Dozens

559 /www/nrich/html/content/98/03/six2/ 1 2011-02-01T00:00:01 <?xml version="1.0" encoding="UTF-8"?> <mdoxml version="1.0"><p>Do you know a quick way to check if a number is a multiple of two? How about three, four or six?<br></br> <br></br> In the interactivity below, the computer generates two random digits.<br></br> Your task is to find the <span style="font-weight: bold;">largest possible three-digit number</span> which uses the computer's digits, and one of your own, to make a multiple of 2, 3, 4 or 6.<br></br> <br></br> <span style="font-weight: bold;">Can you decribe a strategy that ensures your first 'guess' is always correct?</span><br></br> </p> <div style="background-color:#bfe8ff;padding:20px;padding-bottom:20px;width:550px;height:400px;position:relative;"> <div style="color:#000000;font-size:32px;font-family:helvetica,arial,sans-serif">Enter the biggest three-digit multiple of <select id="gamechoice" style="background-color:#ffbff7;color:#000000;font-size:32px;font-family:helvetica,arial,sans-serif"><option value="two">two</option><option value="three">three</option><option value="four">four</option><option value="six">six</option></select> you can think of that uses the digits:</div> <div style="color:#000000;font-size:32px;font-family:helvetica,arial,sans-serif; text-align:center"> <span id="Number1" style="color:#000000;font-size:48px;font-family:helvetica,arial,sans-serif; "></span> and <span id="Number2" style="color:#000000;font-size:48px;font-family:helvetica,arial,sans-serif; "> </span> </div> <input id="Answer" type="text" size="3" style="color:#000000;font-size:32px;font-family:helvetica,arial,sans-serif;" onkeydown="if (event.keyCode !== 13) {blankout();} else {check();}" /> <input id="testButton" name="Test" type="button" value="Check" onClick="check()" style="color:#000000;font-size:28px;font-family:helvetica,arial,sans-serif" /> <br></br> <div id="Feedback" style="color:#000000;font-size:32px;font-family:helvetica,arial,sans-serif"></div> <br></br> <input id="clickhere" name="clickhere" type="button" size="10" value="New digits" onClick="generate()" style="color:#000000;font-size:28px;font-family:helvetica,arial,sans-serif;position:absolute; top:360px;"/> </div> <script src="7410-script.js" type="text/javascript"> </script><br></br> <div style="float: right;"> <mdo:image alt="1, 3, 4 and 5" height="150" src="Dozens.jpg" width="200"></mdo:image></div> <br></br> <br></br> <p><span style="font-weight: bold;">Something else to think about:</span><br></br> <br></br> What is the largest possible five-digit number<br></br> divisible by $12$ that you can make from the digits<br></br> $1$, $3$, $4$, $5$ and one more digit? <br></br> <br></br> <a href="http://nrich.maths.org/8003">Click here for a poster of this problem</a>.</p></mdoxml> <?xml version="1.0" encoding="UTF-8"?> <mdoxml version="1.0"><br></br> <span class="editorial">We received a large number of well considered responses to this problem.</span><br></br> <br></br> <span class="editorial">Ewan from Wilson's School explained his strategy for finding the largest multiple of 2:</span><br></br> <br></br> You put the largest possible combination of three numbers, making sure the two given are there, and the last digit is even.<br></br> For instance, with 9 and 5, the largest would be 958, not 995, because 5 is odd. <br></br> <br></br> <br></br> <span class="editorial">Gabriel, also from Wilson's School, had strategies for identifying multiples of 3, 4 and 6:</span><br></br> <br></br> To find out if your number is a multiple of 3, all you neeed to do is add the digits. If the digits add up to a multiple of 3, the original number is also a multiple of 3.<br></br> For example: 351<br></br> To find out if this is a multiple of 3 using normal division would be quite slow, but if you just add the didgits it takes only a few seconds:<br></br> 3 + 5 + 1 = 9<br></br> 9 is a multiple of 3 so 351 is a multiple of 3.<br></br> <br></br> You can use this for any number: 546822<br></br> 5 + 4 + 6 + 8 + 2 + 2 = 27<br></br> Note: this method only works if you want to find out if your number is a multiple of 3.<br></br> This does not work with any other number.<br></br> <br></br> <span class="editorial">(It actually works for multiples of 9 as well - take a look a</span><a href="http://nrich.maths.org/1308&part=" class="editorial"></a><span class="editorial">t</span> <a class="editorial" href="http://nrich.maths.org/1308&part=">Divisibility Tests</a><span class="editorial">)</span><br></br> <br></br> For the biggest multiple of 4, look at the last two digits. If the number formed by its last two digits is divisible by 4, the original number is a multiple of 4 as well.<br></br> <br></br> For the multiples of 6, just check if the number is divisible by 2 AND 3.<br></br> If it is, then the number is also divisible by 6.<br></br> <br></br> <br></br> <span class="editorial">Abbas from Green Primary School had a different strategy for identifying multiples of 4:</span><br></br> <br></br> You can divide the number by 2 and if the answer is even, that means the number that you first started with is a multiple of 4.<br></br> <br></br> <br></br> <span class="editorial">Patrick and Jenna from Savanna Oaks Middle School, and Thomas from Colet Court School explained how to find the largest multiples in the interactivity challenge.</span><br></br> <br></br> <span class="editorial">Here is Thomas' detailed explanation of his strategy:</span><br></br> <br></br> Multiple of 2:<br></br> <ul> <li>First check if both numbers are even. If they are, then use the highest of the 2 numbers as the tens digit and the lowest as the units digit. Use 9 as the missing number (ie the hundreds digit).</li> <li>If only one number is even, then use the odd number as the tens digit, the even number as the units and use 9 as the missing number (ie the hundreds digit).</li> <li>If both numbers are odd, then choose 8 as the units number (this will be your missing number), then use the biggest odd numbers as the hundreds digit and the smallest odd number as the tens digit.</li> </ul> Multiple of 3:<br></br> <br></br> The sum of the three digits needs to be a multiple of 3.<br></br> Therefore we will derive the missing number by starting from 9 and checking if the sum of the two given digits plus 9 is a multiple of 9.<br></br> If this is the case, then the result will be 9 for the hundreds digit, the highest of the two given digits for the tens and the lowest of the two digits for the unit.<br></br> If 9 does not work, we try with 8 and repeat with 7 then 6 then... if 8 does not work either.<br></br> When we have found the digit, the hundreds digit is the highest of the three numbers, the tens the second highest and the unit the lowest of all the digits.<br></br> <br></br> Multiple of 4:<br></br> <ul> <li>First we try to see if the combination of the 2 given digits is a multiple of 4. By doing so we will try first with the highest of the two digits as the tens digit and the other one as the unit. If it works then we will use 9 as the hundreds digit.</li> <li>If the combination of the highest of the two digits and the lowest does not work, we will try the other way round (lowest of the two given digits as the tens and highest as the unit). If this combination is a multiple of 4 then we will use 9 as the hundreds digit like above.</li> <li>If neither combination of the two digits is a multiple of 4, then we will use the highest of the two given digits as the hundreds digit. Then we will check if the remaining given digit is even. If this is the case we will use this number as the unit and choose as the tens digit the highest digit possible such that the last two digits are a multiple of 4.</li> <li>If the remaining given digit is not even, then we will use this number as the tens digit and choose the highest unit such that the last two digits are a multiple of 4.</li> </ul> <br></br> Multiple of 6:<br></br> <br></br> To be a multiple of 6, it needs to be a multiple of 3 and of 2.<br></br> <br></br> <span class="editorial">Thomas also suggested a solution to the final challenge:</span> <br></br> <br></br> The solution to "Something else to think about" is <span style="font-weight: bold;">53184</span>.<br></br> <br></br> If the number is divisible by 12, it needs to be divisible by 4 and by 3.<br></br> To be divisible by 3 the sum of the digits need to be divisible by 3, and therefore the missing digit can only be 2, 5 or 8.<br></br> The last two digits need to be divisible by 4. Since we have 1, 3, 4 and 5, none of the combination is divisible by 4 and therefore we need to use the highest possible additional number ahead of the 4 divisible by 4, ie 8 (84 is divisible by 4).<br></br> Then for the first three digits, we rank them from the highest down to the lowest.<br></br> <br></br> <br></br> <span class="editorial">This problem caught many people out - we received lots of solutions suggesting that 53184 was the answer (using an 8 as the extra number). </span> <br></br> <br></br> <span class="editorial">Annabel, Harry and Liberty from Hethersett High School managed to improve on Thomas' solution to the final challenge:</span><br></br> <br></br> The largest possible five digit number, including 1, 3, 4, 5 and another digit, which is divisible by 12 is: <span style="font-weight: bold;">54132</span><br></br> <br></br> This is because it is divisible by 3 and 4.<br></br> <br></br> <br></br> <span class="editorial">Ravsimrat from Garden International School managed to improve on that:</span><br></br> <br></br> Largest possible 5 digit number divisible by 12 is: <span style="font-weight: bold;">54312</span><br></br> <br></br> DIVISIBILITY RULES 54312 should be divisible by 12 so...<br></br> The divisibility rule for 12 is: the number should be divisible by 3 and 4<br></br> <br></br> a) Rule for 3: the sum of the digits of the number should be divisible by 3.<br></br> eg: (5 + 4 + 3 + 1 + 2) / 3 = 5 (so...it is divisible by 3)<br></br> <br></br> b) Rule for 4: the last 2 digits of the number should be divisible by 4.<br></br> eg: 12 / 4 = 3 (so...it is divisible by 4)<br></br> <br></br> Hence 54312 is divisible by 12.<br></br> <br></br> <br></br> <span class="editorial">Titus, also from Garden International School, thought about the problem in a similar way:</span><br></br> <br></br> Multiples of 12 are also multiples of 2, 3 and 4.<br></br> Multiples of 3 can be found by the sum of digits which can be divided by 3.<br></br> So the one more digit is limited to 8, 5 or 2.<br></br> Multiples of 4 are obtained if the last two digits together can be divided by 4.<br></br> Multiples of 2 must be even numbers.<br></br> The unit place has to be 2, 4 or 8.<br></br> <br></br> My answer is 54312<br></br> <br></br> <span class="editorial">Luke and Jordan from Breckland Middle School also sent us their clear reasoning:</span><br></br> <br></br> We worked out some facts about this mystery number to help us:<br></br> <br></br> 1. It must end in an even number as 12 is an even number.<br></br> <br></br> 2. To find out if a number is divisible by 3, you can add up all of the digits and see if it is a multiple of 3 that you recognise.<br></br> From this, we know that any number containing only 1, 3, 4 and 5 will not be divisible by 3, and therefore not divisible by 12.<br></br> Therefore, the fifth number MUST be 2, 5 or 8 for the five-digit number to be divisible by 12.<br></br> <br></br> 3. It cannot begin with 8. There are only 6 options beginning with 8 that are divisible by 2 and 3 (81354; 81534; 83154; 83514; 85314; and 85134) and none of these is divisible by 12.<br></br> <br></br> Continued trial and error then showed 54312 to be the biggest number that is divisible by 12.<br></br> <br></br> <span class="editorial">Thomas from A.Y. Jackson School sent us a detailed explanation of his thinking:</span><br></br> <br></br> Using the numbers 1, 3, 4, 5, and another number represented by N, the largest five digit number that is divisible by 12 is 54312. (N=2)<br></br> <br></br> I began by deducing that any number divisible by 12 has to be divisible by 4 and 3 and must be even.<br></br> <br></br> Any number with 3 as a factor must have digits that add up to a multiple of 3. The sum of 1, 3, 4, 5 is 13, therefore, N could be only three numbers: 2, 5, 8.<br></br> <br></br> The first attempt will certainly be to choose the largest possible value of N which is 8, and 8 as the first digit, but I realise that then the last number must be 4, but none of the remaining values (3, 1 or 5) paired with 4 (34, 14, 54) could be divisible by 4 (since if a number is divisible by 4, the last two digits must also be divisible by 4).<br></br> <br></br> Therefore 8 must be stuck to the rear with the number being either 53184 or 53148.<br></br> It then becomes evident that the largest value the ten-thousands digit could hold is 5.<br></br> <br></br> If N=5, the same dilemma returns (3, 1, 5 could not be paired with 4).<br></br> <br></br> The only solution to N then becomes 2, as then 2 could replace 4 as the last digit and thus giving the number a greater thousands place-value.<br></br> <br></br> The template of the number becomes 54 _ _ 2<br></br> Either 12 or 32 is divisible by 4, but 54312 is obviously larger than 54132, and thus the answer is 54312.<br></br> <br></br> <span class="editorial">Harry from The Beacon School also sent us a full and very clear explanation of his thinking:</span><br></br> <br></br> As the number is a multiple of 12, it must be a multiple of 3 also. Therefore its digits must add to a multiple of 3.<br></br> <br></br> 1+3+4+5=13, so the extra digit must be 2, 5, or 8 to make the sum into a multiple of 3.<br></br> <br></br> The number must also be a multiple of 4 (because it is a multiple of 12). For a number to be a multiple of 4, the number formed by its last two digits must be a multiple of 4.<br></br> <br></br> If the extra digit is 5, then no multiples of 4 can be formed.<br></br> The explanation is as follows:<br></br> For the number to be a multiple of 4, it must be even, and the only even number in the selection of 1, 3, 4, 5, 5 is 4, therefore the number must end with 4.<br></br> None of 14, 34 and 54 are multiples of 4. Therefore, if the extra digit is 5, the number cannot be a multiple of 4 and consequently cannot be a multiple of 12.<br></br> <br></br> If the extra digit is 8, the only possible places for it are in the tens and the units.<br></br> This is because the only possible last 2 digits (for the number to be a multiple of 4) are 48 or 84.<br></br> Choosing 84 gives a maximum of 53184 for the solution to the original problem (if the extra digit is 8).<br></br> <br></br> The only other possibility is if the extra digit is 2.<br></br> Now the last two digits can be 12, 24, 32, or 52.<br></br> Choosing 12 is best because this leaves 5, 4 and 3 for the larger places.<br></br> The maximum using an extra digit of 2 is therefore 54312, which is larger than the 53184 produced by using 8.<br></br> <br></br> Therefore the solution is 54312.<br></br> <br></br> This is verified as a multiple of 12 because it is both a multiple of 4 and a multiple of 3.<br></br> <br></br> <br></br> <span class="editorial">Well done to you all.</span><br></br></mdoxml> <?xml version="1.0" encoding="UTF-8"?> <mdoxml version="1.0"><p><em>Do you know a quick way to check if a number is a multiple of two?<br></br> How about three, four or six?</em>... <strong><a href="/559">View the problem</a></strong></p> <h3> </h3> <h3>Why do this problem?</h3> <p>This problem offers a twist on the usual way of thinking about divisibility tests - rather than checking to see if a number is divisible by 2, 3, 4 or 6, students are asked to construct numbers that meet the necessary criteria.</p> <br></br> <h3>Possible approach</h3> <div>Introduce the problem by showing the interactivity with multiples of two selected. Students could write the answer to the computer-generated problem on individual student whiteboards, or they could discuss with their partner before offering their suggestion. If the computer does not agree with the class's suggestion, take some time to discuss why. <span style="font-weight: bold;">Challenge the class to develop a strategy for ALWAYS finding the highest even number on the first attempt.</span></div> <div> </div> <div>Repeat the activity selecting multiples of 3 - this is a good opportunity to introduce the standard divisibility test for multiples of 3 if it has not been met before.</div> <div> </div> <div>"<span style="font-weight: bold;">Your next challenge</span> is to develop a similar strategy for finding the highest multiple of 4, and the highest multiple of 6, on the first attempt."</div> <div><span style="font-style: italic;">In a computer room, pairs of students can work together at a computer to develop their strategy. Alternatively, each pair could be given a set of digit cards and select two at random to model the interactivity.</span></div> <div> </div> <div><span style="font-weight: bold;">Once the class have had time to work on the problem</span> and develop and test their strategies, bring the class together. There are many ways in which this could be done; one possibility is to challenge each pair to construct the solution to a new problem generated by the interactivity.</div> <div>"Will we be able to get all the way round the class without anyone making a mistake?"</div> <div> </div> <div><span style="font-weight: bold;">Finally</span>, the "something else to think about" at the end of the problem could be used as a homework task or in a follow-up lesson.</div> <div> </div> <div>The article <a href="http://nrich.maths.org/1308&part=">Divisibility Tests</a> offers clear explanations of the various divisibility rules and why they work.</div> <br></br> <h3>Key questions</h3> <div>How do you know if a number is a multiple of 4?</div> <div>How do you know if a number is a multiple of 6?</div> <div>How do you know you have found the biggest possible number?</div> <h3>Possible extension</h3> <div>Students could be encouraged to read the article about <a href="http://nrich.maths.org/1308&part=">Divisibility Tests</a>.</div> <div> </div> <div>Ask students to imagine that the interactivity also asks them to construct multiples of 12. Challenge them to come up with a strategy and identify situations when a multiple of 12 cannot be constructed.</div> <br></br> <p><a href="http://nrich.maths.org/public/viewer.php?obj_id=796&part=">American Billions</a> is an engaging extension activity which uses similar ideas to the ones met in this problem. </p> <br></br> <h3>Possible support</h3> <p>The interactivity in the problem <a href="http://nrich.maths.org/7431&part=">Largest Even</a> offers students practice in finding even numbers using just two digits. In a similar way, the rest of the challenges in this problem could be adapted to two-digit numbers.</p></mdoxml> <?xml version="1.0" encoding="UTF-8"?> <mdoxml version="1.0">What is special about multiples of 2? 3? 4? 6?<br></br> You may find the article on <a href="http://nrich.maths.org/public/viewer.php?obj_id=1308&part="> Divisibility Tests</a> helpful.<br></br></mdoxml> <?xml version="1.0" encoding="UTF-8"?> <mdoxml version="1.0"><br></br> <p class="editorial">This problem caught many people out - we received lots of solutions suggesting that 53184 was the answer (using an 8 as the extra number). It is indeed a multiple of 12 but it is not the largest multiple that could be made; that is 54312 (using a 2 as the extra number).</p> <p class="editorial">We received correct solutions from Livvy and Connor, Lily, Maryanne, Adam and Mitchell, Nur, Harry, Thomas, Arpan, Titus, Ed C and Ed O, Charline, Ravsimrat, Luke and Jordan, Andrew, William and Ron. Well done to you all.</p> <p class="editorial">Ravsimrat from Garden International School and Titus from GIS thought about the problem in a similar way. Here is what Titus wrote:</p> Multiples of 12 are also multiples of 2, 3 and 4.<br></br> Multiples of 3 can be found by the sum of digits which can be divided by 3. So the one more digit is limited to 8, 5 or 2.<br></br> Multiples of 4 are obtained if the last two digits together can be divided by 4.<br></br> Multiples of 2 must be even numbers.<br></br> The unit place has to be 2, 4 or 8.<br></br> My answer is 54312<br></br> <p class="editorial">Luke and Jordan from Breckland Middle School also sent us their reasoning:</p> <div>We worked out some facts about this mystery number to help us:</div> <br></br> <div>1. It must end in an even number as 12 is an even number.</div> <br></br> <div>2. To find out if a number is divisible by 3, you can add up all of the digits and see if it is a multiple of 3 that you recognise.</div> <div>From this, we know that any number containing only 1, 3, 4 and 5 will not be divisible by 3, and therefore not divisible by 12.</div> <div>Therefore, the fifth number MUST be 2, 5 or 8 for the five-digit number to be divisible by 12.</div> <br></br> <div>3. It cannot begin with 8. There are only 6 options beginning with 8 that are divisible by 2 and 3 (81354; 81534; 83154; 83514; 85314; and 85134) and none of these are divisible by 12.</div> <br></br> <div>Continued trial and error then showed 54312 to be the biggest number that is divisible by 12.</div> <p><span class="editorial">Thomas from A.Y. Jackson School sent us a detailed explanation of his thinking:</span></p> <div>Using the numbers 1, 3, 4, 5, and another number represented by N, the largest five digit number that is divisible by 12 is 54312. (N=2)</div> <div>I began by deducing any number divisible by 12 has to be divisible by 4 and 3 and must be even.</div> <div>Any number with three as a factor (its digits) must add up to a multiple of three.</div> <div>The sum of 1, 3, 4, 5 is 13, therefore, N could be only three numbers: 2, 5, 8.</div> <div>The first attempt will certainly be to choose the largest possible value of N which is 8, and 8 as the first digit, but I realize that then the last number must be 4, but none of the remaining values (3, 1 or 5) paired with 4 (34, 14, 54) could be divisible by 4 (since if a number is divisible by 4, the last two digits must also be divisible by four)</div> <div>Therefore 8 must be stuck to the rear with the number being either 53184 or 53148.</div> <div>It then becomes evident that the largest value the ten-thousands digit could hold is 5.</div> <div>If N=5, the same dilemma returns (3, 1, 5 could not be paired with 4).</div> <div>The only solution to N then becomes 2, as then 2 could replace 4 as the last digit and thus giving the number a greater thousands place-value.</div> <div>The template of the number becomes 54 _ _ 2 either 12 or 32 is divisible by four, but 54312 is obviously larger than 54132. And thus the number is 54312.</div> <br></br> <div><span class="editorial">Harry from The Beacon School also sent us a full and very clear explanation of his thinking:</span></div> <br></br> <div>As the number is a multiple of 12, it must be a multiple of 3 also. Therefore its digits must add to a multiple of 3.</div> <div>1+3+4+5=13, so the extra digit must be 2, 5, or 8 to make the sum into a multiple of 3.</div> <div>The number must also be a multiple of 4 (because it is a multiple of 12).</div> <div>For a number to be a multiple of 4, the number formed by its last two digits must be a multiple of 4.</div> <br></br> <div>If the extra digit is 5, then no multiples of 4 can be formed.</div> <div>The explanation is as follows:</div> <div>For the number to be a multiple of 4, it must be even, and the only even number in the selection of 1, 3, 4, 5, 5 is 4, therefore the number must end with 4. None of 14, 34 and 54 are multiples of 4. Therefore, if the extra digit is 5, the number cannot be a multiple of 4 and consequently cannot be a multiple of 12.</div> <br></br> <div>If the extra digit is 8, the only possible places for it are in the tens and the units. This is because the only possible last 2 digits (for the number to be a multiple of 4) are 48 or 84.</div> <div>Choosing 84 gives a maximum of 53184 for the solution to the original problem (if the extra digit is 8).</div> <br></br> <div>The only other possibility is if the extra digit is 2. Now the last two digits can be 12, 24, 32, or 52. Choosing 12 is best because this leaves 5, 4 and 3 for the larger places.</div> <br></br> <div>The maximum using an extra digit of 2 is therefore 54312, which is larger than the 53184 produced by using 8. Therefore the solution is 54312.</div> <br></br> <div>This is verified as a multiple of 12 because it is both a multiple of 4 (12 is a multiple of 4) and a multiple of 3 (5+4+3+1+2=15, which is a multiple of 3).</div> <div>The largest possible multiple of 12 that can be made using the digits 5, 4, 3, 1, X is 54312.</div> <br></br> <span style="font-style: italic;">Old solutions: The solution is 54312. Joel from ACS Barker Rd School, Singapore, sent the following solution: A number which is divisible by 12 has to be divisible by 3 and 4. A number which is divisible by 3 has the sum of its digits divisible by 3. A number which is divisible by 4 has its last two digits divisible by 4. The missing number has to be one of the last two digits, since the current 4 digits have no combination of 2 digits divisible by 4. Since the sum of the current digits is 13, the missing number is either 2, 5, or 8. Since 5, 4 and 3 are the largest numbers, we assume that the first 3 digits of our final number are 5, 4 and 3 in respective order. All numbers divisible by 4 are even numbers, and 1 is not, and so the number has to be 5431_. The only digit out of 2, 5 and 8 which will make the last two digits divisible by 4, is 2, so the final number is 54312. Other correct solutions with clear explanations came from Angela of Hethersett High School, Norfolk, Nicola of Madras College, St Andrews and Soon Xing and George from The Chinese High School in Singapore.</span><br></br></mdoxml> 2 3 0 0 1 0 0 Dozens Do you know a quick way to check if a number is a multiple of two? How about three, four or six? Numbers and the Number System Factors and multiples Numbers and the Number System Divisibility Information and Communications Technology smartphone