There are x trees and y parrots. If on each tree only one parrot is sitting,…

Concept

A distribution statement of the form "one item per group with some left over" converts directly into an equation: if each of x trees seats exactly one parrot and c parrots are still left over, then y = x + c. If instead every occupied tree seats k parrots and t trees stay empty, the parrots exactly fill the remaining (x - t) trees, so y = k(x - t). Two such statements about the same pair (x, y) give a linear system in two unknowns, which fixes both values.

Applying it to this problem

1. One parrot on each tree leaves one parrot without a tree, so the parrots outnumber the trees by exactly one: y = x + 1.

2. Two parrots on each tree leaves one tree without a parrot, so the parrots exactly fill the other x - 1 trees at two per tree: y = 2(x - 1).

3. Both expressions describe the same y, so x + 1 = 2(x - 1) = 2x - 2.

4. Solving: x + 1 = 2x - 2 gives 1 + 2 = 2x - x, so x = 3.

5. Substituting back: y = x + 1 = 3 + 1 = 4.

Cross-check

• With 3 trees and 4 parrots, one parrot on each tree seats 3 parrots and leaves 4 - 3 = 1 parrot without a tree, as required.

• With two parrots to a tree, the 4 parrots occupy 4 / 2 = 2 trees, so 3 - 2 = 1 tree is left without a parrot, as required.

So x = 3 and y = 4. In the list given in the question, 3 is labelled C and 4 is labelled D, so x and y are C and D respectively.

Contrast with the near-misses

• Any pair with y = x + 1 meets the first condition on its own; the second condition additionally requires the parrots to fill x - 1 trees exactly two to a tree, which forces x + 1 = 2x - 2 and hence x = 3. For 1 and 2 the two parrots form one pair that fills the only tree, so no tree is left empty; for 2 and 3 the three parrots do not split into complete pairs at all, so one parrot is left unseated.

• A pair with y = x + 2, such as 1 and 3, leaves two parrots without a tree under the one-per-tree seating instead of one.