Highest voted questions tagged magic-square - Puzzling Stack Exchange most recent 30 from puzzling.stackexchange.com 2019-06-17T15:26:49Z https://puzzling.stackexchange.com/feeds/tag?tagnames=magic-square&sort=votes http://www.creativecommons.org/licenses/by-sa/3.0/rdf https://puzzling.stackexchange.com/q/23944 25 Unsolved Mysteries: Magic Square of Squares GentlePurpleRain https://puzzling.stackexchange.com/users/4421 2015-11-12T20:26:33Z 2018-10-22T15:48:16Z <blockquote> <p><sup><em>This is the first in what will hopefully be a series of <strong>Unsolved Mysteries</strong> posts.</em> </sup></p> <p><sup> <em>Note that <strong>this puzzle has no known solution</strong>, nor any proof that a solution is impossible. We will see how smart the denizens of Puzzling.SE actually are...!</em> </sup></p> </blockquote> <hr> <p>Most people are familiar with the concept of a <a href="https://en.wikipedia.org/wiki/Magic_square">Magic Square</a>. (If not, follow the link to read up on it.)</p> <p>There are algorithms available that make it trivial to construct a magic square of almost any size, but by adding a few constraints to the problem, it becomes much more challenging.</p> <p>Consider the following $4\times4$ magic square, where every entry is itself a square number, and the rows, columns and diagonals all sum to $8515$:</p> <p>$$\begin{array}\\ 68^2&amp;29^2&amp;41^2&amp;37^2\\ 17^2&amp;31^2&amp;79^2&amp;32^2\\ 59^2&amp;28^2&amp;23^2&amp;61^2\\ 11^2&amp;77^2&amp;8^2&amp;49^2 \end{array}$$</p> <p><em>Note that</em><br> $68^2 + 29^2 + 41^2 + 37^2 = 17^2 + 31^2 + 79^2 + 32^2$<br> <em>but</em><br> $68 + 29 + 41 + 37 \ne 17 + 31 + 79 + 32$ </p> <p><em>Only the squared values have the properties of a magic square.</em></p> <p>Many such $4\times4$ squares have been constructed, but as of yet, no one has succeeded in constructing a $3\times3$ magic square with the same property, nor in proving that no such magic square exists.</p> <p><strong><em>Your challenge</em></strong>, therefore, is as follows:</p> <blockquote> <p><strong>A)</strong> Build a $3\times3$ magic square where each of the nine entries in the square is itself a square number. </p> <p><strong><em>or</em></strong></p> <p><strong>B)</strong> Prove that no such square exists.</p> </blockquote> <hr> <p>For the pedantic among us (you know who you are), here are a few additional constraints:</p> <ul> <li>Each entry in the square must be unique. (A square consisting entirely of $4$s <em>is not valid</em>.)</li> <li>The definition of "square number" implies this, but I will spell it out here for those who like to quibble: The entries (before squaring) must be integers. Thus a magic square using values $\{ \sqrt1^2, \sqrt2^2, \sqrt3^2, \sqrt4^2, \sqrt5^2, \sqrt6^2, \sqrt7^2, \sqrt8^2, \sqrt9^2\}$ <em>is not valid</em> (although, of course, $\sqrt1^2$, $\sqrt4^2$, and $\sqrt9^2$ can be used in a square, being proper square numbers ($=1^2, 2^2, 3^2$).</li> <li>This also means that using complex numbers, limits, representations of infinity, or any other abstract mathematical concept <em>is not valid</em>. The intent of the question is obvious; please stick to that.</li> </ul> https://puzzling.stackexchange.com/q/22064 20 The 5040 Square Mike Earnest https://puzzling.stackexchange.com/users/10615 2015-09-10T18:06:02Z 2015-09-11T08:03:55Z <p>Fill a $4\times4$ grid with positive integers so that:</p> <ul> <li>Every cell has a different integer</li> <li>The <em>product</em> of the numbers in each row is $5040$, and similarly for the columns</li> </ul> <p><em>Source</em>: This was an NPR weekly listener challenge, aired on 2005-10-09. See <a href="http://www.taterenner.com/nprmps.htm">here</a>.</p> https://puzzling.stackexchange.com/q/10797 19 3x3 "Magic Square" of Prime Numbers LaBird https://puzzling.stackexchange.com/users/9944 2015-03-23T11:22:55Z 2017-11-17T01:11:32Z <p>During the thinking and analysis of some mathematical problems, I came up with this puzzle:</p> <p><img src="https://i.stack.imgur.com/iTc0s.png" alt="enter image description here"></p> <p>Just like any magic square, one has to fill in $9$ different numbers $P_1, P_2, \dots P_9$ to a $3 \times 3$ grid. But this time, all the numbers must be <strong>different prime numbers</strong>. In addition, the $8$ sums ($3$ horizontal, $3$ vertical and $2$ diagonal) must not only be <strong>different prime numbers</strong> among themselves, but also be different from the $9$ numbers in the grid. In other words, $P_1, P_2, \dots, P_9, S_1, S_2, \dots, S_8$ must be <strong>all different prime numbers</strong>.</p> <p>I suppose there are infinitely many solutions, so the challenge is to minimize the sum $S_1 + S_2 + \dots + S_8$. Here is one answer I found:</p> <p><img src="https://i.stack.imgur.com/cj1yE.png" alt="enter image description here"></p> <p>The total of the $8$ sums is $480$. I believe there are very likely solutions that can beat this total. You are welcome to have a try.</p> <p><strong><em>Update (plus Spoiler Alert):</em></strong> It was verified (using computer program) that one of the answers here (the accepted answer) is the optimal solution that cannot be beaten. There are $8$ optimal solutions, but actually they are the same because if you rotate one of the solutions by $90$, $180$ and $270$ degrees, and also horizontally flip each of the resulting grid, you will get all $8$ answers. Hence the "open-ended" tag has been removed.</p> https://puzzling.stackexchange.com/q/73974 19 A riddle that has been killing me the whole day Viktor Jeppesen https://puzzling.stackexchange.com/users/23165 2018-10-16T22:33:21Z 2018-10-24T17:05:39Z <p>So I'm walking around in London and found the following number riddle. The rules say, that what ever pattern you find, must be true for the rows as well as the columns. The answer in level 1 is for example 33 since it follows a+b+2=c. Meaning the third block always is the sum of the two first ones plus two, both in the columns and the rows. </p> <p>I just can't figure out level 5, please help me get some sleep <a href="https://i.stack.imgur.com/V0AGR.jpg" rel="noreferrer"><img src="https://i.stack.imgur.com/V0AGR.jpg" alt="The puzzle"></a></p> https://puzzling.stackexchange.com/q/269 15 Elegant solution to the Magic Hexagon problem durron597 https://puzzling.stackexchange.com/users/42 2014-05-19T18:33:23Z 2014-08-03T23:46:58Z <h2>The Magic Hexagon Problem</h2> <p><img src="https://i.stack.imgur.com/FMgOn.gif" alt="Magic Hexagon"></p> <blockquote> <p>A magic hexagon of order $n$ is an arrangement of close-packed hexagons containing the numbers $1, 2, ..., H_{n-1}$, where $H_n$ is the $n^{th}$ hex number such that the numbers along each straight line add up to the same sum. (Here, the hex numbers are i.e., $1, 7, 19, 37, 61, 91, 127$, ...; <a href="https://oeis.org/A003215%7D" rel="nofollow noreferrer">Sloane's A003215</a>. In the above magic hexagon of order $n=3$, each line (those of lengths $3$, $4$, and $5$) adds up to $38$.</p> </blockquote> <p>From <a href="http://mathworld.wolfram.com/MagicHexagon.html" rel="nofollow noreferrer">Wolfram's Magic Hexagon</a>.</p> <p>Here is the programming puzzles entry to the <a href="https://codegolf.stackexchange.com/questions/6304/code-solution-for-the-magic-hexagon">Magic Hexagon problem</a>. As you can see, the solutions there use appropriate systems of linear equations, and then try all the possibilities until it finds one. Boring, but effective; and can't be done by a human in a reasonable amount of time.</p> <p>This is <a href="http://britton.disted.camosun.bc.ca/magicsq/magic.html" rel="nofollow noreferrer">de la Loubere's method</a> for magic squares. Much cleaner and more beautiful, and a human can do this process easily.</p> <p>I want to know, is there such a solution / methodology to the Magic Hexagon problem that has the elegance of de la Loubere's method? How would you begin the process of coming up with one?</p> https://puzzling.stackexchange.com/q/123 14 Magic square with the position of 8 fixed John Bupit https://puzzling.stackexchange.com/users/129 2014-05-15T16:40:22Z 2015-06-05T12:12:49Z <blockquote> <p>A magic square (of order 3) is a 3x3 matrix consisting of distinct numbers from 1 to 9, where the numbers in each row, column and diagonal add up to 15.</p> </blockquote> <p>For example, the following would be a magic square:</p> <p><img src="https://i.stack.imgur.com/Xlrc1.png" alt="A magic square (Source: Wikipedia)"></p> <p><strong>The problem is to construct a magic square that has the position of the number 8 fixed.</strong></p> <p><img src="https://i.stack.imgur.com/mWm4y.png" alt="A magic square with the position of 8 fixed"></p> https://puzzling.stackexchange.com/q/190 13 What is the fewest number of filled-in squares required to uniquely define a magic square? Joe Z. https://puzzling.stackexchange.com/users/88 2014-05-17T04:46:12Z 2015-06-02T03:11:52Z <p>The magic square is a well-known grid of the numbers from 1 to 9 in which every row, column, and diagonal adds up to 15:</p> <pre><code>4 9 2 3 5 7 8 1 6 </code></pre> <p>But it is also possible to create magic squares using other numbers:</p> <pre><code>24 87 45 73 52 31 59 17 80 </code></pre> <p>It's also known that given just a few filled-in squares, you can determine the rest logically. For example, given the partially filled-in grid:</p> <pre><code>8 9 . . 6 . . 3 . </code></pre> <p>you can immediately infer that the rows, columns, and diagonals add up to 18, and so the bottom-right square is 4 and the top-right square is 1:</p> <pre><code>8 9 1 . 6 . . 3 4 </code></pre> <p>Then the right square is 13 and the bottom-left square is 11:</p> <pre><code> 8 9 1 . 6 13 11 3 4 </code></pre> <p>And finally, in a slightly uncouth twist, the left square is -1.</p> <pre><code> 8 9 1 -1 6 13 11 3 4 </code></pre> <p>But in fact, it's possible to create a set of filled-in numbers that don't have any completed rows at all and still be able to fill the rest of the numbers in:</p> <pre><code>12 .. 27 .. .. 6 .. 18 .. </code></pre> <p>This, as it turns out, has a (unique) solution of:</p> <pre><code>12 24 27 36 21 6 15 18 30 </code></pre> <p>So what is the fewest number of filled-in squares that are actually possible to derive the whole square from, and what arrangement are they in? And what about the case of higher-order magic squares?</p> https://puzzling.stackexchange.com/q/20567 12 Magic square using numbers 0,2,3,4,5,6,7,8,10 tyler https://puzzling.stackexchange.com/users/15971 2015-08-25T22:13:33Z 2018-10-06T17:23:34Z <p>I have to make a 3x3 magic square using the numbers 0-10 without 1 and 9. I have tried various things but am not good at this. The sums of each row, column, and diagonal have to be equal; I added all the numbers up and divided by three so I believe each row should add up to 15?</p> https://puzzling.stackexchange.com/q/41160 12 Put numbers to a star-shaped puzzle Jamal Senjaya https://puzzling.stackexchange.com/users/28756 2016-08-22T03:52:50Z 2016-08-22T11:49:18Z <p><a href="https://i.stack.imgur.com/pxOCs.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/pxOCs.png" alt="enter image description here"></a></p> <p>For users who can not see picture, see description below</p> <pre><code> Put numbers 1,2,3,4,5,6,8,9,10,11,12,13,14 (1 to 14, without 7) to each letter in such a way that the numbers in each row with 3 or 4 letters in all three directions, sum the same constant X M I J K L F G H B C D E A B+C+D+E = F+G+H = I+J+K+L = B+F+J+M = C+G+K = A+D+H+L = I+F+C+A = J+G+D = M+K+H+E </code></pre> https://puzzling.stackexchange.com/q/6195 12 Magic Squares in Sudoku Grids Kevin https://puzzling.stackexchange.com/users/294 2014-12-16T07:07:20Z 2014-12-22T23:28:08Z <p>A 3x3 magic square is a 3x3 grid containing the numbers 1-9 once each, and in which every row, column, and diagonal sums to 15:</p> <pre><code>294 753 618 </code></pre> <p>And I presume we all know the rules to Sudoku.</p> <p>Consider the following completed Sudoku puzzle:</p> <pre> +---+---+---+ |<b><i>294</i></b>|516|738| |<b><i>753</i></b>|842|196| |<b><i>618</i></b>|379|254| +---+---+---+ |571|468|329| |936|125|847| |482|793|561| +---+---+---+ |345|981|672| |827|634|915| |169|257|483| +---+---+---+ </pre> <p>Notice that the top-left 3x3 portion of the Sudoku solution is a magic square! Likewise, consider this Sudoku solution:</p> <pre> +---+---+---+ |8<b><i>29|4</i></b>75|613| |4<b><i>75|3</i></b>61|289| |3<b><i>61|8</i></b>29|574| +---+---+---+ |197|583|426| |643|912|758| |258|647|931| +---+---+---+ |782|194|365| |514|736|892| |936|258|147| +---+---+---+ </pre> <p>This solution also has a magic square in the top-left corner. However, notice that it isn't perfectly aligned with one of the nine 3x3 squares that are considered when solving Sudoku. Instead, it's a column to the right of the edge.</p> <p>Finally, consider this Sudoku solution:</p> <pre> +---+---+---+ |<b><i>294</i></b>|187|365| |<b><i>753</i></b>|269|814| |<b><i>618|34</i></b>5|279| +--<b><i>-+--</i></b>-+---+ |82<b><i>1|59</i></b>4|736| |53<b><i>6|72</i></b>8|941| |947|613|582| +---+---+---+ |165|872|493| |482|936|157| |379|451|628| +---+---+---+ </pre> <p>This one has two overlapping magic squares. The first one is in the top-right corner, like in the first Sudoku example. The second magic square shares the 8 in the bottom-right corner of the first magic square, but the rest of its numbers are below and to the right of the first magic square.</p> <p>Counting all magic squares that show up in a Sudoku solution, including ones that have overlapping numbers and that don't align with the 3x3 squares that are checked in a Sudoku puzzle, what is the maximum number of magic squares that can be present in a single Sudoku solution?</p> https://puzzling.stackexchange.com/q/6250 10 The Quite Unusual Square warspyking https://puzzling.stackexchange.com/users/2383 2014-12-18T13:12:45Z 2019-01-26T18:09:00Z <p>Imagine an $n \times n$ grid filled with the numbers 1, ..., $n$ where $n$ > 3 each number appearing n times, where each row, column, and diagonal all equal the same number. Can you fill grid like this? If so, show an example, and include as much detail as possible (for example if there's only 1 solution). If not, provide mathematical proof it's impossible.</p> <p><strong>I've purposely excluded $1 \times 1$ due to simplicity, $2 \times 2$ due to impossibility and $3 \times 3$ also due to impossibility (for solvers like you!)</strong></p> https://puzzling.stackexchange.com/q/81364 10 Not-Quite-Sufficiently-Advanced-Technology Square Rubio https://puzzling.stackexchange.com/users/30633 2019-04-03T19:40:43Z 2019-04-05T16:29:53Z <p>This was given as an assignment to a group of sixth graders, who were told they could use calculators. Beyond that, no real assistance was provided. They were not working on it as a group, so independent effort was presumably expected.</p> <hr> <blockquote> <p>MAGIC SQUARE </p> <p>Place the numbers provided into the grid so that the numbers ACROSS every row and DOWN every column add up to 200.</p> <p><span class="math-container">$$\begin{array}{|c|c|c|c|c|}\hline27&amp;&amp;&amp;&amp;36\\\hline&amp;40&amp;&amp;19\\\hline&amp;&amp;42\\\hline&amp;47&amp;&amp;42\\\hline32&amp;&amp;&amp;&amp;35\\\hline\end{array}$$</span></p> <p>Numbers:<br> <span class="math-container">$~~~~ ~~~~ \boxed{66} ~~~~ \boxed{46} ~~~~ \boxed{15} ~~~~ \boxed{41} ~~~~ \boxed{30} ~~~~ \boxed{72} ~~~~ \boxed{24} ~~~~ \boxed{28}$</span> <span class="math-container">$~~~~ ~~~~ \boxed{25} ~~~~ \boxed{66} ~~~~ \boxed{39} ~~~~ \boxed{22} ~~~~ \boxed{45} ~~~~ \boxed{54} ~~~~ \boxed{58} ~~~~ \boxed{49} ~~~~$</span></p> </blockquote> <hr> <p>First, this isn't going to be a real "Magic Square" for a number of reasons, but we'll ignore that.<br> The thing is, even with the grid partially filled in, a true unconfined brute-force search for a solution would take a very long time. A more intelligently created exhaustive search program can find a solution in seconds (there are two), but still requires a LOT of permutations to be evaluated. By computer, that's fast. With a hand-held calculator alone, manually keeping track of and testing the various permutations would be hellishly impractical.</p> <p>So -<br> <strong>Is there a way to solve this, using only pencil and paper and a calculator, without sheer luck or exhaustive effort?</strong><br> (Or, to put it another way, is this in any way a fair assignment for a sixth grader?)</p> <hr> <p><strong>Bonus Question</strong> </p> <p>Ok, I'll add this because some of you will want to try it.<br> Go ahead, find the two solutions any way you like. First one to supply them gets the brownie points.<br> But I'm mainly interested in a practical <a href="/questions/tagged/no-computers" class="post-tag" title="show questions tagged &#39;no-computers&#39;" rel="tag">no-computers</a> method, if one exists.</p> https://puzzling.stackexchange.com/q/5934 9 Does a magic rectangle exist? warspyking https://puzzling.stackexchange.com/users/2383 2014-12-05T00:25:21Z 2017-10-31T16:54:51Z <p>My definition of a magic rectangle:</p> <blockquote> <p>Any $m \times n$ rectangle where $m \ne n$ and all the numbers $1, 2, 3,\dots, mn$ fit into the rectangle. All horizontal lines, vertical lines, and diagonal lines (albeit not the same length) add up to the same number called a <em>"magic constant"</em></p> </blockquote> <p>Do any magic rectangles exist? If so, what are some examples? Please include dimensions, magic constant, and if possible, the whole rectangle.</p> <p><strong>BONUS: How can you determine a rectangle's magic constant from its $m \times n$ dimensions?</strong></p> https://puzzling.stackexchange.com/q/41171 8 Modify a magic square Jamal Senjaya https://puzzling.stackexchange.com/users/28756 2016-08-22T07:50:36Z 2016-12-21T04:35:31Z <p>This is a 3x3 magic square of <strong>summation</strong>,<br> in which <strong>sums</strong> of each row, column, and diagonal are equal.</p> <p>$$\begin{array}{c|c|c} 4&amp;9&amp;2\\\hline 3&amp;5&amp;7\\\hline 8&amp;1&amp;6 \end{array}$$</p> <p>Now modify the magic square by defining a simple function $f(x)$,<br> so it becomes a new magic square of <strong>multiplication</strong>,<br> i.e. in which <strong>products</strong> of each row, column, and diagonal are equal.<br></p> <p>$$\begin{array}{c|c|c} f(4)&amp;f(9)&amp;f(2)\\\hline f(3)&amp;f(5)&amp;f(7)\\\hline f(8)&amp;f(1)&amp;f(6) \end{array}$$</p> <p>Note: Create the function as simple as possible.</p> https://puzzling.stackexchange.com/q/23476 8 Complete the magic square! Kay-D Castaneda https://puzzling.stackexchange.com/users/17080 2015-10-23T21:03:45Z 2018-12-24T22:12:29Z <p>So, my math teacher gave us a magic math square with </p> <ul> <li>the 9 in the bottom right corner, </li> <li>the 7 in the left column middle row, and </li> <li>the 1 in the middle column top row.</li> </ul> <p>She said she would give whoever figured this out a can of soda from the teachers' lounge. I can't figure it out. I've tried nonstop for 2 days$\ldots$ I need help :( </p> <p>For those of you who don't know, a magic square is a grid with (in this case) 9 boxes and you have to put the numbers 1-9 in each square. Each column, row, and diagonally have to equal the same number. My math teacher said her and the other math teachers have found a way so it is possible but I'm not the best at math so I really need some help. </p> https://puzzling.stackexchange.com/q/3607 8 Can you fill a 3x3 grid with these numbers so the products of the rows and columns are the same? warspyking https://puzzling.stackexchange.com/users/2383 2014-11-06T00:31:00Z 2018-02-20T19:05:10Z <p>Is it possible to form a $3\mbox{x}3$ grid containing the set of numbers: $${1,2,4,8,16,32,64,128,256}$$</p> <p>in such a way that the product of the numbers in every row, column and diagonal are the same? If so, how?</p> https://puzzling.stackexchange.com/q/9336 8 9-by-9 filled, magic square Daniella https://puzzling.stackexchange.com/users/9688 2015-02-20T18:31:14Z 2018-04-03T11:01:39Z <p>Construct a 9-by-9 filled, magic square using the integers from 0 to 80. The magic square should additionally have the property that when it is divided into ninths according to the picture below, each 3-by-3 subsquare is also magic.</p> <p><img src="https://i.stack.imgur.com/gZirk.png" alt="enter image description here"></p> https://puzzling.stackexchange.com/q/20327 8 The magic of the primes Rand al'Thor https://puzzling.stackexchange.com/users/5373 2015-08-20T15:17:49Z 2015-08-20T16:03:37Z <p>A mathematician, a physicist, and an engineer found themselves caught in an ancient anecdote. Lacking a chemist to brew them an anecdote antidote, they fell to arguing over which of them was to be the butt of the joke ... when suddenly the physicist and the engineer disappeared! Between the spots where they'd been standing, a magician stood glaring at the mathematician.</p> <p>"Long have those of thy profession sought to ridicule my kin," he said menacingly. "Wherefore then, now that thou art in my power, should I release thee? I shall trap thee in unamusing tales with thy fellow <em>scientists</em> for ever."</p> <p>"As I see you are a magician," the mathematician began, "will a <strong>magic square</strong> enable me to avert your power?"</p> <p>The magician scoffed. "What is thy field of expertise within that arid wasteland, Mathematics?"</p> <p>"Number theory," the unperturbed<sup>1</sup> mathematician replied.</p> <p>"Ha! Those fools who seek to discover pattern among the prime numbers! In that case, I challenge thee to find a $2015\times2015$ magic square all of whose entries are <strong>prime</strong>. If thou canst achieve this feat, I shall release thee unharmed; otherwise, thou art my slave for ever!"</p> <hr> <p><strong>Can the mathematician succeed?</strong> Give a proof either that such a magic square exists or that it cannot exist. Unlike the unfortunate mathematician, you don't actually need to construct it!</p> <hr> <p><sup>1</sup><sub>Well, of <em>course</em> he was unperturbed. Perturbation theory is a branch of analysis, not number theory!</sub></p> https://puzzling.stackexchange.com/q/42804 8 How big can a witchcraft square be? Rand al'Thor https://puzzling.stackexchange.com/users/5373 2016-09-19T23:59:24Z 2016-09-20T17:31:12Z <p>A <em>witchcraft square</em> is defined to be an $n\times n$ square of distinct natural numbers such that the row sums and column sums form a set of $2n$ consecutive natural numbers. For example,</p> <pre> 5 11 14 15 8 13 10 17 9 2 22 16 24 21 4 3 </pre> <p>is a witchcraft square because the row and column sums are $45, 46, 47, 48, 49, 50, 51, 52$ in some order.</p> <p><strong>For which values of $n$ does there exist an $n\times n$ witchcraft square?</strong></p> <p><sub><sub>This puzzle was created by Stanley Rabinowitz for the 1982-83 issue of the Journal of Recreational Mathematics. I don't know the answer.</sub></sub></p> https://puzzling.stackexchange.com/q/1957 7 Puzzle of putting numbers 1-9 in 3x3 Grid to add up to 15 Freya https://puzzling.stackexchange.com/users/1935 2014-07-25T22:19:42Z 2016-10-26T13:32:50Z <p>In a 3x3 grid, I'd have to put numbers from 1 to 9 in a manner so that respective row, column and diagonal add up to 15.</p> <p>I have only been able to come up with one solution:</p> <p>$$\begin{array}{ccc} 6 &amp; 1 &amp; 8 \\ 7 &amp; 5 &amp; 3 \\ 2 &amp; 9 &amp; 4 \end{array}$$</p> <p>Through some calculations and trial and error method.</p> <p>Is there any strategy or way of approach to this problem, or is trial and error method the solution to it?</p> https://puzzling.stackexchange.com/q/18500 7 Magic Square Mixups [Challenge] Vincent Tang https://puzzling.stackexchange.com/users/2001 2015-07-30T20:03:05Z 2015-07-31T13:47:01Z <p>This kind of puzzle is different than your normal magic square puzzles. Here are 3, in increasing difficulty. Some numbers have been switched, and you have to find them and swap them around to make the magic square valid again.<br></p> <p>The zeros are for formatting placeholders.</p> <ol> <li><p>The numbers in each of line of five squares across, down, and diagonally should add up to 58, but in every row across and column there is one number of out place. Swap these with one another to make the total correct.<br> 16 16 11 09 14 <br> 17 09 29 13 01 <br> 15 05 06 16 15 <br> 07 03 17 11 12 <br> 14 17 03 08 06 <br></p></li> <li><p>The numbers in each of line of six squares across, down, and diagonally should add up to 122, but in every row across and column there is one number of out place. Swap these with one another to make the total correct.<br> 21 05 14 31 44 15 <br> 30 29 21 09 22 20 <br> 36 29 20 10 06 22 <br> 06 30 22 30 13 17 <br> 10 26 23 17 12 22 <br> 27 12 20 13 21 27 <br></p></li> <li><p>The numbers in each of line of seven squares across, down, and diagonally should add up to 123, but in every row across and column there is one number of out place. Swap these with one another to make the total correct.<br> 31 19 10 13 14 32 15 <br> 06 21 17 22 30 17 07 <br> 17 30 17 24 17 11 08 <br> 07 22 33 13 15 17 11 <br> 14 16 21 22 13 11 29 <br> 16 13 03 19 12 12 43 <br> 27 13 19 13 20 18 11 <br></p></li> </ol> <p>Have fun!!</p> https://puzzling.stackexchange.com/q/35480 7 No ordinary magic square Greg Hastings https://puzzling.stackexchange.com/users/24935 2016-06-09T09:00:25Z 2016-06-09T16:34:33Z <p>Instead of placing every number from 1-9 in the square below such that each column, row and long diagonal has the same sum, do the opposite!</p> <p>Place 1-9 in the squares below in a manner such that none of the columns, rows or long diagonals have the same sum:</p> <p><a href="https://i.stack.imgur.com/x5B0L.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/x5B0L.png" alt="enter image description here"></a></p> <p>Edit: The question above seemed too easy for many of you. Lets take the next step. Please count or calculate the number of possible correct answers to the same question. </p> <p><a href="https://puzzling.stackexchange.com/questions/35488/no-ordinary-magic-square-part-2-how-many-solutions-are-there">No ordinary magic square part 2. How many solutions are there?</a></p> https://puzzling.stackexchange.com/q/22226 7 The magic square with a hole Rohcana https://puzzling.stackexchange.com/users/15544 2015-09-15T22:33:15Z 2015-09-15T23:00:41Z <p>Alice loves magic squares. She has a 4x4 square, where she can put a number in each cell. But alas! Some evil person has poked a hole in her square. Alice is really really sad because she can't make a magic square anymore. Won't you cheer poor Alice up?</p> <blockquote> <p>You are given a 4x4 square with 1 of its cells removed. Put the numbers 1 to 15 in it so that each row, column and diagonal sum to the same.</p> </blockquote> https://puzzling.stackexchange.com/q/40811 7 Arrange the numbers in a 4x4 table Jamal Senjaya https://puzzling.stackexchange.com/users/28756 2016-08-17T22:27:37Z 2016-08-18T16:16:00Z <p>Put these numbers:</p> <blockquote> <p>2, 3, 4, 5, 6, 9, 10, 12, 15, 18, 20, 25, 30, 45, 50, 75 </p> </blockquote> <p>in a 4x4 square table so <strong>the products of all numbers in any given row, column and diagonal are equal.</strong></p> <p><em>Note</em> : There are multiple solutions, it is preferred to find the universal rule to discover all the solutions.</p> https://puzzling.stackexchange.com/q/79519 7 A 4 x 4 Magic Square with Pairwise Relatively Prime Entries Bernardo Recamán Santos https://puzzling.stackexchange.com/users/19026 2019-02-10T14:31:58Z 2019-02-11T00:17:23Z <p>Find a 4 x 4 magic square of positive integers such that any two of its entries are pairwise different and relatively prime, i.e., have no common divisor greater than 1. </p> <p>What is the least that the largest number in such a square can be?</p> https://puzzling.stackexchange.com/q/59876 7 Find the missing numbers in the magic square user2882061 https://puzzling.stackexchange.com/users/39563 2018-01-29T16:27:45Z 2018-06-01T06:50:27Z <p>The following is a magic square: each row, column and diagonal add to 34, all of the numbers 1 to 16 appear exactly once. Find the missing numbers.</p> <p><a href="https://i.stack.imgur.com/xllaS.png" rel="noreferrer"><img src="https://i.stack.imgur.com/xllaS.png" alt="enter image description here"></a></p> https://puzzling.stackexchange.com/q/10868 7 3x3 “Magic Square” of Prime Numbers -- Part II LaBird https://puzzling.stackexchange.com/users/9944 2015-03-24T10:18:43Z 2015-03-24T16:09:41Z <p>Glad to know <a href="https://puzzling.stackexchange.com/questions/10797/3x3-magic-square-of-prime-numbers">the previous puzzle</a>, which was the first puzzle I posted in Puzzling, was warmly welcomed (Thank you!), and an optimal solution was found. Inspired by the comments there, here is the Version $2$ of the puzzle.</p> <p><img src="https://i.stack.imgur.com/urjOD.png" alt="enter image description here"></p> <p>In fact, most of the things are unchanged. We still have this $3 \times 3$ grid, which $9$ distinct prime numbers $P_1, P_2, ..., P_9$ are to be filled in. And there are $8$ sums: $3$ horizontal, $3$ vertical and $2$ diagonal, and they are named $S_1, S_2, ..., S_8$. All the requirements in the <a href="https://puzzling.stackexchange.com/questions/10797/3x3-magic-square-of-prime-numbers">first version</a> still hold here, which mean:</p> <ul> <li>$P_1, P_2, ..., P_9, S_1, S_2, ..., S_8$ are <strong>all distinct prime numbers</strong> (i.e. there are totally $17$ different prime numbers).</li> </ul> <p>But this time, one more additional requirement:</p> <ul> <li>The grand total $P_1 + P_2 + ... + P_9 + S_1 + S_2 + ... + S_8$ also has to be a <strong>prime number</strong>.</li> </ul> <p><strong>The challenge: To minimize the grand total.</strong></p> <p>With the additional requirement, some solutions satisfying the previous puzzle do not satisfy this version. And, the optimal solution will be different.</p> <p>Below is one possible solution I come up with, which has a grand total of $601$, but it is not the optimal solution:</p> <p><img src="https://i.stack.imgur.com/78zyA.png" alt="enter image description here"></p> <p>Feel free to have a try!</p> https://puzzling.stackexchange.com/q/80210 6 I'm trying to create a magic square Lucy https://puzzling.stackexchange.com/users/57524 2019-03-04T08:43:54Z 2019-03-04T14:27:38Z <p>I'm having trouble trying to make a <span class="math-container">$3\times3$</span> magic square with magic number <span class="math-container">$12$</span> and I can't figure it out.</p> <p>Can you please help me?</p> https://puzzling.stackexchange.com/q/14695 6 How do I solve these 3x3 magic squares? [duplicate] user12319 https://puzzling.stackexchange.com/users/12319 2015-05-11T09:07:48Z 2015-05-11T17:53:01Z <div class="question-status question-originals-of-duplicate"> <p>This question already has an answer here:</p> <ul> <li> <a href="/questions/1957/puzzle-of-putting-numbers-1-9-in-3x3-grid-to-add-up-to-15" dir="ltr">Puzzle of putting numbers 1-9 in 3x3 Grid to add up to 15</a> <span class="question-originals-answer-count"> 4 answers </span> </li> </ul> </div> <p>I'm doing 3x3 magic squares. Here are the squares I'm working on:</p> <pre><code>| | 5 | | | | | | | 8 | | | </code></pre> <p>The values must be between 3 and 12, and each line must add to 21.</p> <p>Here's another one: </p> <pre><code>| | 9 | | | | | 3 | | | | | </code></pre> <p>For this one, the boundaries are 3-11. As with the last one, each line total must add to 21.</p> https://puzzling.stackexchange.com/q/1992 6 Magic back yards Lembik https://puzzling.stackexchange.com/users/449 2014-07-30T19:17:38Z 2014-08-02T18:42:37Z <p>My back yard forms a rectangular grid of squares except some of the squares are missing as they are covered by pipes or a small tree. The layout is as follows. A 'x' indicates a square that is free and '.' a covered square.</p> <pre><code>. x x . . x x x . x x x x x x x x x x x x x x x x x x x x x x x . x x x </code></pre> <p>I would like write distinct non-negative numbers in each one of the free squares. However, to make it interesting I would like the sums of all the columns to be the same and the sum of the all the rows to be the same (although the row sum need not equal the column sum). We can assume that the value of a covered square is zero.</p> <blockquote> <p>Is this possible with this layout? If not, why not?</p> </blockquote>