Let $x$, $y$, and $z$ be real numbers such that $x+y+z=20$ and $x+2y+3z=16$. What is the value of $x+3y+5z$?

[i]Proposed by James Lin[/i]
1. We start with the given equations:
   \[
   x + y + z = 20
   \]
   \[
   x + 2y + 3z = 16
   \]

2. Subtract the first equation from the second equation to eliminate \(x\):
   \[
   (x + 2y + 3z) - (x + y + z) = 16 - 20
   \]
   Simplifying, we get:
   \[
   y + 2z = -4
   \]

3. We need to find the value of \(x + 3y + 5z\). Notice that we can use the equation \(y + 2z = -4\) to help us. Let's add this equation to the second given equation:
   \[
   (x + 2y + 3z) + (y + 2z) = 16 + (-4)
   \]
   Simplifying, we get:
   \[
   x + 3y + 5z = 12
   \]

The final answer is \(\boxed{12}\).