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Suppose that there are at most $11 -1 = 10$ cages that hold at least $2$ chickens, then the rest of the $26 - 10 = 16$ cages can hold at most one. This allows us to pack at most $3\times 10 + 16 = 46$ chickens, which doesn't fulfill the conditions.

Conversely, if there are $11$ cages with $3$ chickens and the rest with $1$ chicken, then there are $3 \times 11 + (26-11) = 48$ chickens. Thus, the minimum number of cages that will have at least $2$ chickens is $11$ .

We are minimizing the cages with 2 and 3 chickens which means we have to maximize the cages with 1 chicken and minimize the cages with 2 chicken. First step is to fill all the cages with only 1 chicken. Then, we have 22 chickens remaining which could be filled in just 11 cages.

This question could be further tricky if number of chickens is not a multiple of 3(48).