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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?

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    $\begingroup$ For problems like this, it helps to work backwards. She ate three on the fourth day, so there were three left after the third day. How many were there at the start of the third day? Keep going. $\endgroup$ – Ross Millikan Apr 26 at 0:34
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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 $k$. enter image description here

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

check

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  • $\begingroup$ Thank you for that beginning start! I will see if guess and check can help me work this out $\endgroup$ – sug123 Apr 26 at 0:25
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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 ...

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  • $\begingroup$ Implies $x=128$. $\endgroup$ – dantopa Apr 26 at 22:06
  • $\begingroup$ Your set-up should be: $\frac 45(\frac 23(\frac 12 x -1)-1)-1 = 3$ $\endgroup$ – Doug M Apr 30 at 6:13

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