Let's denote the number of molded veggies in D#1 as \( x \), in D#2 as \( y \), and in D#3 as \( z \).
From the problem:
- \( x = y - 45 \) (D#1 has 45 less molded veggies than D#2)
- \( y = z - 80 \) (D#2 has 80 less molded veggies than D#3)
- \( x + y + z = 236 \) (The total number of molded veggies is 236)
First, substitute \( y \) from the second equation into the first equation:
\[ x = (z - 80) - 45 \] \[ x = z - 125 \]
Now substitute \( x \) and \( y \) into the total equation:
- We already have \( y = z - 80 \).
- We also have \( x = z - 125 \).
Now substituting both into the total:
\[ (z - 125) + (z - 80) + z = 236 \]
Combine like terms:
\[ 3z - 205 = 236 \]
Now solve for \( z \):
\[ 3z = 236 + 205 \] \[ 3z = 441 \] \[ z = 147 \]
Now that we have \( z \), we can find \( y \):
\[ y = z - 80 = 147 - 80 = 67 \]
Now, find \( x \):
\[ x = y - 45 = 67 - 45 = 22 \]
So, D#1 has \( \boxed{22} \) molded veggies.
To check, calculate the total:
- D#1: 22
- D#2: 67
- D#3: 147
Adding these together:
\[ 22 + 67 + 147 = 236 \]
The calculations are consistent. Thus, D#1 indeed has \( \boxed{22} \) molded veggies.