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: Isolate p from each equation separately.
From \(2x+p=2y\): \(p = 2y - 2x\).
From \(p+y=x\): \(p = x - y\).

Step 2: Since both expressions equal the same p, set them equal.
\[ 2y - 2x = x - y \]
\[ 2y + y = x + 2x \]
\[ 3y = 3x \]
\[ y = x \]

Step 3: Substitute back to find p.
Using \(p = x - y\) and \(y=x\):
\[ p = x - x = 0 \]
So whatever \(x\) and \(y\) turn out to be, they must be equal, and this in turn forces \(p\) to be exactly \(0\). There is no freedom to pick \(p=1,2\) or \(3\) and still satisfy both starting equations.

Step 4: Build x + y + z and compare options.
With \(y=x\) and \(z=x+y=2x\), the sum \(x+y+z = 4x\) can be found for \(p=0\), but for any positive \(p\), the equation \(3p=0\) derived the same way is violated, so there is no valid \(x,y,z\) at all, and the sum cannot even be written down.
Between the given choices, only \(p=0\) actually allows the sum to exist, so it is the value that lets \(x+y+z\) attain its value.

Step 5: Double-check by plugging p = 0 back into all three original equations.
With \(p=0\), pick any number for \(x\), say \(x=5\). Then \(y=x=5\) (from Step 2), and \(z=x+y=10\).
Check equation 1: \(2x+p=2(5)+0=10\), and \(2y=2(5)=10\). They match.
Check equation 2: \(p+y=0+5=5\), and \(x=5\). They match.
Check equation 3: \(x+y=5+5=10=z\). This matches too, so all three equations hold together only when \(p=0\), confirming that no other listed value of \(p\) can ever work.
\[ \boxed{p = 0} \]