Maths Olympiad Prep

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Algebra Difficulty 5.5 AIME, harder Prove it United States

Problem:

There are mm friends with nn cupcakes each weighing 1 ounce. They wish to split the cupcakes equally by dividing each cupcake into some number of parts, and allocating some parts to each person.

a) Assume m=3m=3 and n=5n=5. Show that they may divide the cupcakes with all pieces being larger than 13\frac{1}{3} ounces.

b) Assume m=5m=5 and n=3n=3. Show that they may divide the cupcakes with all pieces being larger than 15\frac{1}{5} ounces.

Solution

Solution:

a.
Each person needs 53=2012\frac{5}{3} = \frac{20}{12} of a cupcake. If they cut four of the cupcakes into 512\frac{5}{12} and 712\frac{7}{12} and the last cupcake in half, then one person can take the four 512\frac{5}{12} pieces, giving them
4512=2012 4 \cdot \frac{5}{12} = \frac{20}{12}
and the other two can each take two 712\frac{7}{12} pieces and one 12=612\frac{1}{2} = \frac{6}{12} piece, giving them as well
712+712+612=2012 \frac{7}{12} + \frac{7}{12} + \frac{6}{12} = \frac{20}{12}
All pieces are larger than 13\frac{1}{3} ounces.

b.
Each person needs 35=1220\frac{3}{5} = \frac{12}{20} of a cupcake. We can divide two of the cupcakes into 620\frac{6}{20}, 720\frac{7}{20} and 720\frac{7}{20} pieces and the last cupcake into four 14=520\frac{1}{4} = \frac{5}{20} pieces. Then four of the people can get one 720\frac{7}{20} piece and one 520\frac{5}{20} piece and the fifth person can get the two 620\frac{6}{20} pieces.

All pieces are larger than 15\frac{1}{5} ounces.

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