Macedonian Mathematical Competitions

Let $x$ be the number of candies in the bag in the beginning. Then Ilina ate $\frac{1}{5}x + 3$ during the first day and there were $\frac{4}{5}x - 3$ candies left. During the second day Ilina ate $\frac{1}{5}(\frac{4}{5}x - 3) + 5$ and the third day she ate the remaining 15 candies. Hence
$$\frac{1}{5}x + 3 + \frac{1}{5}(\frac{4}{5}x - 3) + 5 + 15 = x.$$
Expanding:
$$\frac{1}{5}x + 3 + \frac{4}{25}x - \frac{3}{5} + 5 + 15 = x$$
Combine like terms:
$$\left(\frac{1}{5}x + \frac{4}{25}x\right) + (3 - \frac{3}{5} + 5 + 15) = x$$
$$\frac{5}{25}x + \frac{4}{25}x = \frac{9}{25}x$$
$$3 - \frac{3}{5} + 5 + 15 = (3 + 5 + 15) - \frac{3}{5} = 23 - \frac{3}{5}$$
So:
$$\frac{9}{25}x + 23 - \frac{3}{5} = x$$
Move all terms to one side:
$$\frac{9}{25}x + 23 - \frac{3}{5} - x = 0$$
$$\frac{9}{25}x - x = -\frac{16}{25}x$$
So:
$$-\frac{16}{25}x + 23 - \frac{3}{5} = 0$$
$$-\frac{16}{25}x = \frac{3}{5} - 23$$
$$-\frac{16}{25}x = \frac{3 - 115}{5} = -\frac{112}{5}$$
Multiply both sides by $-1$:
$$\frac{16}{25}x = \frac{112}{5}$$
Multiply both sides by $25$:
$$16x = 25 \times \frac{112}{5} = 5 \times 112 = 560$$
$$16x = 560$$
$$x = 35$$
So, there were $35$ candies in the bag in the beginning.