Peaches today, Peaches tomorrow....

2312 /www/nrich/html/content/id/2312/ 1 2011-02-01T00:00:01 <?xml version="1.0" encoding="UTF-8"?> <mdoxml version="1.0"> <div style="text-align: center; float: right;"><mdo:image src="peaches.png"></mdo:image> </div> (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. <div style="text-align: left;">On the second day he decided to keep ${\bf \frac{7}{11}}$ of his peaches. He gave the rest away. Then he ate one.</div> On the third day he decided to keep ${\bf \frac{5}{9}}$ of his peaches. He gave the rest away. Then he ate one. <div style="text-align: left;">On the fourth day he decided to keep ${\bf \frac{2}{7}}$ of his peaches. He gave the rest away. Then he ate one.</div> 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 ${\bf 75}$ 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: ${\bf \frac{1}{2}, \frac{1}{4}, \frac{3}{4}, \frac{3}{5}, \frac{5}{6}, \frac{11}{15}}$ In which order did he use the fractions so that he was left with just one peach at the end? <div style="text-align: left;">(iii) Peach Rationing Whenever the monkey has peaches, he always keeps a fraction of them each day, gives the rest away, and then eats one.</div> <div style="text-align: left;">I wonder how long he could make his peaches last for?</div> <div style="text-align: left;">Here are his rules:</div> <ul> <li style="text-align: left;">Each fraction must be in its simplest form and must be less than $1$.</li> <li style="text-align: left;">The denominator is never the same as the number of peaches left (for example, if there were $45$ peaches left, he would not be allowed to keep $\frac{44}{45}$ of them).</li> </ul> <div style="text-align: left;">Can you start with fewer than 100 peaches and choose fractions so that there is at least one peach left after a week?</div> <div style="text-align: left;">What is the longest that you can make them last, starting with fewer than 100 peaches?</div> <div>Send us your solutions showing how many peaches you started with and the fractions you used each day. </div> <a href="http://nrich.maths.org/7201&part=">Click here for a poster of this problem</a>.</mdoxml> <?xml version="1.0" encoding="UTF-8"?> <mdoxml version="1.0"> 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 $60$ = $45$, subtract $1$ = $44$ $\frac{7}{11}$ of $44$ = $28$, subtract $1$ = $27$ $\frac{5}{9}$ of $27$ = $15$, subtract $1$ = $14$ $\frac{2}{7}$ of $14$ = $4$, subtract $1$ = $3$ $\frac{2}{3}$ of $3$ = $2$, subtract $1$ = $1$ 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 <a class="editorial" href="/content/id/2312/Tree%20diagram%20for%20part%20three.pdf">pdf file of the tree diagram</a>; 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 <a href="/content/id/2312/TreediagramforFrancois.pdf">here</a>), 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.$\frac{32}{33}\ \text{of}\ 99 = 96$; $96- 1 = 95$ 2. $\frac{17}{19}\ \text{of}\ 95 = 85$; $85 - 1 = 84$ 3.$\frac{13}{14}\ \text{of}\ 84 = 78$; $78 - 1 = 77$ 4.$\frac{10}{11}\ \text{of}\ 77 = 70$; $70 - 1 = 69$ 5.$\frac{22}{23}\ \text{of}\ 69 = 66$; $66 - 1 = 65$ 6.$\frac{11}{13}\ \text{of}\ 65 = 55$; $55 - 1 = 54$ 7.$\frac{26}{27}\ \text{of}\ 54 = 52$; $52 - 1 = 51$ 8.$\frac{15}{17}\ \text{of}\ 51 = 45$; $45 - 1 = 44$ 9.$\frac{10}{11}\ \text{of}\ 44 = 40$; $40 - 1 = 39$ 10.$\frac{12}{13}\ \text{of}\ 39$; $39 = 36 - 1 = 35$ 12.$\frac{5}{7}\ \text{of}\ 35 = 25$; $25 - 1 = 24$ 13. $\frac{11}{12}\ \text{of}\ 24$; $24= 22 - 1 = 21$ 14.$\frac{5}{7}\ \text{of}\ 21 = 15$; $15 - 1 = 14$ 15.$\frac{1}{7} \text{of}\ 14 = 2$; $2 - 1 = 1$ 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. $99 \times \frac{32}{33} = 96 - 1 = 95$ 2. $95 \times \frac{17}{19} = 85 - 1 = 84$ 3. $84 \times \frac{41}{42} = 82 - 1 = 81$ 4. $81 \times \frac{26}{27} = 78 - 1 = 77$ 5. $77 \times \frac{10}{11} = 70 - 1 = 69$ 6. $69 \times \frac{22}{23} = 66 - 1 = 65$ 7. $65 \times \frac{11}{13} = 55 - 1 = 54$ 8. $54 \times \frac{26}{27} = 52 - 1 = 51$ 9. $51 \times \frac{15}{17} = 45 - 1 = 44$ 10. $ 44 \times \frac{10}{11} = 40 - 1 = 39$ 11. $39 \times \frac{12}{13} = 36 - 1 = 35$ 12. $35 \times \frac{4}{5} = 28 - 1 = 27$ 13. $27 \times \frac{7}{9} = 21 - 1 = 20$ 14. $20 \times \frac{4}{5} = 16 - 1 = 15$ 15. $15 \times \frac{2}{3} = 10 - 1 = 9$ 16. $9 \times \frac{2}{3} = 6 - 1 = 5$ Francois, from Abingdon School, produced a <a class="editorial" href="/content/id/2312/Peachesfrancoispartiv.pdf">very thorough solution</a>, showing all of the working. Thank you very much to everyone who submitted solutions to this problem. If you enjoyed this, try <a class="editorial" href="http://nrich.maths.org/2382&part=Ben%27s%20Game">Ben's Game</a>, and <a class="editorial" href="http://nrich.maths.org/708&part=Fair%20Shares?">Fair Shares?</a> as follow-up problems. </mdoxml> <?xml version="1.0" encoding="UTF-8"?> <mdoxml version="1.0"> <h3>Why do <a href="http://nrich.maths.org/public/viewer.php?obj_id=2312">this problem</a>?</h3> This problem could replace repetitive textbook work on calculating fractions of integers. It offers plenty of practice of these calculations while requiring students to come up with problem-solving strategies. It offers a good context for thinking about factors of numbers. <h3>Possible approach</h3> <div>Introduce the first part of the problem to the class, and give them a little bit of time to work in pairs to solve it. All three parts of this problem can be displayed on these <a class="powerpoint" href="/content/id/2312/Peaches%20Today%2C%20Peaches%20Tomorrow.ppt">PowerPoint slides</a>. Once students have had a chance to work on the first challenge, share strategies and discuss any difficulties that arose.</div> The second task is much more demanding and interesting, requiring students to work quite systematically and record the steps they take clearly. Challenge students to find the solution and to convince themselves that there is no other way of ordering the fractions. Finally, the last challenge would work very well as a 'simmering activity' set for students to think about beyond a single lesson - perhaps this could be set as a homework. The best solution could be displayed on the classroom wall, and students could be challenged to improve on it. <h3>Key questions</h3> <div>How will you record your work efficiently so that you can keep track of what is happening?</div> <h3>Possible extension</h3> <div>In part (iii) a solution can be found where the peaches last for more than a fortnight! Finding a solution of this magnitude should provide a suitable extension challenge.</div> <div><a href="http://nrich.maths.org/2382&part=">Ben's Game</a> and <a href="http://nrich.maths.org/708&part=">Fair Shares?</a> are both good follow-up fractions problems.</div> <h3>Possible support</h3> <div> </div> <div>For part (ii), suggest to students who are struggling to record their thinking effectively that they could use a tree diagram: at each stage, branch off the fractions it would be possible to try next so that all possibilities are checked.</div> </mdoxml> <?xml version="1.0" encoding="UTF-8"?> <mdoxml version="1.0">The monkey will always keep a whole number of peaches. For Part (ii), why not try a starting value and see what happens? Then decide whether you need to start with a bigger or smaller starting value next time. For part (iii), you could use a tree diagram: at each stage, branch off the fractions it would be possible to try next so that all possibilities are checked. </mdoxml> <?xml version="1.0" encoding="UTF-8"?> <mdoxml version="1.0"> On the 4th day there was 1 left over. So on 3rd day there were 4 left over. So on 2nd day there were 10 left over. So on 1st day there were 22. (iii) If we assume that 1 peach is eaten each day: 11/15, 55 - 1 = 54 5/6, 45 - 1 = 44 3/4, 33 - 1 = 32 1/2, 16 - 1 = 15 3/5, 9 - 1 = 8 1/4, 2 - 1 = 1 (iv) One possibility: 96 47/48 29/31 41/43 26/27 10/11 22/23 11/13 26/27 15/17 10/11 12/13 5/7 11/12 5/7 5/7 1/3 leaving 2 peaches. We received a large number of correct solutions to problem (ii), many with very clear explanations of the successful strategies: Frannie from NGHS reasoned as follows: To end up with 1 peach left, the beginning amount of peaches should be even. There are many ways of trial and error that you can try, and use ways to work it out. At first I picked the random number 12 which turns out to be: *12* 1st day eat 7 leave 5 2nd day eat 3 1/2 leave 1 1/2 3rd day eat 1 3/4 DOESNT WORK! because there aren't enough peaches left. So then I doubled the number and used 24: *24* 1st day eat 13 leave 11 2nd day eat 6 1/2 leave 4 1/2 3rd day eat 3 1/4 leave 1 1/4 4th day there should be 1 left but in this there are 1 1/4, so a 1/4 too many peaches are left. Then I thought there probably wouldnt be 1/2s etc and I was so close (only a 1/4 away) so I went down to 22 (nearest even number) and then that was right! *22* 1st day eat 12 leave 10 2nd day eat 6 leave 4 3rd day eat 3 leave 1 then 4th day would have 1 left! Ian and Charlie from William Lovell School reasoned differently: Working back from the end: There is one left at the end, so you add the one from the third day and then double it because he ate half the peaches plus one: 1 + 1 = 2 2 x 2 = 4 Then you add one from the second day, and then double it: 4 + 1 = 5 5 x 2 = 10 Then you add one more from the first day, and then double it: 10 + 1 = 11 11 x 2 = 22 At the beginning the monkey had 22 peaches. Peter from Lawrence Sherrif School approached it in a similar way: 22 peaches because Day 4: 1 peach Day 3: (1 Peach + 1 Peach) x 2 = 4 Peaches Day 2: (4 Peach + 1 Peach) x 2 = 10 Peaches Day 1: (10 Peach + 1 Peach) x 2 = 22 Peaches Louisa from Nottingham High School for Girls not only showed her stategy, but then also showed that her solution worked: 1+1 = 2 2x2 = 4 4+1 = 5 5x2 = 10 10+1= 11 11x2 = 22 therefore: 22 peaches to start he eats half (11) + 1... this leaves 10 he eats half again (5) + 1... this leaves 4 he eats half (2) + 1... he is left with 1 peach. Matt and Fred from Albion Heights School resorted to some algebraic thinking: The answer to this problem can easily be solved by using one of the simplest rules in algebra - reverse the operations. We began by writing out the problem: "X" equals the original amount of peaches, while "A" represents the number of remaining peaches on day 2, "B" is the amount on day 3, and "C" is what remained on the fourth and final day. (X/2-1) = A (A/2-1) = B (B/2-1) = C If we got "C" by dividing X by 2 and subtracting 1 three times, we must do the opposite to figure out "X". So, that means we must add 1 to "C" and multiply it by 2 three times. (C+1*2) = B (B+1*2) = A (A+1*2) = X C=1 (1+1*2) = 4 (4+1*2) = 10 and (10+1*2) = 22 Therefore, X=22. Ben also resorted to some algebraic reasoning: Taking x as the initial number of peaches, as days go by we end up with Day 1: x/2 - 1 Day 2:1/2(x/2 - 1) - 1 Day 3:1/2(1/2(x/2 - 1) - 1) - 1 Day 4:1/2(1/2(x/2 - 1) - 1) - 1 = 1 Therefore 1/2(1/2(x/2 - 1) - 1) - 1 = 1 1/2(1/2(x/2 - 1) - 1) = 2 1/2(x/2 - 1) - 1 = 4 1/2(x/2 - 1) = 5 x/2 - 1 = 10 x/2 = 11 x = 22 Well done to you all. See <a href="/content/id/2312/Peach%20Problems.pdf">also</a> </mdoxml> 2 4 0 0 1 1 0 Peaches today, Peaches tomorrow.... Whenever a monkey has peaches, he always keeps a fraction of them each day, gives the rest away, and then eats one. How long could he make his peaches last for? 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