Problem 12 · Invariant (conserved cubes) · 4 points
Made by AI and not reviewed by a real person.
A tinkerer has 5 two-cube rods and 7 single cubes; she may split a rod into two singles or join two singles into a rod. Which total number of separate pieces can she NEVER have?
- 8
- 10
- 12
- 11
- 9
Show a hint
- Count cubes rather than pieces.
- How many cubes could 8 pieces hold at most?
Show the answer and solution
Answer: A — 8
The number of cubes stays at 5 × 2 + 7 = 17. Each piece contains at most 2 cubes, so 8 pieces could hold at most 16 cubes: 8 is impossible. Joining zero, one, two, or three pairs of singles gives 12, 11, 10, or 9 pieces. The trap is 9: although 17 cubes cannot all become rods, they can form 8 rods and 1 single.