Stuck on candy bowl fraction I am really stuck on this problem because I'm not even sure where to start.

Larissa has a bowl of candies. On the first day, she eats 1/2 of the candies plus one more. On the second day, she eats 1/3 of the remaining candies plus one more.  On the third day, she eats 1/5 of the remaining candies plus one more.  On the fourth day she eats the three remaining candies.  How many candies did she have at the start?
Hints

Let $N$ be the number of candies.

Day 1: she eats $\frac{1}{2}N + 1$: remainder $N_{1}$

Day 2: she eats $\frac{1}{3}N_{1} + 1$: remainder $N_{2}$

Day 3: she eats $\frac{1}{5}N_{2} + 1$: remainder $N_{3}$

Day 4: she eats $3$: $N_{3} = 3$

Can you go from here? (answer $=20$)



Step 1: English $\to$ algebra

Create a table which defined the key variables of $e_{k}$, the amount of candies eaten on day $k$, and $a_{k}$, the amount available at the end of day $k$.


Step 2: Solution criteria

We need to find $a_{0}$, the initial amount of candies, such that $a_{3}=3$.

Step 3: Solution strategy

We all agree that the solution involves expressing the intermediate variables $a_{k}$, and $e_{k}$, $k=1,2,3$ in terms of $a_{0}$. 

The are a few ways to do this. We chose a top down approach.

Day 1

$$a_{1} = a_{0} - e_{1} = \tfrac{1}{2}a_{0}-1$$

Day 2

$$e_{2} = \tfrac{1}{3}a_{1} + 1 = \tfrac{1}{6}a_{0}+\tfrac{2}{3}$$
$$a_{2} = a_{1} - e_{2} = \tfrac{1}{3}a_{0}-\tfrac{5}{3}$$

Day 3

$$e_{3} = \tfrac{1}{5}a_{2} + 1 = \tfrac{1}{15}a_{0}+\tfrac{2}{3}$$
$$a_{3} = a_{2} - e_{3} = \tfrac{4}{15}a_{0}-\tfrac{7}{3}$$

Step 4: Solve

$$
a_{3} = 3
$$
implies
$$
\tfrac{4}{15}a_{0}-\tfrac{7}{3} = 3 \quad \implies \quad a_{0} = 20
$$

Hopefully, this is reflected in your class notes.


Check the solution
Let $x$ be the number of candies to start with ... then 
\begin{eqnarray*}
\frac{1}{5} \left(\frac{1}{3} \left(\frac{x}{2} -1 \right) -1 \right) -1 =3.
\end{eqnarray*}
So ...