Key idea. Use parity for the even-sized moves; give a construction for five-cup moves.

1. Initially there are 0 upright cups. If a move flips \(k\) cups, of which \(u\) were upright, the upright count changes by \(k-2u\). When \(k\) is even, this change is even.

2. So with either two-cup moves or six-cup moves, the number of upright cups stays even. It cannot become 33. Parts (a) and (b) are impossible.

3. For part (c), put the 33 cups in a circle and number them 1 to 33. Make one move starting at each cup: flip that cup and the next four clockwise.

4. The first moves flip 1-5, then 2-6, and so on. Near the end, wrap around the circle: for example, the move starting at 30 flips 30, 31, 32, 33 and 1.

5. Every cup is in exactly five of these blocks. For example, cup 1 is flipped in the moves starting at 1, 30, 31, 32 and 33. The same count holds for every cup.

6. Each cup is therefore flipped five times. Starting upside down, it ends upright after an odd number of flips. After all 33 moves, every cup is upright.

Remember. A “no” answer needs an obstruction; a “yes” answer needs a method that actually works.