Always a multiple?

7208 /www/nrich/html/content/id/7208/ 1 0000-00-00T00:00:00 <?xml version="1.0" encoding="UTF-8"?> <mdoxml version="1.0">Watch the video to see Charlie's number trick. <video controls="controls" height="315" src="7208%20Always%20a%20multiple.mp4" tabindex="0" width="420"></video> If you can't play the video, you can read a description <a href="/7208/clue">here</a>. Try a few examples for yourself. Do you always get a multiple of 11? Can you explain why? Alison and Charlie came up with their own explanations: <video controls="controls" height="315" src="7208%20Always%20a%20Multiple%20-%20AlisonExplanation.mp4" tabindex="0" width="420"></video> <video controls="controls" height="315" src="7208%20Always%20a%20Multiple%20-%20CharlieExplanation.mp4" tabindex="0" width="420"></video> If you can't play the videos, you can read a description <a href="/7208/clue">here</a>. Here are some similar number tricks. Can you use Charlie's or Alison's representation to explain how they work? <ul> <li>Take any two-digit number. Reverse the digits, and subtract your answer from your original number. What do you notice? </li> <li>Take any two-digit number. Add its digits, and subtract your answer from your original number. What do you notice? </li> <li>Take any three-digit number. Reverse the digits, and subtract your answer from your original number. What do you notice? </li> <li>Take any five-digit number. Reverse the digits, and subtract your answer from your original number. What do you notice?</li> </ul> Once you've been able to explain what is going on above you should be able to explain why many other similar tricks work. Here is a selection you might like to try: <a href="http://nrich.maths.org/2129&part=">Special Numbers</a> <a href="http://nrich.maths.org/1170&part=">Think of Two Numbers</a> <a href="http://nrich.maths.org/564&part=">Legs Eleven</a> </mdoxml> <?xml version="1.0" encoding="UTF-8"?> <mdoxml version="1.0"> Michelle and Cathy, from Mount Waverley, started off by checking a few examples: 21: reverse 12 21+12= 33 33 = a multiple of 11. 25: reverse 52 25+52=77 77 = a multiple of 11. 18: reverse 81 18+81=99 99 = amultiple of 11. 31: reverse 13 31+13=44 44 = a multiple of 11. That's a positive start - it looks like all two-digit numbers are going to behave like this! Hana, from Avenue House School, did a different kind of check: I arranged multilink to show four tens and two units for 42, and two tens and four units for 24. I then put the four units with the four tens, and the two units with the two tens, giving six lots of eleven. Interesting method - does it work for Michelle and Cathy's examples above too? Can we explain why this seems to be true for all two-digit numbers? Mariella, from Westwood Girls' College, wrote the following: The number will always be a multiple of 11 because if (for example) I have the number 13, and I switch the digits, I will then have the number 31. Then I would add the numbers in the tens column, and get the number 4. Then I would add the numbers in the units column, and get the number 4 - again. The reason for this is that the two numbers in the tens and units columns are exactly the same, just swapped around. The first few numbers of the 11 times tables, have the same numbers in the tens and units column making it a member of the 11 times table. If the two numbers add to make a number bigger than 100, they will still make a number in the 11 times tables because; if (for example) I have the number 89 when I switch the digits I will have the number 98, and when I add the numbers I get the answer 187. 187 is made up of 17 units (17) and 17 tens (170): 17x10 +17x1 =17x11 I can see that 17 x 11 (187) will be in the 11 times tables. Great - I'm convinced! What about the next problems? Take any two-digit number. Reverse the digits, and subtract your answer from your original number. What do you notice? Rupa, from Garden International School, solved this using algebra: Using algebra, let a be the tens number and b be the units number. For the number 'ab', the number is worth 10a+b. For 'ba', the number is worth 10b+a. So, (10a+b) - (10b+a) = 9a-9b. Then you can factorize it to get 9(a-b). This shows that 9 is the common factor. For example: 54. Reverse the number, you get 45. And then 54-45= 9. Another example, just to check: 76. Reverse, 67. Then 76-67=9. Great - well spotted! Take any two-digit number. Add its digits, and subtract your answer from your original number. What do you notice? Monel and Rasel, from Globe Academy, gave us their thoughts on this problem: If you have 36 (for example) and you take away the units digit first, then you will always get a multiple of 10. If you then take away the tens digit you will get a multiple of 9 (because 3x10 - 3x1 = 3x9). Nice! Maisy, from England, also solved this using algebra: The number ab (tens: a, units: b) can be written as 10a + b. The digits are a and b, so when you add them you get (a + b). Then (10a + b) - (a + b) = 9a + 0b = 9a (always a multiple of 9). Take any three-digit number. Reverse the digits, and subtract your answer from your original number. What do you notice? Rupa wrote: If a is the hundreds digit, b is the tens digit and c is the units digit, the number would look like this: 100a + 10b + c. After reversing the digits, it would be 100c + 10b + a. Subtracting gives: (100a + 10b + c) - (100c + 10b + a). Simplifying gives: 99(a-c) which means that, for every three digit number, you always get a multiple of 99. Take any five-digit number. Reverse the digits, and subtract your answer from your original number. What do you notice? Maisy wrote: We want to do abcde - edcba: (10,000a + 1000b + 100c + 10d + 1e) - ( 1a + 10b + 100c + 1000d + 10,000e) = 9999a + 990b + 0c - 990d - 9999e. Simplifying gives: 99(101a + 10b - 10d - 101e) which is always a multiple of 99. Rupa also had a go at a couple of the other tricks: <a href="/2129">Special numbers</a> If the tens unit is replaced with a, and the units digit is replaced with b, the value of the original number is 10a+b. The challenge is to arrive at the original number when adding (a+b) to ab. The equation would look like this: (a+b)+ab=10a+b So ab=9a So b=9 <a href="/1170"> Think Of two Numbers</a> Think of two whole numbers under 10.  Take one of them and add 1. Multiply by 5. Add 1 again.  Double your answer.  Subtract 1.  Add your second number.  Add 2.  Double again.  Subtract 8.  Halve this number and tell me your answer. If the two numbers were a and b, each step of the sequence would look like this: a and b a+1 5(a+1)=5a+5 5a+6 10a+12 10a+11 10a+11+b 10a+13+b 20a+26+2b 20a+18+2b 10a+9+b When you are told the answer, subtract 9. You will be left with a two-digit number; the digits of this number will be the numbers that your friend thought of. Fantastic. Thanks!</mdoxml> <?xml version="1.0" encoding="UTF-8"?> <mdoxml version="1.0"><h3>Why do this problem?</h3> This problem introduces students to a useful technique (representing numbers algebraically according to their place value) for solving a wide variety of related problems. <h3>Possible approach</h3> <div> "Think of a two-digit number and write it down." "Reverse the digits and add your answer to your original number." "What answers did you get?" Collect a few students' answers together and write them up on the board. "Does anyone notice anything interesting?" "Multiples of 11." "Does anyone have an answer that isn't a multiple of 11?" "With your partner, without trying all possible two-digit numbers, try to find a convincing explanation why it will always work." Give students some time to explore the problem. While they are working, circulate and listen for useful insights. Then bring the class together and share ideas. If Alison's and Charlie's explanations from the video aren't offered, demonstrate them or show the video. Students could also be invited to work backwards - for example, what two-digit numbers can be reversed and added together to give 154 (a multiple of 11)? "These methods can be used for lots of similar number tricks. Here are <a href="/content/id/7208/Always%20a%20Multiple.pdf">a few more</a>. Work with your partner to figure out what each trick does, and then adapt the methods to explain why the tricks work." If appropriate, bring the class together to share explanations for why each trick works, or ask them to present their clear explanations on a poster to display. Finally, challenge students to devise their own number tricks using similar structures, and to test them out on each other. </div> You may wish to show this <a href="http://www.flashlightcreative.net/swf/mindreader/">online 'mind reader' activity</a> to see if your students can explain how it works. Perhaps you could show it at the start of the lesson (without an explanation) and then again at the end of the lesson once they have the tools to deconstruct it. <h3> </h3> <h3>Possible extension</h3> These problems can all be solved using similar techniques: <a href="http://nrich.maths.org/2129&part=">Special Numbers</a> <a href="http://nrich.maths.org/1170&part=">Think of Two Numbers</a> <a href="http://nrich.maths.org/564&part=">Legs Eleven</a> <h3>Possible support</h3> <a href="/2791">Diagonal Sums</a> provokes a need to use place value to solve the problem, and could be a good foundation for this activity. </mdoxml> <?xml version="1.0" encoding="UTF-8"?> <mdoxml version="1.0">First video: <div class="toggle">Charlie said: "Alison, think of a two-digit number. Reverse the digits and add your answer to your original number. I bet your answer is a multiple of 11." Alison chose 42, added 24 and got the answer 66: "It is! How on earth did you know that?" Charlie said: "I'm not sure. Let's try to work it out."</div> Second video: <div class="toggle">Alison arranged multilink to show four tens and two units for 42, and two tens and four units for 24. She then put the four units with the four tens, and the two units with the two tens, giving six lots of eleven. Charlie imagined a two-digit number $ab$, where $a$ represents the number in the tens column, and $b$ respresents the number in the units. This can be written as $10a+b$. Similarly, $ba$ can be written as $10b+a$. Charlie added these together to get $11a+11b$, which he wrote as $11(a+b)$.</div> </mdoxml> 2 4 0 0 1 0 0 Always a multiple? Think of a two digit number, reverse the digits, and add the numbers together. Something special happens... Numbers and the Number System Place value Algebra Creating expressions/formulae Using, Applying and Reasoning about Mathematics Making and proving conjectures Information and Communications Technology Video