(x\) and \(y\) are positive integers, we can try values for \(x\):
If \(x = 1\), then \(7\times1 + 5y = 36\), \(5y = 29\), \(y\) is not an integer.
If \(x = 2\), then \(7\times2 + 5y = 36\), \(14 + 5y = 36\), \(5y = 22\), \(y\) is not an integer.
If \(x = 3\), then \(7\times3 + 5y = 36\), \(21 + 5y = 36\), \(5y = 15\), \(y = 3\)
Substitute \(x = 3\) and \(y = 3\) into equation \((1)\): \(3 + 3 + z = 36\), \(z = 30\)
So there are 3 men, 3 women, and 30 children.
### Problem 2
Let the number of bullocks be \(x\), the number of cows be \(y\), and the number of calves be \(z\). Then we have:
\[
\begin{cases}
x + y + z = 100 \quad (1) \\
3000x + 4000y + 600z = 100000 \quad (2)
\end{cases}
\]
Simplify equation \((2)\) by dividing by 100: \(30x + 40y + 6z = 1000\), further divide by 2: \(15x + 20y + 3z = 500 \quad (3)\)
Multiply equation \((1)\) by 3: \(3x + 3y + 3z = 300 \quad (4)\)
Subtract equation \((4)\) from equation \((3)\): \((15x + 20y + 3z) - (3x + 3y + 3z) = 500 - 300\)
\(12x + 17y = 200\)
Solve for \(x\): \(x = \frac{200 - 17y}{12}\)
Since \(x\) and \(y\) are positive integers, we try values for \(y\):
If \(y = 4\), then \(x = \frac{200 - 17\times4}{12} = \frac{200 - 68}{12} = \frac{132}{12} = 11\)
Substitute \(x = 11\) and \(y = 4\) into equation \((1)\): \(11 + 4 + z = 100\), \(z = 85\)
So there are 11 bullocks, 4 cows, and 85 calves.
@set c= 不知则觉多,知则觉少,越知越多,便觉越来越少. --- 知多少.
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