Nine boys and three girls agreed to share their cents equally.
Every boy gave an equal sum to every girl, and then every girl gave another equal sum to every boy. Each child then possessed exactly the same amount.
What was the smallest possible amount in cents that each child possessed after the sharing?
Note:
(1) Before the sharing, every boy had the same amount of money, which was
greater than 0. Every girl had the same amount of money, which was greater than 0 too. The money that each boy possessed and the money that each girl possessed were not the same before the sharing.
(2) The amount of money that each boy gave to each girl (which was greater than 0) and the amount of money that each girl gave to each boy (which was greater than 0) were not the same.
Here are some thoughts: Let $b$ the amount of money each boy has, and let $g$ be the amount each girl has.
Let $x$ be the amount the each girl gets from the boys. So $x$ is a multiple of 9.
Let $y$ be the amount each boy gets from the girls. So $y$ is a multiple of 3.
So we have $9b-xg+yb=3g+xg-yb$, which gives use $(9+2y)b=(3+2x)g \implies (9+6y')b=(3+18x')g$
Now we want to find the smallest possible $b$ and $g$.
I still feel like I'm not understanding this question right. Please help.
Thank you!
Best Answer
Using your notation, each boy starts with $b$ and each girl with $g$. Each boy gives $x$ to each girl and each girl gives $y$ to each boy.
Then, at the end of all the sharing, each boy has $$b-3x+3y=b-3(x-y)$$ coins and each girl has $$g+9x-9y=g+9(x-y)$$ coins.
These must be equal so $$b-3(x-y)=g+9(x-y)\quad \text {or}\quad b-g=12(x-y)$$
Thus $b-g$ is divisible by $12$.
Now, the rules appear to exclude $b-g=0$ so, as we are looking for minimal solutions, we need $|b-g|=12$. Now, our goal is to minimize the total number of coins, which is $9b+3g$. To do that, we aim for $b-g=-12$. Noting that $b≥3$, so that the boys can at least give each girl one coin, the minimal solution is $(b,g)=(3,15)$.
We then require $x-y=-1$. We can achieve that with $(x,y)=(1,2)$.
In the end each kid winds up with $$3-3(x-y)=\boxed {6}=15+9(x-y)$$
Note that the "official" solution fails since in that case the kids must start with the same number, counter to assumption.
Note: an earlier version of this solution arbitrarily added the assumption that $b>g$. This error was pointed out by @tomi in a comment.