tag:blogger.com,1999:blog-55453775295449655342017-07-30T17:30:23.286-07:00Math Is CrazyMadhu Sudan Sigdelhttps://plus.google.com/104387178077515574864noreply@blogger.comBlogger27125tag:blogger.com,1999:blog-5545377529544965534.post-49795110079484826332016-02-03T14:04:00.000-08:002016-02-03T14:04:39.713-08:00Monk and the mystery<div dir="ltr" style="text-align: left;" trbidi="on"><span style="background-color: white;"><span style="color: #050505; font-family: 'PT Sans', sans-serif; font-size: 13px; line-height: 19.5px;">300 monks live together in a monastery. They have very strict rules which are followed by all of the monks at all times. One of the rules is, that absolutely no communication between monks is allowed. Another is, that mirrors are forbidden. The monks have their three meals a day together in a large hall, the rest of their day is spent with individual contemplation and chores. One morning, a messenger comes to the monastery and addresses the monks at breakfast. He tells them, that a rare disease is spread throughout the country, and that the monks may have the disease as well. The main symptom of the disease is a large red spot on the head of the afflicted. The disease kills everyone who knows they have it within two hours. The disease was transmitted by a bad shipment of rice, but is not contagious. On the morning of the eleventh day after the messenger arrives, some of the monks don't turn for breakfast and are found dead in their beds. </span></span><br /><span style="background-color: white;"><span style="color: #050505; font-family: 'PT Sans', sans-serif; font-size: 13px; line-height: 19.5px;"><br /></span></span><span style="background-color: white;"><span style="color: #050505; font-family: 'PT Sans', sans-serif; font-size: 13px; line-height: 19.5px;">Question: How many monks died?</span></span></div>Madhu Sudan Sigdelhttps://plus.google.com/104387178077515574864noreply@blogger.com1tag:blogger.com,1999:blog-5545377529544965534.post-42531291103519000122014-05-29T23:19:00.002-07:002014-05-29T23:20:09.125-07:00Cryptarithmetic<div dir="ltr" style="text-align: left;" trbidi="on"><div style="text-align: justify;"><span style="font-family: inherit;">This <span style="background-color: white; color: #252525; font-size: 14px; line-height: 22.399999618530273px;"> is a type of </span>mathematical game<span style="background-color: white; color: #252525; font-size: 14px; line-height: 22.399999618530273px;"> consisting of a mathematical </span>equation<span style="background-color: white; color: #252525; font-size: 14px; line-height: 22.399999618530273px;"> among unknown </span>numbers<span style="background-color: white; color: #252525; font-size: 14px; line-height: 22.399999618530273px;">, whose </span>digits<span style="background-color: white; color: #252525; font-size: 14px; line-height: 22.399999618530273px;"> are represented by </span>letters<span style="background-color: white; color: #252525; font-size: 14px; line-height: 22.399999618530273px;">. The goal is to identify the value of each letter. Lets try this game with the popular one.</span></span></div><div style="text-align: justify;"><span style="font-family: inherit;"><br /></span></div><div style="text-align: justify;"><span style="font-family: inherit;"><img alt="\begin{matrix} & & \text{S} & \text{E} & \text{N} & \text{D} \\ + & & \text{M} & \text{O} & \text{R} & \text{E} \\ \hline = & \text{M} & \text{O} & \text{N} & \text{E} & \text{Y} \\ \end{matrix}" class="mwe-math-fallback-png-inline tex" src="http://upload.wikimedia.org/math/b/4/0/b400a8f1b675bbd3eff8fc8b83691335.png" style="background-color: white; border: none; color: #252525; display: inline; font-size: 14px; line-height: 22.399999618530273px; margin: 0px; vertical-align: middle;" /></span></div><div style="text-align: justify;"><span style="font-family: inherit;"><br /></span></div><div style="text-align: justify;"><span style="font-family: inherit;">You have to find the value of each letter involved in above equation.( S, E, N, D, M, O, R, Y ).</span></div><div style="text-align: justify;"><span style="font-family: inherit;"><br /></span></div><div style="text-align: justify;"><span style="font-family: inherit;">Remember the rules :</span></div><div style="text-align: justify;"><span style="font-family: inherit;">a) Different letters have different value.</span></div><div style="text-align: justify;"><span style="font-family: inherit;">b) Same letters must have same value (i.e. M in MORE and M in MONEY has same value)</span></div><div style="text-align: justify;"><span style="font-family: inherit;">c) The range of value is 0-9.</span></div><div style="text-align: justify;"><span style="font-family: inherit;"><br /></span></div><div style="text-align: justify;"><span style="font-family: inherit;">Refer to comment section if you want to tally your answer.</span></div></div>Madhu Sudan Sigdelhttps://plus.google.com/104387178077515574864noreply@blogger.com1tag:blogger.com,1999:blog-5545377529544965534.post-37184660429759914102014-05-27T21:25:00.001-07:002014-05-27T21:25:15.777-07:00Crocodile Dilemma Paradox <div dir="ltr" style="text-align: left;" trbidi="on"><span style="background-color: white; color: #252525; font-family: sans-serif; font-size: 14px; line-height: 22.399999618530273px;">If a crocodile steals a child and promises its return if the father can correctly guess exactly what the crocodile will do, how should the crocodile respond in the case that the father correctly guesses that the child will not be returned?</span><br /><span style="background-color: white; color: #252525; font-family: sans-serif; font-size: 14px; line-height: 22.399999618530273px;"><br /></span><span style="background-color: white; color: #252525; font-family: sans-serif; font-size: 14px; line-height: 22.399999618530273px;">Don't think that logic is too simple. Think every possible cases in perspective of crocodile. Think hard. If you cannot find the paradox see the comment to find the reason of paradox.</span></div>Madhu Sudan Sigdelhttps://plus.google.com/104387178077515574864noreply@blogger.com1tag:blogger.com,1999:blog-5545377529544965534.post-41642701151123712822014-04-24T12:56:00.003-07:002014-04-24T13:04:00.877-07:00Pirate Gold Problem<div dir="ltr" style="text-align: left;" trbidi="on"><div style="background-color: white; color: #252525; font-family: sans-serif; font-size: 14px; line-height: 22.399999618530273px; margin-bottom: 0.5em; margin-top: 0.5em;">There are 5 rational pirates, A, B, C, D and E. They find 100 gold coins. They must decide how to distribute them.</div><div style="background-color: white; color: #252525; font-family: sans-serif; font-size: 14px; line-height: 22.399999618530273px; margin-bottom: 0.5em; margin-top: 0.5em;">The pirates have a strict order of seniority: A is superior to B, who is superior to C, who is superior to D, who is superior to E.</div><div style="background-color: white; color: #252525; font-family: sans-serif; font-size: 14px; line-height: 22.399999618530273px; margin-bottom: 0.5em; margin-top: 0.5em;">The pirate world's rules of distribution are thus: that the most senior pirate should propose a distribution of coins. The pirates, including the proposer, then vote on whether to accept this distribution. In case of a tie vote the proposer has the casting vote. If not, the proposer is thrown overboard from the pirate ship and dies, and the next most senior pirate makes a new proposal to begin the system again.</div><div style="background-color: white; color: #252525; font-family: sans-serif; font-size: 14px; line-height: 22.399999618530273px; margin-bottom: 0.5em; margin-top: 0.5em;">Pirates base their decisions on three factors. First of all, each pirate wants to survive. Second, given survival, each pirate wants to maximize the number of gold coins he receives. Third, each pirate would prefer to throw another overboard, if all other results would otherwise be equal. The pirates do not trust each other, and will neither make nor honor any promises between pirates apart from the main proposal. </div></div>Madhu Sudan Sigdelhttps://plus.google.com/104387178077515574864noreply@blogger.com1tag:blogger.com,1999:blog-5545377529544965534.post-11007290032009633592011-06-23T11:03:00.001-07:002011-06-23T11:03:58.315-07:00Ones and nines<div dir="ltr" style="text-align: left;" trbidi="on">Show that all the <a class="ml-smartlink" href="http://en.wikipedia.org/wiki/Divisor">divisors</a> of any number of the form 19...9 (with an odd number of nines) end in 1 or 9. For example, the numbers 19, 1999, 199999, and 19999999 are prime (so clearly the property holds), and the (positive) divisors of 1999999999 are 1, 31, 64516129 and 1999999999 itself.<br />Show further that this property continues to hold if we insert an equal number of zeroes before the nines. For example, the numbers 109, 1000999, 10000099999, 100000009999999, and 1000000000999999999 are prime, and the (positive)divisors of 10000000000099999999999 are 1, 19, 62655709, 1190458471, 8400125030569, 159602375580811, 526315789478947368421, and 10000000000099999999999 itself.</div>Madhu Sudan Sigdelhttps://plus.google.com/104387178077515574864noreply@blogger.com0tag:blogger.com,1999:blog-5545377529544965534.post-62084181845864876922011-06-23T10:59:00.001-07:002011-06-23T10:59:40.240-07:00Ant in a field<div dir="ltr" style="text-align: left;" trbidi="on">An ant, located in a square field, is 13 meters from one of the corner posts of the field, 17 meters from the corner post diagonally opposite that one, and 20 meters from a third corner post. Find the area of the field. Assume the land is flat.<br /><img alt="Square field, containing an ant, as described above." class="p" height="215" src="http://www.qbyte.org/puzzles/p113.gif" style="height: 215px; width: 215px;" width="215" /></div>Madhu Sudan Sigdelhttps://plus.google.com/104387178077515574864noreply@blogger.com1tag:blogger.com,1999:blog-5545377529544965534.post-71642111711759438632011-06-23T10:55:00.001-07:002011-06-23T10:55:35.143-07:00Missing digits<div dir="ltr" style="text-align: left;" trbidi="on">Given that 37! = 13763753091226345046315979581abcdefgh0000000, determine, with a minimum of arithmetical effort, the digits a, b, c, d, e, f, g, and h. No calculators or computers allowed!</div>Madhu Sudan Sigdelhttps://plus.google.com/104387178077515574864noreply@blogger.com1tag:blogger.com,1999:blog-5545377529544965534.post-55346169219414233342011-06-23T10:39:00.001-07:002011-06-23T10:39:49.683-07:00Five card trick<div dir="ltr" style="text-align: left;" trbidi="on">Two information theoreticians, A and B, perform a trick with a <a class="ml-smartlink" href="http://en.wikipedia.org/wiki/Shuffling">shuffled deck</a> of cards, jokers removed. A asks a member of the audience to select five cards at random from the deck. The audience member passes the five cards to A, who examines them, and hands one back. A then arranges the remaining four cards in some way and places them face down, in a neat pile.<br />B, who has not witnessed these proceedings, then enters the room, looks at the four cards, and determines the missing fifth card, held by the audience member. How is this trick done?<br />Note: The only communication between A and B is via the arrangement of the four cards. There is no encoded speech or hand signals or ESP, no bent or <a class="ml-smartlink" href="http://en.wikipedia.org/wiki/Marked_Cards">marked cards</a>, no clue in the orientation of the pile of four cards...</div>Madhu Sudan Sigdelhttps://plus.google.com/104387178077515574864noreply@blogger.com2tag:blogger.com,1999:blog-5545377529544965534.post-28669202398856491152011-06-23T10:24:00.000-07:002011-06-23T10:24:41.486-07:00Three children<div dir="ltr" style="text-align: left;" trbidi="on"><h2 id="p17"></h2>On the first day of a new job, a colleague invites you around for a barbecue. As the two of you arrive at his home, a young boy throws open the door to welcome his father. “My other two kids will be home soon!” remarks your colleague.<br />Waiting in the kitchen while your colleague gets some drinks from the basement, you notice a letter from the principal of the local school tacked to the noticeboard. “Dear Parents,” it begins, “This is the time of year when I write to all parents, such as yourselves, who have a girl or girls in the school, asking you to volunteer your time to help the girls' soccer team.” “Hmmm,” you think to yourself, “clearly they have at least one of each!”<br />This, of course, leaves two possibilities: two boys and a girl, or two girls and a boy. Are these two possibilities equally likely, or is one more likely than the other?<br />Note: This is not a trick puzzle. You should assume all things that it seems you're meant to assume, and not assume things that you aren't told to assume. If things can easily be imagined in either of two ways, you should assume that they are equally likely. For example, you may be able to imagine a reason that a colleague with two boys and a girl would be more likely to have invited you to dinner than one with two girls and a boy. If so, this would affect the probabilities of the two possibilities. But if your imagination is that good, you can probably imagine the opposite as well. You should assume that any such extra information not mentioned in the story is not available.<br /><br /></div>Madhu Sudan Sigdelhttps://plus.google.com/104387178077515574864noreply@blogger.com1tag:blogger.com,1999:blog-5545377529544965534.post-17613023024249397402011-06-23T10:14:00.000-07:002011-06-23T10:14:54.277-07:00Dice game<div dir="ltr" style="text-align: left;" trbidi="on"><h2 id="p11"></h2>Two students play a game based on the total roll of two standard dice. Student A says that a 12 will be rolled first. Student B says that two consecutive 7s will be rolled first. The students keep rolling until one of them wins. What is the probability that A will win?<br /><br /><h1 id="main"></h1>Solution:<br /><br />Let p be the probability that student A wins. We consider the possible outcomes of the first two rolls. (Recall that each <em>roll</em> consists of the throw of two dice.) Consider the following <a class="ml-smartlink" href="http://en.wikipedia.org/wiki/Mutually_exclusive_events">mutually exclusive</a> cases, which encompass all possibilities.<br /><ul><li>If the first roll is a 12 (probability 1/36), A wins immediately.</li><li>If the first roll is a 7 and the second roll is a 12 (probability 1/6 · 1/36 = 1/216), A wins immediately.</li><li>If the first and second rolls are both 7 (probability 1/6 · 1/6 = 1/36), A cannot win. (That is, B wins immediately.)</li><li>If the first roll is a 7 and the second roll is neither a 7 nor a 12 (probability <span class="u">1/6 · 29/36 = 29/216),</span> A wins with probability p.</li><li>If the first roll is neither a 7 nor a 12 (probability 29/36), A wins with probability p.</li></ul>Note that in the last two cases we are effectively back at square one; hence the probability that A subsequently wins is p.<br />Probability p is the weighted mean of all of the above possibilities.<br />Hence p = 1/36 + 1/216 + (29/216)p + (29/36)p.<br />Therefore p = 7/13.<br /></div>Madhu Sudan Sigdelhttps://plus.google.com/104387178077515574864noreply@blogger.com3tag:blogger.com,1999:blog-5545377529544965534.post-45139651795820124202011-06-23T10:09:00.000-07:002011-06-23T10:09:10.517-07:00Five men, a monkey, and some coconuts<div dir="ltr" style="text-align: left;" trbidi="on"><h2 id="p7"></h2>Five men crash-land their airplane on a deserted island in the South Pacific. On their first day they gather as many coconuts as they can find into one big pile. They decide that, since it is getting dark, they will wait until the next day to divide the coconuts.<br />That night each man took a turn watching for rescue searchers while the others slept. The first watcher got bored so he decided to divide the coconuts into five equal piles. When he did this, he found he had one remaining coconut. He gave this coconut to a monkey, took one of the piles, and hid it for himself. Then he jumbled up the four other piles into one big pile again.<br />To cut a long story short, each of the five men ended up doing exactly the same thing. They each divided the coconuts into five equal piles and had one extra coconut left over, which they gave to the monkey. They each took one of the five piles and hid those coconuts. They each came back and jumbled up the remaining four piles into one big pile.<br />What is the smallest number of coconuts there could have been in the original pile?</div>Madhu Sudan Sigdelhttps://plus.google.com/104387178077515574864noreply@blogger.com0tag:blogger.com,1999:blog-5545377529544965534.post-9006443901384955982011-06-23T09:32:00.000-07:002011-06-23T09:32:50.757-07:00Cryptogram Puzzle<div dir="ltr" style="text-align: left;" trbidi="on">Cryptograms are word puzzles in which your primary task is to decode a block of cipher-text. You are faced with a sequence of characters that looks like gibberish. The characters can be letters, numbers, or any other symbols you like. The idea is that each character represents a letter of the alphabet, and when all the characters are replaced with their corresponding letters, the gibberish turns into something meaningful and (hopefully!) recognizable.Here's an example of a cryptogram that has been embellished with a theme and an author of the encrypted phrase - just to spice things up a notch... <br /><div style="text-align: center;"> <b>Topic - Realism</b> </div><div align="center"><a href="http://www.word-buff.com/support-files/word-puzzle-cryptogram.pdf" target="_new" title="Click for a Larger Printable Cryptogram Puzzle"><img alt="Word Puzzle Example - Cryptogram" border="0" height="505" src="http://www.word-buff.com/images/word-puzzle-cryptogram.png" style="height: 316px; width: 400px;" width="640" /></a></div>In this example, I've given you a toehold into the grid by telling you that the number 22 represents the letter <b>T</b>. The cryptograms you meet won't necessarily have hints, or themes for that matter (though the themeless cryptograms seem a bit dry and academic for me).As you work your way through the puzzle you make educated guesses at the letters being represented by certain characters by focusing on key word patterns. I'll let you have a go at the one above to get a feel for what I mean by 'key word patterns' here.<br /><br />Download the pdf file from : http://www.word-buff.com/support-files/word-puzzle-cryptogram.pdf<br /><br />Lets not ruin the interest by giving the solution, I will verify if you post the answer.</div>Madhu Sudan Sigdelhttps://plus.google.com/104387178077515574864noreply@blogger.com0tag:blogger.com,1999:blog-5545377529544965534.post-60793799224397574632011-06-19T03:55:00.000-07:002011-06-19T03:55:23.365-07:00Nature's Numbers<div dir="ltr" style="text-align: left;" trbidi="on"><h3 class="post-title entry-title"> <a href="http://musemath.blogspot.com/2007/03/natures-numbers.html"></a> </h3><div class="post-header"> </div>If you've ever looked for a four-leafed clover, you know that nature rarely delivers such a curiosity. Nearly every clover plant you check has the usual three leaves. If you study the flowers in your garden or in the countryside, you'll discover the most common number of petals is five. Buttercups, geraniums, pansies, primroses, rhododendrons, tomato blossoms, and many more all have five petals.<br /><br /><a href="http://1.bp.blogspot.com/_Gb4NmAd9gS8/Rgm6X64iTAI/AAAAAAAAAGI/xyilUHt3gwI/s1600-h/fib2.jpg" onblur="try {parent.deselectBloggerImageGracefully();} catch(e) {}"><img alt="" border="0" id="BLOGGER_PHOTO_ID_5046769777551494146" src="http://1.bp.blogspot.com/_Gb4NmAd9gS8/Rgm6X64iTAI/AAAAAAAAAGI/xyilUHt3gwI/s400/fib2.jpg" style="cursor: pointer; display: block; margin: 0px auto 10px; text-align: center;" /></a><br />Five also shows up in arrangements of seeds. Cut an apple in half across its core (rather than the usual way down the core from the stem), and you'll see the seeds arranged in a beautiful five-pointed star. What numbers do cucumbers, tomatoes, pears, and lemons feature?<br /><br /><a href="http://4.bp.blogspot.com/_Gb4NmAd9gS8/Rgm7Sq4iTCI/AAAAAAAAAGY/zfRH8q4GZM4/s1600-h/fib5.jpg" onblur="try {parent.deselectBloggerImageGracefully();} catch(e) {}"><img alt="" border="0" id="BLOGGER_PHOTO_ID_5046770786868808738" src="http://4.bp.blogspot.com/_Gb4NmAd9gS8/Rgm7Sq4iTCI/AAAAAAAAAGY/zfRH8q4GZM4/s400/fib5.jpg" style="cursor: pointer; display: block; margin: 0px auto 10px; text-align: center;" /></a><br />Pineapples have eight rows of scales, seen as roughly diamond-shaped markings, sloping in one direction and 13 sloping in the other. Pine cones show the same sort of feature.<br /><br /><a href="http://2.bp.blogspot.com/_Gb4NmAd9gS8/Rgm6iK4iTBI/AAAAAAAAAGQ/8aWJVp0_kP0/s1600-h/fib1.gif" onblur="try {parent.deselectBloggerImageGracefully();} catch(e) {}"><img alt="" border="0" id="BLOGGER_PHOTO_ID_5046769953645153298" src="http://2.bp.blogspot.com/_Gb4NmAd9gS8/Rgm6iK4iTBI/AAAAAAAAAGQ/8aWJVp0_kP0/s400/fib1.gif" style="cursor: pointer; display: block; margin: 0px auto 10px; text-align: center;" /></a><br />The head of a sunflower highlights other numbers. In a perfect head, the tiny flowers, or florets, that will become seeds are arranged in two spirals, one winding clockwise and the other counterclockwise. Depending on the species of sunflower, you might find 34 and 55, 55 and 89, or even 89 and 144 florets along a spiral.<br /><br /><a href="http://2.bp.blogspot.com/_Gb4NmAd9gS8/Rgm7pK4iTDI/AAAAAAAAAGg/BKkSYGL1_tw/s1600-h/fib4.jpg" onblur="try {parent.deselectBloggerImageGracefully();} catch(e) {}"><img alt="" border="0" id="BLOGGER_PHOTO_ID_5046771173415865394" src="http://2.bp.blogspot.com/_Gb4NmAd9gS8/Rgm7pK4iTDI/AAAAAAAAAGg/BKkSYGL1_tw/s400/fib4.jpg" style="cursor: pointer; display: block; margin: 0px auto 10px; text-align: center;" /></a><br />Similarly, floret spirals at the center of certain types of daisies feature the numbers 21 and 34. You can look for similar patterns on brocolli or cauliflower.<br /><br />Take a look at the numbers that nature seems to like (at least in plants): 3, 5, 8, 13, 21, 34, 55, 89, and 144. Can you find a pattern?<br /><br /><a href="http://3.bp.blogspot.com/_Gb4NmAd9gS8/Rgm76a4iTEI/AAAAAAAAAGo/TjuePRBfBTw/s1600-h/fib3.jpg" onblur="try {parent.deselectBloggerImageGracefully();} catch(e) {}"><img alt="" border="0" id="BLOGGER_PHOTO_ID_5046771469768608834" src="http://3.bp.blogspot.com/_Gb4NmAd9gS8/Rgm76a4iTEI/AAAAAAAAAGo/TjuePRBfBTw/s400/fib3.jpg" style="cursor: pointer; display: block; margin: 0px auto 10px; text-align: center;" /></a><br />Here's a clue: Start with 1 + 1 = 2, and then add the two numbers on each side of the equal sign. Keep on doing this with each new equation that you get.<br /><br />1 + 1 = 2, 1 + 2 = 3, 2 + 3 = 5, 3 + 5 = 8, 5 + 8 = 13, 8 + 13 = 21, 13 + 21 = 34, and so on.<br /><br />The sums you get are all members of a famous sequence of numbers named for the mathematician <a class="ml-smartlink" href="http://en.wikipedia.org/wiki/Fibonacci">Leonardo of Pisa</a>, also known as Fibonacci, who studied them about 800 years ago. Scientists have long wondered why these number come up in plants. The answer may have something to do with the way plants grow, especially the way petals or buds space themselves to gather the most sunlight and nutrients.<br /><br />Wherever you look, nature certainly has a way with numbers.</div>Madhu Sudan Sigdelhttps://plus.google.com/104387178077515574864noreply@blogger.com0tag:blogger.com,1999:blog-5545377529544965534.post-77770090557141689222011-06-13T13:26:00.000-07:002011-06-13T13:27:54.423-07:00Find Faulty Coin<div dir="ltr" style="text-align: left;" trbidi="on">There are 12 coins. One of them is false; it weights differently. It is not known, if the false coin is heavier or lighter than the right coins. How to find the false coin by three weighs on a simple scale?<br /><br /><br /><b>Solution:</b><br /><br /><a href="http://1.bp.blogspot.com/-h3bWhT02C2o/TfZxedrExjI/AAAAAAAABoA/uQNL82p_kPg/s1600/coin1.gif" imageanchor="1" style="clear: left; float: left; margin-bottom: 1em; margin-right: 1em;"><br /></a><br /><div class="separator" style="clear: both; text-align: center;"><a href="http://1.bp.blogspot.com/-h3bWhT02C2o/TfZxedrExjI/AAAAAAAABoA/uQNL82p_kPg/s1600/coin1.gif" imageanchor="1" style="margin-left: 1em; margin-right: 1em;"><img border="0" src="http://1.bp.blogspot.com/-h3bWhT02C2o/TfZxedrExjI/AAAAAAAABoA/uQNL82p_kPg/s1600/coin1.gif" style="cursor: move;" /></a><a href="http://1.bp.blogspot.com/-h3bWhT02C2o/TfZxedrExjI/AAAAAAAABoA/uQNL82p_kPg/s1600/coin1.gif" imageanchor="1" style="margin-left: 1em; margin-right: 1em;"><img border="0" src="http://1.bp.blogspot.com/-h3bWhT02C2o/TfZxedrExjI/AAAAAAAABoA/uQNL82p_kPg/s1600/coin1.gif" /></a><a href="http://1.bp.blogspot.com/-h3bWhT02C2o/TfZxedrExjI/AAAAAAAABoA/uQNL82p_kPg/s1600/coin1.gif" imageanchor="1" style="margin-left: 1em; margin-right: 1em;"></a><a href="http://1.bp.blogspot.com/-h3bWhT02C2o/TfZxedrExjI/AAAAAAAABoA/uQNL82p_kPg/s1600/coin1.gif" imageanchor="1" style="margin-left: 1em; margin-right: 1em;"><img border="0" src="http://1.bp.blogspot.com/-h3bWhT02C2o/TfZxedrExjI/AAAAAAAABoA/uQNL82p_kPg/s1600/coin1.gif" /></a></div><div class="separator" style="clear: both; text-align: center;"><a href="http://4.bp.blogspot.com/-F140osPD4Ws/TfZxfSXD3BI/AAAAAAAABoE/6Eb3bSvMkYU/s1600/graph1.gif" imageanchor="1" style="margin-left: 1em; margin-right: 1em;"><img border="0" height="166" src="http://4.bp.blogspot.com/-F140osPD4Ws/TfZxfSXD3BI/AAAAAAAABoE/6Eb3bSvMkYU/s400/graph1.gif" width="400" /></a></div><br /></div>Madhu Sudan Sigdelhttps://plus.google.com/104387178077515574864noreply@blogger.com0tag:blogger.com,1999:blog-5545377529544965534.post-11306910607687013272011-06-13T13:09:00.000-07:002011-06-13T13:10:32.375-07:00Two Boys Cross a River Puzzle<div dir="ltr" style="text-align: left;" trbidi="on"><h1 align="center"></h1><h3>The Puzzle:</h3><div style="color: black; float: left; padding: 4px 18px 4px 4px;">Two boys wish to cross a river. The only way to get to the other side is by boat, but that boat can only take one boy at a time. The boat cannot return on its own, there are no ropes or similar tricks, yet both boys manage to cross using the boat. </div><div style="color: black; font-family: Verdana,Georgia,helvetica,serif; font-size-adjust: none; font-size: 15px; font-stretch: normal; font-style: normal; font-variant: normal; font-weight: 300; line-height: 20px;"><br />How?</div><h3>Our Solution:</h3><div style="margin: 0pt 0pt 0pt 50px;"><div style="color: #000044; font-family: Verdana,Georgia,helvetica,serif; font-size-adjust: none; font-size: 15px; font-stretch: normal; font-style: normal; font-variant: normal; font-weight: 300; line-height: 20px;">The boys start on opposite sides of the river.</div></div></div>Madhu Sudan Sigdelhttps://plus.google.com/104387178077515574864noreply@blogger.com0tag:blogger.com,1999:blog-5545377529544965534.post-16953139489311736552011-06-13T12:58:00.000-07:002011-06-13T12:59:34.637-07:00Bridge crossing<div dir="ltr" style="text-align: left;" trbidi="on"> This problem was published in MAA on line: <a 3="" href="http://www.csmonitor.com/ideas/mathchat/chat061198.html" target="page"> Crossing a Rickety Bridge at Night By Flashlight. </a> <br /><ul>A group of four people has to cross a bridge. It is dark, and they have to light the path with a flashlight. No more than two people can cross the bridge simultaneously, and the group has only one flashlight. It takes different time for the people in the group to cross the bridge: <ul><b>A</b>nnie crosses the bridge in 1 minute, </ul><ul><b>B</b>ob crosses the bridge in 2 minutes, </ul><ul>Volodia Mitlin crosses the bridge in 5 minutes,</ul><ul><b>D</b>orothy crosses the bridge in 10 minutes. </ul></ul>How can the group cross the bridge in 17 minutes?<br /><b><span font="" style="color: red;"> <span style="color: black;"> </span></span></b><br /><b><span font="" style="color: red;"><span style="color: black;">Solution: </span></span></b><br /><span font="" style="color: red;"> http://www.math.utah.edu/~cherk/solutions/bridge/bridge.html</span><br /><br /><span font="" style="color: red;"> </span> </div>Madhu Sudan Sigdelhttps://plus.google.com/104387178077515574864noreply@blogger.com0tag:blogger.com,1999:blog-5545377529544965534.post-12114809769382604982011-06-13T11:50:00.001-07:002011-06-13T11:51:30.386-07:00Really InterestingSomebody thought of two natural numbers greater than 1 and calculated their sum and product (both turned out to be less than 100). He told the sum to S. and the product – to P. The following is their discussion:<br />P.: I don’t know the numbers.<br />S.: Me neither.<br />P.: And I knew from the beginning that you would not know.<br />S.: And I knew from the beginning that you would know from the beginning.<br />P.: But I still don’t know them!<br />S.: But now I do.<br /><br />What are the numbers?<br />Try it and post in the comments.....<br />Its interesting one try yourself......<br />Easy But hard.....Madhu Sudan Sigdelhttps://plus.google.com/104387178077515574864noreply@blogger.com0tag:blogger.com,1999:blog-5545377529544965534.post-53980736241986537802011-06-13T11:26:00.000-07:002011-06-13T11:28:42.741-07:00"Light-bulb" Math Riddles<ol class="normal"><li>How many mathematicians does it take to change a light bulb? <span style="font-size:85%;"><span style="font-style: italic;" class="highlight-label" onclick="openClose('ans0')"> <span id="ans0" style="display: block;">None. The answer is intuitively obvious or. It's left to the reader as an exercise. </span></span></span> </li><p> </p><li>How many numerical analysts does it take to change a light bulb? <span style="font-style: italic;font-size:78%;" ><span class="highlight-label" onclick="openClose('ans1')"> <span id="ans1" style="display: block;">3.9967 (after six iterations).</span></span></span> </li><p> </p><li>How many mathematical logicians does it take to change a light bulb? <span style="font-style: italic;font-size:85%;" ><span class="highlight-label" onclick="openClose('ans2')"> <span id="ans2" style="display: block;">None. They can't do it, but they can easily prove that it can be done.</span></span></span> </li><p> </p><li>How many classical geometers does it take to change a light bulb? <span style="font-style: italic;font-size:85%;" ><span class="highlight-label" onclick="openClose('ans3')"> <span id="ans3" style="display: block;">None. You can't do it with a straight edge and a compass.</span></span></span> </li><p> </p><li>How many math analysts does it take to change a light bulb? <span style="font-style: italic;font-size:85%;" ><span class="highlight-label" onclick="openClose('ans4')"> <span id="ans4" style="display: block;">Three. One to prove existence, one to prove uniqueness and one to derive a nonconstructive algorithm to do it.</span></span></span> </li><p> </p><li>How many number theorists does it take to change a light bulb? <span class="highlight-label" onclick="openClose('ans5')"> <span id="ans5" style="display: block;"><span style="font-style: italic;font-size:85%;" >I don't know the exact number, but I am sure it must be some rather elegant prime</span>.</span></span><iframe allowtransparency="true" hspace="0" marginwidth="0" marginheight="0" onload="var i=this.id,s=window.google_iframe_oncopy,H=s&&s.handlers,h=H&&H[i],w=this.contentWindow,d;try{d=w.document}catch(e){}if(h&&d&&(!d.body||!d.body.firstChild)){if(h.call){i+='.call';setTimeout(h,0)}else if(h.match){i+='.nav';w.location.replace(h)}s.log&&s.log.push(i)}" vspace="0" id="aswift_1" name="aswift_1" style="left: 0pt; position: absolute; top: 0pt;" frameborder="0" height="15" scrolling="no" width="468"></iframe></li><ins style="border: medium none ; margin: 0pt; padding: 0pt; display: inline-table; height: 15px; position: relative; visibility: visible; width: 468px;"><ins id="aswift_1_anchor" style="border: medium none ; margin: 0pt; padding: 0pt; display: block; height: 15px; position: relative; visibility: visible; width: 468px;"></ins></ins><li>How many statisticians does it take to screw in a light bulb? <span style="font-style: italic;font-size:85%;" ><span class="highlight-label" onclick="openClose('ans6')"> <span id="ans6" style="display: block;">We really don't know yet. Our entire sample was skewed to the left!</span></span></span> </li><p> </p><li>How many math students does it take to change a light bulb? <span style="font-style: italic;font-size:85%;" ><span class="highlight-label" onclick="openClose('ans7')"> <span id="ans7" style="display: block;">Ten. One to do it and nine to watch.</span></span></span> </li><p> </p><li>How many topologists does it take to change a light bulb? <span style="font-style: italic;font-size:85%;" ><span class="highlight-label" onclick="openClose('ans8')"> <span id="ans8" style="display: block;">Just one. But what will you do with the doughnut?</span></span></span> </li><p> </p><li>How many professors does it take change a light bulb? <span style="font-style: italic;font-size:85%;" ><span class="highlight-label" onclick="openClose('ans10')"> <span id="ans10" style="display: block;">One. With eight research students, two programmers, three post-docs and a secretary to help him.</span></span></span> </li><p> </p><li>How many university lecturers does it take to change a light bulb? <span style="font-style: italic;font-size:85%;" ><span class="highlight-label" onclick="openClose('ans11')"> <span id="ans11" style="display: block;">Four. One to do it and three to co-author the paper.</span></span></span> </li><p> </p><li>How many graduate students does it take to change a light bulb? <span style="font-style: italic;font-size:85%;" ><span class="highlight-label" onclick="openClose('ans12')"> <span id="ans12" style="display: block;">Only one. But it takes nine years.</span></span></span> </li><p> </p><li>How many administrators does it take to change a light bulb? <span style="font-style: italic;font-size:85%;" ><span class="highlight-label" onclick="openClose('ans13')"> <span id="ans13" style="display: block;">None. What was wrong with the old one?</span></span></span> </li></ol>Madhu Sudan Sigdelhttps://plus.google.com/104387178077515574864noreply@blogger.com0tag:blogger.com,1999:blog-5545377529544965534.post-5761905973727009312011-06-13T11:15:00.000-07:002011-06-13T11:20:05.187-07:00A cat has nine tails.<a onblur="try {parent.deselectBloggerImageGracefully();} catch(e) {}" href="http://4.bp.blogspot.com/-uYDpd102Z7A/TfZUZvPVRwI/AAAAAAAABn8/9Du65ViXTPw/s1600/images.jpg"><img style="margin: 0px auto 10px; display: block; text-align: center; cursor: pointer; width: 245px; height: 206px;" src="http://4.bp.blogspot.com/-uYDpd102Z7A/TfZUZvPVRwI/AAAAAAAABn8/9Du65ViXTPw/s320/images.jpg" alt="" id="BLOGGER_PHOTO_ID_5617770386345510658" border="0" /></a><br /><p> <b>Theorem.</b> A cat has nine tails. </p> <p> <i>Proof.</i> No cat has eight tails. Since one cat has one more tail than no cat, it must have nine tails. </p>Madhu Sudan Sigdelhttps://plus.google.com/104387178077515574864noreply@blogger.com0tag:blogger.com,1999:blog-5545377529544965534.post-8450734741910921892011-06-13T10:56:00.000-07:002011-06-13T10:58:18.160-07:00Farmer's Sheep<span style="color: rgb(255, 0, 0);font-family:Arial;" ></span><br /> <div style="text-align: left;"> <span style="color: rgb(255, 255, 255);">.<br /> </span></div> <center> <div style="width: 550px;" class="figcenter"><img style="width: 396px; height: 281px;" title="" alt="" src="http://www.pedagonet.com/mathgenius/images/q120.png" /> </div> </center> <p>Farmer Longmore had a curious aptitude for arithmetic, and was known in his district as the "mathematical farmer."<br />The new vicar was not aware of this fact when, meeting his worthy parishioner one day in the lane, he asked him in the course of a short conversation, "Now, how many sheep have you altogether?"<br />He was therefore rather surprised at Longmore's answer, which was as follows:<br />"You can divide my sheep into two different parts, so that the difference between the two numbers is the same as the difference between their squares.<br />Maybe, Mr. Parson, you will like to work out the little sum for yourself."</p> <p>Can the reader say just how many sheep the farmer had? </p> <p>Supposing he had possessed only twenty sheep, and he divided them into the two parts 12 and 8.<br />Now, the difference between their squares, 144 and 64, is 80.<br />So that will not do, for 4 and 80 are certainly not the same.<br />If you can find numbers that work out correctly, you will know exactly how many sheep Farmer Longmore owned.</p><p><br /></p><p style="font-weight: bold;">Solution:</p><p>If he divided this sheep (which is best done by weight) into two parts, making one part two-thirds and the other part one-third, then the difference between these two numbers is the same as the difference between their squares—that is, one-third.<br /></p> Any two fractions will do if the denominator equals the sum of the two numerators..<br /><br />The farmer had <span style="font-weight: bold;">one</span> sheep only!Madhu Sudan Sigdelhttps://plus.google.com/104387178077515574864noreply@blogger.com0tag:blogger.com,1999:blog-5545377529544965534.post-66298115747155836712011-06-13T10:42:00.000-07:002011-06-13T10:43:58.768-07:00Wizard's Cats<span style="font-family: Arial; color: rgb(255, 0, 0);"></span><br /> <div style="text-align: left;"> <span style="color: rgb(255, 255, 255);">.<br /> </span></div> <center> <div style="width: 400px;" class="figcenter"><img title="" alt="" src="http://www.pedagonet.com/mathgenius/images/q167.png" height="392" width="400" /> </div> </center> <p>A wizard placed ten cats inside a <a href="http://en.wikipedia.org/wiki/Magic_circle" class="ml-smartlink">magic circle</a> as shown in our illustration, and hypnotized them so that they should remain stationary during his pleasure.<br /> He then proposed to draw <a href="http://en.wikipedia.org/wiki/Three_circles" class="ml-smartlink">three circles</a> inside the large one, so that no cat could approach another cat without crossing a <span class=""></span>magic circle.</p><p style="font-weight: bold;"><br /></p><p><span style="font-weight: bold;">Solution</span></p><p>The illustration requires no explanation.<br /> It shows clearly how the <a href="http://en.wikipedia.org/wiki/Three_circles" class="ml-smartlink">three circles</a> may be drawn so that every cat has a separate enclosure, and cannot approach another cat without crossing a line.</p> <center> <div style="width: 400px;" class="figcenter"><img style="width: 59px; height: 59px;" title="" alt="" src="http://www.pedagonet.com/mathgenius/images/a167.png" /> </div> </center>Madhu Sudan Sigdelhttps://plus.google.com/104387178077515574864noreply@blogger.com0tag:blogger.com,1999:blog-5545377529544965534.post-70082625553236008402011-06-13T10:33:00.000-07:002011-06-13T10:35:36.697-07:00The Wizard's Arithmetic<p>Once upon a time a knight went to consult a certain famous wizard.<br /><br /></p> <center> <div class="figcenter" style="width: 500px;"><img style="width: 500px; height: 453px;" src="http://www.pedagonet.com/puzzles/129.png" alt="" /></div> </center> <p>"And art thou learned also in the magic of numbers?" asked the knight.<br />"Show me but one sample of thy wit in these matters."</p> <p>The old wizard took five blocks bearing numbers, and placed them on a shelf, apparently at random, so that they stood in <a id="KonaLink1" class="kLink" style="text-decoration: underline ! important; position: static; font-family: inherit ! important; font-weight: inherit ! important; font-size: inherit ! important;" href="http://www.pedagonet.com/puzzles/wizard.html#"><span style="color: blue ! important; font-family: inherit ! important; font-weight: inherit ! important; font-size: inherit ! important; position: static;color:blue;" ><span class="kLink" style="color: blue ! important; font-family: inherit ! important; font-weight: inherit ! important; font-size: inherit ! important; position: relative;">the </span><span class="kLink" style="color: blue ! important; font-family: inherit ! important; font-weight: inherit ! important; font-size: inherit ! important; position: relative;">order</span></span></a> 41096, as shown in our illustration.<br />He then took in his hands an 8 and a 3, and held them together to form the number 83.<span class=""></span></p> <p>"Sir Knight, tell me," said the wizard, "canst thou multiply one number into the other in thy mind?"</p> <p>"<a id="KonaLink2" class="kLink" style="text-decoration: underline ! important; position: static; font-family: inherit ! important; font-weight: inherit ! important; font-size: inherit ! important;" href="http://www.pedagonet.com/puzzles/wizard.html#"><span style="color: blue ! important; font-family: inherit ! important; font-weight: inherit ! important; font-size: inherit ! important; position: static;color:blue;" ><span class="kLink" style="color: blue ! important; font-family: inherit ! important; font-weight: inherit ! important; font-size: inherit ! important; position: relative;">Nay</span></span></a>, of a truth," the good knight replied.<br />"I should need to set out upon the task with pen and scrip."</p> <p>"Yet mark ye how right easy a thing it is to a man learned in the lore of far Araby, who knoweth all the magic that is hid in the philosophy of numbers!"</p> <p>The wizard simply placed the 3 next to the 4 on the shelf, and the 8 at the other end. It will be found that this gives the answer quite correctly—3410968.<br />Very curious, is it not? How many other two-figure multipliers can you find that will produce the same effect?<br />You may place just as many blocks as you like on the shelf, bearing any figures you choose.</p>Madhu Sudan Sigdelhttps://plus.google.com/104387178077515574864noreply@blogger.com0tag:blogger.com,1999:blog-5545377529544965534.post-265385235822557212011-06-12T09:55:00.001-07:002011-06-12T09:55:24.111-07:00Julius Caesar's Last BreathWhat's the chance that the breath you just inhaled contains at least one air molecule that was in Julius Caesar's last breath--the one in which he said (according to Shakespeare) "Et tu Brute? Then die Caesar"?<br /><br />Assume that the more than two thousand years that have passed have been enough time for all the molecules in Caesar's last breath to mix evenly in the atmosphere, and that only a trivial amount of the molecules have leaked out into the oceans or the ground. Assume further that there are about 1044 molecules of air, and about 2 x 1022 molecules in each breath--yours or Caesar's.<br /><br />That gives a chance of 2 x 1022/1044 = 2x 10-22 that any one particular molecule you breathe in came from Caesar's last breath. This means that the probability that any one particular molecule did not come from Caesar's last breath is [1-2x10-22]. But we want the probability that the first molecule did not come from Caesar's last breath and that the second molecule and that the third molecule and so forth. To determine the probability of not just one thing but of a whole bunch of things that are causally unconnected happening together, we multiply the individual probabilities. Since there are 2x10-22 different molecules, and since each has the same [1-2x10-22] chance of not coming from Caesar's last breath, we need to multiply the probability of any single event--[1-2x10-22]--by itself 2x1022 times. That gives us:<br /><br /> [1-2x10-22][2x10^22]<br /><br />as the probability that none of the molecules in the breath you just inhaled (assuming you are still holding out) came from Julius Caesar's last breath.<br /><br />How to evaluate this? Recall that if a number x is small, then (1-x) is approximately equal to e-x, where e=2.718281828... is the so-called base of the natural logarithms. So we can rewrite the equation above as:<br /><br /> [e[-2x10^(-22)]][2x10^(22)]<br /><br />And remember that when we raise numbers with exponents to further exponents, we simply multiply the exponents together. In this case, one exponent (the chance that a molecule came from Caesar) is very small, and the other (the number of molecules in a breath) is very large. When we multiply them together, we get: [-2x10(-22)] x [2x10(22)] = -4. e-4 is approximately 1/(2.72 x 2.72 x 2.72 x 2.72) = 1/54.7 = 0.018.<br /><br />Thus there is a 1.8% chance that none of the molecules you are (still) holding in your lungs came from Caesar's last breath. And there is a 98.2% chance that at least one of the molecules in your lungs came from Caesar's last breath.<br /><br />From John Allen Paulos's Innumeracy.Madhu Sudan Sigdelhttps://plus.google.com/104387178077515574864noreply@blogger.com0tag:blogger.com,1999:blog-5545377529544965534.post-40014954929031688812011-06-12T09:47:00.001-07:002011-06-12T09:48:25.898-07:00Amazing Mathmatics Number<div style="text-align: center;">Sequential Inputs of numbers with 8<br />1 x 8 + 1 = 9<br />12 x 8 + 2 = 98<br />123 x 8 + 3 = 987<br />1234 x 8 + 4 = 9876<br />12345 x 8 + 5 = 98765<br />123456 x 8 + 6 = 987654<br />1234567 x 8 + 7 = 9876543<br />12345678 x 8 + 8 = 98765432<br />123456789 x 8 + 9 = 987654321<br /><br /><br />Sequential 1's with 9<br />1 x 9 + 2 = 11<br />12 x 9 + 3 = 111<br />123 x 9 + 4 = 1111<br />1234 x 9 + 5 = 11111<br />12345 x 9 + 6 = 111111<br />123456 x 9 + 7 = 1111111<br />1234567 x 9 + 8 = 11111111<br />12345678 x 9 + 9 = 111111111<br />123456789 x 9 + 10 = 1111111111<br /><br /><br />Sequential 8's with 9<br />9 x 9 + 7 = 88<br />98 x 9 + 6 = 888<br />987 x 9 + 5 = 8888<br />9876 x 9 + 4 = 88888<br />98765 x 9 + 3 = 888888<br />987654 x 9 + 2 = 8888888<br />9876543 x 9 + 1 = 88888888<br />98765432 x 9 + 0 = 888888888<br /><br /><br />Numeric Palindrome with 1's<br />1 x 1 = 1<br />11 x 11 = 121<br />111 x 111 = 12321<br />1111 x 1111 = 1234321<br />11111 x 11111 = 123454321<br />111111 x 111111 = 12345654321<br />1111111 x 1111111 = 1234567654321<br />11111111 x 11111111 = 123456787654321<br />111111111 x 111111111 = 12345678987654321<br /><br /><br />Without 8<br />12345679 x 9 = 111111111<br />12345679 x 18 = 222222222<br />12345679 x 27 = 333333333<br />12345679 x 36 = 444444444<br />12345679 x 45 = 555555555<br />12345679 x 54 = 666666666<br />12345679 x 63 = 777777777<br />12345679 x 72 = 888888888<br />12345679 x 81 = 999999999<br /><br /><br />Sequential Inputs of 9<br />9 x 9 = 81<br />99 x 99 = 9801<br />999 x 999 = 998001<br />9999 x 9999 = 99980001<br />99999 x 99999 = 9999800001<br />999999 x 999999 = 999998000001<br />9999999 x 9999999 = 99999980000001<br />99999999 x 99999999 = 9999999800000001<br />999999999 x 999999999 = 999999998000000001<br />......................................<br /><br /><br />Sequential Inputs of 6<br />6 x 7 = 42<br />66 x 67 = 4422<br />666 x 667 = 444222<br />6666 x 6667 = 44442222<br />66666 x 66667 = 4444422222<br />666666 x 666667 = 444444222222<br />6666666 x 6666667 = 44444442222222<br />66666666 x 66666667 = 4444444422222222<br />666666666 x 666666667 = 444444444222222222<br />......................................</div>Madhu Sudan Sigdelhttps://plus.google.com/104387178077515574864noreply@blogger.com0tag:blogger.com,1999:blog-5545377529544965534.post-43137384027023150172011-06-12T09:42:00.000-07:002011-06-12T09:43:27.423-07:00Day of The DateDay of the Week:<br />January has 31 days. It means that every date in February will be 3 days later than the same date in January(28 is 4 weeks exactly). The below table is calculated in such a way. Remember this table which will help you to calculate.<br />January 0<br />February 3<br />March 3<br />April 6<br />May 1<br />June 4<br />July 6<br />August 2<br />September 5<br />October 0<br />November 3<br />December 5<br /><br />Step1: Ask for the Date. Ex: 23rd June 1986<br />Step2: Number of the month on the list, June is 4.<br />Step3: Take the date of the month, that is 23<br />Step4: Take the last 2 digits of the year, that is 86.<br />Step5: Find out the number of leap years. Divide the last 2 digits of the year by 4, 86 divide by 4 is 21.<br />Step6: Now add all the 4 numbers: 4 + 23 + 86 + 21 = 134.<br />Step7: Divide 134 by 7 = 19 remainder 1.<br />The reminder tells you the day.<br />Sunday 0<br />Monday 1<br />Tuesday 2<br />Wednesday 3<br />Thursday 4<br />Friday 5<br />Saturday 6<br />Answer: MondayMadhu Sudan Sigdelhttps://plus.google.com/104387178077515574864noreply@blogger.com0