Maths Olympiad Prep

Library / Netherlands — First Round

Netherlands — First Round

Combinatorics Difficulty 4.7 AIME Prove it Netherlands

At a congress all attendees are either a mathematician or a biologist and there is no one that is both. The mathematicians all know each other and each of them knows four of the biologists. The biologists also all know each other and each of them knows nine of the mathematicians. It turns out that every mathematician knows twice as many people as every biologist. (If person AA knows person BB, then person BB also knows person AA.)
How many mathematicians are at the congress?

Solution

Let mm be the number of mathematicians and bb the number of biologists.

Each mathematician knows all other mathematicians (m−1m-1 people) and 4 biologists, so each mathematician knows m−1+4=m+3m-1+4 = m+3 people.

Each biologist knows all other biologists (b−1b-1 people) and 9 mathematicians, so each biologist knows b−1+9=b+8b-1+9 = b+8 people.

We are told that every mathematician knows twice as many people as every biologist:
m+3=2(b+8) m+3 = 2(b+8)

Also, the number of mathematician-biologist acquaintances can be counted in two ways:
- Each mathematician knows 4 biologists: m×4m \times 4
- Each biologist knows 9 mathematicians: b×9b \times 9
So:
m×4=b×9 m \times 4 = b \times 9

Now solve the system:

From the second equation:
m×4=b×9  ⟹  m=94b m \times 4 = b \times 9 \implies m = \frac{9}{4}b

Substitute mm into the first equation:
m+3=2(b+8)94b+3=2b+1694b−2b=16−394b−84b=1314b=13b=52 m+3 = 2(b+8) \\ \frac{9}{4}b + 3 = 2b + 16 \\ \frac{9}{4}b - 2b = 16 - 3 \\ \frac{9}{4}b - \frac{8}{4}b = 13 \\ \frac{1}{4}b = 13 \\ b = 52

Now, m=94b=94×52=9×13=117m = \frac{9}{4}b = \frac{9}{4} \times 52 = 9 \times 13 = 117

Answer:

There are 117117 mathematicians at the congress.

Want a route through all this instead of an archive? The track puts 2,604 problems in a working order, from Junior Challenge level to the IMO shortlist.

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