Competition problem — Combinatorics

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Combinatorics Difficulty 5.7 AIME, harder Prove it Brazil

When two red amoebas join, the result in one blue amoeba; when a red amoeba and a blue amoeba join, they turn into three red amoeba; and when two blue amoeba join, they become four red amoeba. Fernando observes a test tube with initially $201$ blue amoebas and $112$ red amoebas.

a. Is it possible that after some amoebas transform the test tube contains $100$ blue amoebas and $314$ red amoebas?

b. Is it possible that after some amoebas transform the test tube contains $99$ blue amoebas and $314$ red amoebas?

Solution

If the number of blue amoebas is $b$ and the number of red amoebas is $r$ then $2b + r$ is invariant: indeed, whenever one blue amoeba appears/disappears, two red amoebas disappear/appear. In the problem, such number is $2 \cdot 201 + 112 = 514$ .

a. Since $2 \cdot 100 + 314 = 514$ , it can be possible. Indeed, if $50$ pairs of blue amoebas turn into $200$ red amoebas and a pair of one amoeba of each color turn into $3$ red amoebas, we have $201 - 2 \cdot 50 - 1 = 100$ blue amoebas and $112 + 200 - 1 + 3 = 314$ red amoebas.

b. Since $2 \cdot 99 + 314 = 512$ , it is not possible.

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Source: MathNet, licensed CC-BY-4.0. Statement reproduced verbatim; metadata (topic, difficulty) added by this project.