The Oran Berry Passing Circle

Problem 11: The Oran Berry Passing Circle

Answer

The smallest possible starting total is 24 Oran Berries.

• Pikachu starts with 11.
• Eevee starts with 7.
• Bulbasaur starts with 6.

Check the three passes

| Moment | Pikachu | Eevee | Bulbasaur | | --- | ---: | ---: | ---: | | Start | 11 | 7 | 6 | | Pikachu gives Eevee 7 | 4 | 14 | 6 | | Eevee gives Bulbasaur 6 | 4 | 8 | 12 | | Bulbasaur gives Pikachu 4 | 8 | 8 | 8 |

Every pass gives the receiver exactly the number that receiver already has, so the receiver's pile doubles. The ending piles are equal, and the starting total is 11 + 7 + 6 = 24.

Why no smaller total works

Passing berries does not change the total. If each Pokémon ends with k berries, the total is k + k + k.

In the last pass, Pikachu's pile doubles to k. That means k must be even. If the total were smaller than 24, then k would be smaller than 8. The only positive even choices are 2, 4, and 6.

Now undo the passes, starting at the end:

| Equal ending pile | Undo the last pass | Keep undoing | Result | | ---: | --- | --- | --- | | 2 each | (1, 2, 3) | Bulbasaur's 3 cannot be halved into whole berries. | Does not work | | 4 each | (2, 4, 6) | Undo Eevee's pass: (2, 7, 3). Eevee's 7 cannot be halved into whole berries. | Does not work | | 6 each | (3, 6, 9) | Bulbasaur's 9 cannot be halved into whole berries. | Does not work | | 8 each | (4, 8, 12) | (4, 14, 6) → (11, 7, 6) | Works |

So 2, 4, and 6 berries each at the end are impossible. 8 each is the first ending amount that works, which makes 24 the smallest possible starting total.