Question

Let \(p\) be any positive integer, and let
\[ 2x + p = 2y, \quad p + y = x, \quad x + y = z \]
For what value of \(p\) would \(x + y + z\) attain its maximum value?

• A. 0
• B. 1
• C. 2
• D. 3

Hint:

Substitute p + y = x into 2x + p = 2y and see what value of p keeps the system from contradicting itself.

Answer 1

The Correct Option is A

Step 1: Write the two equations that involve only x, y and p.
From \(p + y = x\), we get \(x = y + p\).
Substitute this into \(2x + p = 2y\):
\[ 2(y+p) + p = 2y \]
\[ 2y + 2p + p = 2y \]
\[ 3p = 0 \]

Step 2: Interpret this result.
The two equations \(2x+p=2y\) and \(p+y=x\) together force \(3p = 0\), which means \(p\) must equal \(0\).
For any other value of \(p\) (in particular any positive integer such as 1, 2 or 3), these two equations contradict each other, so no real numbers \(x\) and \(y\) can satisfy both at once. That means for \(p=1,2,3\) the system has no solution at all, so \(x+y+z\) is not even defined for those choices of \(p\).

Step 3: See what happens when p = 0.
When \(p=0\): from \(p+y=x\) we get \(x=y\). Both original equations then hold for any value of \(y\), so \(x\) and \(y\) are free (they simply have to be equal), and \(z = x+y = 2x\).
So \(x+y+z = x + x + 2x = 4x\), which is defined for every choice of \(x\), unlike the other values of \(p\) where the system breaks down completely.

Step 4: Compare with the options.
\(p=0\) is the only value among the choices for which the system is consistent and \(x+y+z\) actually exists, so it is the value that lets the sum attain its (only attainable, hence maximum) value. Options 2, 3 and 4 (\(p=1,2,3\)) all make the equations contradictory, so they can never be right.

Final Answer:
\[ \boxed{p = 0} \]