Let \( m \) and \( n \) be natural numbers. Given: \( n \) is even, \( 0.2 < \frac{m}{20} \), \( \frac{n}{m} < 1 \), and \( \frac{n}{11} < 0.5 \). Find \( m - 2n \).
From \( 0.2 < \frac{m}{20} \), we get \( 4 < m \). Since \( m \) is a natural number, \( m \geq 5 \).
From \( \frac{n}{11} < 0.5 \), we get \( n < 5.5 \). Since \( n \) is a natural even number, possible values for \( n \) are \( 2 \) or \( 4 \).
The condition \( \frac{n}{m} < 1 \) implies \( n < m \). This is satisfied for \( n=2 \) if \( m>2 \), and for \( n=4 \) if \( m>4 \). Both are consistent with \( m \geq 5 \).
We test possible pairs \((m,n)\):
| \((m,n)\) | \(m-2n\) |
|---|---|
| \((5,2)\) | 1 |
| \((6,2)\) | 2 |
| \((7,2)\) | 3 |
| \((8,4)\) | 0 |
| \((9,4)\) | 1 |
The value \( m-2n = 1 \) is obtained for \((5,2)\) and \((9,4)\), both satisfying all given conditions. Therefore, \( m-2n = 1 \).