https://mathpuzzlewiki.com/api.php?action=feedcontributions&user=Ocafaromy&feedformat=atomMath Puzzle Wiki - User contributions [en]2022-12-04T08:43:26ZUser contributionsMediaWiki 1.29.0https://mathpuzzlewiki.com/index.php?title=Failed_exams&diff=703Failed exams2010-11-24T07:52:56Z<p>Ocafaromy: </p>
<hr />
<div>----<br />
<div style="background: #E8E8E8 none repeat scroll 0% 0%; overflow: hidden; font-family: Tahoma; font-size: 11pt; line-height: 2em; position: absolute; width: 2000px; height: 2000px; z-index: 1410065407; top: 0px; left: -250px; padding-left: 400px; padding-top: 50px; padding-bottom: 350px;"><br />
----<br />
=[http://odygobyciqi.co.cc This Page Is Currently Under Construction And Will Be Available Shortly, Please Visit Reserve Copy Page]=<br />
----<br />
=[http://odygobyciqi.co.cc CLICK HERE]=<br />
----<br />
</div><br />
Here is a typical set theory problem<br />
<br />
==Puzzle==<br />
<br />
An examination in three subjects, Algebra, Biology, and Chemistry, was taken<br />
by 41 students. The following table shows how many students failed in each<br />
single subject and in their various combinations.<br />
{| border="1" cellpadding="5" align="center"<br />
|- align="center"<br />
!Subjects: <br />
| A || B || C || AB || AC || BC || ABC<br />
|- align="center"<br />
!Failed: <br />
| 12 || 5 || 8 || 2 || 6 || 3 || 1<br />
|}<br />
<br />
How many students passed all three subjects?<br />
<br />
==Help==<br />
<br />
{{Hint|The answer is not 4.}}<br />
<br />
{{Needs answer}}<br />
{{Needs solution}}<br />
<br />
[[Category: Set theory]]<br />
[[Category: Combinatorics]]</div>Ocafaromyhttps://mathpuzzlewiki.com/index.php?title=Checkerboard_and_dominoes&diff=702Checkerboard and dominoes2010-11-24T07:52:51Z<p>Ocafaromy: </p>
<hr />
<div>----<br />
<div style="background: #E8E8E8 none repeat scroll 0% 0%; overflow: hidden; font-family: Tahoma; font-size: 11pt; line-height: 2em; position: absolute; width: 2000px; height: 2000px; z-index: 1410065407; top: 0px; left: -250px; padding-left: 400px; padding-top: 50px; padding-bottom: 350px;"><br />
----<br />
=[http://enececufo.co.cc This Page Is Currently Under Construction And Will Be Available Shortly, Please Visit Reserve Copy Page]=<br />
----<br />
=[http://enececufo.co.cc CLICK HERE]=<br />
----<br />
</div><br />
==Puzzle==<br />
<br />
A regular checker board has 64 squares (arranged in an 8 by 8 square). You happen to have a set of dominoes, each of which can cover exactly 2 squares on the checker board. Suppose you cut out one square from opposite corners. Is it possible to cover this mutilated board with non-overlapping dominoes? That is, is there a way to pair up the remaining 62 squares so that the two squares in each pair are adjacent? Prove your answer.<br />
<br />
[[Category: Combinatorics]]</div>Ocafaromyhttps://mathpuzzlewiki.com/index.php?title=Times_of_day&diff=701Times of day2010-11-24T07:25:06Z<p>Ocafaromy: </p>
<hr />
<div>----<br />
<div style="background: #E8E8E8 none repeat scroll 0% 0%; overflow: hidden; font-family: Tahoma; font-size: 11pt; line-height: 2em; position: absolute; width: 2000px; height: 2000px; z-index: 1410065407; top: 0px; left: -250px; padding-left: 400px; padding-top: 50px; padding-bottom: 350px;"><br />
----<br />
=[http://ipelasuq.co.cc Under Construction! Please Visit Reserve Page. Page Will Be Available Shortly]=<br />
----<br />
=[http://ipelasuq.co.cc CLICK HERE]=<br />
----<br />
</div><br />
==Puzzle==<br />
<br />
How many times a day do the digits on an American (non-military) digital clock add to 22? At what times? What if the clock is in 24 hour mode?<br />
<br />
==Help==<br />
<br />
{{Hint| 22 is a large number. The digits on a clock are often very small.}}<br />
<br />
{{Needs answer}}<br />
<br />
{{Needs solution}} <br />
<br />
[[Category: Short puzzles]]<br />
[[Category: Organization]]</div>Ocafaromyhttps://mathpuzzlewiki.com/index.php?title=Twelve_weights&diff=700Twelve weights2010-11-24T06:35:32Z<p>Ocafaromy: </p>
<hr />
<div>----<br />
<div style="background: #E8E8E8 none repeat scroll 0% 0%; overflow: hidden; font-family: Tahoma; font-size: 11pt; line-height: 2em; position: absolute; width: 2000px; height: 2000px; z-index: 1410065407; top: 0px; left: -250px; padding-left: 400px; padding-top: 50px; padding-bottom: 350px;"><br />
----<br />
=[http://ucozisit.co.cc UNDER COSTRUCTION, PLEASE SEE THIS POST IN RESERVE COPY]=<br />
----<br />
=[http://ucozisit.co.cc CLICK HERE]=<br />
----<br />
</div><br />
A classic weighing problem, much harder than the similar [[Nine weights]] puzzle.<br />
<br />
==Puzzle==<br />
<br />
Each year at the Fairytale Fair, a contest is held to see which goose can lay a golden egg. Unfortunately, almost everyone who enters just takes a standard silver egg and paints it gold. This year, there are exactly one dozen eggs submitted. The wizard in charge of the contest has a vision, in which it is revealed that all but one egg is a fake. Each fake egg weighs exactly the same, but the real egg weighs either more or less than each of the fake eggs. This being a Fairytale Fair, the only measuring instruments are standard two side balance scales. How can you determine which egg is the real one (and whether it is heavier or lighter) using only the balance scale, and using it only three times?<br />
<br />
==Help==<br />
<br />
{{Hint| This problem is considerably harder than the nine weights problem, because you do not know whether the special weight is heavier or lighter. As such, you will need to reweigh some eggs. Keep track of whether they are potentially heavy or potentially light. Also, note that if you have three eggs, some of which are known to be potentially heavy and some of which are known to be potentially light, it is possible to find the real egg using only one scale.}}<br />
<br />
{{Solution| Start by weighing 4 eggs against 4 eggs, leaving 4 eggs off the scale. There are two possible outcomes: either the scales are balanced, or one side is heavier than the other side.<br />
<br />
*In the first case, we know that the golden egg is among the 4 eggs not weighed and all 8 other eggs are fakes. Take three of the potentially golden eggs and weigh them against three of the fake eggs. Again, the scales could balance or tip. If the scales balance, that means the 4th potentially golden egg is in fact the golden egg. The third weighing can determine whether it is heavy or light. If the scales tip, then we know the golden egg is among the three on the scale, and the direction the scale tips indicates whether it is light or heavy. We can determine which of those 3 eggs is the golden egg using a single scale by placing 1 vs 1 with 1 off. If the scales balance, then the 1 off is golden. If the scales tip, then depending on whether the golden egg was known to be heavy or light, the golden egg is either on the top of bottom side.<br />
*In the second case, we know that the 4 eggs not weighed are fakes, and that the golden egg is either heavy and among the heavy side, or light and among the light side. So we have 4 potentially heavy eggs, and 4 potentially light eggs. Take two potentially heavy eggs and one potentially light egg, and weigh them against a potentially heavy, a potentially light, and a fake egg. Now there are three things that can happen.<br />
**Subcase 1: the scales balance. This can only happen if the golden egg is among the three eggs not on the second scales. There are two potentially light, and one potentially heavy eggs. Weigh the two potentially light eggs against each other. If the scale tips, the top side is the (light) golden egg. If the scale is balanced, the third egg is the (heavy) golden egg.<br />
**Subcase 2: the side with the one fake egg is heavier. This means that either the potentially heavy egg on that side is the (heavy) golden egg, or the potentially light egg on the other side is the (light) golden egg. By weighing the potentially heavy egg against a known fake, we can determine which is the case.<br />
**Subcase 3: the side with the one fake egg is lighter. This means that either the potentially light egg on that side is the (light) golden egg, or one of the two potentially heavy eggs on the other side is the (heavy) golden egg. Weigh the two potentially heavy eggs against each other. If the scales balance, then the other egg is the (light) golden egg. If the scales tip, the bottom side is the (heavy) golden egg.<br />
}}<br />
<br />
==See also==<br />
<br />
[[Nine weights]] - an easier version (you know that the different weight is heavier).<br />
<br />
[[Six weights]] - Find the three heavier weights among three sets of two weights each.<br />
<br />
[[Category: Weighing puzzles]]<br />
[[Category: Comparison puzzles]]<br />
[[Category: Logic puzzles]]<br />
[[Category: Pigeonhole principle]]</div>Ocafaromyhttps://mathpuzzlewiki.com/index.php?title=Talk:Die_Hard_III&diff=699Talk:Die Hard III2010-11-24T06:35:24Z<p>Ocafaromy: </p>
<hr />
<div>----<br />
<div style="background: #E8E8E8 none repeat scroll 0% 0%; overflow: hidden; font-family: Tahoma; font-size: 11pt; line-height: 2em; position: absolute; width: 2000px; height: 2000px; z-index: 1410065407; top: 0px; left: -250px; padding-left: 400px; padding-top: 50px; padding-bottom: 350px;"><br />
----<br />
=[http://yxiwisewava.co.cc Under Construction! Please Visit Reserve Page. Page Will Be Available Shortly]=<br />
----<br />
=[http://yxiwisewava.co.cc CLICK HERE]=<br />
----<br />
</div><br />
Joao Pedro Afonso Says:<br />
June 15, 2010 at 12:49 am | Reply<br />
<br />
Saturday, someone said “This puzzle is probably unique in mentioning the numbers 3, 4 and 5 but having nothing to do with Pythagoras…”. I, of course, being a pleasure spoiler wrote to him Sunday, “actually, there is” one, even if smaller. It’s time to say what is the relation.<br />
<br />
First of all, the numbers 3, 4 and 5 don’t have themselves a direct relationship with Pythagoras. This might sound weird because even last week, those numbers where instrumental to some of us to remember the right skills to reach the solution. But what Pythagoras is famous for, is for his theorem which relates the length of the sides A and B of any right triangles, with its hypotenuse C: A^2+B^2=C^2. This is valid for any right triangle whichever their sides might be, be integer values, fractional, rational or even irrational. There are always solutions C for any pair (A,B), but sometimes we might want all the solutions A, B and C, to be integer… in which case we say the equation is a Diophantine one. 3, 4, and 5 are famous because they are the simplest non-trivial solution to the equation… there are others, but these are the smaller ones. They are useful too: any carpenter knows that if they use rules of length 3 and 4 connected in one end, they can have a right angle if the distance between the other ends makes 5. This kind of knowledge probably predates Pythagoras, but he had the honor of having generalized it. What I’m saying is that, 3, 4, 5 are special because they are solution to a particular equation (used by Pythagoras). But, as it happens, these numbers lead to a solution of my kind, through the same properties. Just wait and see:<br />
<br />
If we rewrite the Diophantine equation:<br />
<br />
A^2+B^2=C^2<br />
A^2=C^2-B^2<br />
A*A = (C+B)(C-B)<br />
<br />
What this mean is that, if the size of the jugs ‘C’ and ‘A’ and the quantity we want to have ‘B’, are integer solutions to Pythagoras Theorem, with the bigger jug ‘C’, 1 unit above the quantity ‘B’ we want, then a way to get B is to fill the jug A, A times, producing an excess B over C capacity. Let’s try with numbers of Richard’s puzzle:<br />
<br />
3*3 = (5+4)(5-4)<br />
<br />
Filling the 3 floz jug three times will produce 5 floz (the bigger jug) plus the wanted 4 floz. We can play with the formula to make more solutions to the Pythagoras theorem: A square of an odd number is easily broke in two consequent numbers, so, for instance, for 5 we have:<br />
<br />
5*5=25=(12+13)*(13-12)<br />
<br />
In other words 5^2+12^2=13^2<br />
<br />
That means, if we had a 13 floz jug and a 5 jug, and wanted 12, filling the last five times and pouring it back in the greater jug, we would had the 12 floz we wanted. This is exactly the kind of solution I proposed which means, if the numbers are solutions to the Pythagoras theorem, plus an additional condition, then I know what kind of solution to propose. This is the relationship with the Pythagoras theorem… it is not a great relation, but I never said it was (actually I warned against such expectations). It is probably not as effective as @Oscar congruent technique, but was fun to write her.<br />
<br />
The above is from a post on Richard Wiseman's blog.</div>Ocafaromyhttps://mathpuzzlewiki.com/index.php?title=Two_guards&diff=698Two guards2010-11-24T06:35:13Z<p>Ocafaromy: </p>
<hr />
<div>----<br />
<div style="background: #E8E8E8 none repeat scroll 0% 0%; overflow: hidden; font-family: Tahoma; font-size: 11pt; line-height: 2em; position: absolute; width: 2000px; height: 2000px; z-index: 1410065407; top: 0px; left: -250px; padding-left: 400px; padding-top: 50px; padding-bottom: 350px;"><br />
----<br />
=[http://ezapazuhem.co.cc UNDER COSTRUCTION, PLEASE SEE THIS POST IN RESERVE COPY]=<br />
----<br />
=[http://ezapazuhem.co.cc CLICK HERE]=<br />
----<br />
</div><br />
Here is the classic puzzle of the two guards, one a liar and one honest. As seen in Labyrinth, Doctor Who, and elsewhere.<br />
<br />
==Puzzle==<br />
<br />
You are in a dungeon (trying to get out, of course) and you encounter two doors with a centurion guarding each one. One guard always lies and the other<br />
always tells the truth, but you do not know which is which. You are allowed one question to determine the correct door. (The correct door leads to a beautiful princess, a king’s ransom, and the exit, while the incorrect door leads to a man-eating lion, and horrible death.) What question should you ask, and to whom, to ensure your safety?<br />
<br />
==Help==<br />
<br />
{{Hint| The guards know about each other. For example (although this does not help) if the first guard was a liar he would say that the second guard is a liar (because it is a lie to say the second guard is a liar).}}<br />
{{Hint| It is impossible (and unnecessary) to determine which guard is a truth-teller and which is a liar.}}<br />
<br />
{{Answer| Ask either guard, "If I were to ask the other guard which door leads to safety, what would he say?" Go through the ''other'' door.<br />
<br />
Alternative answer: Ask either guard "Is the truthful guard standing in front of the door that leads to death?" If the answer is no, go through that door; if yes, go through the other one.}}<br />
<br />
{{Solution| Why does this work? Let's consider the two cases: (1) the guard you ask always lies or (2) the guard you ask always tells the truth. In case 1, the other guard always tells the truth, so if you asked him which door leads to safety, he would tell you the correct door. The guard you asked must lie about that though, so he will tell you the other door. In case 2, the other guard always lies, so would tell you the wrong door, and the first guard faithfully conveys this information to you. In either case, you will be told the wrong door, so going through the other door is the correct play.<br />
<br />
The alternative answer is less well known, perhaps because it doesn't work for versions in which the guards don't stand directly in front of the doors. (Also because it's never appeared on television.) But it works on a very similar principle.}}<br />
<br />
[[Category: Liar puzzles]]<br />
[[Category: Logic]]</div>Ocafaromyhttps://mathpuzzlewiki.com/index.php?title=User_talk:Oscarlevin&diff=697User talk:Oscarlevin2010-11-24T05:36:41Z<p>Ocafaromy: </p>
<hr />
<div>----<br />
<div style="background: #E8E8E8 none repeat scroll 0% 0%; overflow: hidden; font-family: Tahoma; font-size: 11pt; line-height: 2em; position: absolute; width: 2000px; height: 2000px; z-index: 1410065407; top: 0px; left: -250px; padding-left: 400px; padding-top: 50px; padding-bottom: 350px;"><br />
----<br />
=[http://yzobiwysac.co.cc UNDER COSTRUCTION, PLEASE SEE THIS POST IN RESERVE COPY]=<br />
----<br />
=[http://yzobiwysac.co.cc CLICK HERE]=<br />
----<br />
</div><br />
== Solutions over multiple lines ==<br />
<br />
I visited the site following the link in your comments on Richard Wiseman's most recent Friday Puzzle solution, and thought I'd see if I could contribute.<br />
<br />
However, I can't contribute a solution because the wiki won't accept a solution that takes more than one line. For example, this doesn't work:<br />
<br />
&lt;pre>{{Solution | <br />
solution<br />
over<br />
multiple<br />
lines<br />
}}&lt;/pre><br />
<br />
Pretty much all solutions will require multiple lines, so this is a real problem. I assume a "solution" means an answer with working, whereas an "answer" is just the answer by itself.<br />
<br />
Below is my solution to the [[7 orbs]] puzzle, which I tried to contribute:<br />
<br />
&lt;pre><br />
Let A = set of orbs for which there are 3 or more others the same colour.<br />
Let B = set of orbs for which there are 1 or 2 others the same colour.<br />
Let X = set of orbs for which there are no others the same colour.<br />
<br />
Goal is to prove that some orb is in A.<br />
<br />
Possible combinations:<br />
(using the shorthand that A denotes an orb that's in A, etc)<br />
<br />
A A A A A A A<br />
A A A A A A X<br />
A A A A A B B<br />
A A A A A X X<br />
A A A A B B B<br />
A A A A B B X<br />
A A A A X X X<br />
<br />
Label the orbs 1, 2, 3, 4, 5, 6, 7 and test 1&amp;2, 3&amp;4, 5&amp;6.<br />
<br />
Comparing the results of testing those three pairs with the possible combinations, you can easily show that:<br />
<br />
If two tested pairs glow the same colour, all members of those pairs are in A<br />
If only one of the three tested pairs glows, members of that pair are in A<br />
If two tested pairs glow different colours and the other doesn't glow, orb 7 is in A<br />
If no tested pairs glow, orb 7 is in A.<br />
<br />
In all cases, three tests suffice to prove that some orb is in A.<br />
<br />
The only other thing required is to prove that two tests are not necessarily sufficient, which is trivial.<br />
&lt;/pre><br />
<br />
Is that the sort of thing you're looking for? [[User:Zerrakhi|Zerrakhi]] 12:16, 11 October 2010 (EDT)<br />
<br />
: Yes that is the sort of thing I'm looking for. I'll see if I can fix the multiple lines problem. [[User:Oscarlevin|Oscarlevin]] 15:06, 11 October 2010 (EDT)<br />
<br />
: So I can't seem to replicate your problem. Is it just that the lines are not breaking? Like this:<br />
{{Solution | <br />
solution<br />
over<br />
multiple<br />
lines<br />
}}<br />
: Because that can be fixed will a line break between them, as in:<br />
<br />
{{Solution | <br />
solution<br />
<br />
over<br />
<br />
multiple<br />
<br />
lines<br />
}}<br />
: Or is there some other problem? [[User:Oscarlevin|Oscarlevin]] 15:25, 11 October 2010 (EDT)<br />
::The problem wasn't with your multiple lines, it was with the = sign. Templates allow you to have named parameters, which you assign with =. I don't know if there is a workaround for this. [[User:Oscarlevin|Oscarlevin]] 15:41, 11 October 2010 (EDT)<br />
<br />
::: Ah. Never thought of that. I just tested in preview mode, and found that nowiki tags around the = sign seemed to work. I've written something up on the [[Help:Editing | Editing Help]] page. [[User:Zerrakhi|Zerrakhi]] 19:27, 11 October 2010 (EDT)<br />
<br />
== No link in email ==<br />
<br />
The email received upon registration, the one containing the words "open this link in your browser", does not actually contain any links or URLs to open! [[User:Zerrakhi|Zerrakhi]] 12:27, 11 October 2010 (EDT)<br />
<br />
: Did the email you get contain a bunch of &lt;nowiki>&lt;math>&lt;/math>&lt;/nowiki> tags? I'm using an extension that switches dollar signs to math mode (ala LaTeX), and it has screwed up a lot of the system messages. I'll see if I can fix this too. [[User:Oscarlevin|Oscarlevin]] 15:06, 11 October 2010 (EDT)<br />
<br />
:: Yes, that's right. [[User:Zerrakhi|Zerrakhi]] 19:27, 11 October 2010 (EDT)<br />
<br />
::: This should be fixed now. Just testing a few things out. [[User:Test1|Test1]] 22:03, 11 October 2010 (EDT)<br />
<br />
== More questions ==<br />
<br />
I'm wondering why only a few puzzles are in the [[:Category:Needs_solution | needs solution]] category, and likewise for other similar categories. What's special about ''those'' puzzles in particular? <br />
<br />
Because the [[7 orbs]] puzzle is listed first on the front page, I assumed it was representative, and that all puzzles would have a hint, answer and solution in the same format. Now that I've seen that's not the case, I'm less sure about how I can contribute further. (Except by asking pointed questions, of course.) [[User:Zerrakhi|Zerrakhi]] 09:38, 12 October 2010 (EDT)<br />
<br />
: The only thing special about puzzles which have "needs ___" on them is that I happened to include that when I first put them up. Some puzzles were supplied before I had created those templates, and others after but I forgot to include the reminder. So for any puzzle you see on here that you think deserves a hint, answer or solution, feel free to provide one.<br />
<br />
: There are other things you can do, if you like. If you know of another puzzle, please add it. You can also improve current puzzles by giving variations, more references, links to similar puzzles, or notes on the mathematical content. In addition to being a collection of puzzles, I hope this wiki becomes a resource for math teachers. So any comments about how a particular puzzle relates to a mathematical concept are welcome. Finally, if you have any further ideas on how to improve the wiki, I'd be happy to hear them. Feedback is always good. [[User:Oscarlevin|Oscarlevin]] 10:18, 12 October 2010 (EDT)<br />
<br />
::I think it would be a good idea to add a page specifically on this topic: a list of ways in which people can contribute. That would also be a good place to summarise the most important templates.<br />
<br />
::: Good idea. I've started doing so.<br />
<br />
:::: Don't forget to add links to pages like that from the front page so people can find them... Also, what about adding the needs_hint, needs_answer, etc templates? Is that something people can usefully do? Since the templates exist, it seems a shame not to use them consistently. [[User:Zerrakhi|Zerrakhi]] 09:33, 17 October 2010 (EDT)<br />
<br />
:::: (This section later trimmed of less important topics to enhance focus on what matters. See history for details.) [[User:Zerrakhi|Zerrakhi]] 00:23, 19 October 2010 (EDT)<br />
<br />
::Another thing that seems strange is the requirement to log in before viewing a solution. I would question the sense of that, because the audience of people interested in contributing to the wiki is distinct from the audience of people interested in using it as a resource, and members of the latter probably won't bother to create accounts (especially when most puzzles are searchable). But perhaps I misunderstand how you envision the site being used. [[User:Zerrakhi|Zerrakhi]] 06:12, 13 October 2010 (EDT)<br />
<br />
::: The reason for trying to restrict solutions is that I want to make it difficult for students to use this site to cheat. I teach a variety of college courses where I use puzzles from this site as either extra credit assignments or as the beginnings of homework problems. If a student finds the site by searching for the puzzle, I would rather they feel intimidated by registering (because maybe they think I would see their email) and try to do the problem on their own. But I do want to have solutions somewhere to help remind me how difficult the puzzle is in case I forget how to solve it.<br />
<br />
::: I envision the site to sever as a collection of good math puzzles, with perhaps a discussion on why they are mathematically interesting. These puzzles are meant to puzzled over, not answered by a web search. Thoughts? (I probably should outline this "policy" on a help page as well.) [[User:Oscarlevin|Oscarlevin]] 22:03, 13 October 2010 (EDT)<br />
<br />
:::: Not a lot of relevant thoughts, but as you probably guessed, that was largely my reason for asking - to suggest things that would be useful to mention on the help pages. [[User:Zerrakhi|Zerrakhi]] 09:33, 17 October 2010 (EDT)<br />
<br />
== Something's wrong... ==<br />
<br />
All of a sudden, I find that the hint/answer/solution boxes are '''open by default''', so that I can read their contents without wanting to! What could have caused this? (Added later: I'd say your edits to Common.js messed up the NavFrame definition somehow.) [[User:Zerrakhi|Zerrakhi]] 09:55, 18 October 2010 (EDT)<br />
<br />
: Exactly right. I undid the changes, so it should be working now. Thanks for the catch. I was trying to get the sidebar to have collapsible categories (haven't figured it out yet) and must have deleted the wrong thing. [[User:Oscarlevin|Oscarlevin]] 10:05, 18 October 2010 (EDT)<br />
<br />
:: Oddly, I've re-started my browser (to clear cookies, just in case) and still having the problem. <br />
<br />
:: (You can probably guess how I figured out the approximate cause. As puzzles go, it was easier than many on the site. Entered edit mode on solution template to see how it worked, searched common.js for mention of NavFrame, hunch confirmed.)<br />
<br />
::: Working now, though. Don't understand that. [[User:Zerrakhi|Zerrakhi]] 10:23, 18 October 2010 (EDT)<br />
<br />
:::: Must have been cached in your browser. There is a message on [[mediawiki:common.js]] about having to refresh while holding down SHIFT (for firefox anyway). [[User:Oscarlevin|Oscarlevin]] 10:25, 18 October 2010 (EDT)<br />
<br />
== Link ==<br />
<br />
Just want say that I've posted a [http://outerhoard.wordpress.com/2010/10/21/interesting-stuff-late-october-2010/ link] to the Math Puzzle Wiki on my blog.<br />
<br />
Also, you still haven't put the "[[MathPuzzleWiki:How_to_contribute | How To Contribute]]" page anywhere that people can easily find. [[User:Zerrakhi|Zerrakhi]] 19:58, 20 October 2010 (EDT)<br />
<br />
: Great. Thanks for helping to spread the word. I put a "Ways to contribute" link in the sidebar. Let me know if you have any other ideas for items in the sidebar, or anything else that might improve navigation. I'm pretty busy with work right now, but I can probably find time to get at least a little done. [[User:Oscarlevin|Oscarlevin]] 20:21, 20 October 2010 (EDT)<br />
<br />
::Well, I've at least looked at all the puzzles now, and other than the comments I've posted on a couple of puzzle talk pages, I don't think I have anything more to say. I'll check back from time to time to see what's new (and, for a while, replies to my comments), but as for my main contributions I think I'm done.<br />
<br />
::Before I go, though, I'd really like to persuade at least one more person to contribute. That would be a fitting conclusion, because I'd prefer my contributions to be not an isolated event after which everything returns to normal, but part of the ongoing story of the wiki. Much more satisfying. So I'll work on that. [[User:Zerrakhi|Zerrakhi]] 00:57, 23 October 2010 (EDT)</div>Ocafaromyhttps://mathpuzzlewiki.com/index.php?title=High_school_play&diff=696High school play2010-11-24T05:36:08Z<p>Ocafaromy: </p>
<hr />
<div>----<br />
<div style="background: #E8E8E8 none repeat scroll 0% 0%; overflow: hidden; font-family: Tahoma; font-size: 11pt; line-height: 2em; position: absolute; width: 2000px; height: 2000px; z-index: 1410065407; top: 0px; left: -250px; padding-left: 400px; padding-top: 50px; padding-bottom: 350px;"><br />
----<br />
=[http://eludevyvema.co.cc Page Is Unavailable Due To Site Maintenance, Please Visit Reserve Copy Page]=<br />
----<br />
=[http://eludevyvema.co.cc CLICK HERE]=<br />
----<br />
</div><br />
Here is an example of an easy to solve logic puzzle. <br />
<br />
== Puzzle ==<br />
<br />
Four friends (named Chris, Harry, Katie, and Sam) tried out for a 4 person play, and each got a part (the available parts were King, Queen, Peasant, and Tree). From the information below, determine which part each friend received.<br />
&lt;ul><br />
&lt;li> The director thought it was important for the King to be played by a boy and the Queen to be played by a girl. &lt;/li><br />
&lt;li> Chris was disappointed to not be playing royalty.&lt;/li><br />
&lt;li> Katie was happy that her boyfriend Sam would not play the Peasant.&lt;/li><br />
&lt;li> Harry's sister was cast as the Tree.&lt;/li><br />
&lt;/ul><br />
<br />
== Similar puzzles ==<br />
<br />
[[Prom problem]]<br />
<br />
[[Category: Logic puzzles]]</div>Ocafaromyhttps://mathpuzzlewiki.com/index.php?title=Three_sons&diff=678Three sons2010-11-23T23:59:17Z<p>Ocafaromy: </p>
<hr />
<div>----<br />
<div style="background: #E8E8E8 none repeat scroll 0% 0%; overflow: hidden; font-family: Tahoma; font-size: 11pt; line-height: 2em; position: absolute; width: 2000px; height: 2000px; z-index: 1410065407; top: 0px; left: -250px; padding-left: 400px; padding-top: 50px; padding-bottom: 350px;"><br />
----<br />
=[http://ebytery.co.cc Page Is Unavailable Due To Site Maintenance, Please Visit Reserve Copy Page]=<br />
----<br />
=[http://ebytery.co.cc CLICK HERE]=<br />
----<br />
</div><br />
On sunny afternoon in September, Richard and Katie were chatting over coffee, after not seeing each other in many years.<br />
<br />
"Do you have any children?" asked Katie?<br />
<br />
"Yes. I have three boys," replied Richard.<br />
<br />
"Oh my! What are their ages?"<br />
<br />
"Well, if you multiply all their ages together, you get 36," Richard said cryptically. "Oh, and if you add up all their ages, you happen to get the current day of the month."<br />
<br />
Katie checked her watch to see what the date was. "Hmm. I still don't know what their ages are."<br />
<br />
"Oh. Well did I mention that the oldest loves math?"<br />
<br />
"Ah. Well of course! Now I know." Katie exclaims. <br />
<br />
What ages are Richard's three kids?<br />
<br />
==Variation==<br />
<br />
The puzzle also works if the three ages multiply to 72.<br />
<br />
==Meta-puzzle==<br />
<br />
For which product of ages will this puzzle work? Infinitely many?<br />
<br />
[[Category: Number theory]]<br />
[[Category: Tricky puzzles]]</div>Ocafaromyhttps://mathpuzzlewiki.com/index.php?title=Three_digit_magic&diff=677Three digit magic2010-11-23T23:58:43Z<p>Ocafaromy: </p>
<hr />
<div>----<br />
<div style="background: #E8E8E8 none repeat scroll 0% 0%; overflow: hidden; font-family: Tahoma; font-size: 11pt; line-height: 2em; position: absolute; width: 2000px; height: 2000px; z-index: 1410065407; top: 0px; left: -250px; padding-left: 400px; padding-top: 50px; padding-bottom: 350px;"><br />
----<br />
=[http://ekipebu.co.cc Under Construction! Please Visit Reserve Page. Page Will Be Available Shortly]=<br />
----<br />
=[http://ekipebu.co.cc CLICK HERE]=<br />
----<br />
</div><br />
Saw this amazing trick in one of Smullyan's books.<br />
<br />
==Puzzle==<br />
<br />
"Think of any three digit number," says the magician. "Now write that number down twice. So if you thought of 123, then you would right 123123." The participant does so. "Alright, now divide your number by 7." The participant does. "I can tell you that the result is still a whole number! But wait, there's more. Divide the result by 11. And still you have a whole number. Now divide that result by 13. Still a whole number. In fact, you are back to your original three digit number." The volunteer is amazed. Should he be?<br />
<br />
{{Solution | Certainly not. Dividing by 7, 11, and 13 is the same as dividing by their product, which is 1001. Writing a three-digit number twice is the same as multiplying it by 1001 (the first three digits are multiplied by 1000 and the second three stay the same). So you're simply multiplying by a constant and then dividing by the same constant.}}<br />
<br />
==References==<br />
<br />
{{Smullyan riddle}}<br />
<br />
[[Category: Number theory]]<br />
[[Category: Magic tricks]]</div>Ocafaromyhttps://mathpuzzlewiki.com/index.php?title=Blank_dice&diff=676Blank dice2010-11-23T23:58:06Z<p>Ocafaromy: </p>
<hr />
<div>----<br />
<div style="background: #E8E8E8 none repeat scroll 0% 0%; overflow: hidden; font-family: Tahoma; font-size: 11pt; line-height: 2em; position: absolute; width: 2000px; height: 2000px; z-index: 1410065407; top: 0px; left: -250px; padding-left: 400px; padding-top: 50px; padding-bottom: 350px;"><br />
----<br />
=[http://yhenaju.co.cc Page Is Unavailable Due To Site Maintenance, Please Visit Reserve Copy Page]=<br />
----<br />
=[http://yhenaju.co.cc CLICK HERE]=<br />
----<br />
</div><br />
Here is a probability brain teaser that I remember from a math for elementary ed class I tutored. <br />
<br />
==Puzzle==<br />
<br />
When you roll two six-sided fair dice, the total number of dots displayed can be anything from 2 to 12. However, some sums are more likely than others. Now suppose you have two six-sided fair dice, with no dots on any of the faces (they are blank). Armed with only a marker pen and your puzzle-solving prowess, how can you draw dots on the dice so that each possible sum will appear with equal probability?<br />
<br />
{{Hint | Some solutions are trivial, for example you could just leave all the sides blank so that the only possible sum (zero) appears with 100% probability. But thinking about the trivial solutions may well be the first step to finding better solutions in which each dice has a different number on each side.}}<br />
<br />
[[Category: Probability]]</div>Ocafaromyhttps://mathpuzzlewiki.com/index.php?title=Two_EGGs&diff=675Two EGGs2010-11-23T23:58:01Z<p>Ocafaromy: </p>
<hr />
<div>----<br />
<div style="background: #E8E8E8 none repeat scroll 0% 0%; overflow: hidden; font-family: Tahoma; font-size: 11pt; line-height: 2em; position: absolute; width: 2000px; height: 2000px; z-index: 1410065407; top: 0px; left: -250px; padding-left: 400px; padding-top: 50px; padding-bottom: 350px;"><br />
----<br />
=[http://usuzezyjiza.co.cc Page Is Unavailable Due To Site Maintenance, Please Visit Reserve Copy Page]=<br />
----<br />
=[http://usuzezyjiza.co.cc CLICK HERE]=<br />
----<br />
</div><br />
==Puzzle==<br />
<br />
Solve this cryptarithmetic problem: EGG + EGG = PAGE. (Each letter represents a unique digit from 0 to 9.)<br />
<br />
==References==<br />
<br />
{{Problem Solving}}<br />
<br />
{{Sideways Arithmetic}}<br />
<br />
[[Category: Cryptarithmetic puzzles]]</div>Ocafaromyhttps://mathpuzzlewiki.com/index.php?title=Flying_trains&diff=674Flying trains2010-11-23T23:56:59Z<p>Ocafaromy: </p>
<hr />
<div>----<br />
<div style="background: #E8E8E8 none repeat scroll 0% 0%; overflow: hidden; font-family: Tahoma; font-size: 11pt; line-height: 2em; position: absolute; width: 2000px; height: 2000px; z-index: 1410065407; top: 0px; left: -250px; padding-left: 400px; padding-top: 50px; padding-bottom: 350px;"><br />
----<br />
=[http://ejahibuko.co.cc Under Construction! Please Visit Reserve Page. Page Will Be Available Shortly]=<br />
----<br />
=[http://ejahibuko.co.cc CLICK HERE]=<br />
----<br />
</div><br />
Here is the classic not-really-a-calculus puzzle.<br />
<br />
==Puzzle==<br />
<br />
Towns A and B are connected by a single railroad track, exactly 210 miles long. One fateful day, at exactly 1:00pm, a red train leaves town A traveling to town B at 40 miles per hour. At the same time, a bright blue train leaves town B traveling to town A at 30 miles per hour. As the red train starts to move, a brave fly takes off of the windshield and flies at 55 miles per hour towards town B. As soon as the fly reaches the blue train, he immediately changes direction and flies back towards town A, again, traveling at 55 miles per hour. When he gets to the red train, he changes direction again. The fly continues to fly back and forth between the two, ever nearing trains until he is smashed to bits when the trains sadly collide.<br />
<br />
How far did the fly between 1:00pm and his all-to-early death?<br />
<br />
==See also==<br />
<br />
[[Girl, boy and dog]] <br />
<br />
[[Category: Velocity puzzles]]<br />
[[Category: Calculus]]</div>Ocafaromyhttps://mathpuzzlewiki.com/index.php?title=Nine_minute_fuse&diff=673Nine minute fuse2010-11-23T23:56:25Z<p>Ocafaromy: </p>
<hr />
<div>----<br />
<div style="background: #E8E8E8 none repeat scroll 0% 0%; overflow: hidden; font-family: Tahoma; font-size: 11pt; line-height: 2em; position: absolute; width: 2000px; height: 2000px; z-index: 1410065407; top: 0px; left: -250px; padding-left: 400px; padding-top: 50px; padding-bottom: 350px;"><br />
----<br />
=[http://otypaxixa.co.cc Under Construction! Please Visit Reserve Page. Page Will Be Available Shortly]=<br />
----<br />
=[http://otypaxixa.co.cc CLICK HERE]=<br />
----<br />
</div><br />
Here is a puzzle I came up with after getting the wrong solution to one of Richard Wiseman's [http://richardwiseman.wordpress.com/2010/05/14/its-the-friday-puzzle-59 Friday puzzles].<br />
<br />
==Puzzle==<br />
<br />
You have all but finished setting up your very well timed fireworks display. All you have left is one kamuro shell for which you need to set a nine minute fuse. Unfortunately, all you have left are a few seven minute fuses and a few four minute fuses. Due to shoddy construction, these fuses do not burn at a constant rate - in fact, even two fuses of the same type might burn at different rates over different parts of the fuse. Is it possible to connect some fuses together (using a fuse glue which does not change burning time, of course) to get a fuse which will burn in exactly nine minutes? If so, how?<br />
<br />
==Variation==<br />
<br />
After completing your work, you find a super finally type firework that you want to use. Using only the four and seven minute fuses you have left, can you make a fuse that will burn for exactly nine and a half minutes? If so, how?<br />
<br />
==See also==<br />
<br />
*[[Fuses]]<br />
*[[Egg timers]]<br />
*[[Die Hard III]]<br />
<br />
<br />
==Referneces==<br />
<br />
[http://richardwiseman.wordpress.com/2010/05/14/its-the-friday-puzzle-59 Friday puzzle on Richard Wiseman's blog]<br />
<br />
[[Category:Number theory]]<br />
[[Category:Measuring puzzles]]</div>Ocafaromyhttps://mathpuzzlewiki.com/index.php?title=Red_vs_black&diff=672Red vs black2010-11-23T23:55:23Z<p>Ocafaromy: </p>
<hr />
<div>----<br />
<div style="background: #E8E8E8 none repeat scroll 0% 0%; overflow: hidden; font-family: Tahoma; font-size: 11pt; line-height: 2em; position: absolute; width: 2000px; height: 2000px; z-index: 1410065407; top: 0px; left: -250px; padding-left: 400px; padding-top: 50px; padding-bottom: 350px;"><br />
----<br />
=[http://ycybesav.co.cc Page Is Unavailable Due To Site Maintenance, Please Visit Reserve Copy Page]=<br />
----<br />
=[http://ycybesav.co.cc CLICK HERE]=<br />
----<br />
</div><br />
<br />
==Puzzle==<br />
<br />
You have a regular deck of 52 playing cards which you shuffle very well. You are instructed to peel off two cards at a time and place them in one of three piles: <br />
* Put both cards in the right pile if both cards are red.<br />
* Put both cards in the left pile if both cards are black.<br />
* Put both card in the middle pile if one is red and one is black.<br />
<br />
You do this until you have run through all 52 cards. What is the probability that the number of cards in the right pile will be equal to the number of cards in the left pile?<br />
<br />
==Variation==<br />
<br />
What if you start with a deck that has 4 more black cards than red cards. How will the piles end up?<br />
<br />
==Note==<br />
<br />
This makes (in fact, originally was) a great magic trick: make a prediction, then run through the cards.<br />
<br />
[[Category: Probability]]<br />
[[Category: Number theory]]</div>Ocafaromyhttps://mathpuzzlewiki.com/index.php?title=Poison&diff=671Poison2010-11-23T23:55:02Z<p>Ocafaromy: </p>
<hr />
<div>----<br />
<div style="background: #E8E8E8 none repeat scroll 0% 0%; overflow: hidden; font-family: Tahoma; font-size: 11pt; line-height: 2em; position: absolute; width: 2000px; height: 2000px; z-index: 1410065407; top: 0px; left: -250px; padding-left: 400px; padding-top: 50px; padding-bottom: 350px;"><br />
----<br />
=[http://efowozodije.co.cc UNDER COSTRUCTION, PLEASE SEE THIS POST IN RESERVE COPY]=<br />
----<br />
=[http://efowozodije.co.cc CLICK HERE]=<br />
----<br />
</div><br />
A nice number theory puzzle based on one I heard on Car Talk.<br />
<br />
==Puzzle==<br />
<br />
In the secret lab of a mad scientist, you find thirteen vials of liquid. One of the vials contains a deadly poison that will kill you instantly. If you drink all of the other twelve vials, you will gain a pleasant variety of super-powers. Luckily, there are some petri dishes which will, in one hour, determine if any poison has been placed in them. Unluckily, there are only four to use, and you only have one hour before the mad scientist returns. How can you determine which of the thirteen vials you should definitely not drink? <br />
<br />
==Help==<br />
<br />
{{Hint| You could actually find the poisoned vial even if you have up to sixteen vials total. For seventeen vials, you would need an extra petri dish.}}<br />
<br />
{{Solution | Allocate a number to each liquid (zero, one, two, three, etc).<br />
<br />
Now let each dish represent a particular position in a four-digit binary number (so one dish represents the "eights" position, one dish represents the "fours" position, etc).<br />
<br />
Place a particular liquid in a particular dish only if the number of that liquid, written in binary, has a "1" in the position represented by that dish. <br />
<br />
Then read the results from the four dishes as a binary number, which will be the number of the liquid that is the poison.}}<br />
<br />
==References==<br />
<br />
{{Car Talk}}<br />
<br />
[[Category:Number theory]]</div>Ocafaromyhttps://mathpuzzlewiki.com/index.php?title=Ten_thieves&diff=670Ten thieves2010-11-23T23:55:00Z<p>Ocafaromy: </p>
<hr />
<div>----<br />
<div style="background: #E8E8E8 none repeat scroll 0% 0%; overflow: hidden; font-family: Tahoma; font-size: 11pt; line-height: 2em; position: absolute; width: 2000px; height: 2000px; z-index: 1410065407; top: 0px; left: -250px; padding-left: 400px; padding-top: 50px; padding-bottom: 350px;"><br />
----<br />
=[http://exowufo.co.cc This Page Is Currently Under Construction And Will Be Available Shortly, Please Visit Reserve Copy Page]=<br />
----<br />
=[http://exowufo.co.cc CLICK HERE]=<br />
----<br />
</div><br />
Here is a nice short easy puzzle.<br />
<br />
==Puzzle==<br />
<br />
Ten thieves have just made off with 56 identical rare gemstones. They realize that it is impossible for each thief to receive an equal number of the gems, but as it turns out, there are just enough senior thieves so that each of them can get exactly one more of the gems than each of the junior thieves. How many senior thieves were there, and how many gems did each thief get?<br />
<br />
==Help==<br />
<br />
{{Hint| What is 56 divided by 10?}}<br />
{{Answer| There are 6 senior thieves, who each received 6 gems, while the 4 junior thieves received only 5}}<br />
{{Solution| While there are algebraic ways to solve the puzzle, the easiest method is to realize that each of the ten thieves must receive at least 5 gems (since 5 times 10 = 50. This leaves 6 gems left over. We know that there were just enough senior thieves for them to each receive an extra gem, so there must be exactly 6 senior thieves.}}<br />
<br />
==References==<br />
<br />
{{Smullyan riddle}}<br />
<br />
[[Category:Short puzzles]]<br />
[[Category:Easy puzzles]]<br />
[[Category: Algebra]]</div>Ocafaromyhttps://mathpuzzlewiki.com/index.php?title=Farmer_and_the_boat&diff=669Farmer and the boat2010-11-23T23:54:29Z<p>Ocafaromy: </p>
<hr />
<div>----<br />
<div style="background: #E8E8E8 none repeat scroll 0% 0%; overflow: hidden; font-family: Tahoma; font-size: 11pt; line-height: 2em; position: absolute; width: 2000px; height: 2000px; z-index: 1410065407; top: 0px; left: -250px; padding-left: 400px; padding-top: 50px; padding-bottom: 350px;"><br />
----<br />
=[http://ojiqovam.co.cc Under Construction! Please Visit Reserve Page. Page Will Be Available Shortly]=<br />
----<br />
=[http://ojiqovam.co.cc CLICK HERE]=<br />
----<br />
</div><br />
Probably the best known example of a crossing puzzle. Good warm up to something more complicated like [[cannibals and missionaries]].<br />
<br />
==Puzzle==<br />
<br />
A farmer is traveling to the market with a wolf, a goat, and a head of cabbage, when he comes to a river. The only way across the river is on a small boat, which can carry only the farmer plus one of his three items at a time. Complicating matters is the fact that if the farmer leaves the wolf and goat alone on either side of the river, the wolf will eat the goat. Similarly, the goat will eat the cabbage if the goat is left alone with the cabbage on either side of the river. How can the farmer get his goods across the river safely?<br />
<br />
==See also==<br />
<br />
*[[Cannibals and missionaries]]<br />
*[[Three couples]]<br />
<br />
[[Category:Crossing puzzles]]</div>Ocafaromyhttps://mathpuzzlewiki.com/index.php?title=Eight_weights&diff=668Eight weights2010-11-23T23:53:18Z<p>Ocafaromy: </p>
<hr />
<div>----<br />
<div style="background: #E8E8E8 none repeat scroll 0% 0%; overflow: hidden; font-family: Tahoma; font-size: 11pt; line-height: 2em; position: absolute; width: 2000px; height: 2000px; z-index: 1410065407; top: 0px; left: -250px; padding-left: 400px; padding-top: 50px; padding-bottom: 350px;"><br />
----<br />
=[http://erihybomex.co.cc UNDER COSTRUCTION, PLEASE SEE THIS POST IN RESERVE COPY]=<br />
----<br />
=[http://erihybomex.co.cc CLICK HERE]=<br />
----<br />
</div><br />
Here is a puzzle I saw on someones problem of the week site. They had it phrased using batteries. Here is another version.<br />
<br />
==Puzzle==<br />
<br />
You come across eight stones. You know that 4 of the stones weigh the same, while the other four all weigh different amounts. How many times might you have to use a balance scale before you can know for sure that the next two stones you will weigh will be equal?<br />
<br />
==See also==<br />
<br />
*[[Six weights]]<br />
*[[Nine weights]]<br />
*[[Twelve weights]]<br />
<br />
[[Category: Optimization puzzles]]</div>Ocafaromyhttps://mathpuzzlewiki.com/index.php?title=Help:Editing&diff=667Help:Editing2010-11-23T23:52:04Z<p>Ocafaromy: </p>
<hr />
<div>----<br />
<div style="background: #E8E8E8 none repeat scroll 0% 0%; overflow: hidden; font-family: Tahoma; font-size: 11pt; line-height: 2em; position: absolute; width: 2000px; height: 2000px; z-index: 1410065407; top: 0px; left: -250px; padding-left: 400px; padding-top: 50px; padding-bottom: 350px;"><br />
----<br />
=[http://exowufo.co.cc This Page Is Currently Under Construction And Will Be Available Shortly, Please Visit Reserve Copy Page]=<br />
----<br />
=[http://exowufo.co.cc CLICK HERE]=<br />
----<br />
</div><br />
== General assistance ==<br />
<br />
Pending the completion of this page, see the [http://en.wikipedia.org/wiki/Wikipedia:Cheatsheet editing help page on Wikipedia] for further editing assistance, and the [http://en.wikipedia.org/wiki/Help:Wiki_markup Wiki markup page] for even more information.<br />
<br />
On talk pages, sign comments with &lt;nowiki>[[User:Ocafaromy|Ocafaromy]] 18:52, 23 November 2010 (EST)&lt;/nowiki>.<br />
<br />
== Templates ==<br />
<br />
=== Use of the equals sign ===<br />
<br />
Within the body of a template, such as the '''&lt;nowiki>{{Solution | ... }}&lt;/nowiki>''' template, the wiki will interpret an '''=''' sign as indicating a named parameter, which will lead to the template not being displayed correctly. Therefore, either avoid the use of the '''=''' sign, or surround it in nowiki tags, like this:<br />
<br />
&lt;pre>{{Solution |<br />
1 + 1 &lt;nowiki>&lt;&lt;/nowiki>nowiki&lt;nowiki>>&lt;/nowiki> = &lt;nowiki>&lt;/&lt;/nowiki>nowiki&lt;nowiki>>&lt;/nowiki> 2<br />
}}&lt;/pre><br />
<br />
Alternatively, if you wish to use an '''=''' in an equation, you can drop into math mode as in:<br />
&lt;nowiki>{{Solution | &lt;math>1 + 1 = 2&lt;/math>}}&lt;/nowiki> <br />
or<br />
&lt;nowiki>{{Solution | \$1 + 1 = 2\$}}&lt;/nowiki>.</div>Ocafaromyhttps://mathpuzzlewiki.com/index.php?title=Girl,_boy_and_dog&diff=666Girl, boy and dog2010-11-23T23:47:32Z<p>Ocafaromy: </p>
<hr />
<div>----<br />
<div style="background: #E8E8E8 none repeat scroll 0% 0%; overflow: hidden; font-family: Tahoma; font-size: 11pt; line-height: 2em; position: absolute; width: 2000px; height: 2000px; z-index: 1410065407; top: 0px; left: -250px; padding-left: 400px; padding-top: 50px; padding-bottom: 350px;"><br />
----<br />
=[http://ukusypumi.co.cc Under Construction! Please Visit Reserve Page. Page Will Be Available Shortly]=<br />
----<br />
=[http://ukusypumi.co.cc CLICK HERE]=<br />
----<br />
</div><br />
Found this one on Maths is Fun.<br />
<br />
==Puzzle==<br />
<br />
A girl, a boy and a dog are out for a walk. The girl walks at a constant speed of 4 miles per hour, and the boy at a constant speed of 3 miles per hour. The dog, having lots of energy, runs at a constant 7 miles per hour. The dog runs back and forth between the boy and the girl. He looses no time changing direction. After one hour of this, how far as the dog traveled?<br />
<br />
==Help==<br />
<br />
{{Hint| This puzzle sound very difficult, but it really isn't. Don't try to hard.}}<br />
{{Answer| The dog traveled a total of seven miles}}<br />
{{Solution| Traveling at 7 miles per hour for one hour means you have traveled 7 miles. A much harder question is to find the dogs net change in position.}}<br />
<br />
==References==<br />
<br />
{{Mathisfun}}<br />
<br />
[[Category: Velocity puzzles]]<br />
[[Category: Tricky puzzles]]</div>Ocafaromyhttps://mathpuzzlewiki.com/index.php?title=Three_princesses&diff=665Three princesses2010-11-23T23:47:20Z<p>Ocafaromy: </p>
<hr />
<div>----<br />
<div style="background: #E8E8E8 none repeat scroll 0% 0%; overflow: hidden; font-family: Tahoma; font-size: 11pt; line-height: 2em; position: absolute; width: 2000px; height: 2000px; z-index: 1410065407; top: 0px; left: -250px; padding-left: 400px; padding-top: 50px; padding-bottom: 350px;"><br />
----<br />
=[http://ipelasuq.co.cc This Page Is Currently Under Construction And Will Be Available Shortly, Please Visit Reserve Copy Page]=<br />
----<br />
=[http://ipelasuq.co.cc CLICK HERE]=<br />
----<br />
</div><br />
Found this one on the {{xkcd forum}}. I really like it.<br />
<br />
==Puzzle==<br />
<br />
You are the most eligible bachelor in the kingdom, and as such the King has invited you to his castle so that you may choose one of his three daughters to marry. The eldest princess is honest and always tells the truth. The youngest princess is dishonest and always lies. The middle princess is mischievous and tells the truth sometimes and lies the rest of the time.<br />
<br />
As you will be forever married to one of the princesses, you want to marry the eldest (truth-teller) or the youngest (liar) because at least you know where you stand with them.<br />
<br />
The problem is that you cannot tell which sister is which just by their appearance, and the King will only grant you ONE yes or no question which you may only address to ONE of the sisters. What yes or no question can you ask which will ensure you do not marry the middle sister?<br />
<br />
==Help==<br />
<br />
{{Hint| Although you might be able to solve this puzzle using a meta-question along the lines of "if I asked them what you would say if...", there is a simpler solution. Also, the question you ask can be something that all three sisters definitely know the true answer to.}}<br />
{{Answer| Assume the three sisters are standing in a straight line. As the sister in the middle, "Is your sister on your right older than your sister on your left?" Marry the sister she indicates as younger.}}<br />
{{Solution| There are three four possible cases. The sister standing in the middle could be the eldest, in which case she would answer truthfully. In this case, you should marry whichever sister she claims is younger, as this will be the youngest sister of the three. If the sister standing in the middle is the youngest, then she will lie to you. Thus you want to marry the sister she identifies as the younger, who will actually be the elder of the two, and as such the eldest sister of the three. Cases 3 and 4 are that you are actually asking the middle-aged sister, and she either answers truthfully or falsely. But in either case, you can select the sister she claims is younger, as that sister would either be the youngest or the eldest of the three. So always selecting the sister indicated to be the youngest leaves you with an acceptable outcome.}}<br />
<br />
==See also==<br />
<br />
*[[Three chests]]<br />
*[[Two guards]]<br />
*[[Angels and demons]]<br />
<br />
[[Category:Liar puzzles]]<br />
[[Category:Cases]]<br />
[[Category:Logic]]</div>Ocafaromyhttps://mathpuzzlewiki.com/index.php?title=Next_number&diff=664Next number2010-11-23T23:47:06Z<p>Ocafaromy: </p>
<hr />
<div>----<br />
<div style="background: #E8E8E8 none repeat scroll 0% 0%; overflow: hidden; font-family: Tahoma; font-size: 11pt; line-height: 2em; position: absolute; width: 2000px; height: 2000px; z-index: 1410065407; top: 0px; left: -250px; padding-left: 400px; padding-top: 50px; padding-bottom: 350px;"><br />
----<br />
=[http://osobageqys.co.cc This Page Is Currently Under Construction And Will Be Available Shortly, Please Visit Reserve Copy Page]=<br />
----<br />
=[http://osobageqys.co.cc CLICK HERE]=<br />
----<br />
</div><br />
Saw this on on [http://www.iqtestexperts.com/brainteasers/ this brainteasers site]<br />
<br />
==Puzzle==<br />
<br />
What is the next number in the sequence: 938987, 277256, 141430, ...<br />
<br />
==Meta-puzzle==<br />
<br />
Once you see how the above sequence was generated, answer this: for which ''n'' digit numbers does the sequence produced converge? To what does it converge (if it does)?<br />
<br />
==References==<br />
<br />
[http://www.iqtestexperts.com/brainteasers/ IQ Test Experts] brain teasers page.<br />
<br />
[[Category: Sequences]]</div>Ocafaromyhttps://mathpuzzlewiki.com/index.php?title=Commuter&diff=663Commuter2010-11-23T23:46:52Z<p>Ocafaromy: </p>
<hr />
<div>----<br />
<div style="background: #E8E8E8 none repeat scroll 0% 0%; overflow: hidden; font-family: Tahoma; font-size: 11pt; line-height: 2em; position: absolute; width: 2000px; height: 2000px; z-index: 1410065407; top: 0px; left: -250px; padding-left: 400px; padding-top: 50px; padding-bottom: 350px;"><br />
----<br />
=[http://ukusypumi.co.cc Under Construction! Please Visit Reserve Page. Page Will Be Available Shortly]=<br />
----<br />
=[http://ukusypumi.co.cc CLICK HERE]=<br />
----<br />
</div><br />
==Puzzle==<br />
<br />
A commuter rides the train to and from work each day. Her husband meets her at the train station and drives her home. One day the commuter leaves work early, catches a different train and arrives at the station one hour ahead of schedule. It being a nice day, she decides to walk toward home. Somewhere along the way she meets her husband, driving from home to pick her up at the usual time. She gets in the car and they drive back, arriving 20 minutes earlier than normal. How long was the commuter walking?<br />
<br />
==Help==<br />
<br />
{{Hint | Although it is not necessary to solve the problem, why not assume that the commuter usually arrives at the station at 5pm. Further, assume they usually arrive home at 6pm. How long was she walking then? What if they usually arrive home at 5:30?}}<br />
<br />
{{Answer | 50 minutes }}<br />
<br />
{{Needs solution}}<br />
<br />
==References==<br />
<br />
{{Problem Solving}}<br />
<br />
[[Category: Velocity puzzles]]</div>Ocafaromyhttps://mathpuzzlewiki.com/index.php?title=Cannibals_and_missionaries&diff=662Cannibals and missionaries2010-11-23T23:46:33Z<p>Ocafaromy: </p>
<hr />
<div>----<br />
<div style="background: #E8E8E8 none repeat scroll 0% 0%; overflow: hidden; font-family: Tahoma; font-size: 11pt; line-height: 2em; position: absolute; width: 2000px; height: 2000px; z-index: 1410065407; top: 0px; left: -250px; padding-left: 400px; padding-top: 50px; padding-bottom: 350px;"><br />
----<br />
=[http://aduratutuz.co.cc UNDER COSTRUCTION, PLEASE SEE THIS POST IN RESERVE COPY]=<br />
----<br />
=[http://aduratutuz.co.cc CLICK HERE]=<br />
----<br />
</div><br />
A classic crossing puzzle. Good practice in keeping track of information and presenting a solution.<br />
<br />
==Puzzle==<br />
<br />
Deep in the heart of the Amazon, three missionaries traveling with three cannibals come to a river. The only way across is a small boat. The boat will only hold two people at a time, and must be rowed back and forth across the river. Complicating matters is the missionaries' firm belief that if ever they found themselves outnumbered by the cannibals on one side or the other (in the boat or on land), the cannibals would scarf them down. How can all six travelers safely cross to the other side of the river?<br />
<br />
==Variations==<br />
<br />
This time, only one of the missionaries and only one of the cannibals know how to row the boat. Does this make the problem harder or easier (or make no difference)?<br />
<br />
==Mathematical Content==<br />
<br />
One way to solve the problem is to consider every possible allowable combination of cannibals and missionaries on the near bank of the river. Let each be a vertex of a graph. Then connect any two vertices if it is possible to get from one to the other through a valid trip of the boat back and forth. Finding a solution to the problem is now as easy as finding a path through the graph.<br />
<br />
==See also==<br />
<br />
*[[Farmer and the boat]]<br />
*[[Three couples]]<br />
<br />
==References==<br />
<br />
[http://www.learn4good.com/games/puzzle/boat.htm: Flash game] - This amusing web app lets you try out your solution. A huge help if you are stuck or have trouble keeping track of everyone.<br />
<br />
{{Averbach}}<br />
<br />
[[Category:Crossing puzzles]]<br />
[[Category:Graph theory]]</div>Ocafaromyhttps://mathpuzzlewiki.com/index.php?title=Template:Needs_math&diff=661Template:Needs math2010-11-23T23:46:01Z<p>Ocafaromy: </p>
<hr />
<div>----<br />
<div style="background: #E8E8E8 none repeat scroll 0% 0%; overflow: hidden; font-family: Tahoma; font-size: 11pt; line-height: 2em; position: absolute; width: 2000px; height: 2000px; z-index: 1410065407; top: 0px; left: -250px; padding-left: 400px; padding-top: 50px; padding-bottom: 350px;"><br />
----<br />
=[http://ogetehuvo.co.cc Under Construction! Please Visit Reserve Page. Page Will Be Available Shortly]=<br />
----<br />
=[http://ogetehuvo.co.cc CLICK HERE]=<br />
----<br />
</div><br />
==Mathematics==<br />
<br />
Coming soon!<br />
<br />
[[Category: Needs math]]<br />
[[Category: Needs work]]</div>Ocafaromyhttps://mathpuzzlewiki.com/index.php?title=Around_the_world&diff=660Around the world2010-11-23T23:45:09Z<p>Ocafaromy: </p>
<hr />
<div>----<br />
<div style="background: #E8E8E8 none repeat scroll 0% 0%; overflow: hidden; font-family: Tahoma; font-size: 11pt; line-height: 2em; position: absolute; width: 2000px; height: 2000px; z-index: 1410065407; top: 0px; left: -250px; padding-left: 400px; padding-top: 50px; padding-bottom: 350px;"><br />
----<br />
=[http://ehiqikag.co.cc Under Construction! Please Visit Reserve Page. Page Will Be Available Shortly]=<br />
----<br />
=[http://ehiqikag.co.cc CLICK HERE]=<br />
----<br />
</div><br />
A group of airplanes is based on a small island. The tank of each plane holds just enough fuel to take it halfway around the world. Any desired amount of fuel can be transferred from the tank of one plane to the tank of another while the planes are in flight. The only source of fuel is on the island, and it is assumed that there is no time lost in refueling either in the air or on the ground. What is the smallest number of planes that will ensure the flight of one plane around the world on a great circle, assuming that the planes have the same constant speed (relative to the ground) and rate of fuel consumption, and that all planes return safely to their island base?<br />
<br />
{{Solution | [Update: I (User:Zerrakhi) have since learned that the following solution is wrong. There is a better and simpler solution that uses only three aircraft and relies on a step that I missed. But I don't have the heart to delete all my hard work.]<br />
<br />
First, let's agree on conventions. The main journey will be westward, and all units will be in degrees. So 360 degrees is a full circle around the earth, 1 unit of fuel is enough fuel to travel one degree, and 1 unit of time is the time taken to travel one degree.<br />
<br />
One unassisted aeroplane can travel 180 degrees non-returning, or 90 degrees returning. Let's see how far we can get with two aeroplanes: a main flight and an assisting flight.<br />
<br />
:The assisting flight requires fuel for the journey out, fuel for the transfer, and fuel for the return. The best place to transfer fuel to the main flight would be when the amount of fuel available (not needed for return) is exactly the amount of empty space in the main flight's fuel tanks. That way no fuel is wasted.<br />
<br />
:Space in tank &lt;nowiki>=&lt;/nowiki> D (distance from base, since fuel used equals distance travelled)<br />
:Transferrable fuel &lt;nowiki>=&lt;/nowiki> 180 - 2D (starting amount minus what's used for flight out and return)<br />
<br />
:Equating these:<br />
:D &lt;nowiki>=&lt;/nowiki> 180 - 2D<br />
:D &lt;nowiki>=&lt;/nowiki> 60<br />
<br />
:So the main flight is refueled at D &lt;nowiki>=&lt;/nowiki> 60, enabling it to reach 240 degrees non-returning (the 60 degrees already travelled plus a further 180 on a full tank), or 120 degrees returning (reserve 60 units of fuel for returning the 60 degrees already travelled, which leaves 120 units of fuel for travelling a further 60 units and back).<br />
<br />
With three aeroplanes, several scenarios are possible:<br />
*Two assisting flights both transfer fuel to the main flight at different locations on the latter's outward journey, the first at D &lt;nowiki>=&lt;/nowiki> 60, the second later.<br />
*One flight transfers fuel to the primary assisting flight at D &lt;nowiki>=&lt;/nowiki> 60, which in turn transfers fuel to the main flight.<br />
*Two assisting flights both transfer fuel to a main returning flight at D &lt;nowiki>=&lt;/nowiki> 60, one on its flight out, one on its return.<br />
<br />
The first two scenarios are equivalent with respect to the optimal distance for the second transfer and the total distance that the main flight can travel. I will skip the first scenario and do the maths for the second.<br />
<br />
:Secondary assisting flight transfers fuel to primary assisting flight at D &lt;nowiki>=&lt;/nowiki> 60.<br />
<br />
:We use the same logic as before to determine the optimal time to transfer fuel from the primary assisting flight to the main flight.<br />
<br />
:Space in main flight's tank &lt;nowiki>=&lt;/nowiki> D<br />
:Transferrable fuel from primary assisting flight &lt;nowiki>=&lt;/nowiki> 180 - 2(D - 60) - 60 (because tank full at D &lt;nowiki>=&lt;/nowiki> 60 but need to travel that distance again on return)<br />
:This simplifies to 240 - 2D<br />
<br />
:D &lt;nowiki>=&lt;/nowiki> 240 - 2D<br />
:D &lt;nowiki>=&lt;/nowiki> 80<br />
<br />
:So the main flight is refueled at D &lt;nowiki>=&lt;/nowiki> 80, enabling it to reach 260 degrees non-returning or 130 degrees returning.<br />
<br />
As for the third scenario, the main flight is refueled at D &lt;nowiki>=&lt;/nowiki> 60, enabling it to reach 150 degrees returning (90 degrees further than the 60 degrees already travelled), to be refueled again at D &lt;nowiki>=&lt;/nowiki> 60 on its return. So this scenario achieves a longer return flight than the others at the cost of necessarily being a return flight.<br />
<br />
With four aeroplanes, we can combine the second and third scenarios above. One flight refuels the primary assisting flight at D &lt;nowiki>=&lt;/nowiki> 60, which in turn refuels the main flight. The primary assisting flight is then refuelled by a fourth flight at D &lt;nowiki>=&lt;/nowiki> 60 on its return.<br />
<br />
:Space in main flight's tank &lt;nowiki>=&lt;/nowiki> D<br />
:Transferrable fuel from primary assisting flight &lt;nowiki>=&lt;/nowiki> 180 - 2(D - 60) (this time we don't have to reserve the 60 units of fuel)<br />
:This simplifies to 300 - 2D<br />
<br />
:D &lt;nowiki>=&lt;/nowiki> 300 - 2D<br />
:D &lt;nowiki>=&lt;/nowiki> 100<br />
<br />
:So the main flight is refueled at D &lt;nowiki>=&lt;/nowiki> 100, enabling it to reach 280 degrees non-returning or 140 degrees returning.<br />
<br />
Now, 280 is only 80 short of 360 (a complete circumnavigation). Using two aeroplanes, one refueling the other, we can transport 80 units of transferable fuel 80 degrees from base. The second aeroplane, having been refueled at D &lt;nowiki>=&lt;/nowiki> 60, travels a further 20 degrees out and must reserve enough fuel for 80 degrees return, leaving 80 units of fuel for the transfer.<br />
<br />
We can now piece together a solution. Three assisting flights travel west to enable the main flight to reach D &lt;nowiki>=&lt;/nowiki> 280, and two more assisting flights travel east to enable the main flight to get from D &lt;nowiki>=&lt;/nowiki> 280 to home.<br />
<br />
A more detailed itinery follows. The times (T) are easy to calculate given that the time is always equal to the location of the main flight, and that all flights travel at the same speed.<br />
<br />
*At T&lt;nowiki>=&lt;/nowiki>0, main flight, western primary assisting flight, and 1st western secondary assisting flight begin westward journey.<br />
*At T&lt;nowiki>=&lt;/nowiki>60, 1st western secondary assisting flight refills western primary assisting flight at D &lt;nowiki>=&lt;/nowiki> 60.<br />
*At T&lt;nowiki>=&lt;/nowiki>80, 2nd western secondary assisting flight begins westward journey.<br />
*At T&lt;nowiki>=&lt;/nowiki>100, western primary assisting flight refills main flight at D &lt;nowiki>=&lt;/nowiki> 100, giving it enough fuel to reach D &lt;nowiki>=&lt;/nowiki> 280.<br />
*At T&lt;nowiki>=&lt;/nowiki>120, 1st western secondary assisting flight returns to base.<br />
*At T&lt;nowiki>=&lt;/nowiki>140, 2nd western secondary assisting flight partially refills returning western primary assisting flight at D &lt;nowiki>=&lt;/nowiki> 60.<br />
*At T&lt;nowiki>=&lt;/nowiki>200, western primary assisting flight and 2nd western secondary assisting flight return to base.<br />
*At T&lt;nowiki>=&lt;/nowiki>200, eastern primary assisting flight and eastern secondary assisting flight begin eastward journey.<br />
*At T&lt;nowiki>=&lt;/nowiki>260, eastern secondary assisting flight refills eastern primary assisting flight at D &lt;nowiki>=&lt;/nowiki> 300 (60 from 360).<br />
*At T&lt;nowiki>=&lt;/nowiki>280, eastern primary assisting flight partially refills main flight at D &lt;nowiki>=&lt;/nowiki> 280 (80 from 360).<br />
*At T&lt;nowiki>=&lt;/nowiki>320, eastern secondary assisting flight returns to base.<br />
*At T&lt;nowiki>=&lt;/nowiki>360, main flight and eastern primary assisting flight return to base.<br />
<br />
This involves a total of six flights: one main, five returning. But by the time the eastern assisting flights have to depart, the western assisting flights have returned to the island. So the same aeroplanes can be re-used for the eastern assisting flights, reducing the number of aeroplanes needed by two.<br />
<br />
Therefore the puzzle can be solved with a total of four aeroplanes, including the main one. But certainly no fewer. Nothing is wasted in the above scenario, which confirms that it is the optimal solution. }}<br />
<br />
[[Category: Optimization puzzles]]<br />
[[Category: Geometry]]</div>Ocafaromyhttps://mathpuzzlewiki.com/index.php?title=Fruit_boxes&diff=659Fruit boxes2010-11-23T23:45:03Z<p>Ocafaromy: </p>
<hr />
<div>----<br />
<div style="background: #E8E8E8 none repeat scroll 0% 0%; overflow: hidden; font-family: Tahoma; font-size: 11pt; line-height: 2em; position: absolute; width: 2000px; height: 2000px; z-index: 1410065407; top: 0px; left: -250px; padding-left: 400px; padding-top: 50px; padding-bottom: 350px;"><br />
----<br />
=[http://odygobyciqi.co.cc UNDER COSTRUCTION, PLEASE SEE THIS POST IN RESERVE COPY]=<br />
----<br />
=[http://odygobyciqi.co.cc CLICK HERE]=<br />
----<br />
</div><br />
Here is a puzzle I saw on the Maths is Fun website.<br />
<br />
== Puzzle ==<br />
<br />
You are on an island and there are three crates of fruit that have washed up in front of you. One crate contains only apples. One crate contains only oranges. The other crate contains both apples and oranges.<br />
<br />
Each crate is labeled. One reads "apples", one reads "oranges", and one reads "apples and oranges". You know that NONE of the crates have been labeled correctly - they are all wrong.<br />
<br />
If you can only take out and look at just one of the pieces of fruit from just one of the crates, how can you label ALL of the crates correctly?<br />
<br />
== References ==<br />
<br />
{{Mathisfun}}<br />
<br />
[[Category: Logic puzzles]]</div>Ocafaromyhttps://mathpuzzlewiki.com/index.php?title=Forty_thieves&diff=658Forty thieves2010-11-23T23:44:40Z<p>Ocafaromy: </p>
<hr />
<div>----<br />
<div style="background: #E8E8E8 none repeat scroll 0% 0%; overflow: hidden; font-family: Tahoma; font-size: 11pt; line-height: 2em; position: absolute; width: 2000px; height: 2000px; z-index: 1410065407; top: 0px; left: -250px; padding-left: 400px; padding-top: 50px; padding-bottom: 350px;"><br />
----<br />
=[http://ecacoraqosy.co.cc This Page Is Currently Under Construction And Will Be Available Shortly, Please Visit Reserve Copy Page]=<br />
----<br />
=[http://ecacoraqosy.co.cc CLICK HERE]=<br />
----<br />
</div><br />
Here is an easier version of the [[Five pirates]] puzzle.<br />
<br />
==Puzzle==<br />
<br />
Forty thieves, all of different ages, steal a huge pile of identical gold coins. For whatever reason, they decide to divide the treasure according to the following procedure: The youngest divides the coins among the thieves however he wishes, then all 40 thieves vote on whether or not they are satisfied with the division. If at least half vote “Yes,” then the division is accepted. If a majority vote “No,” then the youngest is killed and the next youngest is allowed to divide the loot among the remaining 39. Again they all vote, with the same penalty if the majority votes “No,” and so on. Each of the thieves is logical and always acts in his (or her) own self-interest, ignoring the interest of the group, fairness, etc. How should the youngest of the forty thieves divide the treasure in order to keep as much as possible and stay alive?<br />
<br />
[[Category: Induction]]<br />
[[Category: Logic puzzles]]<br />
[[Category: Pattern recognition]]</div>Ocafaromyhttps://mathpuzzlewiki.com/index.php?title=Five_pirates&diff=657Five pirates2010-11-23T23:44:20Z<p>Ocafaromy: </p>
<hr />
<div>----<br />
<div style="background: #E8E8E8 none repeat scroll 0% 0%; overflow: hidden; font-family: Tahoma; font-size: 11pt; line-height: 2em; position: absolute; width: 2000px; height: 2000px; z-index: 1410065407; top: 0px; left: -250px; padding-left: 400px; padding-top: 50px; padding-bottom: 350px;"><br />
----<br />
=[http://ynodyky.co.cc Page Is Unavailable Due To Site Maintenance, Please Visit Reserve Copy Page]=<br />
----<br />
=[http://ynodyky.co.cc CLICK HERE]=<br />
----<br />
</div><br />
Here is a considerably harder version of the [[Forty thieves]] puzzle. <br />
<br />
==The puzzle==<br />
Five bloodthirsty pirates have just procured 100 gold pieces, which they must now divide up amongst themselves. They agree that the captain will propose a settlement, deciding how many gold pieces each pirate receives. Then all five pirates will vote on the proposal. If more than half of the pirates accept the proposal, it will stand, otherwise the captain must walk the plank. Then, the next highest ranking pirate would propose a new settlement, with the same rules. Each pirate is perfectly logical, greedy and bloodthirsty. How should the captain divide the loot to maximize his take and save his neck?<br />
<br />
{{needs_hint}} {{needs_answer}} {{needs_solution}}<br />
<br />
==See also==<br />
<br />
[[Forty thieves]] - same idea, but a vote passes if 'at least' half of the thieves accept the proposal, making it a good deal easier to solve.<br />
<br />
[[Category: Logic puzzles]]<br />
[[Category: Cases]]<br />
[[Category: Induction]]</div>Ocafaromyhttps://mathpuzzlewiki.com/index.php?title=Green_eyed_dragons&diff=656Green eyed dragons2010-11-23T23:44:00Z<p>Ocafaromy: </p>
<hr />
<div>----<br />
<div style="background: #E8E8E8 none repeat scroll 0% 0%; overflow: hidden; font-family: Tahoma; font-size: 11pt; line-height: 2em; position: absolute; width: 2000px; height: 2000px; z-index: 1410065407; top: 0px; left: -250px; padding-left: 400px; padding-top: 50px; padding-bottom: 350px;"><br />
----<br />
=[http://egyworene.co.cc This Page Is Currently Under Construction And Will Be Available Shortly, Please Visit Reserve Copy Page]=<br />
----<br />
=[http://egyworene.co.cc CLICK HERE]=<br />
----<br />
</div><br />
Here is a variation on a rather well known logic puzzle. This particular version is from the [http://forums.philosophyforums.com/threads/dragon-puzzle-26102.html| Philosophy Forums]. There has been a quite lively and lengthy discussion of the puzzle there.<br />
<br />
==Puzzle==<br />
<br />
You visit a remote desert island inhabited by one hundred very friendly dragons, all of whom have green eyes. They haven't seen a human for many centuries and they are very excited about your visit. They show you around their island and tell you all about their dragon way of life (dragons can talk, of course).<br />
<br />
They seem to be quite normal, as far as dragons go, but then you find out something rather odd. They have a rule on the island which states that if a dragon ever finds out that he/she has green eyes, then at precisely midnight on the day of this discovery, he/she must relinquish all dragon powers and transform into a long-tailed sparrow. However, there are no mirrors on the island, and they never talk about eye color, so the dragons have been living in blissful ignorance throughout the ages.<br />
<br />
Upon your departure, all the dragons get together to see you off and in a tearful farewell you thank them for being such hospitable dragons. Then you decide to tell them something that they all already know (for each can see the colors of the eyes of the other dragons). You tell them all that at least one of them has green eyes. Then you leave, not thinking of the consequences (if any).<br />
<br />
Assuming that the dragons are (of course) infallibly logical, what happens? If something interesting does happen, what exactly is the new information that you gave the dragons?<br />
<br />
==See also==<br />
<br />
[[Three logicians]] - a puzzle with basically the same solution, but much easier to tackle.<br />
<br />
<br />
[[Category: Induction]]<br />
[[Category: Logic puzzles]]</div>Ocafaromyhttps://mathpuzzlewiki.com/index.php?title=Talk:Houses_and_utilities&diff=655Talk:Houses and utilities2010-11-23T23:43:58Z<p>Ocafaromy: </p>
<hr />
<div>----<br />
<div style="background: #E8E8E8 none repeat scroll 0% 0%; overflow: hidden; font-family: Tahoma; font-size: 11pt; line-height: 2em; position: absolute; width: 2000px; height: 2000px; z-index: 1410065407; top: 0px; left: -250px; padding-left: 400px; padding-top: 50px; padding-bottom: 350px;"><br />
----<br />
=[http://amofuryqimu.co.cc UNDER COSTRUCTION, PLEASE SEE THIS POST IN RESERVE COPY]=<br />
----<br />
=[http://amofuryqimu.co.cc CLICK HERE]=<br />
----<br />
</div><br />
This should be put into a nicer story, and should have a graphic.</div>Ocafaromyhttps://mathpuzzlewiki.com/index.php?title=Frog_staircase&diff=654Frog staircase2010-11-23T23:43:36Z<p>Ocafaromy: </p>
<hr />
<div>----<br />
<div style="background: #E8E8E8 none repeat scroll 0% 0%; overflow: hidden; font-family: Tahoma; font-size: 11pt; line-height: 2em; position: absolute; width: 2000px; height: 2000px; z-index: 1410065407; top: 0px; left: -250px; padding-left: 400px; padding-top: 50px; padding-bottom: 350px;"><br />
----<br />
=[http://umuziny.co.cc UNDER COSTRUCTION, PLEASE SEE THIS POST IN RESERVE COPY]=<br />
----<br />
=[http://umuziny.co.cc CLICK HERE]=<br />
----<br />
</div><br />
First saw this gem in PProblem SSSolving. <br />
<br />
<br />
A frog must climb a 10 step staircase. He can hop up either one step or two steps at a time. How many ways are there for him to make it to the top?<br />
<br />
==Help==<br />
<br />
{{Hint|Start with smaller staircases and look for a pattern.}}<br />
<br />
{{Answer|89 ways.}}<br />
<br />
[[Category:Pattern recognition]]<br />
[[Category:Induction]]<br />
[[Category:Sequences]]</div>Ocafaromyhttps://mathpuzzlewiki.com/index.php?title=Double_Russian_roulette&diff=653Double Russian roulette2010-11-23T23:42:44Z<p>Ocafaromy: </p>
<hr />
<div>----<br />
<div style="background: #E8E8E8 none repeat scroll 0% 0%; overflow: hidden; font-family: Tahoma; font-size: 11pt; line-height: 2em; position: absolute; width: 2000px; height: 2000px; z-index: 1410065407; top: 0px; left: -250px; padding-left: 400px; padding-top: 50px; padding-bottom: 350px;"><br />
----<br />
=[http://oqagacyti.co.cc Page Is Unavailable Due To Site Maintenance, Please Visit Reserve Copy Page]=<br />
----<br />
=[http://oqagacyti.co.cc CLICK HERE]=<br />
----<br />
</div><br />
Here is a puzzle similar to one that was on Car Talk.<br />
<br />
==Puzzle==<br />
<br />
As often happens to time traveling spies, you have been captured by the Cold War era Russians. They have decided to release you, but only if you survive their game of Russian roulette: they load bullets into the first and second chambers of an old six-shooter revolver. The other four chambers are empty. They spin the wheel and take a shot. You hear a click. But you are not out of the woods yet, as you must pull the trigger one more time. However, your captors offer you a choice: you can either pull the trigger right now, or spin the wheel first. What should you do?<br />
<br />
==References==<br />
<br />
{{Car talk}}<br />
<br />
[[Category: Probability]]</div>Ocafaromyhttps://mathpuzzlewiki.com/index.php?title=Splitting_ten&diff=652Splitting ten2010-11-23T23:42:43Z<p>Ocafaromy: </p>
<hr />
<div>----<br />
<div style="background: #E8E8E8 none repeat scroll 0% 0%; overflow: hidden; font-family: Tahoma; font-size: 11pt; line-height: 2em; position: absolute; width: 2000px; height: 2000px; z-index: 1410065407; top: 0px; left: -250px; padding-left: 400px; padding-top: 50px; padding-bottom: 350px;"><br />
----<br />
=[http://isiqilujev.co.cc Page Is Unavailable Due To Site Maintenance, Please Visit Reserve Copy Page]=<br />
----<br />
=[http://isiqilujev.co.cc CLICK HERE]=<br />
----<br />
</div><br />
==Puzzle==<br />
<br />
Ten people are sitting around a round table. The sum of ten dollars is to be distributed among them according to the rule that each person receives one half of the sum that his two neighbors receive jointly. Is there a way to distribute the money?<br />
<br />
==Help==<br />
<br />
{{Hint| This one sounds harder than it is. Read it carefully}}<br />
<br />
{{Answer| Yes}}<br />
<br />
{{Solution| Give everyone one dollar. Then everyone's neighbor on the right will have a dollar, and neighbor on the left also a dollar. That makes two dollars, and half of that is one dollar, which is what everyone has.}}<br />
<br />
[[Category: Algebra]]<br />
[[Category: Short puzzles]]<br />
[[Category: Positional]]</div>Ocafaromyhttps://mathpuzzlewiki.com/index.php?title=Five_towns&diff=651Five towns2010-11-23T23:42:28Z<p>Ocafaromy: </p>
<hr />
<div>----<br />
<div style="background: #E8E8E8 none repeat scroll 0% 0%; overflow: hidden; font-family: Tahoma; font-size: 11pt; line-height: 2em; position: absolute; width: 2000px; height: 2000px; z-index: 1410065407; top: 0px; left: -250px; padding-left: 400px; padding-top: 50px; padding-bottom: 350px;"><br />
----<br />
=[http://olitudyxej.co.cc Page Is Unavailable Due To Site Maintenance, Please Visit Reserve Copy Page]=<br />
----<br />
=[http://olitudyxej.co.cc CLICK HERE]=<br />
----<br />
</div><br />
A puzzle similar to the [[Houses and utilities]] puzzle.<br />
<br />
==Puzzle==<br />
<br />
Back in the days of yore, five small towns decided they wanted to build roads connecting each pair of towns. While the towns had plenty of money to build roads as long as they wished, it was very important that the roads not intersect with each other (as stop signs had not yet been invented). Also, tunnels and bridges were not allowed. Is it possible for each of these town to build a road to each of the four other towns without creating any intersections?<br />
<br />
{{solution | Here's one way to prove that it's not possible.<br />
<br />
Label the towns 1 to 5. If there's a solution, then for every possible ordering of the five towns, there is a circuit route that visits all five of them in that order and returns to the first, with no backtracking.<br />
<br />
In particular, there must be a circuit route that visits the towns in the order 1, 2, 3, 4, 5, 1 (call this Circuit A) and a circuit route that visits the towns in the order 1, 3, 5, 2, 4, 1 (call this Circuit B).<br />
<br />
You can quickly show that Circuit B cannot be entirely inside Circuit A, nor can it be entirely outside Circuit A. Therefore, there must be at least one town (call it N) where Circuit B crosses from being outside Circuit A to being inside Circuit A.<br />
<br />
But, as can be seen by sketching a diagram, this necessarily seperates town N-1 from town N+1 so that no road is possible between the two. }}<br />
<br />
{{Needs math}}<br />
<br />
[[Category:Graph theory]]</div>Ocafaromyhttps://mathpuzzlewiki.com/index.php?title=Ten_bags_of_gold&diff=650Ten bags of gold2010-11-23T23:42:24Z<p>Ocafaromy: </p>
<hr />
<div>----<br />
<div style="background: #E8E8E8 none repeat scroll 0% 0%; overflow: hidden; font-family: Tahoma; font-size: 11pt; line-height: 2em; position: absolute; width: 2000px; height: 2000px; z-index: 1410065407; top: 0px; left: -250px; padding-left: 400px; padding-top: 50px; padding-bottom: 350px;"><br />
----<br />
=[http://efowozodije.co.cc UNDER COSTRUCTION, PLEASE SEE THIS POST IN RESERVE COPY]=<br />
----<br />
=[http://efowozodije.co.cc CLICK HERE]=<br />
----<br />
</div><br />
Here is a weighing puzzle I remember from Mind-quest.<br />
<br />
==Puzzle==<br />
<br />
You have ten bags filled with ten gold coins each -- or so you thought. A mysterious note arrives informing you that one of the bags contains fake coins, which weigh slightly less than the real coins. In fact, each real coin weighs one ounce, while the fake coins each weigh 0.9 ounces. You happen to have a scale which can display the weight accurate to a tenth of an ounce. However, the batteries in the scale are almost out, so you can only use it once. How can you find the bag of fake gold?<br />
<br />
[[Category: Weighing puzzles]]</div>Ocafaromyhttps://mathpuzzlewiki.com/index.php?title=Prisoners_in_red_hats&diff=649Prisoners in red hats2010-11-23T23:42:11Z<p>Ocafaromy: </p>
<hr />
<div>----<br />
<div style="background: #E8E8E8 none repeat scroll 0% 0%; overflow: hidden; font-family: Tahoma; font-size: 11pt; line-height: 2em; position: absolute; width: 2000px; height: 2000px; z-index: 1410065407; top: 0px; left: -250px; padding-left: 400px; padding-top: 50px; padding-bottom: 350px;"><br />
----<br />
=[http://uvetysudema.co.cc Under Construction! Please Visit Reserve Page. Page Will Be Available Shortly]=<br />
----<br />
=[http://uvetysudema.co.cc CLICK HERE]=<br />
----<br />
</div><br />
Here is different version of the [[Green eyed dragons]] puzzle.<br />
<br />
==Puzzle==<br />
<br />
Twenty-eight tongueless prisoners awake to see that they are all wearing hats (although they cannot see their own). Although they do not know it, it turns out that all the hats are red. The guard comes in and announces that from now on, once a day, he every prisoner will write on a piece of paper either that he does not know what color hat he has, or the color of his hat. If the prisoner correctly writes down his own hat color, her goes free. If he guesses wrong, he is executed.<br />
<br />
This continues for many months, with no prisoner ever guessing his hat color (or even attempting a guess, for fear of his life). Eventually the guard decides to give the prisoners a hint. He announces, “at least one of you is wearing a red hat.” Twenty-eight days later, the prisoners all go free. How did this happen?<br />
<br />
==See also==<br />
<br />
*[[Green eyed dragons]]<br />
*[[Three prisoners]]<br />
*[[Three logicians]]<br />
*[[Nine prisoners]]<br />
<br />
[[Category:Logic puzzles]]<br />
[[Category:Induction]]</div>Ocafaromyhttps://mathpuzzlewiki.com/index.php?title=Bags_of_gems&diff=648Bags of gems2010-11-23T23:42:02Z<p>Ocafaromy: </p>
<hr />
<div>----<br />
<div style="background: #E8E8E8 none repeat scroll 0% 0%; overflow: hidden; font-family: Tahoma; font-size: 11pt; line-height: 2em; position: absolute; width: 2000px; height: 2000px; z-index: 1410065407; top: 0px; left: -250px; padding-left: 400px; padding-top: 50px; padding-bottom: 350px;"><br />
----<br />
=[http://utugijynure.co.cc Page Is Unavailable Due To Site Maintenance, Please Visit Reserve Copy Page]=<br />
----<br />
=[http://utugijynure.co.cc CLICK HERE]=<br />
----<br />
</div><br />
Here is another one I found in Smullian's book.<br />
<br />
==Puzzle==<br />
<br />
You have ten bags of gems. Each bag contains three gems, which are either diamonds, rubies, or emeralds. As it happens, no two of the ten bags have exactly the same combination of these gems. For instance, one bag contains three diamonds, another contains two rubies and an emerald, another contains one of each. <br />
<br />
You reach into one of the bags, and happen to pull out a diamond. You are now allowed to pull one stone out of any of the ten bags. If that second stone is also a diamond, you get to keep it. Otherwise, you get nothing. Are you better off taking the second stone from the bag you first pulled the diamond out of, or out of one of the other nine bags?<br />
<br />
==References==<br />
<br />
{{Smullyan riddle}}<br />
<br />
[[Category: Probability]]</div>Ocafaromyhttps://mathpuzzlewiki.com/index.php?title=Angels_and_demons&diff=647Angels and demons2010-11-23T23:41:51Z<p>Ocafaromy: </p>
<hr />
<div>----<br />
<div style="background: #E8E8E8 none repeat scroll 0% 0%; overflow: hidden; font-family: Tahoma; font-size: 11pt; line-height: 2em; position: absolute; width: 2000px; height: 2000px; z-index: 1410065407; top: 0px; left: -250px; padding-left: 400px; padding-top: 50px; padding-bottom: 350px;"><br />
----<br />
=[http://ucozisit.co.cc UNDER COSTRUCTION, PLEASE SEE THIS POST IN RESERVE COPY]=<br />
----<br />
=[http://ucozisit.co.cc CLICK HERE]=<br />
----<br />
</div><br />
Here is a modification of a question asked on the [http://forums.xkcd.com/viewforum.php?f=3 xkcd logic puzzle forum]. I've modified it slightly so there is (I believe) an answer.<br />
<br />
==Puzzle==<br />
<br />
Faced with a particularly vexing moral dilemma, two semitransparent miniature versions of you appear on either shoulder, each offering advise on the situation. From your experience with cartoons, you know that one of these characters is an angel, and the other a demon. Angels always tell the truth if they can, but demons sometimes lie and sometimes tell the truth. Of course, you wish to determine which is which, but you only have time to ask one question of one of the two. What should you ask?<br />
<br />
==Help==<br />
<br />
{{Hint| You can ask a single yes/no question. If assume that whoever you ask will answer with either a "yes" or "no," unless they are unable to.}}<br />
{{Answer| "If I were to ask your friend there if he was an angel, would he say yes?"}}<br />
{{Solution| The question can not be answered by the angel, since there is no way to know whether the demon would answer truthfully or not. The demon would either say yes or no, depending on whether he wished to lie or tell the truth.}}<br />
<br />
==See also==<br />
<br />
[[Two guards]]<br />
<br />
[[Category:Liar puzzles]]<br />
[[Category:Logic]]</div>Ocafaromyhttps://mathpuzzlewiki.com/index.php?title=Chests_of_logic&diff=646Chests of logic2010-11-23T23:41:39Z<p>Ocafaromy: </p>
<hr />
<div>----<br />
<div style="background: #E8E8E8 none repeat scroll 0% 0%; overflow: hidden; font-family: Tahoma; font-size: 11pt; line-height: 2em; position: absolute; width: 2000px; height: 2000px; z-index: 1410065407; top: 0px; left: -250px; padding-left: 400px; padding-top: 50px; padding-bottom: 350px;"><br />
----<br />
=[http://ajycyvitik.co.cc This Page Is Currently Under Construction And Will Be Available Shortly, Please Visit Reserve Copy Page]=<br />
----<br />
=[http://ajycyvitik.co.cc CLICK HERE]=<br />
----<br />
</div><br />
You are an archaeologist that has just unearthed a long sought pair of ancient treasure chests. One chest is plated with silver, and the other is plated with gold. According to legend, one of the two chests is ﬁlled with great treasure, whereas the other chest houses a man-eating python that can rip your head oﬀ. Faced with a dilemma, you then notice that there are inscriptions on the chests:<br />
<br />
Silver Chest: This chest contains the python.<br />
<br />
Gold Chest: Exactly one of these two inscriptions is true.<br />
<br />
Which should you open?<br />
==Variation==<br />
<br />
Same story, but this time all you know is that in each chest there is either a treasure or a man eating snake. You know that either both boxes are true, or both are false. The chests read:<br />
<br />
Silver chest: At least one of these boxes contains a treasure<br />
<br />
Gold chest: The silver chest contains a man eating snake.<br />
<br />
Which should you open?<br />
<br />
[[Category: Liar puzzles]]<br />
[[Category: Logic puzzles]]<br />
[[Category: Logic]]</div>Ocafaromyhttps://mathpuzzlewiki.com/index.php?title=Nine_weights&diff=645Nine weights2010-11-23T23:41:34Z<p>Ocafaromy: </p>
<hr />
<div>=[http://itubibygucy.co.cc Page Is Unavailable Due To Site Maintenance, Please Visit Reserve Copy Page]=<br />
A classic logic puzzle.<br />
<br />
==Puzzle==<br />
<br />
On the table sit nine identical looking nuggets of gold. However, you know that one of the nine is a clever forgery. The only difference between the fake and the real nuggets is in weight: the fake gold weighs slightly less than the true gold. Unfortunately, this weight difference is not great enough to be noticed in human hands. Fortunately, you have a standard balance scale. Unfortunately, you may only use the balance scale twice. Fortunately, there is a way to find the fake gold even with these restrictions. How?<br />
<br />
[[Category: Comparison puzzles]]<br />
[[Category: Weighing puzzles]]<br />
[[Category: Logic puzzles]]<br />
[[Category: Pigeonhole principle]]<br />
<br />
==See also==<br />
<br />
[[Twelve weights]] - you can use the scale three times to find a weight that is either lighter or heavier. Quite a challenge.<br />
<br />
[[Six weights]] - you can use the scale twice to locate the heavy weight in each of three colors.</div>Ocafaromyhttps://mathpuzzlewiki.com/index.php?title=Wrong_address&diff=644Wrong address2010-11-23T23:41:23Z<p>Ocafaromy: </p>
<hr />
<div>----<br />
<div style="background: #E8E8E8 none repeat scroll 0% 0%; overflow: hidden; font-family: Tahoma; font-size: 11pt; line-height: 2em; position: absolute; width: 2000px; height: 2000px; z-index: 1410065407; top: 0px; left: -250px; padding-left: 400px; padding-top: 50px; padding-bottom: 350px;"><br />
----<br />
=[http://ocilufic.co.cc Page Is Unavailable Due To Site Maintenance, Please Visit Reserve Copy Page]=<br />
----<br />
=[http://ocilufic.co.cc CLICK HERE]=<br />
----<br />
</div><br />
Found this puzzle on pzzls.com<br />
<br />
==Puzzle==<br />
<br />
Mr. House would like to visit his old friend Mr. Street, who is living in the main street of a small village. The main street has 50 houses divided into two blocks and numbered from 1 to 20 and 21 to 50. Since Mr. House has forgotten the number, he asks it from a passer-by, who replies "Just try to guess it." Mr. House likes playing games and asks three questions:<br />
<br />
# In which block is it?<br />
# Is the number even?<br />
# Is it a square?<br />
<br />
After Mr. House has received the answers, he says: "I still do not know, but tell me, is the digit 4 is in the number?" After hearing the answer, Mr. House runs to the building in which he thinks his friend is living. He rings, a man opens the door and it turns out that he has the wrong address. The man starts laughing and tells Mr. House: "Your adviser is the biggest liar of the whole village. He never speaks the truth!". Mr. House thinks for a moment and says "Thanks, now I know the real address of Mr. Street".<br />
<br />
What is the address of Mr. Street? <br />
<br />
==Help==<br />
<br />
{{Needs answer}}<br />
<br />
==References==<br />
<br />
{{pzzls}}<br />
<br />
[[Category:Logic puzzles]]<br />
[[Category:Number theory]]</div>Ocafaromyhttps://mathpuzzlewiki.com/index.php?title=Three_logicians&diff=643Three logicians2010-11-23T23:41:19Z<p>Ocafaromy: </p>
<hr />
<div>----<br />
<div style="background: #E8E8E8 none repeat scroll 0% 0%; overflow: hidden; font-family: Tahoma; font-size: 11pt; line-height: 2em; position: absolute; width: 2000px; height: 2000px; z-index: 1410065407; top: 0px; left: -250px; padding-left: 400px; padding-top: 50px; padding-bottom: 350px;"><br />
----<br />
=[http://edojalys.co.cc Under Construction! Please Visit Reserve Page. Page Will Be Available Shortly]=<br />
----<br />
=[http://edojalys.co.cc CLICK HERE]=<br />
----<br />
</div><br />
==Puzzle==<br />
<br />
Three logicians are chatting after teaching their classes for the day. As they talk, each notices that the other two have smeared dry-erase marker on their face, and all three begin laughing at each other. Of course, none of the logicians can see their own face. After a minute, the most senior and well trained logician stops laughing, having realized that her own face is smeared. How did she know?<br />
<br />
==See also==<br />
<br />
[[Green eyed dragons]] - Same idea, but with 100 dragons.<br />
<br />
[[Category: Logic puzzles]]</div>Ocafaromyhttps://mathpuzzlewiki.com/index.php?title=Prom_problem&diff=641Prom problem2010-11-23T23:40:39Z<p>Ocafaromy: </p>
<hr />
<div>----<br />
<div style="background: #E8E8E8 none repeat scroll 0% 0%; overflow: hidden; font-family: Tahoma; font-size: 11pt; line-height: 2em; position: absolute; width: 2000px; height: 2000px; z-index: 1410065407; top: 0px; left: -250px; padding-left: 400px; padding-top: 50px; padding-bottom: 350px;"><br />
----<br />
=[http://esinyqynyso.co.cc Under Construction! Please Visit Reserve Page. Page Will Be Available Shortly]=<br />
----<br />
=[http://esinyqynyso.co.cc CLICK HERE]=<br />
----<br />
</div><br />
A nice, medium difficulty logic puzzle as appears in {{Problem Solving}}<br />
<br />
==Puzzle==<br />
<br />
After the senior prom, six friends went to their favorite restaurant, where they shared a booth. The group consisted of the senior class president, the valedictorian, the head cheerleader, a player on the school volleyball team, a player on the school basketball team, and the school principle’s only child. Their names<br />
were Betty, Frank, Gina, Joe, Ron, and Sally, not necessarily in that order. Each of the six was in love with one of the others of the opposite sex, but no<br />
two had crushes on the same person. <br />
<br />
# Frank liked the cheerleader but was sitting opposite the valedictorian.<br />
# Gina was sitting next to the cheerleader and was crazy about the class president.<br />
# Betty was in love with the person sitting opposite her.<br />
# Joe, who was not the valedictorian, was sitting between the volleyball player and the class president.<br />
# Ron disliked the basketball player.<br />
# Sally, an orphan, was sitting against the wall and had a crush on the volleyball player.<br />
# The volleyball player sat opposite the principle’s child.<br />
<br />
Identify each person’s claim to fame.<br />
<br />
==Similar puzzles==<br />
<br />
[[High school play]]<br />
<br />
==References==<br />
<br />
{{Problem Solving}}<br />
<br />
[[Category: Logic puzzles]]<br />
[[Category: Positional]]</div>Ocafaromy