Baby Shower Problem. Too hard for 1st grader but got parents thinking

Solution 1:

$$(P+B-1)(P+B)=4PB$$ is equivalent to

$$(P-B)^2=P+B $$

Let $x:=P-B$ then $P+B=x^2$.

Solving Yields $P= \frac{x^2+x}{2}$ and $B= \frac{x^2-x}{2}$, where $x$ needs to be an integer....(Note that $x$ can also be negative). Note that this generates all solutions.

Thanks to Henning who pointed the small mistake.

Solution 2:

Say there are $p$ pink balls and $b$ blue balls; then the number of pairs of balls of the same color is $p(p-1)+b(b-1)$ and the total number of pairs of balls is $(p+b)(p+b-1)$. We want the second of these to be twice the first, giving

$$ 2 [p(p-1)+b(b-1)] = (p+b)(p+b-1). $$

After much rearrangement, this is equivalent to

$$ (p-b)^2 = p+b $$.

So the total number of balls must be the square of the difference between the number of balls of each color. Let $p + b = n^2$, and assume $p>b$; then we have $p+b = n^2, p-b = n$ which gives $p = (n^2+n)/2, b = (n^2-n)/2$; all the solutions are of this form or of the reverse form. Furthermore we must have $n \ge 2$; if $n = 0$ or $n = 1$ there aren't enough tiles to play the game. In particular the solutions are

$$ (p,b) = (3,1), (6,3), (10,6), (15,10), \cdots $$

and their reversals $(1,3), (3,6), \cdots$. I agree that it would be quite difficult for a first grader to come up with any of these solutions.

Solution 3:

Probably it's supposed to be solved by trial and error with small numbers. Empirically, I find quickly that $(P,B)=(1,3)$ works, and no combinations of 2 or 3 balls in total do.

If we number the blue balls B1, B2, B3, it is fairly simple to write down all ways to pick two balls:

P+B1    B1+B3
P+B2    B2+B3
P+B3    B3+B1

and these all have to be equally possible. It is just barely conceivable that a 1st-grader would be able to follow that argument, but I'm not sure I would have the balls (so to say) to depend on it.