A little light thinking

7016 /www/nrich/html/content/id/7016/ 1 2011-02-01T00:00:01 <?xml version="1.0" encoding="UTF-8"?> <mdoxml version="1.0"> This problem follows on from <a href="http://nrich.maths.org/7024&part=" style="font-style: italic;">Charlie's Delightful Machine</a>, so take a look at that first. The rules for turning on the lights of Charlie's Delightful Machine are all given by linear sequences (like those found in <a href="/6713">Shifting Times Tables</a>). In the first problem, you found efficient strategies for working out the rules controlling each light. Now try to make two lights light up at once. <div style="background-color:#000;padding:20px;padding-bottom:20px;width:260px;height:280px"> <div style="position:relative;"><label style="position:absolute;top:10;color:#fff;font-size:12px;font-family:helvetica,arial,sans-serif">Type in a number</label> <input id="numberText" maxlength="8" name="number" size="8" style="position:absolute;top:-10px;left:130px;font-size:1.2em" tabindex="1" type="text" value="0"></input> <mdo:image id="red" src="red.png" style="position:absolute;left:20px;top:30px;opacity:0;z-index:1"></mdo:image> <mdo:image id="redOff" src="redOff.png" style="position:absolute;left:20px;top:30px;opacity:1;z-index:0"></mdo:image> <mdo:image id="yellow" src="yellow.png" style="position:absolute;left:120px;top:30px;opacity:0;z-index:1"></mdo:image> <mdo:image id="yellowOff" src="yellowOff.png" style="position:absolute;left:120px;top:30px;opacity:1;z-index:0"></mdo:image> <mdo:image id="green" src="green.png" style="position:absolute;left:20px;top:130px;opacity:0;z-index:1"></mdo:image> <mdo:image id="greenOff" src="greenOff.png" style="position:absolute;left:20px;top:130px;opacity:1;z-index:0"></mdo:image> <mdo:image id="blue" src="blue.png" style="position:absolute;left:120px;top:130px;opacity:0;z-index:1"></mdo:image> <mdo:image id="blueOff" src="blueOff.png" style="position:absolute;left:120px;top:130px;opacity:1;z-index:0"></mdo:image> <input id="restartButton" name="restartButton" style="position:absolute;top:260px" type="button" value="Restart"></input> </div> </div> Once you have made a pair of lights light up simultaneously, can you find another number that will light them both up? And another? And another? ... Is there a connection between the rules that light up each individual light and the rule that lights up the pair? Sometimes it's impossible to switch a pair of lights on simultaneously. How can you decide whether it is possible to switch a pair of lights on simultaneously? Now explore turning on three or even all four lights. If you find an example where it's impossible to light them all up, try to explain why it's not possible. If you find an example where it is possible to light them all up, work out what is special about the sequence of numbers that light up all four lights simultaneously.</mdoxml> <?xml version="1.0" encoding="UTF-8"?> <mdoxml version="1.0"> Zafirah and Folashade suggested some ideas for attacking the problem: Do not use random numbers when solving this problem go up in a strategic method. Before trying to work out anything at least get three numbers which light up each light to see if there is any pattern occouring. Look closely at the relationship between the numbers for each colour. Once you find out the relationship between the numbers, start to look if you are able to put the method into a formula. Isaac from Bryn Offa school used this method and tried numbers in ascending order until he had enough to spot a pattern. Natalie explained how she worked out the rules that turned on the lights: The sequences and rules for each light follows: Red: $7, 15, 23, 31$ The difference between $7$ and $15$ is $8$ so the beginning of the rule is $8n$ and then $1 \times 8$ is $8$, but the first term of the sequence is $7$. To change the $8$ to $7$ you subtract $1$. So the rule is $8n - 1$. Yellow: $5, 15, 25$ The difference between $5$ and $15$ is $10$ so the beginning of the rule is $10n$. However, the first term of the sequence is $5$ so to change the $5$ into 10, you add $5$. The rule is $10n + 5$. Green: $10, 20, 30$ The difference between $10$ and $20$ is $10$ so the beginning of the rule is $10n$. the first term of the sequence is $10$, so the rule doesn't need to change. The rule is therefore $10n$ Blue: $2, 6, 10, 14, 17, 22, 26, 30, 34$ The difference between $6$ and $10$ is $4$ so the beginning of the rule is $4n$. The first term of the sequence is $2$, so to change the $4$ to a $2$ you subtract $2$. The rule is $4n - 2$. The numbers that light 2 lights are: 10, 15 and 30. Some of you found sequences which didn't ever light both lights. Josh, Christian, Pavan, Alex and Vishaal found one such scenario: We found the rules: $3n+1$ and $6n+2$. We have decided that the two are impossible to light up together, as it will never both have the same number. The reason for that is because $3n+1$ will always be one more than a multiple of three whereas $6n+2$ will always two more then a multiple of three. Rajeev gave some examples when both lights could be lit: $8n+3$ and $6n+5$... the number which would turn both lights on is $11$ because $8+3$ and $6+5$ both equal $11$. The other numbers which would work are $35, 83,131$. The formula is $24n-13$. $11n+10$ and $10n+1$... the number which would turn both lights on is $21$ because $11+10$ and $20+1$ both equal $21$. The other numbers are $131,241,351$ and so on. The formula is $110n-89$. Jessica and Sam suggested some other combined sequences: $4n+1$ and $5m-2$ have a combined rule of $20k-7$ $10n+4$ and $3m-1$ have a combine rule of $30k+14$ $11n-5$ and $6m-5$ have a combined rule of $66k-5$ A pattern we can see for the combined rules is that to find the combined rules times the coefficients of $n$ together and then find the common differences in the new sequence. Finally Herschel gave an insight into the method known as the Chinese Remainder Theorem: We want to find a number that is part of two Arithmetic Progressions (APs): $an+b$ and $cn+d$. To put it another way, we are searching for a number $x$ that satisfies the congruences $x = b$ (mod$ a$) and $x = d$ (mod $c$). (The "mod $a$" part means that you look at the remainder on each side of the equation after dividing by "$a$".) Finding this value of $x$ that will satisfy both equations and hence turn on both lights can be done using Chinese Remainder Theorem, which deals precisely with this problem - solving congruences. For more information on the Chinese Remainder Theorem please look at this <a class="editorial" href="http://nrich.maths.org/5466&part=">article</a>. </mdoxml> <?xml version="1.0" encoding="UTF-8"?> <mdoxml version="1.0"> <div class="embed"> <h2>A Little Light Thinking</h2> The rules for turning on the lights of Charlie's Delightful Machine are all given by linear sequences (like those found in <a href="/6713">Shifting Times Tables</a>). In the first problem, you found efficient strategies for working out the rules controlling each light. Now try to make two lights light up at once. <div style="background-color:#000;padding:20px;padding-bottom:20px;width:260px;height:280px"> <div style="position:relative;"><label style="position:absolute;top:10;color:#fff;font-size:12px;font-family:helvetica,arial,sans-serif">Type in a number</label> <input id="numberText" maxlength="8" name="number" size="8" style="position:absolute;top:-10px;left:130px;font-size:1.2em" tabindex="1" type="text" value="0"></input> <mdo:image id="red" src="red.png" style="position:absolute;left:20px;top:30px;opacity:0;z-index:1"></mdo:image> <mdo:image id="redOff" src="redOff.png" style="position:absolute;left:20px;top:30px;opacity:1;z-index:0"></mdo:image> <mdo:image id="yellow" src="yellow.png" style="position:absolute;left:120px;top:30px;opacity:0;z-index:1"></mdo:image> <mdo:image id="yellowOff" src="yellowOff.png" style="position:absolute;left:120px;top:30px;opacity:1;z-index:0"></mdo:image> <mdo:image id="green" src="green.png" style="position:absolute;left:20px;top:130px;opacity:0;z-index:1"></mdo:image> <mdo:image id="greenOff" src="greenOff.png" style="position:absolute;left:20px;top:130px;opacity:1;z-index:0"></mdo:image> <mdo:image id="blue" src="blue.png" style="position:absolute;left:120px;top:130px;opacity:0;z-index:1"></mdo:image> <mdo:image id="blueOff" src="blueOff.png" style="position:absolute;left:120px;top:130px;opacity:1;z-index:0"></mdo:image> <input id="restartButton" name="restartButton" style="position:absolute;top:260px" type="button" value="Restart"></input> </div> </div> </div> <h3>Why do this problem?</h3> This problem encourages students to think about the properties of numbers. It could be used to consolidate work on linear sequences. The interactivity encourages learners to begin the problem experimentally before working more theoretically as they engage with the main ideas. <h3>Possible approach</h3> <div> This problem follows on from <a href="http://nrich.maths.org/7024&part=note" style="font-style: italic;">Charlie's Delightful Machine</a>, so we assume students have a strategy for determining the rule to switch individual lights on. "Did anyone find examples where two or more lights switched on simultaneously?" "Our challenge now is to determine how to turn on two lights (or more) simultaneously, if we know the rules for each individual light." If computers are available, students could work in pairs determining the rules for individual lights, then finding the rule for switching on pairs of lights simultaneously. Remind them of the importance of recording their findings to share with the class. If computers are not available, this <a class="pdflink" href="/content/id/7016/LightThinking.pdf">worksheet</a> contains the rules for a collection of sequences that students could explore, searching for examples where two sequences have numbers in common. </div> One way they could record their work is by creating a table with sequence rules along the top and down the side, and indicating with a tick or a cross in each cell whether both lights could light up simultaneously. When appropriate, they could also indicate numbers which successfully switch on both lights. <div style="font-weight: bold;">Remind the class that the task is not simply to find examples, but to find a way of determining, by just looking at the sequence rules, whether or not a pair of lights will ever light up simultaneously.</div> <div>When trying to produce convincing arguments, ideas from the problem <a href="http://nrich.maths.org/2669&part=">Stars</a> might be useful.</div> Leave some time for the class to come together to share examples of rules where more than one light could be switched on simultaneously, and examples where it was impossible, together with their reasoning. <h3>Key questions</h3> What is true about any pair of rules where it is not possible to light up both lights? If two sequences are described by the rules $an+b$ and $cn+d$, can you explain the conditions for determining whether the lights will ever switch on together? <h3>Possible extension</h3> Some students may wish to use ideas of modular arithmetic to prove their findings; reading <a href="http://nrich.maths.org/4350&part=">this article</a> first may help. <h3>Possible support</h3> The problem <a href="http://nrich.maths.org/6713&part=">Shifting Times Tables</a> offers an introductory challenge for exploring linear sequences. In the <a href="/7016/clue">Hint</a> there is a version of the interactivity with just two lights which students might find more accessible. Students could use a 100 square (<a href="/content/id/7016/1-100_NumberGrid.doc">Word</a>, <a href="/content/id/7016/1-100_NumberGrid.pdf">pdf</a>) as a visual way to record sequences and see where (if) they coincide. </mdoxml> <?xml version="1.0" encoding="UTF-8"?> <mdoxml version="1.0"> The problem <a href="http://nrich.maths.org/2669&part=">Stars</a> has a visual way of looking at relationships between numbers which might be useful in this problem. The problem <a href="http://nrich.maths.org/6713&part=">Shifting Times Tables</a> offers an introductory challenge for exploring linear sequences. You may prefer to experiment with this two-light version of the interactivity first. <div style="background-color:#000;padding:20px;padding-bottom:20px;width:260px;height:180px"> <div style="position:relative;"><label style="position:absolute;top:10;color:#fff;font-size:12px;font-family:helvetica,arial,sans-serif">Type in a number</label> <input id="numberText" maxlength="8" name="number" size="8" style="position:absolute;top:-10px;left:130px;font-size:1.2em" tabindex="1" type="text" value="0"></input> <mdo:image id="red" src="red.png" style="position:absolute;left:20px;top:30px;opacity:0;z-index:1"></mdo:image> <mdo:image id="redOff" src="redOff.png" style="position:absolute;left:20px;top:30px;opacity:1;z-index:0"></mdo:image> <mdo:image id="blue" src="blue.png" style="position:absolute;left:120px;top:30px;opacity:0;z-index:1"></mdo:image> <mdo:image id="blueOff" src="blueOff.png" style="position:absolute;left:120px;top:30px;opacity:1;z-index:0"></mdo:image> <input id="restartButton" name="restart" onclick="restart()" style="position:absolute;top:160px" type="button" value="Restart"></input> </div> </div> <script src="Twolights.js" type="text/javascript"> </script></mdoxml> <?xml version="1.0" encoding="UTF-8"?> <mdoxml version="1.0"> Zafirah and Folashade suggested some ideas for attacking the problem: Do not use random numbers when solving this problem go up in a strategic method. Before trying to work out anything at least get three numbers which light up each light to see if there is any pattern occouring. Look closely at the relationship between the numbers for each colour. Once you find out the relationship between the numbers, start to look if you are able to put the method into a formula. Dylan and Ashara followed on with these thoughts: Try to display the rules algebraicly. Our sequence which goes $3, 9, 15$ could be described as "going up by $6$ from the number $3$", or could be described as $6n - 3$, $n$ being the number in the sequence. If you're having difficulty displaying your results algebraically, why not try to imagine a numberline. $+ 1$ merely describes you shifting all the numbers up by one. Once you have found out the rules for Red & Blue, finding out the combined rule shouldn't prove difficult. At times, the combined rule could also be the rule for either Red or Blue! Just try loads of number, and don't think too hard. Sometimes the easiest, most simple answers are also the correct ones. Some of you found sequences which didn't ever light both lights. Josh, Christian, Pavan, Alex and Vishaal found one such scenario: We concluded to the rules: $3n+1$ and $6n+2$. We have decided that the two are impossible to solve, as it will never both have the same number. The reason for that is because $3n+1$ will always be one more than a multiple of three whereas $6n+2$ will always two more then a multiple of three. Rajeev of Fairfield Junior School gave some examples when both lights could be lit: $8n+3$ and $6n+5$... the number which would turn both lights on is $11$ because $8+3$ and $6+5$ both equal $11$. The other numbers which would work are $35, 83,131$. The formula is $24n-13$. $11n+10$ and $10n+1$... the number which would turn both lights on is $21$ because $11+10$ and $20+1$ both equal $21$. The other numbers are $131,241,351$ and so on. The formula is $110n-89$. In the above you have to find the lowest common multiple of both sequences and that number minus another number so the resulting number would turn both lights on. Jessica and Sam suggested some other combined sequences: $4n+1$ and $5m-2$ have a combined rule of $20k-7$ $10n+4$ and $3m-1$ have a combine rule of $30k+14$ $11n-5$ and $6m-5$ have a combined rule of $66k-5$ A pattern we can see for the combined rules is that to find the combined rules times the gradients together and then find the common differences in the new sequence. Finally Herschel of the European School of Varese gave an insight into the method known as the Chinese Remainder Theorem: We want to find a number that is part of two Arithmetic Progressions (APs): $an+b$ and $cn+d$. To put it another way, we are searching for a number $x$ that satisfies the congruences $x = b$ (mod$ a$) and $x = d$ (mod $c$). (The "mod $a$" part means that you look at the remainder on each side of the equation after dividing by "$a$".) Finding this value of $x$ that will satisfy both equations and hence turn on both lights can be done using Chinese Remainder Theorem, which deals precisely with this problem - solving congruences. For more information on the Chinese Remainder Theorem please look at this <a class="editorial" href="http://nrich.maths.org/5466&part=">article</a>. </mdoxml> 2 3 0 0 0 1 0 A little light thinking Here is a machine with four coloured lights. Can you make two lights switch on at once? Three lights? All four lights? Numbers and the Number System Properties of numbers Numbers and the Number System Number theory Sequences, Functions and Graphs Arithmetic sequence Sequences, Functions and Graphs Sequences Using, Applying and Reasoning about Mathematics Making and proving conjectures Secondary Mapping Document Patterns and sequences US Secondary Mapping Document DisplayCabinet ajk44 solution needs editing