Multiplication square

2821 /www/nrich/html/content/id/2821/ 1 2011-02-01T00:00:01 <?xml version="1.0" encoding="UTF-8"?> <mdoxml version="1.0"> Take a look at the multiplication square below: <mdo:image alt="times table" height="657" src="times%20table.gif" width="612"></mdo:image> Pick any 2 by 2 square and add the numbers on each diagonal. For example, if you take: <div style="margin-left: 40px;"><mdo:image alt="part of times table" height="82" src="small%20times%20table.gif" width="72"></mdo:image></div> the numbers along one diagonal add up to $77$ ($32 + 45$) and the numbers along the other diagonal add up to $76$ ($36 + 40$). Try a few more examples. What do you notice? Can you show (prove) that this will always be true? Now pick any 3 by 3 square and add the numbers on each diagonal. For example, if you take: <div style="margin-left: 40px;"><mdo:image alt="part of times table" height="124" src="larger%20times%20table.gif" width="115"></mdo:image></div> the numbers along one diagonal add up to $275$ ($72 + 91 + 112$) and the numbers along the other diagonal add up to $271$ ($84 + 91 + 96$). Try a few more examples. What do you notice this time? Can you show (prove) that this will always be true? Now pick any 4 by 4 square and add the numbers on each diagonal. For example, if you take: <div style="margin-left: 40px;"><mdo:image alt="part of times table" height="162" src="largest%20times%20table.gif" width="148"></mdo:image></div> the numbers along one diagonal add up to $176$ ($24 + 36 + 50 + 66$) and the numbers along the other diagonal add up to $166$ ($33 + 40 + 45 + 48$). Try a few more examples. What do you notice now? Can you show (prove) that this will always be true? Can you predict what will happen if you pick a 5 by 5 square, a 6 by 6 square ... an n by n square, and add the numbers on each diagonal? Can you prove your prediction? <a href="http://nrich.maths.org/public/viewer.php?obj_id=6057&part=">Click here for a poster of this problem</a>. </mdoxml> <?xml version="1.0" encoding="UTF-8"?> <mdoxml version="1.0"> Well done to Maulik aged 11 who sent in some nice work on this problem. Neil's solution is given below. For a 2 by 2 square with column headings of x and x+1, and row headings of y and y+1, Neil says that: For the two by two square you can always express it algebraically like this: <mdo:image height="44" width="128" alt="2x2" src="multiplication%20sq1.JPG"></mdo:image> So the diagonal from top right to bottom left is: $(x+1)y+x(y+1) = xy+y+xy+x = 2xy+x+y$ Lets call that Z. The diagonal from top left to bottom right is: $xy+(x+1)(y+1) = xy+xy+x+y+1 = 2xy+x+y+1$ So the first diagonal is Z and the second Z+1 so the diagonal from top left to bottom right is always 1 more than the diagonal from top right to bottom left. For a 3 by 3 square with column headings of x, x+1 and x+2, and row headings of y, y+1 and y+2, Neil says that: The 3 by 3 square looks like this: <mdo:image height="76" width="194" alt="3x3" src="multiplication%20sq2.JPG"></mdo:image> The diagonal from top right to bottom left is: $x(y+2)+(x+1)(y+1)+(x+2)y = xy+2x+xy+x+y+1+xy+2y = 3xy+3x+3y+1$ The diagonal from top left to bottom right is: $ xy+(x+1)(y+1)+(x+2)(y+2) = xy+xy+x+y+1+xy+2x+2y+4 = 3xy+3x+3y+5 $ Let's say that $3xy+3x+3y = W$ The diagonal from top right to bottom left is W+1. The diagonal from top left to bottom right is W+5. So the difference between the diagonals is 4. More generally Continuing with the same method: <mdo:image height="195" width="224" alt="summary of results" src="multiplication%20sq3.JPG"></mdo:image> Note that 1 = 1² 4 = 2² 10 = 1² + 3² 20 = 2² + 4² 35 =1² + 3² + 5² 56 = 2² + 4² + 6² 84 = 1² + 3² + 5² + 7² Neil goes on to point out that the differences are the <a href="http://thesaurus.maths.org/mmkb/entry.html?action=entryById&id=3118"> tetrahedral numbers.</a> Tetrahedral numbers are the sum of consecutive triangular numbers. The formula is 1/6n(n+1)(n+2). The first few tetrahedral numbers are 1, 4, 10, 20, 35, 56, 84, 120, ... The tetrahedral numbers are found in the fourth diagonal of Pascal's triangle: <mdo:image height="307" width="475" alt="pascal's triangle" src="multiplication%20square%204.JPG"></mdo:image> </mdoxml> <?xml version="1.0" encoding="UTF-8"?> <mdoxml version="1.0"> This problem offers an opportunity: <ul><li> to notice and explain patterns in a multiplication square </li> </ul> <ul><li>to use some algebraic techniques, in particular multiplying out brackets. </li> </ul> </mdoxml> <?xml version="1.0" encoding="UTF-8"?> <mdoxml version="1.0"> For a general $2$ by $2$ square, the column headings could be $x$ and $(x + 1)$, and the row headings could be $y$ and $(y + 1)$. For a general $3$ by $3$ square, the column headings could be $x$, $(x + 1)$ and $(x + 2)$, and the row headings could be $y$, $(y + 1)$ and $(y + 2)$. If you are struggling with the algebra take a look at <a href="http://nrich.maths.org/public/viewer.php?obj_id=2823&part=index"> Partitioning revisited.</a> </mdoxml> 2 4 0 0 1 0 0 Multiplication square Pick a square within a multiplication square and add the numbers on each diagonal. What do you notice? Algebra Creating expressions/formulae Algebra Manipulating algebraic expressions/formulae Using, Applying and Reasoning about Mathematics Mathematical reasoning & proof Using, Applying and Reasoning about Mathematics Generalising Using, Applying and Reasoning about Mathematics Making and proving conjectures Secondary Mapping Document Expanding and factorising quadratics