Giving Candies

Assume that at first, Andrew has $3y$ candies ( $y$ can be any positive integer).

"If Andrew gives Cindy one-third of his candies, all of us will have the same number of candies."

If Andrew gives Cindy that many of his candies, Andrew will have $2y$ candies remaining because he will have given $y$ candies away, which means all the other kids will also have $2y$ candies in their hand, including Cindy.

This implies that in the first place, Andrew has $3y$ candies (as assumed), Cindy has $y$ candies and all the other kids have $2y$ candies each (including Becky).

"If Andrew gives Becky all of his candies, the number of Becky's candies will be equal to the total number of candies of everyone else in the group."

If Andrew gives Becky all of his candies, which is $3y$ candies, Andrew will have no candies and Becky will have $5y$ candies. Cindy will have $y$ candies, because the number of candies she's holding does not change. Each of the other (unnamed) kids will still have $2y$ candies in their hand.

Becky now has $5y$ candies, Andrew and Cindy have $y$ candies together, so all the other (unnamed) kids should have $5y-y=4y$ candies together. Because each unnamed kid has $2y$ candies in their hand, the number of unnamed kids should be: $4y\div 2y =2$ .

We conclude that there are $\boxed{5}$ kids in total.

Let $a$ , $b$ and $c$ be the number of candies that Andrew, Becky and Cindy have, respectively. From the first statement, we can deduce that all the kids except for Andrew and Cindy have the same number of candies; let this number be $n$ (note that $c=n$ ). Then $n=a-\frac{1}{3}a = \frac{2}{3}a$ , so $a=\frac{3}{2}n$

From the second statement, we know that $c+a = \frac{1}{2}xn$ . Substituting what we learned from the first statement, $n+\frac{3}{2}n = \frac{1}{2}xn \implies x=\boxed{5}$ .