How old am I?

631 /www/nrich/html/content/99/03/six2/ 1 2011-02-01T00:00:01 <?xml version="1.0" encoding="UTF-8"?> <mdoxml version="1.0"><div style="float: right;"><mdo:image alt="birthday cake" height="100" src="icon.jpg" width="100"></mdo:image></div> On my last birthday, my friend said to me: "In 15 years' time, your age will be the square of your age 15 years ago!" Can you work out how old I am? This got me thinking... Was there ever a time in my life when I had other birthdays that were special in this way? Could I have said: "In 3 years' time, my age will be the square of my age 3 years ago" or: "In 4 years' time, my age will be the square of my age 4 years ago" or: "In 5 years' time, my age will be the square of my age 5 years ago" or: "In 6 years' time, my age will be the square of my age 6 years ago" or...? Can you make any generalisations about which birthdays are special in this way? Can you prove your findings? <a href="http://nrich.maths.org/8007">Click here for a poster of this problem</a>.</mdoxml> <?xml version="1.0" encoding="UTF-8"?> <mdoxml version="1.0"> We received lots of correct answers to the first part of the problem. An anonymous solver used a trial-and-improvement method: First I added the two 15s which gave me a total of 30 so I knew that the square number and its square root had to have a difference of 30. So I tried out 5 squared = 25 but there was a difference of only 20 so next I tried 6 squared which gave me 36 and that had a difference of 30. So I halved 30 which gave me 15 and I added that to 6 and the answer was 21. So he/she is 21 years old. Aporva used a neat numerical method: To solve this problem, I wrote down the square numbers: 1, 4, 9, 16, 25, 36, 49, 64, 81, 100 Then, I crossed out all the numbers under 30, since my age couldn't be negative. This left: 36, 49, 64, 81, 100 Then, I took away 30 from all the numbers, squared it, and looked if it was the same number. If it was, I took away 15 from the square number to find my age. $36 - 30 = 6, 6^2$ DOES equal $36, 36 - 15 = 21$ Therefore, my age must be 21 $49 - 30 = 19, 19^2$ DOES NOT equal $49$. I realised that it was not logical to go on since I had the answer and since the the numbers where getting too big. Kamal from St John's and Milena from Colegio Desafio in Brazil used algebra to solve the problem. Here is Milena's solution: We can solve the problems which are presented to us through quadratic equations. For the first one, for instance: $x + 15=(x - 15)^2$ $x + 15 = x^2 - 30x + 225$ $x^2 - 31x + 210 = 0$ Using the quadratic equation formula, $x= \frac{31 \pm \sqrt{31^2 - 4\times1\times210}}{2 \times 1}$ $x = \frac{(31 +11)}{2} = 21$ or $x = \frac{(31 - 11)}{2} = 10$ As he had a birthday 15 years ago, he cannot be 10 years old so his age is 21. It can be verified: $21 - 15 =6, 21 + 15 = 36, 6^2 = 36$ In order to find out if we can do with a number what was done with 15, we have to first think of a number and its square. Then, we have to subtract them; the result will always be even, for the square of an even number is always even and the square of an odd number is always odd. After that, we divide the result by two and add this quantity to the number with which we started. For an age $y$ and an integer $x$, $y = x+ \frac{x^2 - x}{2}$ For example, $6 + \frac{6^2 - 6}{2} = 6 + \frac{36 - 6}{2} = 6 + 15 = 21$ I calculated it for different numbers: $2 + \frac{2^2 - 2}{2} = 2 + \frac{4 - 2}{2} = 2 + 1 = 3$ $3 + \frac{3^2 - 3}{2} = 3 + \frac{9 - 3}{2} = 3 + 3 = 6$ $4 + \frac{4^2 - 4}{2} = 4 + \frac{16 - 4}{2} = 4 + 6 = 10$ $5 + \frac{5^2 - 5}{2} = 5 + \frac{25 - 5}{2} = 5 + 10 = 15$ Interestingly, I noted that the age y is always the sum of all integers from 1 up to the number which is represented by x in the equation: $3 = 1 + 2 $ $6 = 1 + 2 + 3$ $10 = 1 + 2 + 3 + 4 $ $15 = 1 + 2 + 3 + 4 + 5 $ $21 = 1 + 2 + 3 + 4 + 5 + 6 $ Aurimas from Chatham Grammar School, Rajeev from Haberdashers' Aske's School, Ben from Hethersett High School and Shivin from the British School in Manila, also spotted that special birthdays occur when your age is a triangular number. Connor from Gladesmore School described a way to work out the special birthdays: Find the age squared, for example $6^2 = 36$, and then subract your original number: $36 - 6=30$. Then halve this and add it to your original number: $\frac{30}{2}=15, 6+15=21$. Using algebra to express this gives: $$\frac{n^2-n}{2} + n = \frac{1}{2}n^2 + \frac{1}{2}n$$ Muntej from Wilson's School explained how he worked out that the age had to be a triangular number: This problem will only work if your age is a triangular number. Why? If I start by doing the squares of various numbers with the difference between the square number and its root: $3\times3=9$ difference $6$ $4\times4=16$ difference $12$ $5\times5=25$ difference $20$ Now halve the differences: $\frac{6}{2}=3, \frac{12}{2}=6, \frac{20}{2}10$ We can see a pattern emerging. When the difference is halved, the sequence of triangular numbers appears. We must use the difference as the age is halfway between the square and the square root. The triangular numbers are always halfway between a square number and its root, therefore "in N years time my age will be the square of how old I was N years ago" only works if N is a triangular number. Toby from Repton School used algebra to explain: Using the quadratic equation formula: $$x=\frac{-b \pm \sqrt{b^2-4ac}}{2a}$$ $x$ is a whole number if $b^2-4ac$ is a square number. $(x-n)^2=x+n$ rearranges to give $x^2-(2n+1)x+(n^2-n)$. So $x$ is a whole number if $(2n+1)^2-4(n^2-n)$ is a square number. So $8n+1$ must be a square number. As $8n+1$ is clearly odd, our square number must be odd, $2r+1$, say. $$8n+1 = (2r+1)^2$$ $$\Rightarrow 8n = (2r+1)^2-1$$ $$\Rightarrow 8n = (2r+1+1)(2r+1-1)$$ $$\Rightarrow 8n = (2r+2)(2r)$$ $$\Rightarrow n=\frac{4r(r+1)}{8} = \frac{r(r+1)}{2}$$ Therefore the birthdays that are special are triangular numbers. Suzie from St Saviours & St Olaves also used algebra and imagined the three ages in the problem as the ages of three different people: My age is $21$. $(21 - 15 = 6, 21 + 15 = 36)$ $A$, $B$ and $C$ are three people that fill the same rules, so if $A$ is $6$, $B$ is $21$ and $C$ is $36$ then these three people could have stood in for me in the initial problem Looked at from $B$'s point of view this problem always looks difficult, but from $A$'s point of view $A$'s age is $n$ (any number); $C$'s age is $n \times n$ and $B$'s age is $\frac{C - A}{2} + A = \frac{n^2 - n}{2} + n = \frac{n(n+1)}{2}$ So, for every age of $A$, $n$, there is a person aged $B$, $\frac{n(n+1)}{2}$, and a person aged $C$, $n^2$, and the differences between the ages is $\frac{n(n-1)}{2}$. Finally, Daniel from Savile Park Primary sent us <a href="/content/99/03/six2/Danielsol.pdf" class="editorial">this solution</a>. </mdoxml> <?xml version="1.0" encoding="UTF-8"?> <mdoxml version="1.0"> <h3>Why do this problem?</h3> In <a href="/631">this problem</a> students have the opportunity to create quadratic equations and solve them by factorisation or by using the quadratic formula. The follow-up questions offer the chance for some interesting generalisations and justifications. <h3>Posssible approach</h3> This task is ideal for students who have already been introduced to solving quadratic equations. Set the initial challenge to the whole class to work on individually or in pairs: "On my last birthday, my friend said to me: 'In 15 years' time, your age will be the square of your age 15 years ago!' Can you work out how old I am?" As they are working, move around the class to identify the different methods students are using to solve it (eg trial and improvement, algebraically using factorisation, algebraically using the quadratic formula). Once students have had a chance to make some progress with the problem, bring the class together and share approaches. If nobody has tried an algebraic approach, model it with the class on the board. "I wonder whether this sort of thing could happen for other ages." Suggest that they choose a statement of the form "In n years' time, my age will be the square of my age n years ago" for values of n between 2 and 30. Those who are confident might be encouraged to choose larger values of n. Ask students to report, as they go along, which values of n lead to a special age, and which don't - these could be collected on the board for all to see. Once results have been collected, a pattern should emerge as to what is special about the ages that work. "Can anyone suggest a large value of n that you think might work?" (For example, if I am 210, my age in 190 years' time is the square of my age 190 years ago!) To prove that the pattern will continue to hold, it may be necessary to introduce an algebraic representation for triangular numbers. Expressions for two consecutive triangle numbers can be used as the basis for the proof. <a href="http://nrich.maths.org/2274&part=">Picturing Triangle Numbers</a> provides a nice way to introduce the expression $\frac{n(n+1)}{2}$. Alternatively, the pictorial representation of triangle numbers can be used to construct a visual proof. <h3>Key Questions</h3> If I am $x$ years old now how old was I 15 years ago? If I am $x$ years old now how old will I be in 15 years' time? Can you use these expressions to form an equation to solve? <h3>Possible extension</h3> The proof that has been suggested shows that triangle numbers are always special ages; to prove that it is only the triangle numbers that have this property is a suitable challenging extension. <h3>Possible support</h3> A simpler route into the task could be to start by considering someone whose age in 3 years' time is the square of their age 3 years ago. An alternative strategy to solve the problem without needing to solve quadratic equations involves thinking about the difference between numbers and their squares. </mdoxml> <?xml version="1.0" encoding="UTF-8"?> <mdoxml version="1.0">If I am $x$ years old now how old was I 15 years ago? If I am $x$ years old now how old will I be in 15 years' time? Can you use these expressions to form an equation to solve? </mdoxml> <?xml version="1.0" encoding="UTF-8"?> <mdoxml version="1.0"> This one produced a bumper crop! Well done the following who touched first base: KS3 Maths Club, Strabane Grammar School; Lucinda, Stamford High School, Jack, Hethersett High School; Claire, Madras College; Arwa, The International School, Brunei and Jonathan & Charlie from St Peter's College, Adelaide, Austral ia. Kang of The Chinese High School of Singapore used algebra as follows: Let Reza's present age be $x$ $x+3 = (x-3)^2$ $\Rightarrow x+3 = x^2 - 6x+9$ $\Rightarrow x^2-7x+6=0$ $\Rightarrow (x-1)(x-6)=0$ Therefore either $x = 1$ or $6$ Since Reza is at least $3$ years old, he must be $6$. Renate and Amy of Hethersett High School suggested a very good method (see the table below). They noted that you can start with column (iii), then fill column (iv) with the squares of the numbers in (iii), then the age in column (i) is the mean of columns (iii) and (iv). Similar tables came from Y8, Y9 and Y10 The Mount School York where they noticed that the ages are triangle numbers, as did Ying of Tao Nan School, Singapore, and also Mark ofKing Edward VI High School, Birmingham and Steven ofBedlington High School, Nothumberland. <table border="1"> <tbody> <tr> <th width="100">(i)</th> <th width="100">(ii)</th> <th width="100">(iii)</th> <th width="100">(iv)</th> </tr> <tr> <th>Age</th> <th>$k$</th> <th> <div>Age $k$ years ago</div> </th> <th>Age $k$ years ahead</th> </tr> <tr> <td>$1$</td> <td>$0$</td> <td>$1$</td> <td>$1$</td> </tr> <tr> <td>$3$</td> <td>$1$</td> <td>$2$</td> <td>$4$</td> </tr> <tr> <td>$6$</td> <td>$3$</td> <td>$3$</td> <td>$9$</td> </tr> <tr> <td>$10$</td> <td>$6$</td> <td>$4$</td> <td>$16$</td> </tr> <tr> <td>$15$</td> <td>$10$</td> <td>$5$</td> <td>$25$</td> </tr> <tr> <td>$21$</td> <td>$15$</td> <td>$6$</td> <td>$36$</td> </tr> <tr> <td>$28$</td> <td>$21$</td> <td>$7$</td> <td>$49$</td> </tr> <tr> <td>$36$</td> <td>$28$</td> <td>$8$</td> <td>$64$</td> </tr> <tr> <td>$45$</td> <td>$36$</td> <td>$9$</td> <td>$81$</td> </tr> <tr> <td>$55$</td> <td>$45$</td> <td>$10$</td> <td>$100$</td> </tr> </tbody> </table> If you look at the first two columns, another pretty pattern emerges from the fact that every square number splits into two consecutive triangle numbers. Try investigating this with dot patterns and algebra. </mdoxml> 2 4 0 0 0 1 0 How old am I? In 15 years' time my age will be the square of my age 15 years ago. Can you work out my age, and when I had other special birthdays? Using, Applying and Reasoning about Mathematics Making and proving conjectures Numbers and the Number System Patterned numbers Algebra Quadratic equations Using, Applying and Reasoning about Mathematics Trial and improvement Information and Communications Technology smartphone Admin Stage 3&4 Investigation Secondary Mapping Document Equations and formulae US Secondary Mapping Document DisplayCabinet