22. Members of the "Handy Hands" club had to cut a piece of wire 102 cm long into parts 15 cm and 12 cm long, and so that there were no scraps. The children found all possible solutions. Find them as well.
22. Since the total length of the wire is an even number (102 cm) and the sum of the lengths of the segments that are 12 cm each is even, the sum of the lengths of the segments that are 15 cm each must also be even. Therefore, the number of pieces that are 15 cm each must be even. Let's consider all possible cases:
1) If there are 2 pieces that are 15 cm each, then \(15 \cdot 2 = 30\) and \(102 - 30 = 72\), and \(72 \div 12 = 6\). So, there will be 6 pieces that are 12 cm each.
2) If there are 4 pieces that are 15 cm each, then \(15 \cdot 4 = 60\) and \(102 - 60 = 42\), but 42 is not divisible by 12 without a remainder.
3) If there are 6 pieces that are 15 cm each, then \(15 \cdot 6 = 90\) and \(102 - 90 = 12\). Therefore, only 2 cases are possible:

a) 2 pieces that are 15 cm each and 6 pieces that are 12 cm each;

b) 6 pieces that are 15 cm each and 1 piece that is 12 cm.