The Oran Berry Passing Circle

At a glance

• Part 1: The smallest possible starting total is 24 Oran Berries.
• Part 2: Pikachu starts with 11, Eevee with 7, and Bulbasaur with 6.
• Part 3: No smaller total works because the only smaller possible equal ending piles are 2, 4, and 6 berries each, and every one fails when the passes are undone.

Key idea: undo a doubling pass

When one Pokémon gives another exactly as many berries as the receiver already has, the receiver’s pile doubles. To undo that pass, halve the receiver’s doubled pile and move the other half back to the giver.

The states below always list the piles in this order: (Pikachu, Eevee, Bulbasaur).

1. Find the smallest working total

First find a working ending amount. Suppose all three Pokémon finish with 8 berries.

1. End: (8,8,8).
2. Undo Bulbasaur’s pass to Pikachu: Pikachu’s 8 came from doubling 4. Move 4 back to Bulbasaur. Just before pass 3, the state was (4,8,12).
3. Undo Eevee’s pass to Bulbasaur: Bulbasaur’s 12 came from doubling 6. Move 6 back to Eevee. Just before pass 2, the state was (4,14,6).
4. Undo Pikachu’s pass to Eevee: Eevee’s 14 came from doubling 7. Move 7 back to Pikachu. At the start, the state was (11,7,6).

The starting total is 11 +7 +6 =24 berries.

2. Show and check all three passes

Start with Pikachu 11, Eevee 7, and Bulbasaur 6.

Every giver has enough berries, each receiver gets exactly the size of that receiver’s existing pile, and the three ending piles are equal.

3. Prove no smaller total works

Passing berries changes who has the berries but never changes the total. Because the three ending piles are equal, the total is three copies of the ending amount.

The last pass doubles Pikachu’s pile, so the equal ending amount must be even. To make a total smaller than 24, each equal ending pile must be smaller than 8. The complete list of positive even choices below 8 is 2, 4, and 6.

• Try 2 each: Undo the last pass to get (1,2,3). Undoing the previous pass would require splitting Bulbasaur’s 3 into two equal whole-number piles, which is impossible.
• Try 4 each: Undo the last pass to get (2,4,6), then undo the previous pass to get (2,7,3). Undoing the first pass would require splitting Eevee’s 7 into two equal whole-number piles, which is impossible.
• Try 6 each: Undo the last pass to get (3,6,9). Undoing the previous pass would require splitting Bulbasaur’s 9 into two equal whole-number piles, which is impossible.

All possible smaller ending amounts fail, while 8 each works. Therefore 24 is the smallest possible starting total.

Check

The forward table verifies every pass. It also preserves the total: the start has 11 +7 +6 =24 berries and the end has 8 +8 +8 =24 berries.