Competition problem — Combinatorics

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Combinatorics Difficulty 6.5 National Olympiad 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 $19$ blue amoebas and $95$ red amoebas.

a. He observes that all the amoebas join in pairs, giving birth to a new generation of amoebas. Find the maximum number of amoebas in this generation.

b. There is, at some point, $100$ amoebas. How many of them are blue?

Solution

a. Suppose there are $x$ pairs of red amoebas, $y$ pairs of blue amoebas and $z$ pairs of one amoeba of each color. Then $2x + z = 95 \iff x = (95 - z)/2$ and $2y + z = 19 \iff y = (19 - z)/2$ . The number of amoebas in the next generation is $x + 4y + 3z = (95 - z)/2 + 4 \cdot (19 - z)/2 + 3z = (171 + z)/2$ . So we need to maximize $z$ , that is, take $z = 19$ , and the maximum number of amoebas in the next generation is $(171 + 19)/2 = 95$ .

b. 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. So if $b + r = 100$ then $2b + r = 2 \cdot 19 + 95 \iff b + 100 = 133 \iff b = 33$ . So there were $33$ blue amoebas.

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