# How many apples did he have when he began his deliveries?

## A farmer has to make 8 stops in delivering apples. He begins with exactly the number of apples he needs for these 8 deliveries. At the first stop, he delivers half of the apples he has plus 1/2 of an apple. At each of the next 7 stops, he delivers half of the remaining apples plus 1/2 of an apple. When he is finished he has no apples left, and none have been lost/damaged when making the deliveries.

Mar 11, 2017

$\text{255 apples}$

#### Explanation:

The trick here is actually the last delivery that the farmer makes.

You know that at each delivery, the farmer delivers half of the number of apples that he had after the previous delivery and $\textcolor{b l u e}{\frac{1}{2}}$ of an apple.

This means that he must end up with $\textcolor{red}{1}$ whole apple before his ${8}^{\text{th}}$ delivery, since

$\frac{\textcolor{red}{1}}{2} - \textcolor{b l u e}{\frac{1}{2}} = 0$

Half of the whole apple leaves him with $\frac{1}{2}$ of an apple, which he then delivers as the $\frac{1}{2}$ of an apple

Moreover, you can say that he was left with $\textcolor{red}{3}$ whole apples before his ${7}^{\text{th}}$ delviery, since

$\frac{\textcolor{red}{3}}{2} - \textcolor{b l u e}{\frac{1}{2}} = 1$

Half of the $3$ whole apples leaves him with $1$ whole apple and $\frac{1}{2}$ of an apple, which he then delivers as the $\frac{1}{2}$ of apple

How about before his ${6}^{\text{th}}$ delivery?

Following the same pattern, you can say that he was left with $\textcolor{red}{7}$ whole apples before his sixth delivery, since

$\frac{\textcolor{red}{7}}{2} - \textcolor{b l u e}{\frac{1}{2}} = 3$

Half of the $7$ whole apples leaves him with $3$ whole apples and $\frac{1}{2}$ of an apple, which he then delivers as the $\frac{1}{2}$ of apple

Can you see the pattern?

You get the number of apples he had before his previous delivery by doubling what he has now and adding $1$.

You can thus say that he has

$7 \times 2 + 1 = \text{15 apples } \to$ before his ${5}^{\text{th}}$ delivery

$15 \times 2 + 1 = \text{31 apples } \to$ before his ${4}^{\text{th}}$ delivery

$31 \times 2 + 1 = \text{63 apples } \to$ before his ${3}^{\text{rd}}$ delivery

$63 \times 2 + 1 = \text{127 apples } \to$ before his ${2}^{\text{nd}}$ delivery

$127 \times 2 + 1 = \text{255 apples } \to$ before his ${1}^{\text{st}}$ delivery

Therefore, you can say that the farmer started with $255$ apples.

$\textcolor{w h i t e}{.}$
ALTERNATIVE APPROACH

Let's assume that the farmer did not deliver $\frac{1}{2}$ of an apple at every stop. In this case, he would simply deliver half of the number of apples he has left at every stop.

In this case, the number of apples he has left would be halved with every stop. Let's say he starts with $x$ apples. He would have

• $x \cdot \frac{1}{2} = \frac{x}{2} \to$ after the ${1}^{\text{st}}$ delivery

• $\frac{x}{2} \cdot \frac{1}{2} = \frac{x}{4} \to$ after the ${2}^{\text{nd}}$ delivery

• $\frac{x}{4} \cdot \frac{1}{2} = \frac{x}{8} \to$ after the ${3}^{\text{rd}}$ delivery

• $\frac{x}{8} \cdot \frac{1}{2} = \frac{x}{16} \to$ after the ${4}^{\text{th}}$ delivery

• $\vdots$

and so on. After his ${8}^{\text{th}}$ delivery, he would be left with

$\frac{x}{2} ^ 8 = \frac{x}{256}$

apples. However, this number cannot be equal to $0$ because that would imply that he started with $0$ apples, which is not the case here.

We know that he scheduled the number of deliveries to ensure that he delivers half of what he had at every delivery, so the maximum number of apples that he can start with is $256$, since

$\frac{256}{2} ^ 8 = \frac{256}{256} = 1$

But since he must be left with $0$ apples after his ${8}^{\text{th}}$ delviery, it follows that he must have started with $1$ less apple than the maximum number of apples, and so

$256 - 1 = \text{255 apples}$

Thefore, you can say that if he starts with $255$ apples and adjusts his deliveries from just half of what he has to half of what he has and $\frac{1}{2}$ of an apple, he will manage to deliver all the apples in $8$ deliveries.