If there are x large boxes and y small boxes, there is 12x+5y= 103, and it can be concluded that only the conditions of x=4 and y= 1 1 can be satisfied.

4 boxes 12.

1 1 boxed 5.

Specific algorithm

There is an X box and another Y box.

12X+5Y= 103

103 is odd.

Y is odd.

The value of 12X ends with 8.

100, when the last number of 12X is 8.

12X=48

∴ X=4。

Y= 1 1