Question

The number of solutions \((x, y, z)\) to the equation \(x\ –\  y \ –\ z = 25\), where x, y, and z are positive integers such that \(x ≤ 40, y ≤ 12\), and \(z ≤ 12\) is

• A. 101
• B. 99
• C. 87
• D. 105

Answer 1

The Correct Option is B

The equation \(x−y−z=25\) can be expressed as \(x=25+y+z. \)
Given that y and z are positive integers with \(y≤12\) and \(z≤12\), the range for \(y+z\) is \(2≤(x+y)≤15\) when \(27≤x≤40. \)
The minimum value for x is 27. 
For \(y=1, z\) can take 12 values. 
Similarly, for \(y=2, z\) can take 12 values, and so on, until \(y=12\) where z can take 10 values. 
Therefore, the total number of solutions is \(3+4+5+6+7+8+9+10+11+12+12+12=99. \)
Hence, the required result is \(99.\)