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

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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.

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.

##### 1 Answer

#### 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**

This means that he must end up with **whole apple** before his

#color(red)(1)/2 - color(blue)(1/2) = 0#

Half of the whole apple leaves him with#1/2# of an apple, which he then delivers as the#1/2# of an apple

Moreover, you can say that he was left with **whole apples** before his

#color(red)(3)/2 - color(blue)(1/2) = 1#

Half of the#3# whole apples leaves him with#1# whole appleand#1/2# of an apple, which he then delivers as the#1/2# of apple

How about before his

Following the same pattern, you can say that he was left with **whole apples** before his sixth delivery, since

#color(red)(7)/2 - color(blue)(1/2) = 3#

Half of the#7# whole apples leaves him with#3# whole applesand#1/2# of an apple, which he then delivers as the#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**

You can thus say that he has

#7 xx 2 + 1 = "15 apples " -># before his#5^"th"# delivery

#15 xx 2 + 1 = "31 apples " -># before his#4^"th"# delivery

#31 xx 2 + 1 = "63 apples " -># before his#3^"rd"# delivery

#63 xx 2 + 1 = "127 apples " -> # before his#2^"nd"# delivery

#127 xx 2 + 1 = "255 apples " -> # before his#1^"st"# delivery

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

**ALTERNATIVE APPROACH**

Here's a cool trick to use to double-check your answer.

Let's *assume* that the farmer **did not** 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 * 1/2 = x/2 -># after the#1^"st"# delivery

#x/2 * 1/2 = x/4 -># after the#2^"nd"# delivery

#x/4 * 1/2 = x/8 -># after the#3^"rd"# delivery

#x/8 * 1/2 = x/16 -># after the#4^"th"# delivery

#vdots#

and so on. **After** his

#x/2^8 = x/256#

apples. However, this number **cannot** be equal to

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

#256/2^8 = 256/256 = 1#

But since he must be left with **apples** after his **less apple** than the *maximum number of apples*, and so

#256 - 1= "255 apples"#

Thefore, you can say that if he starts with **apples** and adjusts his deliveries from just half of what he has to half of what he has **and**