781 /www/nrich/html/content/01/06/six2/ 1 2011-02-01T00:00:01 <?xml version="1.0" encoding="UTF-8"?> <mdoxml version="1.0"><br></br> <p>Each of the different letters below stands for a different number.<br></br> How many solutions can you find to this alphanumeric?<br></br> How can you be sure you found them all?<br></br>  </p> <div style="margin-left: 80px;"><mdo:image src="TwoplusTwo.png"></mdo:image></div> <div style="text-align: center;"><br></br>  </div> <div><a href="http://nrich.maths.org/public/viewer.php?obj_id=6324&amp;part=">Click here for a poster of this problem</a>.</div> <br></br></mdoxml> <?xml version="1.0" encoding="UTF-8"?> <mdoxml version="1.0"><br></br> <p>There were seven solutions to this alphanumeric, and four people or groups found all seven: Andrei (School 205, Bucharest, Romania), Sim (Raffles Girls' Primary, Singapore), Prateek (Riccarton High School, New Zealand) and Damen, Katrina and Oliver (Crownfield Junior School, England). A number of pupils from Crownfield Junior School produced good work on this problem.</p> <p>The key to this sort of problem is to be systematic, and Andrei explained very clearly how he worked through every possibility. Here is his solution:</p> <p>I started from the observation that I must add two 3-digit numbers, and the result is 4-digit one, so that I must assign the value 1 to the letter F.</p> <p>My method is to give values to the letters from right to left, the way additions are performed. <em>[O is also a good place to start as it occurs three times. - Ed.]</em></p> <ul> <li>If O=0, then R must be 0 too. As each letter stands for a different number, it isn't possible.</li> <li>O can't be 1, as F=1.</li> <li>If O=2, then <table cellspacing="0" cellpadding="4"> <tbody> <tr> <td></td> <td></td> <td>T</td> <td>W</td> <td>2</td> </tr> <tr> <td>+</td> <td></td> <td>T</td> <td>W</td> <td>2</td> </tr> <tr> <td>-</td> <td>-</td> <td>-</td> <td>-</td> <td>-</td> </tr> <tr> <td></td> <td>1</td> <td>2</td> <td>U</td> <td>R</td> </tr> </tbody> </table> This means R=4 and T=6. In this case W+W=U, and I analyse the numbers for W one by one: <ul> <li>W=0 NO, because 0+0=0, so U must be 0 too.</li> <li>W=1 NO, because F=1.</li> <li>W=2 NO, because O=2.</li> <li>W=3, so U=6. NO, because T=6.</li> <li>W=4 NO, because R=4.</li> <li>W=5 NO, because this would make U greater than 9.</li> </ul> </li> <li>If O=3, then <table cellspacing="0" cellpadding="4"> <tbody> <tr> <td></td> <td></td> <td>T</td> <td>W</td> <td>3</td> </tr> <tr> <td>+</td> <td></td> <td>T</td> <td>W</td> <td>3</td> </tr> <tr> <td>-</td> <td>-</td> <td>-</td> <td>-</td> <td>-</td> </tr> <tr> <td></td> <td>1</td> <td>3</td> <td>U</td> <td>R</td> </tr> </tbody> </table> There is no solution, because both R and T must correspond to 6.</li> <li>If O=4, then <table cellspacing="0" cellpadding="4"> <tbody> <tr> <td></td> <td></td> <td>T</td> <td>W</td> <td>4</td> </tr> <tr> <td>+</td> <td></td> <td>T</td> <td>W</td> <td>4</td> </tr> <tr> <td>-</td> <td>-</td> <td>-</td> <td>-</td> <td>-</td> </tr> <tr> <td></td> <td>1</td> <td>4</td> <td>U</td> <td>R</td> </tr> </tbody> </table> This means R=8 and T=7. In this case, W+W=U, and analysing as previously, I find that W cannot be 0, 1, 2, 4, and cannot be greater than or equal to 5. There is a solution with W=3 and U=6.</li> <li>If O=5, then <table cellspacing="0" cellpadding="4"> <tbody> <tr> <td></td> <td></td> <td>T</td> <td>W</td> <td>5</td> </tr> <tr> <td>+</td> <td></td> <td>T</td> <td>W</td> <td>5</td> </tr> <tr> <td>-</td> <td>-</td> <td>-</td> <td>-</td> <td>-</td> </tr> <tr> <td></td> <td>1</td> <td>5</td> <td>U</td> <td>R</td> </tr> </tbody> </table> This means T=7 and R=0. Now, W must be at least 5, because W+W+1=10+U. <ul> <li>W=5 NO, because O=5.</li> <li>W=6, U=3 - SOLUTION</li> <li>W=7 NO, because T=7.</li> <li>W=8, U=7 NO, because T=7.</li> </ul> </li> <li>If O=6, then <table cellspacing="0" cellpadding="4"> <tbody> <tr> <td></td> <td></td> <td>T</td> <td>W</td> <td>6</td> </tr> <tr> <td>+</td> <td></td> <td>T</td> <td>W</td> <td>6</td> </tr> <tr> <td>-</td> <td>-</td> <td>-</td> <td>-</td> <td>-</td> </tr> <tr> <td></td> <td>1</td> <td>6</td> <td>U</td> <td>R</td> </tr> </tbody> </table> This means R=2 and T=8. W+W+1=U, so W=4. <ul> <li>W=0, so U=1. NO, because F=1.</li> <li>W=1 NO, because F=1.</li> <li>W=2 NO, because R=2.</li> <li>W=3, U=7 - SOLUTION</li> <li>W=4, U=9 - SOLUTION</li> </ul> </li> <li>If O=7, then <table cellspacing="0" cellpadding="4"> <tbody> <tr> <td></td> <td></td> <td>T</td> <td>W</td> <td>7</td> </tr> <tr> <td>+</td> <td></td> <td>T</td> <td>W</td> <td>7</td> </tr> <tr> <td>-</td> <td>-</td> <td>-</td> <td>-</td> <td>-</td> </tr> <tr> <td></td> <td>1</td> <td>7</td> <td>U</td> <td>R</td> </tr> </tbody> </table> This means R=4 and T=8. W+W+1=10+U. So, W=5. <ul> <li>W=5, so U=1. NO, because F=1.</li> <li>W=6, so U=3 - SOLUTION</li> <li>W=7 NO, because O=7.</li> <li>W=8 NO, because T=8.</li> <li>W=9, so U=9 - NO</li> </ul> </li> <li>If O=8, then <table cellspacing="0" cellpadding="4"> <tbody> <tr> <td></td> <td></td> <td>T</td> <td>W</td> <td>8</td> </tr> <tr> <td>+</td> <td></td> <td>T</td> <td>W</td> <td>8</td> </tr> <tr> <td>-</td> <td>-</td> <td>-</td> <td>-</td> <td>-</td> </tr> <tr> <td></td> <td>1</td> <td>8</td> <td>U</td> <td>R</td> </tr> </tbody> </table> This means R=6 and T=9. W+W+1=U. So, W=4. <ul> <li>W=0, so U=1. NO, because F=1.</li> <li>W=1 NO, because F=1.</li> <li>W=2, U=5 - SOLUTION</li> <li>W=3, U=7 - SOLUTION</li> <li>W=4, so U=9. NO, becuase T=9.</li> </ul> </li> <li>If O=9, then <table cellspacing="0" cellpadding="4"> <tbody> <tr> <td></td> <td></td> <td>T</td> <td>W</td> <td>9</td> </tr> <tr> <td>+</td> <td></td> <td>T</td> <td>W</td> <td>9</td> </tr> <tr> <td>-</td> <td>-</td> <td>-</td> <td>-</td> <td>-</td> </tr> <tr> <td></td> <td>1</td> <td>9</td> <td>U</td> <td>R</td> </tr> </tbody> </table> This means R=8 and T=9, but this gives no solutions because O=9.</li> </ul> <br></br> <p>So, the only solutions are:</p> <table cellspacing="0" cellpadding="4"> <tbody> <tr> <td></td> <td></td> <td>7</td> <td>3</td> <td>4</td> <td></td> <td></td> <td></td> <td>7</td> <td>6</td> <td>5</td> <td></td> <td></td> <td></td> <td>8</td> <td>3</td> <td>6</td> <td></td> <td></td> <td></td> <td>8</td> <td>4</td> <td>6</td> </tr> <tr> <td>+</td> <td></td> <td>7</td> <td>3</td> <td>4</td> <td></td> <td>+</td> <td></td> <td>7</td> <td>6</td> <td>5</td> <td></td> <td>+</td> <td></td> <td>8</td> <td>3</td> <td>6</td> <td></td> <td>+</td> <td></td> <td>8</td> <td>4</td> <td>6</td> </tr> <tr> <td>-</td> <td>-</td> <td>-</td> <td>-</td> <td>-</td> <td></td> <td>-</td> <td>-</td> <td>-</td> <td>-</td> <td>-</td> <td></td> <td>-</td> <td>-</td> <td>-</td> <td>-</td> <td>-</td> <td></td> <td>-</td> <td>-</td> <td>-</td> <td>-</td> <td>-</td> </tr> <tr> <td></td> <td>1</td> <td>4</td> <td>6</td> <td>8</td> <td></td> <td></td> <td>1</td> <td>5</td> <td>3</td> <td>0</td> <td></td> <td></td> <td>1</td> <td>6</td> <td>7</td> <td>2</td> <td></td> <td></td> <td>1</td> <td>6</td> <td>9</td> <td>2</td> </tr> <tr> <td></td> </tr> <tr> <td></td> <td></td> <td>8</td> <td>6</td> <td>7</td> <td></td> <td></td> <td></td> <td>9</td> <td>2</td> <td>8</td> <td></td> <td></td> <td></td> <td>9</td> <td>3</td> <td>8</td> </tr> <tr> <td>+</td> <td></td> <td>8</td> <td>6</td> <td>7</td> <td></td> <td>+</td> <td></td> <td>9</td> <td>2</td> <td>8</td> <td></td> <td>+</td> <td></td> <td>9</td> <td>3</td> <td>8</td> </tr> <tr> <td>-</td> <td>-</td> <td>-</td> <td>-</td> <td>-</td> <td></td> <td>-</td> <td>-</td> <td>-</td> <td>-</td> <td>-</td> <td></td> <td>-</td> <td>-</td> <td>-</td> <td>-</td> <td>-</td> </tr> <tr> <td></td> <td>1</td> <td>7</td> <td>3</td> <td>4</td> <td></td> <td></td> <td>1</td> <td>8</td> <td>5</td> <td>6</td> <td></td> <td></td> <td>1</td> <td>8</td> <td>7</td> <td>6</td> </tr> </tbody> </table> <br></br></mdoxml> <?xml version="1.0" encoding="UTF-8"?> <mdoxml version="1.0"><div class="embed"> <h2>Two and Two</h2> <br></br> <p>How many solutions can you find to this alphanumeric? Each of the different letters stands for a different number.</p> <table style="" border="1"> <tbody> <tr> <td> </td> <td> </td> <td>T</td> <td>W</td> <td>O</td> </tr> <tr> <td>+</td> <td> </td> <td>T</td> <td>W</td> <td>O</td> </tr> <tr> <td>--</td> <td>--</td> <td>--</td> <td>--</td> <td>--</td> </tr> <tr> <td> </td> <td>F</td> <td>O</td> <td>U</td> <td>R</td> </tr> </tbody> </table> <br></br> <br></br> <div><a href="http://nrich.maths.org/public/viewer.php?obj_id=6324&amp;part=">Click here for a poster of this problem</a>.</div> </div> <br></br> <h3>Why do this problem?</h3> <div><a href="http://nrich.maths.org/public/viewer.php?obj_id=781&amp;part=">This problem</a> offers an opportunity to practise addition in a more interesting and challenging context than is usual. It requires students to work systematically, record their progress efficiently and apply their understanding of place value.</div> <h3>Possible approach</h3> <div>Write the following sum on the board and ask students to complete it.</div> <br></br> <div>$\; \; \; 8 \; 6 \; 1$</div> <div style="text-decoration: none;"><span style="text-decoration: underline;">$+ \; ? \; ? \; ?$</span></div> <div>$1 \; 4 \; 2 \; 7$</div> <br></br> <div>Expect justifications for any suggestions.</div> <div>This should be unproblematic so move onto a more challenging problem.</div> <br></br> <div>If each letter stands for a distinct digit what are the values of $a$, $b$ and $c$?</div> <br></br> <div>$\; \; \; b \; b$</div> <div><span style="text-decoration: underline;">$+ \; c \; b$</span></div> <div>$a \; c \; c$</div> <br></br> <div>"How can we approach this?"</div> <div>Expect fairly random suggestions to start with but aim to use the discussion as an opportunity to model working systematically.</div> <div>"What happens if $b$ is $1, 2, 3, \ldots$ ?" - rejecting values as soon as it is apparent they do not work and discussing how you know.</div> <div>"How can we be sure we have all the solutions?"</div> <br></br> <div>Set the students off in pairs to work on the problem <span style="font-weight: bold;">TWO + TWO = FOUR</span> .</div> <div>Establish that you are not going to announce how many solutions there are and that you will expect students to work systematically and be able to justify that they have all the possible solutions.</div> <br></br> <h3>Key questions</h3> Are we certain we have considered all the possibilities?<br></br> <br></br> <h3>Possible extension</h3> <div>Which other word sums work? - eg</div> <div style="margin-left: 80px; font-weight: bold;">ONE + ONE = TWO</div> <div style="margin-left: 80px; font-weight: bold;">FOUR + FIVE = NINE</div> <div style="margin-left: 40px;"> </div> <br></br> <div>Which definitely don&#39;t work? - eg</div> <div style="margin-left: 80px; font-weight: bold;">ONE + TWO = THREE</div> <div style="margin-left: 80px; font-weight: bold;">ONE + THREE = FOUR</div> <div style="margin-left: 40px;"> </div> <br></br> <div>Suggest students try <a href="http://nrich.maths.org/public/viewer.php?obj_id=5001&amp;part=">Fly Fly Away</a> and <a href="http://nrich.maths.org/public/viewer.php?obj_id=654&amp;part=">Alphabet Soup</a></div> <br></br> <h3>Possible support</h3> Suggest students start with <a href="http://nrich.maths.org/public/viewer.php?obj_id=1071&amp;part=">Spell by Numbers</a><br></br> <br></br></mdoxml> 2 5 0 1 1 0 0 Two and Two How many solutions can you find to this sum? Each of the different letters stands for a different number. Using, Applying and Reasoning about Mathematics Working systematically Numbers and the Number System Place value Calculations and Numerical Methods Addition & subtraction Information and Communications Technology smartphone Secondary Mapping Document Number operations and calculation methods Secondary Mapping Document DisplayCabinet Admin Upper primary mapping document Secondary processes PM - Working Systematically