At certain store, a package of 3 apples and 12 oranges costs 5 dollars, and a package of 20 apples and 5 oranges costs 13 dollars. Given that apples and oranges can only be bought in these two packages, what is the minimum nonzero amount of dollars that must be spent to have an equal number of apples and oranges?

[i]Ray Li[/i]
1. Let us denote the first package as \( P_1 \) and the second package as \( P_2 \).
   - \( P_1 \) contains 3 apples and 12 oranges and costs 5 dollars.
   - \( P_2 \) contains 20 apples and 5 oranges and costs 13 dollars.

2. We need to find the minimum nonzero amount of dollars that must be spent to have an equal number of apples and oranges. Let \( x \) be the number of \( P_1 \) packages and \( y \) be the number of \( P_2 \) packages.

3. The total number of apples and oranges can be expressed as:
   \[
   \text{Total apples} = 3x + 20y
   \]
   \[
   \text{Total oranges} = 12x + 5y
   \]

4. We need the number of apples to be equal to the number of oranges:
   \[
   3x + 20y = 12x + 5y
   \]

5. Simplifying the equation:
   \[
   3x + 20y = 12x + 5y
   \]
   \[
   20y - 5y = 12x - 3x
   \]
   \[
   15y = 9x
   \]
   \[
   5y = 3x
   \]
   \[
   y = \frac{3x}{5}
   \]

6. Since \( x \) and \( y \) must be integers, \( x \) must be a multiple of 5. Let \( x = 5k \) for some integer \( k \):
   \[
   y = \frac{3(5k)}{5} = 3k
   \]

7. The total cost for \( x = 5k \) packages of \( P_1 \) and \( y = 3k \) packages of \( P_2 \) is:
   \[
   \text{Total cost} = 5(5k) + 13(3k) = 25k + 39k = 64k
   \]

8. The minimum nonzero amount of dollars is when \( k = 1 \):
   \[
   \text{Total cost} = 64 \times 1 = 64 \text{ dollars}
   \]

9. To verify, we check the number of apples and oranges:
   \[
   \text{Total apples} = 3(5) + 20(3) = 15 + 60 = 75
   \]
   \[
   \text{Total oranges} = 12(5) + 5(3) = 60 + 15 = 75
   \]

   Both are equal, confirming our solution.

The final answer is \(\boxed{64}\) dollars.