### Chocolate

There are three tables in a room with blocks of chocolate on each. Where would be the best place for each child in the class to sit if they came in one at a time?

### Rationals Between

What fractions can you find between the square roots of 56 and 58?

### F'arc'tion

At the corner of the cube circular arcs are drawn and the area enclosed shaded. What fraction of the surface area of the cube is shaded? Try working out the answer without recourse to pencil and paper.

# Peaches Today, Peaches Tomorrow....

##### Stage: 3 and 4 Challenge Level:

Lots of great solutions were submitted to this problem about the clever, and also generous little monkey. The problem encouraged you to practise calculating fractions, and come up with problem-solving strategies. In addition, as several students found, it higlighted the importance of efficently recording your working.

Part (i) required some calculations. Thomas, from Wharncliffe Side Primary, explained his answer:

The answer is $1$ left.

This is how I found it:
$\frac{3}{4}$ of $60$ is $45$, subtract $1$ to get $44$
$\frac{7}{11}$ of $44$ is $28$, subtract $1$ to get $27$
$\frac{5}{9}$ of $27$ is $15$, subtract $1$ to get $14$
$\frac{2}{7}$ of $14$ is $4$, subtract $1$ to get $3$
$\frac{2}{3}$ of $3$ is $2$, subtract $1$ to get $1$ left

For part (ii), the fractions are given, but the order has to be worked out. Many people noticed that the trial and improvement method would work here; try out the fractions in a different order, until the monkey is left with one after six days. However, several students also noticed that there could be a systematic way of working this out, in order to make the process more efficent. Recording the working clearly was particularly helpful.

Prevenia, from Crest Girls' Academy, noted that the peaches had to last the monkey for six days, and so could not run out too quickly. Therefore, it seemed likely that the monkey would keep larger fractions initially. For example, if the little monkey only kept half on the first day, the peaches would be dwindling rapidly!

Andrew, on behalf of his Year 8 Maths class at Holmemead Middle School, submitted the following solution:

We decided, that to be thorough, we would use a tree method. This would enable us to prove that we had pursued all the different lines of inquiry within the problem.

Have a look at the pdf file of the tree diagram; it is a very good and clear explanation! The circles with dots indicate a number which cannot be divided by any of the remaining fractions The tree was produced by Hugh from the maths group.

Francois, from Abingdon School, also produced a lovely solution to part (iii):

Each day the little monkey kept a fraction of his peaches. To do so, the denominator of the fraction must be a factor of the number of peaches at the beginning of the day and must not be the number 1. At the beginning of the first day, he has $75$ peaches. The factors of $75$ are $1, 3, 5, 15, 25, 75$. Therefore on the first day, the only fractions available are $\frac{3}{5}$ and $\frac{11}{15}$. Using this method and a tree diagram (see here), I found that the order of the fractions which the monkey used are:

$\frac{11}{15}$, $\frac{5}{6}$, $\frac{3}{4}$, $\frac{1}{2}$, $\frac{3}{5}$ and $\frac{1}{4}$.

Other students who submitted correct solutions included: Thomas from Wharncliffe Side Primary, Philip and Paul from Wilson's Junior School, Gwyneth and Benedict from A.D.J.S., and Alvina from South Island School. Thank you also to Tom, Alyssa, Ruth, and Ed, who are all from Whitby Community College.

Part (iii) was a problem-solving exercise where you were trying to make the peaches last as long as possible for the generous little monkey. Some people managed to make the peaches last for seven days, and other students made them last even longer. The maximum was sixteen days! As with part (iii), the main method used was "trial and improvement". Again, however, there could be some strategy to this.

Emily, Tom, Kendal, Ella, Luke and Alice, from John Ray Junior School, submitted a solution to this problem. They realised that if the peaches were to last for a long time, they should choose fractions that left the monkey with as many peaches as possible:

Starting with $98$ peaches:

Day $1$ = He keeps $\frac{48}{49}$, which means he has $96$. But he eats $1$ so he has $95$.

Day $2$ = He keeps $\frac{4}{5}$, which means he has $76$. But he eats $1$ so he has $75$.

Day $3$ = He keeps $\frac{14}{15}$, which means he has $70$. But he eats $1$ so he has $69$.

Day $4$ = He keeps $\frac{22}{23}$ which means he has $66$. But he eats $1$ so he has $65$.

Day $5$ = He keeps $\frac{4}{5}$ which means he has $52$. But he eats $1$ so he has $51$.

Day $6$ = He keeps $\frac{2}{3}$ which means he has $34$. But he eats $1$ so he has $33$.

Day $7$ = He keeps $\frac{10}{11}$ which means he has $30$. But he eats $1$ so he has $29$.

Since $29$ is a prime number and they weren't allowed to use $29$ as a denominator, they had to stop here. What would have happened if, on day seven, the monkey had kept $\frac{9}{11}$? Could the peaches have lasted longer?

Aporva considered this, and worked out a general method to help lengthen the number of days that the peaches could last for:

For this problem I started with the highest number of peaches, worked out the highest factor of the number and used this number as the denominator.
Then let $x$ = numerator, and $y$ = denominator.
Make the numerator equal to the denominator minus $1$
i.e. ($x = y-1$).
If $\frac{x}{y}$-$1$ of $n = \text{prime number}$ then make $x = y - 2$.
If this was also a prime number, I just repeated it (ie. let $x = y - 3$).

Next, I made the fractions into their simplest form if they weren't already.

Starting with $99$ peaches:

1.$\frac{32}{33}\ \text{of}\ 99\ \text{is}\ 96$; $96- 1 = 95$

2. $\frac{17}{19}\ \text{of}\ 95\ \text{is}\ 85$; $85 - 1 = 84$

3.$\frac{13}{14}\ \text{of}\ 84\ \text{is}\ 78$; $78 - 1 = 77$

4.$\frac{10}{11}\ \text{of}\ 77\ \text{is}\ 70$; $70 - 1 = 69$

5.$\frac{22}{23}\ \text{of}\ 69\ \text{is}\ 66$; $66 - 1 = 65$

6.$\frac{11}{13}\ \text{of}\ 65\ \text{is}\ 55$; $55 - 1 = 54$

7.$\frac{26}{27}\ \text{of}\ 54\ \text{is}\ 52$; $52 - 1 = 51$

8.$\frac{15}{17}\ \text{of}\ 51\ \text{is}\ 45$; $45 - 1 = 44$

9.$\frac{10}{11}\ \text{of}\ 44\ \text{is}\ 40$; $40 - 1 = 39$

10.$\frac{12}{13}\ \text{of}\ 39\ \text{is}\ 36$; $36 - 1 = 35$

12.$\frac{5}{7}\ \text{of}\ 35\ \text{is}\ 25$; $25 - 1 = 24$

13. $\frac{11}{12}\ \text{of}\ 24\ \text{is}\ 22$; $22 - 1 = 21$

14.$\frac{5}{7}\ \text{of}\ 21\ \text{is}\ 15$; $15 - 1 = 14$

15.$\frac{1}{7} \text{of}\ 14\ \text{is}\ 2$; $2 - 1 = 1$

What would have happened if, on day $15$, the monkey had kept $\frac{5}{7}$ of $14$? Could the peaches have lasted even longer?

Genie, from Putney High School also managed to make the peaches last for $15$ days. Katie and Jade from Pudsey Grangefield School, and Max from St . Peter's Primary School made the peaches last for $16$ days, which was the record time submitted. Hye Soo Kwon from Kings Norton Girls also got $16$, and explains her working:

1) Start with $99$ peaches because it is the biggest number lower than $100$

2) Now a fraction of the peaches is needed, the fraction has to be as large as possible.

- Find the factors of 99 to try as denominators because they are the only numbers that $99$ can be divided into. They are: $3, 5, 9, 11, 33$ ($1$ and $99$ cannot be the denominators accorsing to the rules.

- The numerator has to be $1$ less than its denominator in order to make the largest fraction.

- So the possible fractions of the peaches are $\frac{2}{3}, \frac{4}{5}, \frac{8}{9}, \frac{32}{33}$

3) Multiply each of these fractions with $99$ to see which one has the biggest answer.

4) Take away $1$ from the biggest answer. If it is prime, try the next biggest fraction. If it is not then this is the answer.

5) Carry out the same procedure again.

So I got this solution, which lasts $16$ days.

1. $99 \times \frac{32}{33}-1 = 96 - 1 = 95$

2. $95 \times \frac{17}{19}-1 = 85 - 1 = 84$

3. $84 \times \frac{41}{42}-1 = 82 - 1 = 81$

4. $81 \times \frac{26}{27}-1 = 78 - 1 = 77$

5. $77 \times \frac{10}{11}-1 = 70 - 1 = 69$

6. $69 \times \frac{22}{23}-1 = 66 - 1 = 65$

7. $65 \times \frac{11}{13}-1 = 55 - 1 = 54$

8. $54 \times \frac{26}{27}-1 = 52 - 1 = 51$

9. $51 \times \frac{15}{17}-1 = 45 - 1 = 44$

10. $44 \times \frac{10}{11}-1 = 40 - 1 = 39$

11. $39 \times \frac{12}{13}-1 = 36 - 1 = 35$

12. $35 \times \frac{4}{5}-1 = 28 - 1 = 27$

13. $27 \times \frac{7}{9}-1 = 21 - 1 = 20$

14. $20 \times \frac{4}{5}-1 = 16 - 1 = 15$

15. $15 \times \frac{2}{3}-1 = 10 - 1 = 9$

16. $9 \times \frac{2}{3}-1 = 6 - 1 = 5$

Francois, from Abingdon School, produced a very thorough solution, showing all of the working.

Thank you very much to everyone who submitted solutions to this problem. If you enjoyed this, try Ben's Game, and Fair Shares? as follow-up problems.