470 /www/nrich/html/content/03/05/15plus1/ 1 2011-02-01T00:00:01 <mdoxml version="1.0"><p><br></br> This problem is now part of the collection <a href="/7821">Magic Letters</a>.</p></mdoxml> <mdoxml version="1.0"><br></br> <span class="editorial">Bob has sent us one solution:</span> <br></br> <br></br> $6\quad\quad 4\quad\quad 9$<br></br> $\ 5\quad 7\quad 8\quad 3$<br></br> $\quad 2\quad \quad 1$ <br></br> <div><span class="editorial">George used algebra to tackle the problem:</span></div> <br></br> <div>Call the numbers in the different dots $a, b, \ldots, i$, so the W looks like</div> <br></br> <div>$a\quad\quad e\quad\quad i$</div> <div>$\ b\quad d\quad f\quad h$</div> <div>$\quad c\quad \quad g$</div> <br></br> <div>Then $a+b+c=13$, $c+d+e$=13, $e+f+g=13$, and $g+h+i=13$. Adding these equations together, we get $a+b+2c+d+2e+f+2g+h+i=52$. But we know that the letters are just the numbers $1, 2, \ldots, 9$ in some order, so $a+b+c+d+e+f+g+h+i=45$, so $c+e+g=7$. Now $c$, $e$ and $g$ are all different, so the only possibility is that they are 1, 2 and 4 in some order. If $c$=1 and $e=2$, then we'd have $d=10$, and that's not possible. So of the three dots $c$, $e$ and $g$, the 1 and the 2 must be in ones that aren't next to each other. So we could have $c=1$, $e=4$ and $g=2$. Now we know that $d=8$ and $f=7$, and $a+b=12$ and $h+i=11$, and $a$, $b$, $h$ and $i$ are $3, 5, 6$ and $9$ in some order. So $a$ and $b$ are $3$ and $9$, and $h$ and $i$ are $5$ and $6$, but it could be either way round. So there are four different solutions, not counting reflections. One is shown below, and you get the others by switching the two numbers on one end or the other (or both!).</div> <br></br> <div>$3\quad\quad 4\quad\quad 5$</div> <div>$\ 9\quad 8\quad 7\quad 6$</div> <div>$\quad 1\quad \quad 2$</div> <br></br> <div>We can do something very similar if we want the sums to be $14$. This time, we get $c+e+g+45=4\times 14=56$, so $c+e+g=11$. There are several possibilities, which we'll consider separately.</div> <br></br> <div>Case $1$: ($1$,$2$,$8$). As above, we can't have $1$ and $2$ next to each other, so the only possibility (apart from reflection) is $c=1$, $e=8$, $g=2$. So $d=5$ and $f=4$, and $a+b=13$ and $h+i=12$, with $\{a,b,h,i\}=\{3,6,7,9\}$. So we get $\{a,b\}=\{6,7\}$ and $\{h,i\}=\{3,9\}$.</div> <br></br> <div>Case $2$: ($1$,$3$,$7$). This time, we can't have $1$ and $3$ next to each other either (as $14-1-3=10$), so the only possibility (again not counting reflection) is $c=1$, $d=6$, $e=7$, $f=4$, $g=3$, and we have $a+b=13$, $h+i=11$, with $\{a,b,h,i\}=\{2,5,8,9\}$. So $\{a,b\}=\{5,8\}$ and $\{h,i\}=\{2,9\}$, and there are four possibilities (by swapping ends), of which one is</div> <br></br> <div>$5\quad\quad 7\quad\quad 2$</div> <div>$\ 8\quad 6\quad 4\quad 9$</div> <div>$\quad 1\quad \quad 3$</div> <br></br> <div>Case $3$: ($1$,$4$,$6$). Now we can't have $4$ and $6$ next to each other, or we'd need another $4$ in the gap between them (to make $14$), so the only possibility is $c=4$, $d=9$, $e=1$, $f=7$, $g=6$, $a+b=10$, $h+i=8$, and so we must have $\{a,b\}=\{2,8\}$ and $\{h,i\}=\{3,5\}$.</div> <br></br> <div>Case $4$: ($2$,$3$,$6$). This time we can't have $2$ and $6$ next to each other, so the only possibilitiy is $c=2$, $d=9$, $e=3$, $f=5$, $g=6$, $a+b=12$, $h+i=8$, and so we get $\{a,b\}=\{4,8\}$ and $\{h,i\}=\{1,7\}$.</div> <br></br> <div>Case $5$: ($2$,$4$,$5$). Now we can't have $4$ and $5$ next to each other, so the only possibility is $c=5$, $d=7$, $e=2$, $f=8$, $g=4$, $a+b=9$, $h+i=10$, and we get the only solution as $\{a,b\}=\{3,6\}$, $\{h,i\}=\{1,9\}$.</div> <br></br></mdoxml> <mdoxml version="1.0">Even though this subject is accessible to school students, Magic Graphs is currently an active research field (see Joseph Gallian's website <a href="http://citeseer.ist.psu.edu/gallian00dynamic.html">http://citeseer.ist.psu.edu/gallian00dynamic.html</a> which lists all the papers published on the subject).<br></br></mdoxml> <mdoxml version="1.0"><br></br> The sum of the numbers 1 to 9 is 45 and four rows of 13 add up to 52. How do you make up the extra 7? <br></br> <br></br> See also <a href="http://nrich.maths.org/public/viewer.php?obj_id=460">Olympic Magic</a>.<br></br> <br></br></mdoxml> <mdoxml version="1.0"><br></br> <p>The numbers in the 4 lines making the W shape add up to 14, 14, 13 and 13. Can you make a Magic W with each line adding up to 13? Can you find more than one solution? How many ways can you make a Magic W total of 14?</p> <mdo:flash height="430" width="580"><param name="allowfullscreen" value="true" ></param><param value="Magic_w.swf" name="movie" ></param><param value="high" name="quality" ></param><param value="#FFFFFF" name="bgcolor" ></param></mdo:flash><br></br></mdoxml> 2 4 0 0 0 1 0 Find all the ways of placing the numbers 1 to 9 on a W shape, with 3 numbers on each leg, so that each set of 3 numbers has the same total. Algebra Creating expressions/formulae Using, Applying and Reasoning about Mathematics Working systematically Decision Mathematics and Combinatorics Combinatorics Decision Mathematics and Combinatorics Networks/Graph Theory Information and Communications Technology Interactivities