Special Numbers

2129 /www/nrich/html/content/02/08/penta3/ 1 2011-02-01T00:00:01 <mdoxml version="1.0">One day our teacher asked us a puzzling question: <mdo:image alt="" height="427" src="Special.png" width="771"></mdo:image> Can you find a special number? Can you find more than one? Can you find them all? One way to convince yourself that you've found all the special numbers is to use a little algebra... Any two-digit number can be represented algebraically as $10a + b$. $a$ is the tens digit and $b$ is the units digit. Write down the algebraic expressions for the sum and the product of the digits. Now write an equation that all special numbers will satisfy. See the <a href="http://nrich.maths.org/2129&part=clue">hint</a> if you're not sure how to do this. Solve the equation, and use it to convince yourself that you have found all the special numbers. There are other sorts of special two-digit numbers... <ul> <li>I add twice the tens digit to the units digit, then add this to the product of the digits. I get back to my original number. </li> <li>I add three times the tens digit to the units digit, then add this to the product of the digits. I get back to my original number. </li> <li>I add four times... or five times... or...</li> </ul> <div>Can you use algebra to help you to find these special numbers?</div> <a href="http://nrich.maths.org/8015">Click here for a poster of this problem</a>.</mdoxml> <mdoxml version="1.0"> We received a mass of good solutions to this problem. Aileen, Becca, and Paulina from Walter J. Paton School described how they arrived at the solution: The problem "Special Numbers" was really fun to figure out. At first, we didn't understand what it meant. But after a little figuring we noticed a pattern. Here is what we did: We just tried random numbers and came across the number 18. It was only 1 off, so we tried 19. The sum of 1+9 was 10 and the product of 1x9 was 9. After that, we added 10+9 and it equaled 19! That was our original number! We found a special number! But, there was still another question: is there more than one special numbers? As we tried to figure out more we tried the number 29. That worked too! After a little observation, we noticed 9 was in the ones (units) place both times! We tried more numbers such as 39 and 49 and so on. Those all worked! We had found a pattern! That was how we figured out that problem. Joseph, from the same school, and May, from Heartlands High School, showed some examples that worked: 29: 2 + 9 = 11 2 x 9 = 18 11 + 18 = 29 39: 3 + 9 = 12 3 x 9 = 27 12 + 27 = 39 49: 4 + 9 = 13 4 x 9 = 36 13 + 36 = 49 59: 5 + 9 = 14 5 x 9 = 45 14 + 45 = 59 69: 6 + 9 = 15 6 x 9 = 54 15 + 54 = 69 Joanna, Sophie and Gaby from Kings Sutton Primary School discovered these numbers and then moved on to other sets of special two-digit numbers: On the first problem we divided up the numbers between the 3 of us and worked on different sections. Joanna started with 0 to 33, Sophie did 34 to 66 and Gaby did the rest. Gaby found 69 worked and at about the same time Joanna found that 19 worked. So then we decided to try every 2 digit number ending in a 9. They all worked. The algebra was a bit tricky so we looked at the second part and this is what we found: We found that if you double the tens then the units have to be 8, and if you triple the tens it has to end in 7. We then noticed a pattern, as the number we multiplied by increases the units value decreases. We tested out our hypothesis and we found it worked. Eddie, from Wilson's School, did use some algebra to explain his results: Firstly, to make this problem simpler to solve, we can use algebra. This means substituting the first digit of the special number (the tens digit) for 'a', and substituting the second digit (the units digit) for 'b'. So, the product of the digits (ab) + the sum of the digits (a + b) must equal the original number (10a + b). As an equation, this is written as: ab + a + b = 10a + b. If you take away 'b' from both sides this equates to: ab + a = 10a Then, if you take away one 'a' from both sides you get: ab = 9a This is because 10 lots of 'a' take away one lot of 'a' is 9 lots (9a). This can be written as: b x a = 9 x a. Therefore, when we divide both sides by 'a' we get: b = 9 This means the second digit (b) of the special number must equal 9. Now we can check if this works. Let's substitute 'a' for 1 and 'b' for 9: 1 x 9 + 1 + 9 = 10 x 1 + 9 19 = 19 It works. Now substitute 'a' for 2: 2 x 9 + 2 + 9 = 10 x 2 + 9 29 = 29 Does 3 work? 3 x 9 + 3 + 9 = 10 x 3 + 9 39 = 39 And so on. 4, 5, 6, 7, 8 and 9 work too. Now I will provide a reason for why the special numbers go up by 10: If 'a' is 1, and 'b' is 9, then the number is 19. The substitutions above show that this works. If 'a' is 2, then the number is 29 (10 more than 19). The difference between 1 x 9 and 2 x 9 is +9, and the difference between 1 + 9 and 2 + 9 is +1. Add the differences to get a total difference of +10. Repeat this paragraph with the successive two numbers in the pattern and you will find that it works. Ben Page from Hethersett High School also used some algebra to explain his results: 1x9 + 1 + 9 = 19 2x9 + 2 + 9 = 29 3x9 + 3 + 9 = 39 4x9 + 4 + 9 = 49 5x9 + 5 + 9 = 59 6x9 + 6 + 9 = 69 7x9 + 7 + 9 = 79 8x9 + 8 + 9 = 89 9x9 + 9 + 9 = 99 ab + a + b = 10a + b ab + a = 10a ab = 9a b= 9 So all the two digit "special numbers" end in 9. Jamie from Mold Alun School explained why the special numbers increased by 10: Any 2-digit number ending in 9 is one of the special numbers. Starting with 19 you have 1 + 9 = 10 and 1x9 = 9 10 + 9 = 19 If you increase 19 by 10, the sum increases by 1 and the product increases by 9, giving you the extra 10 you need to make up the next number, 29. Every time you increase the previous special number by 10, the sum increases by 1 and the product increases by 9, giving you the extra 10 you need to make up the next number. e.g. for 29: 2 + 9 = 11, 1 more than the sum of the digits of 19, and 2 x 9 = 18, 9 more than the product of the digits in 19. Lena from SHHS found two sets of special numbers: The expression for a two digit number is 10a+b. To solve the first special number you create an algebraic equation. You need to add the sum of the two digits to the product of the two digits to get the original number. The equation looks like this: 10a + b = a + b + ab 9a + b = b + ab 9a = ab b = 9 So now you know the units part of the two digit number will always be 9. A different special number could be to add twice the tens digit to the units digit, then add this to the product of the digits and this gives back to the original number. The equation would look like this: 10a + b = 2a + b + ab 8a + b = b + ab 8a = ab b = 8 So you know the the second digit in the two digit number will always be 8. For example 28 is a special number because 2x2 + 8 + 2x8 = 12 + 16 = 28 Joseph, from Wilson's Grammar School, chose ab to represent a two digit number and showed how to find special numbers that satisfied the different criteria: a + b + ab = 10a + b a + b - b + ab = 10a a + ab = 10a ab = 10a - a ab = 9a b = 9 So the special numbers are 19, 29, 39, 49, 59, 69, 79, 89, 99 2a + b + ab =10a + b 2a + ab = 10a ab = 10a - 2a ab = 8a b = 8 So in this case the special numbers are 18, 28, 38, 48, 58, 68, 78, 88, 98 3a + b + ab = 10a + b 3a + ab = 10a ab = 10a - 3a ab = 7a b = 7 So in this case the special numbers are 17, 27, 37, 47, 57, 67, 77, 87, 97 4a + b + ab = 10a + b 4a + ab = 10a ab = 10a - 4a ab = 6a b = 6 So in this case the special numbers are 16, 26, 36, 46, 56, 66, 76, 86, 96 5a + b + ab = 10a + b 5a + ab = 10a ab = 10a - 5a ab = 5a b = 5 So in this case the special numbers are 15, 25, 35, 45, 55, 65, 75, 85, 95 Rajeev from Haberdashers' Aske's Boys' School found a general rule: When the tens digit is multiplied n times, the units digit of the special numbers will be 10-n. Max from Twyford C of E High School came to the same conclusion: When it's 2 times the tens digit you take 2 away from 10 and that gives you the value of the units digit of the special numbers. When its 3 times, or 4 times, etc, you just subtract it from 10 to give you the value of the units digit of the special numbers. Well done to you all. </mdoxml> <mdoxml version="1.0"> <h3>Why do this problem?</h3> This problem provides a great opportunity to introduce a very useful way of representing numbers algebraically. <h3>Possible approach</h3> <div>Present the learners with the problem displayed in the speech bubbles. Give the class some time to look for special numbers - once a few have been found, more will quickly follow. Once there is a consensus that a family of special numbers has been found, bring the class together and share what they think is going on. Ask them "Are you sure you've found all the special numbers? How could we be certain?" Suggest that an exhaustive search would be exhausting!</div> <div> </div> <div>Introduce the class to the algebraic representation of two-digit numbers and work through the reasoning together to establish that there are only nine solutions.</div> <div> </div> <div>Then present the suggestions for other types of special numbers, and challenge learners to predict, using algebra, which numbers will be special for each type.</div> <h3>Key questions</h3> <div>Can we use algebra to help us to understand what's going on?</div> <h3>Possible extension</h3> <div><a href="http://nrich.maths.org/1170&part=">Think of Two Numbers</a> requires some of the same algebraic skills.</div> <div><a href="http://nrich.maths.org/564&part=">Legs Eleven</a> is a very challenging extension.</div> <h3>Possible support</h3> For students who find the leap into algebra too difficult, the problem can be used to offer an interesting numerical challenge. </mdoxml> <mdoxml version="1.0">Try some different numbers to get a feel for the problem. Special numbers all satisfy the following equation: $$a + b + ab = 10a + b$$ </mdoxml> <mdoxml version="1.0"> Lots of you sent in particular numbers that you spotted. Kristian noticed that $19$ works: $1+9=10$ $1 \times 9=9$ $10+9=19$ Daniel found that $29$ works: $(2+9)+(2 \times 9) = 11+18=29$ Bill and Ben discovered that $39$ works: Because $3 + 9 = 12$ and $3 \times 9 = 27$ and $27 + 12 = 39$ Class 5AA at Raglan Junior sent us something interesting: Any two digit number which ends in $9$ will give you the solution. However you can't use $99$ as the digits are the same. But why does this work? Daniel, from Bacons College, sent us his explanation: The $2$ digit number has $a$ tens and $b$ units, so I can write the equation for this question like this: $a + b + a \times b = 10 \times a + b$ so: $a + a \times b = 10 \times a$ so: $ a \times (1 + b) = 10 \times a$ so: $b = 9$ and it turns out that $a$ can be anything! I think that what Daniel means is that a can be any of $1, 2, 3, 4, 5, 6, 7, 8,$ because it can't be $9$ (as Class 5AA pointed out, $99$ isn't allowed). But the idea of using algebra is a good one, if you've met it. Otherwise, you could just try all $8$ numbers. </mdoxml> 2 3 0 0 1 0 0 Special Numbers My two digit number is special because adding the sum of its digits to the product of its digits gives me my original number. What could my number be? Algebra Introducing algebra Numbers and the Number System Properties of numbers Using, Applying and Reasoning about Mathematics Working systematically Algebra Creating expressions/formulae Information and Communications Technology smartphone