This problem is in three parts. If you are feeling confident about working with fractions, you might want to skip straight ahead to part (ii) or part (iii).

(i) A little monkey had 60 peaches.

**Can you start with fewer than 100 peaches and choose fractions so that there is at least one peach left after a week?**

**Starting with fewer than 100, what is the longest you can make the peaches last?**

(i) A little monkey had 60 peaches.

On the first day he decided to keep ${\bf \frac{3}{4}}$ of his peaches.

He gave the rest away. Then he ate one.

On the second day he decided to keep ${\bf \frac{7}{11}}$ of his peaches.

He gave the rest away. Then he ate one.

He gave the rest away. Then he ate one.

On the third day he decided to keep ${\bf \frac{5}{9}}$ of his peaches.

He gave the rest away. Then he ate one.

On the fourth day he decided to keep ${\bf \frac{2}{7}}$ of his peaches.

He gave the rest away. Then he ate one.

He gave the rest away. Then he ate one.

On the fifth day he decided to keep ${\bf \frac{2}{3}}$ of his peaches.

He gave the rest away. Then he ate one.

**How many did he have left at the end?**

(ii) A little monkey had 75 peaches.

Each day, he kept a fraction of his peaches, gave the rest away, and then ate one.

These are the fractions he decided to keep: $$ \frac{1}{2} \qquad \frac{1}{4} \qquad \frac{3}{4} \qquad \frac{3}{5} \qquad \frac{5}{6} \qquad \frac{11}{15}$$

**In which order did he use the fractions so that he was left with just one peach at the end?**

(iii) Whenever the monkey has some peaches, he always keeps a fraction of them each day, gives the rest away, and then eats one.

I wonder how long he could make his peaches last for...

Here are his rules:

- Each fraction must be in its simplest form and must be less than 1.
- The denominator is never the same as the number of peaches left.

For example, if there were 45 peaches left, he would not choose to keep $\frac{44}{45}$ of them.

Click here for a poster of part (ii) of this problem.