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?
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 $=20$)
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 $k$.
We need to find $a_{0}$, the initial amount of candies, such that $a_{3}=3$.
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.
$$a_{1} = a_{0} - e_{1} = \tfrac{1}{2}a_{0}-1$$
$$e_{2} = \tfrac{1}{3}a_{1} + 1 = \tfrac{1}{6}a_{0}+\tfrac{2}{3}$$ $$a_{2} = a_{1} - e_{2} = \tfrac{1}{3}a_{0}-\tfrac{5}{3}$$
$$e_{3} = \tfrac{1}{5}a_{2} + 1 = \tfrac{1}{15}a_{0}+\tfrac{2}{3}$$ $$a_{3} = a_{2} - e_{3} = \tfrac{4}{15}a_{0}-\tfrac{7}{3}$$
$$ a_{3} = 3 $$ implies $$ \tfrac{4}{15}a_{0}-\tfrac{7}{3} = 3 \quad \implies \quad a_{0} = 20 $$
Hopefully, this is reflected in your class notes.
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 ...
