This year we have to make our school assignments in pairs.
With each classmate must be made exactly one of those assignments.
Exactly 30% of the assignments will be made by a pair of girls.

How many assignments do I have to make?


Say we have $b$ boys and $g$ girls, then we have $${g\choose 2} = {3\over 10}{b+g\choose 2}$$ so we have this equation $$3b^2 +3b(2g-1)= 7g(g-1)$$

Now for each $g$ find $b$. If we put $x = 2(b+g)-1$ and $y= 2g-1$ we get $$\boxed{10y^2-7=3x^2}$$

Here are some values

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  • 3
    $\begingroup$ Looks like a variant of Pell's equation, so I would try to build infinite solutions with the same algorithm used in Pell's equation. I don't know the exact theory for this variant, though. $\endgroup$ Apr 24 at 18:36
  • 4
    $\begingroup$ This doesn’t directly answer the original question. Also it doesn’t work for y=x=1 since that would mean g=1 and b=0, which would mean there were zero assignments. (I guess technically 30% of zero is zero, but it still seems to go against the spirit of the question.) $\endgroup$ Apr 24 at 18:57
  • $\begingroup$ @Greedoid much better, though strictly speaking the question was “how many assignments do I have?” $\endgroup$ Apr 25 at 15:20
  • $\begingroup$ I agree with Nick (both times). An addition that the most reasonable class size is 16 (at least I think so), and esp. that the (probably intended) solution is thus 15 would really make it an answer to my question. Still it seems complete enough to already accept it. $\endgroup$
    – Retudin
    Apr 25 at 20:28

I'm not sure if this is the only possible solution to the puzzle, but I found that if:

You have 2 boys and 3 girls in the class, then 10 total assignments will be made where 3 are made by solely girls. This means that you have to make four assignments in total.

Again, there may be more solutions based on the rules outlined in the puzzle.

  • $\begingroup$ How could I have missed that one.. I was looking for the other answer and maybe a proof that there is no solution with very large classes. I guess 4 would already be too many for many students. $\endgroup$
    – Retudin
    Apr 24 at 12:11

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