Going round in circles

6651 /www/nrich/html/content/id/6651/ 1 2011-02-01T00:00:01 <?xml version="1.0" encoding="UTF-8"?> <mdoxml version="1.0"><br></br> <h4><em><em><em><em>If you are a teacher, click <a href="http://nrich.maths.org/6651&part=note">here</a> for a version of the problem suitable for classroom use, together with supporting materials. Otherwise, read on...</em></em></em></em></h4> <p> </p> <p style="float: right;"><mdo:image src="November.png" style="padding-top: 50px;"></mdo:image></p> <p><span style="font-weight: bold;">Charlie said:</span> "It's Monday today, so it will be Monday again in $7$ days..</p> <p>and in $770$ days...</p> <p>and in $140$ days...</p> <p>and in $35 035$ days...</p> <p>and in $14 000 000 007$ days!"<br></br> <br></br> <strong>Alison said:</strong> "and it will be Wednesday in $2$ days...</p> <p>and in $72$ days...</p> <p>and in $702$ days...</p> <p>and in $779$ days...</p> <p>and in $14 777 002$ days!"<br></br> </p> <p>Do you agree with all of Charlie's and Alison's statements?</p> <p>Charlie and Alison chose numbers that were easy to work with. Can you see why they were chosen?</p> <p>Can you make up some similar statements of your own?</p> <p> </p> <p><strong>If today is Monday, what day will it be in $1000$ days' time?</strong></p> <p> </p> <p>Once you've had a go, have a look at how two students got started on this question:</p> <p>Ann's Method:</p> <p> </p> <div class="toggle">"It will be Monday in $700$ days, $770$ days, $840$ days... "<br></br> Can you suggest how Ann might continue?</div> <p> </p> <p>Luke's Method:</p> <p> </p> <div class="toggle">"On my calculator, I can work out that $1000 \div 7 = 142.8571429$.<br></br> Then I can work out $142 \times 7$..."<br></br> Can you suggest how Luke might continue?</div> <p> </p> <p><strong>Can you suggest any other methods for solving the problem?</strong></p> <p> </p> <p>Now try to suggest efficient methods to answer the following questions.</p> <p>You could use mental strategies, pencil and paper, or calculator methods.<br></br> </p> <ul> <li>If it is autumn now, what season will it be in $100$ seasons?<br></br> </li> <li>If it is November, what month will it be in $1000$ months?<br></br> </li> <li>A railway line has $27$ stations on a circular loop. If I fall asleep and travel through $312$ stations, where will I end up in relation to where I started?<br></br> </li> <li>If it is midday now, will it be light or dark in $539$ hours?<br></br> </li> <li>If a running track is $400$ metres around, where will I be in relation to the start after running 6 miles (approximately $9656$ metres)?<br></br> </li> <li>I was facing North and then spun around through $945 ^\circ$ clockwise. In what direction was I facing at the end?<br></br> </li> <li>If I get on at the bottom of a fairground wheel and the wheel turns through $5000$ degrees, whereabouts on the wheel will I be?<br></br> <br></br> </li> </ul> <p> </p> <div class="framework"><strong>Notes and Background</strong><br></br> <br></br> For more information on calendars and how mathematics can be used to work out quickly days of the week far in the past and future, take a look at the Plus article <a href="http://plus.maths.org/content/what-day-week-were-you-born">On What Day Of The Week Were You Born?</a></div> <p><br></br> <br></br> </p></mdoxml> <?xml version="1.0" encoding="UTF-8"?> <mdoxml version="1.0"><br></br> <p><span class="editorial">We received lots of great solutions to this question. One student noticed:</span></p> <p>If it is Monday today, then in 7 days it will be Monday again, then in 14 days, than 21 and so on. This goes on like the 7 times table: any number divisible by 7 will be a Monday.<br></br> If today is Monday, in 29 days what day will it be?<br></br> Well, 29/7 = 4 with remainder 1, so in 29 days it will be 1 day after Monday,<br></br> i.e. Tuesday. So it matters what the remainder is.</p> <p><span class="editorial">Charlie and Alison's statements were all correct, and we received great explanations of this from Jebin and Matthew from Junior King's School Canterbury, Alastair from St Mary's Primary School in Yorkshire, Cherice from Taipei European School and Sam from Shrewsbury House School. Alastair wrote:</span></p> <p>Charlie is right with his numbers because they are all multiples of 7; therefore all will be Mondays. Alison is right because her numbers are all multiples of 7 <em>plus two</em>; therefore all will be Wednesdays.</p> <p><span class="editorial">Lots of students correctly pointed out that, if today is Monday, then in 1000 days it will be Sunday. For example, Nicholas from Brazil used Ann's method and added on multiples of 7 to get closer to 1000:</span></p> <p>It will be Monday in 700 days, 770 days, 840 days, 910 days, 980 days, 987 days and 994 days, and then 6 days later it is Sunday.</p> <p><span class="editorial">Many students also commented on Luke's method. Here is one very clear explanation we received from Yasmine at Garden School, Kuala Lumpur:</span></p> <p>First I divided 1000 by 7, which gave 142.8571429. I rounded this down to 142, then multiplied 142 by 7, to give me 994. This means that 994 days is exactly 142 weeks - so in 994 days, it's still Monday. Since we want to know what happens in 1000 days, we just add 6 days to Monday, and find that the answer is Sunday.</p> <p><span class="editorial">George from England pointed out that this is the same as saying that 1000/7 = 142 remainder 6.</span></p> <p><span class="editorial">We received responses to the later questions from Tom from Monmouth School, Noor-ul-Ain from Westfield Middle School, Sam from Shrewsbury House School, Alastair from St Mary's Primary, and Katherine, Jack, Fred, Megan, Amani, Jasmine, Fiona, Michael, Jet, Markus and Gaeti from Garden School, Kuala Lumpur. Here are some of them:</span></p> <ul> <li><span class="editorial">If it is autumn now, what season will it be in 100 seasons?</span><br></br> <span class="editorial">Tom said:</span> there are four seasons, and 4 is a factor of 100, so whatever season it starts off as, it will be the same after 100 seasons. So in 100 seasons it will be autumn again.</li> <li><span class="editorial">If it is November, what month will it be in 1000 months?<br></br> Sam answered:</span> on my calculator, 1000/12 = 83.33333333. If I multiply 12 by 83 I get 996. Therefore in 996 months it will be November, so if we add 4 months to that we get March in 1000 months.</li> <li><span class="editorial">A railway line has 27 stations on a circular loop. If I fall asleep and travel through 312 stations, where will I end up in relation to where I started?<br></br> Fred and Megan gave this solution:</span><br></br> First you find out how many complete circuits there are in 312 stops: 312/27 = 11.55 loops.<br></br> Then you find out how many stops there are in 11 loops: 11×27 = 297 stops.<br></br> Next you find out how many further stops they have travelled: 312-297 = 15.<br></br> They woke up 15 stations further on than their starting station.</li> <li><span class="editorial">If it is midday now, will it be light or dark in 539 hours?<br></br> Sam said:</span> on my calculator 539/24 = 22.4583333. Then 22 × 24 = 528, therefore in 528 hours it will be midday, and 11 hours on from that it will be 539 hours from midday now, so it will be 11 pm. Therefore it will be dark.</li> <li><span class="editorial">If a running track is 400 metres around, where will I be in relation to the start after running 6 miles (approximately 9656 metres)?<br></br> Alastair noted:</span> 9656 / 400 = 24.14, and 24 × 400 = 9600, so after 9600 metres I will be back where I started, and after 9656 metres I will be 56 metres around.</li> <li><span class="editorial">I was facing North and then spun around through 945$^\circ$ clockwise. In what direction was I facing at the end?<br></br> Garden School sent us this helpful explanation and picture:</span><br></br> <mdo:image alt="" src="compass.png" style="width: 350px; height: 296px;"></mdo:image><br></br> Spinning around twice is the same as spinning through 720$^\circ$, and 945 - 720 = 225, so the required direction is 225 degrees from North, i.e. South-West.</li> <li><span class="editorial">If I get on at the bottom of a fairground wheel and the wheel turns through 5000 degrees, whereabouts on the wheel will I be?<br></br> Markus and Gaeti suggested:</span><br></br> First you have to try as get as close to 5000 as you can by multiplying 360 by 13, which is 4680. After 4680$^\circ$ you have come back to where you started. But you still have 320$^\circ$ left to spin through. On a clock face, if the hour hand started off pointing downwards (at 6:00) and spun clockwise through 320$^\circ$ it would be pointing to approximately 4:23.</li> </ul> <p><br></br> <br></br> </p></mdoxml> <?xml version="1.0" encoding="UTF-8"?> <mdoxml version="1.0"><br></br> <h3>Why do this problem?</h3> <p><a href="http://nrich.maths.org/6651">This problem</a> encourages students to consider a range of alternative methods of calculation. There is an opportunity for students to appreciate the importance of the quotient and the remainder when using division to solve a problem. When using calculators, students can explore the relationship between the remainder and the decimal part of the answer.</p> <br></br> <br></br> <h3>Possible approach</h3> <div><br></br> Work on this problem complements the ideas introduced in <a href="http://nrich.maths.org/6713&part=">Shifting Times Tables</a>.</div> <br></br> <div>"If today is Monday, it will be Monday again in 7 days, 770 days, 140 days."</div> <div>Pause, and ask students to suggest other numbers of days. Keep a record on the board as they are suggested, without any comment. Encourage them to suggest some large numbers. If none are forthcoming, perhaps ask "Does 35035 belong here?" and other similar questions.</div> <br></br> <div>"It will be Wednesday in 2 days, 72 days, 702 days, 772 days."</div> <div>Pause, and ask students to suggest other numbers of days. Keep a record on the board as they are suggested, without any comment.</div> <br></br> <div>Once a large collection of correct examples have been recorded, point to some of them and ask the class to explain how they know they are correct.</div> <br></br> <div>Ask the class to devise a way of working out what day it will be in 1000 days' time, initially without a calculator and then with a calculator. Share strategies, and if Ann's or Luke's methods (below) have not been suggested, show them the half-completed methods and ask them to finish them off.</div> <div> <p>Ann's Method:</p> <div class="toggle">"It will be Monday in $700$ days, $770$ days, $840$ days... "<br></br> Can you suggest how Ann might continue?</div> <p>Luke's Method:</p> <div class="toggle">"On my calculator, I can work out that $1000 \div 7 = 142.8571429$.<br></br> Then I can work out $142 \times 7$..."<br></br> Can you suggest how Luke might continue?</div> </div> <br></br> <div>Hand out <a href="/content/id/6651/Going%20Round%20In%20Circles.pdf">this worksheet</a>, which has space for students to keep a record of their methods. Perhaps organise students to work in pairs, and challenge them to find the most efficient mental, written, and calculator methods for each question.</div> <br></br> <div>Once the seven set questions have been answered, ask students to generate similar questions (perhaps using random numbers between 100 and 10000), and to suggest efficient methods for answering their questions.</div> <br></br> <br></br> <div><a href="http://nrich.maths.org/5617&part=">Round and Round and Round</a> encourages similar ways of working with division and remainders on a calculator, in the context of positions around a circle.</div> <div> </div> <div>Students could use the insights from this problem to analyse the game <a href="/1272">GOT IT</a>.</div> <div> </div> <h3>Key questions</h3> <p>How can I use what I know (small multiples of 7, 12, 360...) to generate number facts that are not at my fingertips (large multiples of 7, 12, 360...)?</p> <div> </div> <div>When I divide using a calculator, how does the answer on the screen help me to work out the quotient and remainder?</div> <div> <h3> </h3> </div> <h3>Possible extension</h3> <p><a href="http://nrich.maths.org/308&part=">Days and Dates</a> begins in the same way as this problem, but then encourages exploration of the algebra behind modular arithmetic.</p> <div> </div> <div><a href="http://nrich.maths.org/4350&part=">Modular Arithmetic</a> is an article introducing the notation and uses of modular arithmetic.</div> <br></br> <h3>Possible support</h3> <p><a href="http://nrich.maths.org/2220&part=">Colour Wheels</a> provides a simple context for exploring repeating patterns.<br></br> <br></br> </p> <br></br></mdoxml> <?xml version="1.0" encoding="UTF-8"?> <mdoxml version="1.0"><br></br> You might find it helpful to start by taking a look at the Level 1 and 2 problems in <a href="http://nrich.maths.org/6713&part=">Shifting Times Tables</a>.<br></br></mdoxml> <?xml version="1.0" encoding="UTF-8"?> <mdoxml version="1.0"><br></br> <p class="editorial">Lots of really great solutions to this problem were submitted. It is good to see a variety of methods used to solve the questions; have a look at the range below to see the different ways students have approached the task. Although there are several ways of working out the answers, some are more efficient than others, and some can be done mentally, whilst others require a calculator. Also, as several people noticed, you can apply the same method to the different problems. This is nice as it shows that you can solve many real-life problems with a few methods that you learn at school.</p> <p class="editorial">Some people used Anita's method; others used Luke's. Some students used a combination of the two, and others devised formulae as well. Some preferred to use remainders, and others worked with decimals or fractions. Have a look at the different ways, and think about the pros and cons of each of them.</p> <p class="editorial">The first question asked you to suggest other numbers that we could have chosen. John, from CGSB, gave some examples:</p> <br></br> <div>Monday: $35, 350, 3500, 35000, 21, 210, 2100, 21000$</div> <br></br> <br></br> Wednesday: $352, 3502, 37, 35002, 23, 212, 2102, 21002$<br></br> <br></br> <p><span class="editorial">Sam, from Charters Sunningdale, explained that you could choose any number in the seven times table eg.$182, 14,406$ ... Lizzie and Becca pointed out that, for Wednesdays, you could choose any number that was $2$ more than any multiple of $7$.</span></p> <p class="editorial">Siddhartha, from the Garden International School, expressed this in an algebraic form:</p> <br></br> <div>Let $x$ equal the number of days since Monday.</div> <div>For the day to be a Monday, the answer to $x\div7$ has to be a multiple of seven to be correct</div> <div>i.e. the answer must be a whole number when divided by $7$.</div> <br></br> <p class="editorial">Have a go at devising a formula for Wednesday.</p> <p class="editorial">Now, let us look at the next questions. James, from Wilson's School, explained how Anita could continue with her working. He also described how Luke's method helped him:</p> <br></br> <div>Going on from Anita's problem, she should continue adding on seventies until she's one $70$ away from going over $1000$, bringing her to the number $980$.</div> <div>Then add on sevens for the days of the week which should make $994$.</div> <div>Add on the extra six days to get to $1000$. Therefore in $1000$ days, it will be a Sunday (Monday + $6$ days is Sunday).</div> <br></br> Luke's method helps him because he divides $1000$ by $7$ (because there are $7$ days in a week), which will give him the exact number of weeks in the $1000$ days. He got the figure $142.8571429$<br></br> He then multiplies $142$ (which is the number of whole weeks in $1000$ days) by $7$ to get $994$. This brings him to the nearest multiple of seven below $1000$.<br></br> Then do what Anita did by adding on the six days, and Luke will also know that Sunday is the $1000$th day.<br></br> However Luke could have rounded up the number $142.8571429$ to $143$, and multiplied that by $7$ to get $1001$. Then he could have just taken off the one day to get Sunday.<br></br> <br></br> <p><span class="editorial">Several others got the correct answer, using Anita's method, Luke's method or a combination of the two: Patrick and William from St. Hughs, Christina from Tudhoe Grange, Nicole and Ann from York House School, and Natasha from Wilmslow High School.</span></p> <p class="editorial">Now let us look at the additional questions, starting with: "If it is November, what month will it be in $1000$ months?"</p> <p class="editorial">Shona from Tudhoe Grange applied Anita's method to this problem:</p> <br></br> <div>If it is November now, in $1000$ months' time it will be March.</div> <div>I worked this out because in $12$ months' time it will be November.</div> <div>It will also be November in $120$ months, $240$ months, $360$ months, $480$ months, $600$ months, $720$ months, $840$ months and $960$ months.</div> <div>Then if you keep adding $12$ months at a time the highest number you can get without going over $1000$ months is $996$ months. So in $996$ months' time it will be November. Then if you add one month at a time then you will get to $1000$ months and it will therefore be March in $1000$ months' time.</div> <br></br> <p class="editorial">Nicole and Ann, from York House School, used Luke's method:</p> <br></br> <div>To know what month it would be in $1000$ months if it is now November:</div> $1000\div12 = 83.333$.<br></br> That would be exactly the number of years in $1000$ months.<br></br> $83 \times 12$ would be how many months in $83$ whole years, which would be $996$.<br></br> There are $4$ more months until the $1000$th month. We know that the $996$th month is November, so $4$ more months would be March. (One month until January and then $3$ months into the next year).<br></br> <br></br> <p><span class="editorial">William, from Toot Hill School, used Luke's method but with remainders:</span></p> <br></br> <div>$1000\div12$ months = $83$ remainder $4$</div> So, you add $4$ months on to November which leaves you with the answer March<br></br> <br></br> <p><span class="editorial">Jenni began using Luke's method. She noticed:</span></p> <div><br></br> $1000\div12= 83\frac{1}{3}$.</div> <div>This means that there are $83$ whole more years to get back to November. Then another third of a year must be added on.</div> <div>One third of a year is $4$ months ($12\times \frac{1}{3}$ or $\frac{12}{3}$).</div> <div>Therefore, we add $4$ months to November, which gives March.</div> <br></br> <p class="editorial">Other students who solved this problem correctly include: Thomas and Laura from Tudhoe Grange, Lizzie and Becca, and Elli from Aylesbury High.</p> <p class="editorial">Flora and Lydia, from Cottenham Village College, noticed that the same method can be applied to similar questions, but in a different context. They came up with a general set of formulae for "circular loop" problems:</p> <br></br> We came up with a formula:<br></br> $x =$ the amount you want to add on. (in the months example, $1000$ months)<br></br> $y =$ the amount in the circular loop. (in the months example, $12$ months, which is the amount of months in a year)<br></br> $t =$ the starting point (in the example, November)<br></br> Integer part of $\frac{x}{y} = n$ (in the example, $83$ years)<br></br> $yn = z$ (in the example, $996$)<br></br> $x -z = p$ (in the example, $4$ months)<br></br> $t +p =$ the answer.<br></br> <br></br> <p><span class="editorial">We will finish by briefly looking at the other questions; the methods shown above can be applied here too. Try the different methods for yourself.</span></p> <p class="editorial">"A railway line has $27$ stations on a circular loop. If I fall asleep and travel through $312$ stations, where will I end up in relation to where I started?"</p> <p class="editorial">Elli, from Aylesbury High, used Luke's method, but with fractions:</p> <div><br></br> The first thing we do is determine how many loops we go through if we travel through $312$ stations.</div> <div>$312\div27 = 11 \frac{5}{9}$</div> <div>$\frac{5}{9}$ of $27$ stations is $15$</div> <div>So we will be $15$ stations ahead, in relation to where we started.</div> <br></br> <br></br> <p><span class="editorial">Josh, from Wilmslow High School used the formulae described by Flora and Lydia above:</span></p> <br></br> $X=312$<br></br> $Y=27$<br></br> $N=312\div27 = 11$<br></br> $Z=11\times27 = 297$<br></br> $P= 312 - 297 = 15$<br></br> Therefore the answer = $15$ Stations later<br></br> <br></br> <p><span class="editorial">Natasha, also from Wilmslow High School got this right too, as did Lizzie and Becca, and Siddhartha, from the Garden International School.</span></p> <p class="editorial">Now for the next question: "If it is midday now, will it be light or dark in $539$ hours?"</p> <p class="editorial">Elli, from Aylesbury High, used Luke's method:</p> <br></br> $539\div24 = 22.4583333$ This calculation gives us the exact number of days.<br></br> $0.458333$ of one day is $11$ hours.<br></br> Midday + $11$ hours = $11$pm.<br></br> Therefore it will be dark.<br></br> <br></br> <p><span class="editorial">Others who were also right included: Laura from Tudhoe Grange, Nicole and Ann from York House School, Josh and Natasha from Wilmslow High School, and Siddhartha from the Garden International School.</span></p> <p class="editorial">The answer is indeed $11$pm. However, an interesting point is that there are some places in the world where it is still light at $11$pm at certain times of the year! So "light" could also be correct in certain special circumstances. Can you think of where these places may be?</p> <p class="editorial">Let us go running now ...</p> <p class="editorial">"If a running track is $400$ metres around, where will I be in relation to the start after running $6$ miles (approximately $9656$ metres)?"</p> <p class="editorial">Those who submitted the correct answer include: John from CGSB, Nicole and Ann from York House School, Luke from Charters, Elli from Aylesbury High, and Siddhartha from the Garden International School.</p> <p class="editorial">Elli showed us one way of working out the answer:</p> <br></br> $9656$ metres $\div$ $400$ metres = $24.14$ circuits of the entire track.<br></br> $0.14 \times 400$ metres = $56$ metres.<br></br> So we'll be $56$ metres ahead<br></br> <br></br> <p><span class="editorial">Can you use the other methods to work out the answer?</span></p> <p class="editorial">The final question was about a fairground wheel: "If I get on at the bottom of a fairground wheel and the wheel turns through 5000 degrees, whereabouts on the wheel will I be?"</p> <p class="editorial">Elli explained how she solved the problem, using Luke's method, but with fractions rather than decimals:</p> <br></br> A fairground wheel is circular, and so going around once will take you through $360^\circ$.<br></br> $5000 \div 360 = 13 \frac{8}{9}$<br></br> So, if the wheel turns around $5000^\circ$, we go around in a circle $13$ and $\frac{8}{9}$ times.<br></br> $\frac{8}{9} \times 360 = 320^\circ$<br></br> So we will be $320^\circ$ around the wheel.<br></br> <br></br> <p><span class="editorial">Laurence, Olivia, Abi, Jessie, Frank, Joe all attend Over Primary School. They used Luke's method to work out the correct answer:</span></p> <br></br> $5000^\circ$ divided by $360^\circ$ (because there are $360^\circ$ in a full circle)<br></br> This makes $13.888$<br></br> Round down to $13$<br></br> Multiply $13$ by $360$ to get $4680^\circ$ which is the nearest multiple of $360$ below $5000$.<br></br> Find the difference between $4680$ and $5000$ which is $320^\circ$<br></br> So, after turning through $5000^\circ$ the wheel would end up $320^\circ$ from the bottom.<br></br> <br></br> <p><span class="editorial">John, from CGSB, pointed out that the answer could also be $40^\circ$ from the bottom ($360-320=40$). Nicole and Ann, from York House School, noticed this too.</span></p> <p class="editorial">Amy from Charters used Anita's method:</p> <div><br></br> A full turn is $360^\circ$, so after $3600^\circ$, you will also be at the bottom.</div> <div>After $3960^\circ$ ($3600+360$), you will again be at the bottom.</div> <div>You can keep on adding $360^\circ$, and still be at the bottom: $4320, 4680, 5040$ ...</div> <div>So, you are $40^\circ$ from the bottom (or $320^\circ$, depending on the way you look at it).</div> <br></br> <p><span class="editorial">Thank you to everyone who submitted solutions! If you want to have a go at a similar problem, take a look at</span> <a class="editorial" href="http://nrich.maths.org/5617&part=Round and Round and Round"> Round and Round and Round</a> <span class="editorial">which encourages similar ways of working with division and remainders on a calculator.</span><br></br> <br></br> <span class="editorial">If you want to do an extension of this problem, try</span> <a class="editorial" href="http://nrich.maths.org/308&part=Days and Dates">Days and Dates</a> <span class="editorial">and</span> <a class="editorial" href="http://nrich.maths.org/4350&part=Modular Arithmetic">Modular Arithmetic</a><span class="editorial">. Have fun!</span></p> <br></br></mdoxml> 2 3 0 0 1 0 0 Going round in circles Mathematicians are always looking for efficient methods for solving problems. How efficient can you be? 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