Candy store with eggs

Solution 1:

UPDATE: here's a two-use strategy that doesn't involve breaking the bag: The customer bags together α/β/γ/δ and shows that the bag is intact. They then bag together α/ε/ζ and show that the bag is intact.

Now, let's set Ω=α+β+γ+δ+ε+ζ. On the one hand, Ω is the sum of six different numbers, so it must be $\geq 21 (=1+2+3+4+5+6)$. On the other hand, since α+β+γ+δ $\leq11$ (from the first bagging) and ε+ζ+α $\leq11$ (from the second bagging), we have α+β+γ+δ+ε+ζ+α = Ω+α $\leq22$. But Ω$\geq21$ and Ω+α$\leq22$ together imply α $=1$. (Many thanks to Joffan in the comments for this easier argument!)

(For posterity, here's my initial four-use strategy: our customer bags together α/β/γ/δ and shows that the bag is intact, then bags together α/β/γ/ε and shows that the bag is intact. Since the only two four-egg bags that don't break the bag are 1/2/3/4 and 1/2/3/5, we now have two known sets: we know that {α, β, γ} are {1,2,3} in some order, and we know that {δ, ε} are {4,5} in some order (though we haven't differentiated members of those sets).

Next, the customer bags together δ/ζ and shows that the bag remains intact; since δ is either 4 or 5, then ζ must be either 6 or 7.

Finally, the customer tries to bag β/γ/ζ and shows that the bag breaks. Any combination of two eggs from the set {1,2,3} with 6 would be 'good' and not break the bag; only the combination {2,3,7} can break it. This proves that ζ is 7 and that {β, γ} are {2,3} in some order, and therefore that α is egg 1.)