Pair Products

2278 /www/nrich/html/content/id/2278/ 1 2011-02-01T00:00:01 <?xml version="1.0" encoding="UTF-8"?> <mdoxml version="1.0"><mdo:image align="top" alt="A pair of juicy green pears." bgcolor="" src="pears.gif" style="width: 104px; height: 120px; float: right;"></mdo:image> Choose four consecutive whole numbers. Multiply the first and last numbers together. Multiply the middle pair together. Choose several different sets of four consecutive whole numbers and do the same. What do you notice? Can you explain what you have noticed? Will it always happen? Click below to see how Charlie and Alison explained what they noticed. Charlie said: <div class="toggle">I noticed that the product of the outer pair was always $2$ less than the product of the inner pair. I can explain this by labelling the four consecutive numbers $n, n+1, n+2, n+3$. Outer pair: $n(n+3) = n^2 + 3n$ Inner pair: $(n+1)(n+2) = n^2 + 3n + 2$ </div> Alison said: <div class="toggle"> <mdo:image src="Alison.png" style="float: right;"></mdo:image> I drew a diagram, in which the product of each pair is represented by the area of a rectangle: The outer pair is represented by the red rectangle. The inner pair is represented by the blue rectangle. The purple area is common to both. The area of the red strip will always be two units less than the area of the blue strip. Therefore, the product of the outer pair is always two less than the product of the inner pair. </div> Instead of doing lots of calculations, can you use these representations to compare the product of the first and last numbers with the product of the second and penultimate numbers, when you have: <ul> <li>$5$ consecutive whole numbers </li> <li>$6, 7, 8, \ldots x$ consecutive whole numbers </li> <li>$4$ consecutive even numbers </li> <li>$4$ consecutive odd numbers </li> <li>$5, 6, 7, 8, \ldots x$ consecutive even or odd numbers </li> <li>$4$ consecutive multiples of $3, 4, 5 \ldots $</li> </ul> <ul> <li>$1.2, 2.2, 3.2, 4.2$ </li> <li>$2, 5, 8, 11$ </li> <li>$4, 4\frac{1}{2}, 5, 5\frac{1}{2}$ </li> </ul> Make up a few similar questions of your own. Impress your friends by giving them a calculator and 'predicting' what will happen! <a href="http://nrich.maths.org/public/viewer.php?obj_id=5978&part=">Click here for a poster of this problem</a>.</mdoxml> <?xml version="1.0" encoding="UTF-8"?> <mdoxml version="1.0">Kira, from the UK, started off by doing some calculations: Say your numbers were 4, 5, 6, 7. When you times the outer numbers 4 and 7 it will equal 28, and when you times the two middle numbers 5 and 6 it will equal 30. Let's try it again on four new consecutive numbers, such as 1, 2, 3, 4: when you times the outer numbers 1 and 4 it equals 4, and when you times the two middle numbers 2 and 3 it equals 6. Conclusion: the product of the outer numbers will always be 2 less than the product of the numbers in the middle. Well spotted! A student from Germany drew rectangles, like in Alison's explanation, to see why this was true. The class at Elgin Academy had the clever idea of testing this for some negative numbers too! Sam, from Park Grove Primary, gave the following explanation: If you choose the first number of the set to be n, then the outside multiplication will be n(n+3), which simplifies to $n^2+3n$. The inside multiplication will be (n+1)(n+2), which simplifies to $n^2+3n+2$. If you look back at the outside multiplication, you can see that it is 2 less than the inside multiplication, proving the answer. Great! Nathan also gave a convincing explanation: Call the number halfway in between the middle two x. This is the second number in our list plus 0.5, or the third number minus 0.5, so if we multiply the second and third numbers together we will get $x^2-0.25$. But x is also the first number plus 1.5, or the fourth number minus 1.5, so if we multiply them together we will get $x^2-2.25$. Nice - here Nathan is making use of the identity: $(x-a)(x+a) = x^2 - a^2$. Well done! Keone, from Sage Ridge School in Reno, started with four consecutive whole numbers: The product of the first and last numbers is always $2$ less than the product of the middle two numbers. Explanation: Suppose the first number is $x$. Then the second number is $x+1$, the third is $x+2$, and the fourth is $x+3$. So the product of the first and fourth numbers is $x(x+3) = x^2 + 3x$. Also, the product of the second and third numbers is $(x+1)(x+2) = x^2 + 3x + 2$. So $(x+1)(x+2) = x(x+3) + 2$ for any chosen value of $x$. When he considered five consecutive whole numbers he found that: The product of the first and last numbers is always $3$ less than the product of the second and fourth numbers. Explanation: Again, let the first number be $x$; then the second number is $x+1$, the third is $x+2$, the fourth is $x+3$, and the fifth is $x+4$. So the product of the first and last numbers is $x(x+4) = x^2 + 4x$. Also, the product of the second and fourth numbers is $(x+1)(x+3) = x^2 + 4x + 3$. So $(x+1)(x+3) = x(x+4) + 3$ for any chosen value of $x$. And with $n$ consecutive whole numbers he found that: The product of the first and last numbers is always $n-2$ less than the product of the second and penultimate numbers. Let us consider the general case where there are n consecutive whole numbers. As before, let the first number be $x$; then the second number is x+1, the third is x+2, and so on. The last number (the $n$th number) will be $x+n-1$. Thus, the penultimate (second-to-last) number will be $x+n-2$. So the product of the first number and the last number will be $x(x+n-1) = x^2 + nx - x$ The product of the second and the penultimate numbers will be $(x+1)(x+n-2) = x^2 + nx - 2x + x + n - 2 = x^2 + nx - x + n - 2$ So $(x+1)(x+n-2) = x(x+n-1) + n - 2$; that is, the product of the second and penultimate numbers will always exceed the product of the first and last numbers by exactly $n - 2$. For example, if we take 6 numbers, the product of the $2$nd and $5$th numbers will be $4$ more than the product of the $1$st and last numbers. Natasha, from the European School, generalised her findings in a similar way and then went on to check her conclusion: We can generalise this problem by substituting particular numbers with letters. Let the first number be $a$. If there are $n$ numbers, where a is the first, then the last number is $(a+n-1)$. The second number is $(a+1)$, the penultimate number $(a+n-2)$. By multiplying the first and last numbers together, we get $a(a+n-1) = a^2 +an-a$ Multiplication of the second and penultimate numbers gives $(a+1)(a+n-2) = a^2+an-a+n-2$ The difference therefore in the product of the first and last numbers and the product of the second and penultimate numbers is always $n-2$. For added confirmation we can take a random example: Consider the numbers $56, 57, 58, 59, 60, 61, 62$ where $n = 7$ and $a = 56$ Our general formula tells us that the difference in the product pairs should be $n-2$ (i.e. $5$). When we do the calculation, we get $$57 \times61 - 56 \times62 = 3477 - 3472 = 5$$ This result corresponds with the general one established above. Tom explained it a different way: <mdo:image alt="" height="106" src="rectangles.JPG" width="420"></mdo:image> If we overlap the two rectangles we cut a piece off at the bottom and create a new piece at the right hand side. The difference between the two small rectangles is always two. This works for all numbers, not just $9, 10, 11, 12$. We received a slightly different summary of findings from Aisling, from Grand Avenue Primary School: The product of the first and last numbers of a series of consecutive whole numbers, is the same as the product of the second and penultimate numbers of that series, minus the number of numbers separating the first and last numbers. Well done to you all for such clear reasoning. </mdoxml> <?xml version="1.0" encoding="UTF-8"?> <mdoxml version="1.0"> <h3>Why do this problem?</h3> <div> </div> <div><a href="/2278">This problem</a> provides a purpose for practising the routine algebraic procedure of expanding brackets.</div> <div>For students who are unfamiliar with algebra, this problem is an excellent context for observing, conjecturing and thinking about proof. It can be a good introduction to the power of algebra, and to a related geometrical interpretation.</div> <div> </div> <h3>Possible approach</h3> <div> </div> <div>"Choose four consecutive numbers, multiply the outer pair and the inner pair. What were your two answers?"</div> <div>Write a selection of students' responses on the board.</div> <div>"What do you notice?" The inner pair product is always two more than the product of the outer pair.</div> <div> </div> <div>"Will this always happen? Can you explain why?"</div> <div>Give students some time to discuss with their partner why the answers always differ by two. Circulate and listen out for interesting insights.</div> <div>Bring the class together and share any explanations they have found. Perhaps share Charlie's and Alison's representations from the problem if they haven't emerged.</div> <div>"We've worked out what happens when you find the product of the inner and outer pair of a set of four consecutive numbers. What questions do you think a mathematician might ask next?"</div> <div>Write students' suggestions up on the board. If suggestions are not forthcoming, introduce some of the ideas listed in the problem.</div> <div> </div> <div>"You should be able to work out what will happen in the situations you've suggested using one of the powerful representations we've looked at, without having to try out lots of numerical examples first. Of course, if you want, you can use a numerical example to verify what you've done."</div> <div> </div> <div>Students could be offered a choice of which situations to work on. Alternatively, you may want everybody to work on a series of related problems (5, 6, 7 ... n consecutive numbers, for example) that will lead to a generalisation.</div> <div>One nice plenary activity is to challenge students to work out quickly what the difference in pair products will be for a randomly chosen sequence of numbers.</div> <div> </div> <div>You can read on the <a href="https://www.ncetm.org.uk/resources/34137">NCETM site</a> how one teacher used this task in the classroom (requires free registration). One blogger wrote a piece on his views of the problem, which you can read <a href="http://casmusings.wordpress.com/2012/05/13/multiplication-puzzle-for-the-very-young/">here</a>. </div> <h3>Key question</h3> Is there a way to represent the pair products that will explain the patterns you noticed? <ul class="noindent"></ul> <h3>Possible extension</h3> <div> </div> <div>This problem only operated on the end numbers and the 'end but one' numbers. Invite students to generalise further by looking at other pairs within the sequence.</div> <div>For example, if you have an odd number of consecutive numbers, what's the difference between the product of the end numbers and the square of the middle number? </div> <h3>Possible support</h3> <div>This problem could also be approached purely numerically, as an exercise in developing fluency with multiplication tables while looking for pattern and structure.</div> <div> </div> <div>In the <a href="/2278/clue">hint</a>, there is a dynamic image of Alison's representation that could be used to help students to see why the difference will always be 2.</div> <div> </div></mdoxml> <?xml version="1.0" encoding="UTF-8"?> <mdoxml version="1.0">The GeoGebra applet below allows you to see how Alison's representation changes as the consecutive numbers increase. Move the slider to change the numbers. <mdo:applet archive="http://jars.geogebra.org/webstart/4.0/geogebra.jar" code="geogebra.GeoGebraApplet" codebase="/content/id/2278/" datafile="" height="420" width="400"><param name="filename" value="http://nrich.maths.org/content/id/2278/pair.ggb"></param> <param name="framePossible" value="true"></param> <param name="showResetIcon" value="true"></param> <param name="enableRightClick" value="true"></param> <param name="showMenuBar" value="false"></param> <param name="showToolBar" value="false"></param> <param name="showToolBarHelp" value="true"></param> <param name="showAlgebraInput" value="false"></param></mdo:applet> If the GeoGebra applet does not load correctly you can save the <a href="http://nrich.maths.org/content/id/2278/pair.ggb">GeoGebra file</a> and open it using the free to download <a href="http://www.geogebra.org">GeoGebra</a> software.</mdoxml> <?xml version="1.0" encoding="UTF-8"?> <mdoxml version="1.0">The difference is 2 when we have 4 consecutive numbers: x (x+3) = x2 + 3x (x+1) (x+2) = x2 + 3x + 2 The difference is 3 when we have 5 consecutive numbers: x (x+4) = x2 + 4x (x+1) (x+3) = x2 + 4x + 3 The difference is 4 when we have 6 consecutive numbers: x (x+5) = x2 + 5x (x+1) (x+4) = x2 + 5x + 4 The difference is (n-2) when we have n consecutive numbers: x (x+n-1) = x2 + nx - x (x+1) (x+n-2) = x2 + nx - x + n - 2 We received many correct solutions to Pair Products, including some with clear explanations from Jiaixang from St George's School in Harpenden, Keone from Sage Ridge School (Reno, NV), Gabriel from the British School of Brussels, Natasha from the European School, Aisling from Grand Avenue Primary School, Benjamin from Woodbridge School and Rebekah from Bayhouse School in Hampshire. Emily tried out some examples: <mdo:image alt="" height="151" src="Table.JPG" width="293"></mdo:image> There are two patterns going on. The $2$nd $\times$ $3$rd is two more than the $1$st $\times$ $4$th and this always happens. This is because both of those columns are going up in the same way. You do $+6$, then $+8$, then $+10$ and so on, adding the next even number. Keone found that with four consecutive whole numbers: The product of the first and last numbers is always $2$ less than the product of the middle two numbers. Explanation: Suppose the first number is $x$. Then the second number is $x+1$, the third is $x+2$, and the fourth is $x+3$. So the product of the first and fourth numbers is $x(x+3) = x^2 + 3x$. Also, the product of the second and third numbers is $(x+1)(x+2) = x^2 + 3x + 2$. So $(x+1)(x+2) = x(x+3) + 2$ for any chosen value of $x$. With five consecutive whole numbers he found that: The product of the first and last numbers is always $3$ less than the product of the second and fourth numbers. Explanation: Again, let the first number be $x$; then the second number is $x+1$, the third is $x+2$, the fourth is $x+3$, and the fifth is $x+4$. So the product of the first and last numbers is $x(x+4) = x^2 + 4x$. Also, the product of the second and fourth numbers is $(x+1)(x+3) = x^2 + 4x + 3$. So $(x+1)(x+3) = x(x+4) + 3$ for any chosen value of $x$. And with $n$ consecutive whole numbers he found that: The product of the first and last numbers is always $n-2$ less than the product of the second and penultimate numbers. Let us consider the general case where there are n consecutive whole numbers. As before, let the first number be $x$; then the second number is x+1, the third is x+2, and so on. The last number (the $n$th number) will be $x+n-1$. Thus, the penultimate (second-to-last) number will be $x+n-2$. So the product of the first number and the last number will be $x(x+n-1) = x^2 + nx - x$ The product of the second and the penultimate numbers will be $(x+1)(x+n-2) = x^2 + nx - 2x + x + n - 2 = x^2 + nx - x + n - 2$ So $(x+1)(x+n-2) = x(x+n-1) + n - 2$; that is, the product of the second and penultimate numbers will always exceed the product of the first and last numbers by exactly $n - 2$. For example, if we take 6 numbers, the product of the $2$nd and $5$th numbers will be $4$ more than the product of the $1$st and last numbers. Natasha generalised her findings in a similar way and then went on to check her conclusion: We can generalise this problem by substituting particular numbers with letters. Let the first number be $a$. If there are $n$ numbers, where a is the first, then the last number is $(a+n-1)$. The second number is $(a+1)$, the penultimate number $(a+n-2)$. By multiplying the first and last numbers together, we get $a(a+n-1) = a^2 +an-a$ Multiplication of the second and penultimate numbers gives $(a+1)(a+n-2) = a^2+an-a+n-2$ The difference therefore in the product of the first and last numbers and the product of the second and penultimate numbers is always $n-2$. For added confirmation we can take a random example: Consider the numbers $56, 57, 58, 59, 60, 61, 62$ where $n = 7$ and $a = 56$ Our general formula tells us that the difference in the product pairs should be $n-2$ (i.e. $5$). When we do the calculation, we get $$57 \times61 - 56 \times62 = 3477 - 3472 = 5$$ This result corresponds with the general one established above. Tom explained it a different way: <mdo:image alt="" height="106" src="rectangles.JPG" width="420"></mdo:image> If we overlap the two rectangles we cut a piece off at the bottom and create a new piece at the right hand side. The difference between the two small rectangles is always two. This works for all numbers, not just $9, 10, 11, 12$. We received a slightly different summary of findings from Aisling: The multiple of the first and last numbers of a series of consecutive whole numbers, is the same as the multiple of the second and penultimate numbers of that series, minus the number of numbers separating the first and last numbers. Well done to you all for such clear reasoning. </mdoxml> 2 3 0 0 0 1 0 Pair Products Choose four consecutive whole numbers. Multiply the first and last numbers together. Multiply the middle pair together. What do you notice? Numbers and the Number System Properties of numbers Algebra Creating expressions/formulae Algebra Manipulating algebraic expressions/formulae Using, Applying and Reasoning about Mathematics Generalising Information and Communications Technology smartphone Admin Stage 3&4 Investigation Secondary Mapping Document Expanding and factorising quadratics