Let $a, b$, and $c$ be integers that satisfy $2a + 3b = 52$, $3b + c = 41$, and $bc = 60$. Find $a + b + c$
1. We start with the given equations:
   \[
   2a + 3b = 52
   \]
   \[
   3b + c = 41
   \]
   \[
   bc = 60
   \]

2. From the second equation, solve for \( c \):
   \[
   c = 41 - 3b
   \]

3. Substitute \( c = 41 - 3b \) into the third equation:
   \[
   b(41 - 3b) = 60
   \]
   \[
   41b - 3b^2 = 60
   \]
   \[
   3b^2 - 41b + 60 = 0
   \]

4. Solve the quadratic equation \( 3b^2 - 41b + 60 = 0 \) using the quadratic formula \( b = \frac{-B \pm \sqrt{B^2 - 4AC}}{2A} \), where \( A = 3 \), \( B = -41 \), and \( C = 60 \):
   \[
   b = \frac{41 \pm \sqrt{(-41)^2 - 4 \cdot 3 \cdot 60}}{2 \cdot 3}
   \]
   \[
   b = \frac{41 \pm \sqrt{1681 - 720}}{6}
   \]
   \[
   b = \frac{41 \pm \sqrt{961}}{6}
   \]
   \[
   b = \frac{41 \pm 31}{6}
   \]

5. This gives us two possible values for \( b \):
   \[
   b = \frac{41 + 31}{6} = \frac{72}{6} = 12
   \]
   \[
   b = \frac{41 - 31}{6} = \frac{10}{6} = \frac{5}{3}
   \]

6. Since \( b \) must be an integer, we discard \( b = \frac{5}{3} \) and take \( b = 12 \).

7. Substitute \( b = 12 \) back into the equation \( c = 41 - 3b \):
   \[
   c = 41 - 3 \cdot 12
   \]
   \[
   c = 41 - 36
   \]
   \[
   c = 5
   \]

8. Substitute \( b = 12 \) into the first equation to solve for \( a \):
   \[
   2a + 3 \cdot 12 = 52
   \]
   \[
   2a + 36 = 52
   \]
   \[
   2a = 16
   \]
   \[
   a = 8
   \]

9. Finally, calculate \( a + b + c \):
   \[
   a + b + c = 8 + 12 + 5 = 25
   \]

The final answer is \(\boxed{25}\)