1134 /www/nrich/html/content/02/10/penta1/ 1 2011-02-01T00:00:01 <?xml version="1.0" encoding="UTF-8"?> <mdoxml version="1.0"><br></br> <p>In the $2 \times 2$ multiplication square below, the boxes at the end of each row and the foot of each column give the result of multiplying the two numbers in that row or column.</p> <table cellspacing="0" border="1" align="center" summary="" id="tableedit1"> <tbody> <tr> <td bgcolor="#FF0000" align="center">7</td> <td bgcolor="#FF0000" align="center">5</td> <td bgcolor="#FFFFFF" align="center">35</td> </tr> <tr> <td bgcolor="#FF0000" align="center">3</td> <td bgcolor="#FF0000" align="center">4</td> <td bgcolor="#FFFFFF" align="center">12</td> </tr> <tr> <td bgcolor="#FFFFFF" align="center">21</td> <td bgcolor="#FFFFFF" align="center">20</td> <td></td> </tr> </tbody> </table> <p>The $3 \times3$ multiplication square below works in the same way. The boxes at the end of each row and the foot of each column give the result of multiplying the three numbers in that row or column.</p> <table cellspacing="0" border="1" align="center" summary="" id="tableedit2"> <tbody> <tr align="center"> <td bgcolor="#FF0000"></td> <td bgcolor="#FF0000"></td> <td bgcolor="#FF0000"></td> <td bgcolor="#FFFFFF">15</td> </tr> <tr align="center"> <td bgcolor="#FF0000"></td> <td bgcolor="#FF0000"></td> <td bgcolor="#FF0000"></td> <td bgcolor="#FFFFFF">108</td> </tr> <tr align="center"> <td bgcolor="#FF0000"></td> <td bgcolor="#FF0000"></td> <td bgcolor="#FF0000"></td> <td bgcolor="#FFFFFF">224</td> </tr> <tr align="center"> <td bgcolor="#FFFFFF">144</td> <td bgcolor="#FFFFFF">8</td> <td bgcolor="#FFFFFF">315</td> <td></td> </tr> </tbody> </table> <p>The numbers $1 - 9$ may be used once and once only.</p> <p>Can you work out the arrangement of the digits in the square so that the given products are correct?</p> <br></br></mdoxml> <?xml version="1.0" encoding="UTF-8"?> <mdoxml version="1.0"><br></br> <p class="editorial">This solution came from Adam from Moorfield Junior School.</p> <p>This is how I arranged the numbers from one to nine in a multiplication square once and once only:</p> <table cellspacing="0" cellpadding="8" border="0" summary=""> <tbody> <tr> <td>1st row: 3x1x5=15</td> <td>2nd row: 6x2x9=108</td> <td>3rd row: 8x4x7=224</td> </tr> <tr> <td>1st column: 3x6x8=144</td> <td>2nd column: 1x2x4=8</td> <td>3rd column: 5x9x7=315</td> </tr> </tbody> </table> <p class="editorial">He is right but doesn't tell us how he worked it out. This solution is from James, Terry, John and Indiya from Sabden Primary School, Clitheroe, Lancashire.</p> <p><mdo:image width="419" height="611" src="solution.jpg" alt="Scan of Sabden Primary School proof."></mdo:image></p> <p class="editorial">This solution is from Ailsa from Romsey Abbey Primary School, Romsey, Hampshire. She says:</p> <p>I managed to crack the Multiplication Square, below, here is how I did it:</p> <table cellspacing="0" cellpadding="8" border="1" summary=""> <tbody> <tr> <td>3</td> <td>1</td> <td>5</td> <td>15</td> </tr> <tr> <td>6</td> <td>2</td> <td>9</td> <td>108</td> </tr> <tr> <td>8</td> <td>4</td> <td>7</td> <td>224</td> </tr> <tr> <td>144</td> <td>8</td> <td>315</td> <td></td> </tr> </tbody> </table> <p>First I worked out that the only way to make 15 was to use the numbers 1, 3 and 5. I then discovered that the only way to make 8 was to use the numbers 1, 2 and 4. Therefore the 1 had to go in the top row and second column. I then worked out 5 had to go in the top row and third column because 315 is a multiple of 5 and 144 is not, so that meant that the 3 had to go in the top row in the first column.</p> <p>Then I divided 108 by 2, the answer was 54, and then made 54 using the numbers I had left, 6, 7, 8 and 9, the answer was 9 x 6, so I knew that 9 and 6 were the 2 other numbers that went in the second row. 6 is not a multiple of 315 so 6 went in the first column and 9 went in the third column.</p> <p>So now the only 2 numbers left were 7 and 8 so I worked the answer out by discovering how to make 315. I multiplied 5 by 9 to give me the answer 45 and then tried multiplying it by both 7 and 8. 7 was the correct number so 7 went in the third column and 8 went in the first.</p> <p class="editorial">This is a really clear description of your method, well done!</p> <p class="editorial">Kaden from Coleridge Community College, Cambridge also sent a correct solution. Well done!</p> <p class="editorial">We have had some other solutions, from a different school, that had got the 6 and the 9 swapped over. Will it work then?</p> <br></br></mdoxml> <?xml version="1.0" encoding="UTF-8"?> <mdoxml version="1.0"><div class="embed"> <h2>Multiplication Squares</h2> <br></br> <p>In the $2 \times 2$ multiplication square below, the boxes at the end of each row and the foot of each column give the result of multiplying the two numbers in that row or column.</p> <table style="" border="1"> <tbody> <tr> <td align="center" bgcolor="#FF0000" style="">7</td> <td align="center" bgcolor="#FF0000" style="">5</td> <td align="center" bgcolor="#FFFFFF" style="">35</td> </tr> <tr> <td align="center" bgcolor="#FF0000" style="">3</td> <td align="center" bgcolor="#FF0000" style="">4</td> <td align="center" bgcolor="#FFFFFF" style="">12</td> </tr> <tr> <td align="center" bgcolor="#FFFFFF" style="">21</td> <td align="center" bgcolor="#FFFFFF" style="">20</td> <td> </td> </tr> </tbody> </table> <p>The $3 \times3$ multiplication square below works in the same way. The boxes at the end of each row and the foot of each column give the result of multiplying the three numbers in that row or column.</p> <table style="" border="1"> <tbody> <tr align="center"> <td bgcolor="#FF0000" style=""> </td> <td bgcolor="#FF0000" style=""> </td> <td bgcolor="#FF0000" style=""> </td> <td bgcolor="#FFFFFF" style="">15</td> </tr> <tr align="center"> <td bgcolor="#FF0000" style=""> </td> <td bgcolor="#FF0000" style=""> </td> <td bgcolor="#FF0000" style=""> </td> <td bgcolor="#FFFFFF" style="">108</td> </tr> <tr align="center"> <td bgcolor="#FF0000" style=""> </td> <td bgcolor="#FF0000" style=""> </td> <td bgcolor="#FF0000" style=""> </td> <td bgcolor="#FFFFFF" style="">224</td> </tr> <tr align="center"> <td bgcolor="#FFFFFF" style="">144</td> <td bgcolor="#FFFFFF" style="">8</td> <td bgcolor="#FFFFFF" style="">315</td> <td> </td> </tr> </tbody> </table> <p>The numbers $1 - 9$ may be used once and once only.</p> <p>Can you work out the arrangement of the digits in the square so that the given products are correct?</p> </div> <br></br> <h3><span style="font-weight: bold;">Why do this problem?</span></h3> <div><a href="http://nrich.maths.org/public/viewer.php?obj_id=1134&amp;part=index">This problem</a> requires learners to find common factors and offers a context in which they can explain their reasoning.</div> <br></br> <h3>Possible approach</h3> <div>You could begin by having a partially completed multiplication square (or squares) on the board, with just a few gaps, for example:</div> <br></br> <div><mdo:image alt="" height="137" src="examples.gif" width="300"></mdo:image></div> <div>Ask the class to describe what they see. Can they fill in the spaces? Invite some children to explain how they know what goes in each empty cell and encourage use of appropriate vocabulary.</div> <br></br> <div>Then show them the multiplication square in the problem and give them some time to begin to work on it. It would be a good idea to have a mini-plenary after a few minutes to share insights so far - where did they start? Not all learners will necessarily have begun in the same way so it would be valuable to talk about the possibilities, focusing on good explanations of the reasoning involved.</div> <br></br> <div>Pupils could then work in pairs to complete the square. Is there only one solution? How do they know? In a final plenary, you could specifically introduce the language of common factors if it has not come up already.</div> <br></br> <h3>Key questions</h3> <div>Which products are odd? How will this help?</div> <div>What are the factors of this number?</div> <br></br> <br></br> <h3>Possible extension</h3> <div><a href="http://nrich.maths.org/public/viewer.php?obj_id=5429&amp;part=index">Multiples Grid</a> would be a good challenge for some children once they have tried this problem.</div> <br></br> <h3>Possible support</h3> <div>Calculators might be useful.</div> <br></br> <br></br></mdoxml> <?xml version="1.0" encoding="UTF-8"?> <mdoxml version="1.0"><br></br> <p>Try placing the numbers $1$, $5$ and $7$ first.</p> <p>Which rows/columns have an odd product?</p> <p>Which are not divisible by $3$?</p> <br></br></mdoxml> 2 3 0 1 0 0 0 Can you work out the arrangement of the digits in the square so that the given products are correct? The numbers 1 - 9 may be used once and once only. Using, Applying and Reasoning about Mathematics Working systematically Numbers and the Number System Factors and multiples Calculations and Numerical Methods Multiplication & division Numbers and the Number System Common factors Admin Upper primary mapping document