tag:blogger.com,1999:blog-57303912018-01-20T11:02:05.829-08:00Blaine's Puzzle BlogWeekly discussion on the NPR puzzler, brain teasers, math problems and more.Blainehttp://www.blogger.com/profile/06379274325110866036noreply@blogger.comBlogger50125tag:blogger.com,1999:blog-5730391.post-41840008057151061442017-12-18T14:51:00.000-08:002017-12-18T14:51:15.676-08:00Christmas Puzzle for 2017<div class="separator" style="clear: both; text-align: center;"><a href="http://home.wavecable.com/~blaineblog@astound.net/xmas/2017/XmasPuzzle2017.pdf" imageanchor="1" style="clear: left; float: left; margin-bottom: 1em; margin-right: 1em;"><img border="0" src="https://3.bp.blogspot.com/-6jp8yVnxBC0/Wjg2kalgbxI/AAAAAAAC1vg/1WzzVG6cTTgWSjvxOaunhcF4wmcw3VTiwCLcBGAs/s400/Christmas%2BPresent%2BGrid.png" width="311" height="400" data-original-width="871" data-original-height="1120" /></a></div>Our annual Christmas puzzle is available now. <br />As in prior years, the reward for solving is a <a href="https://vimeo.com/blainefelicia/xmas2017">video Christmas card</a>, but you'll need to figure out the password by <a href="http://home.wavecable.com/~blaineblog@astound.net/xmas/2017/XmasPuzzle2017.pdf">solving the puzzle first</a>.<br /><br /><i>Note: If you need some help, the full answer is posted <a href="http://home.wavecable.com/~blaineblog@astound.net/xmas/2017/">here</a>, but try solving it <b>without</b> help first... it's more fun that way.</i><br /><br />Feel free to add a comment below to let us know that you successfully figured it out (without giving away the answer to others). We are always looking for new ideas for next year's <a href="http://family.blainesville.com/p/christmas-puzzles.html">Christmas puzzle</a>, so submit those too.Blainehttp://www.blogger.com/profile/06379274325110866036noreply@blogger.com11tag:blogger.com,1999:blog-5730391.post-24283616507326882372016-12-26T21:58:00.002-08:002016-12-26T21:58:48.850-08:00Christmas Snowflake Matching Puzzle for 2016<div class="separator" style="clear: both; text-align: center;"><a href="http://home.wavecable.com/~blaineblog@astound.net/xmas/2016/XmasPuzzle2016.pdf" imageanchor="1" style="clear: left; float: left; margin-bottom: 1em; margin-right: 1em;"><img border="0" src="https://2.bp.blogspot.com/-3Utn0Cw6PzU/WGIB3UH80fI/AAAAAAACgCg/1MmbibWBArk9g_bNh1J5LRSFT2TfElQVACLcB/s320/Snowflakes.png" width="311" height="320" /></a></div>Our annual Christmas puzzle is available now. <br />As in prior years, the reward for solving is a <a href="http://vimeo.com/blainefelicia/xmas2016">video Christmas card</a>, but you'll need to figure out the password by <a href="http://home.wavecable.com/~blaineblog@astound.net/xmas/2016/XmasPuzzle2016.pdf">solving the puzzle first</a>.<br /><br /><i>Note: If you need some help, the full answer is posted <a href="http://home.wavecable.com/~blaineblog@astound.net/xmas/2016/">here</a>, but try solving it <b>without</b> help first... it's more fun that way.</i><br /><br />Feel free to add a comment below to let us know that you successfully figured it out (without giving away the answer to others). We are always looking for new ideas for next year's <a href="http://family.blainesville.com/p/christmas-puzzles.html">Christmas puzzle</a>, so submit those too.Blainehttp://www.blogger.com/profile/06379274325110866036noreply@blogger.com22tag:blogger.com,1999:blog-5730391.post-70784294975436994142015-12-17T12:29:00.001-08:002016-11-18T09:11:54.640-08:00Christmas Puzzle for 2015<div class="separator" style="clear: both; text-align: center;"><a href="http://home.wavecable.com/~blaineblog@astound.net/xmas/2015/XmasPuzzle2015.pdf" imageanchor="1" style="clear: left; float: left; margin-bottom: 1em; margin-right: 1em;"><img border="0" src="http://3.bp.blogspot.com/-AcSqzAuw0qI/VnMZdJt2AwI/AAAAAAACPio/dFM7WYJakts/s320/Christmas%2BTree%2B%2528small%2529.png" /></a></div>Our annual Christmas puzzle is available now. <br />As in prior years, the reward for solving is a <a href="http://vimeo.com/blainefelicia/xmas2015">video Christmas card</a>, but you'll need to figure out the password by <a href="http://home.wavecable.com/~blaineblog@astound.net/xmas/2015/XmasPuzzle2015.pdf">solving the puzzle first</a>.<br /><br /><i>Note: If you need some help, the full answer is posted <a href="http://home.wavecable.com/~blaineblog@astound.net/xmas/2015/">here</a>, but try solving it <b>without</b> help first... it's more fun that way.</i><br /><br />Feel free to add a comment below to let us know that you successfully figured it out (without giving away the answer to others). We are always looking for new ideas for next year's <a href="http://family.blainesville.com/p/christmas-puzzles.html">Christmas puzzle</a>, so submit those too.Blainehttp://www.blogger.com/profile/06379274325110866036noreply@blogger.com14tag:blogger.com,1999:blog-5730391.post-79095674523750351652014-12-17T10:42:00.001-08:002016-11-18T09:12:02.113-08:00Christmas Puzzle 2014 - Power of TEN<div class="separator" style="clear: both; text-align: center;"><a href="http://home.wavecable.com/~blaineblog@astound.net/xmas/2014/XmasPuzzle2014.pdf" imageanchor="1" style="clear: left; float: left; margin-bottom: 1em; margin-right: 1em;"><img border="0" src="http://2.bp.blogspot.com/-Xuxw-KkFHOU/VJHMoBGe3vI/AAAAAAAAMWM/33Z7gxwPyjM/s320/ChristmasPuzzle2014.PNG" /></a></div>Our annual Christmas puzzle "Power of TEN" is available now. Every answer is a word containing the letters TEN. For example, if the clue were <i>"These keep your hands warm when playing in the snow"</i>, the answer would be <i>mit<u>ten</u>s</i>.<br /><br />As in prior years, the reward for solving is a <a href="http://vimeo.com/blainefelicia/xmas2014">video Christmas card</a>, but you'll need to figure out the password by <a href="http://home.wavecable.com/~blaineblog@astound.net/xmas/2014/XmasPuzzle2014.pdf">using the first letter of each word</a>. As a hint, the password could be used to describe an occasion such as the new year.<br /><br /><i>Note: If you need some help, the full answer is posted <a href="http://home.wavecable.com/~blaineblog@astound.net/xmas/2014/">here</a>, but try solving it <b>without</b> help first... it's more fun that way.</i><br /><br />Feel free to add a comment below to let us know that you successfully figured it out (without giving away the answer to others). We are always looking for new ideas for next year's <a href="http://family.blainesville.com/p/christmas-puzzles.html">Christmas puzzle</a>, so submit those too.Blainehttp://www.blogger.com/profile/06379274325110866036noreply@blogger.com14tag:blogger.com,1999:blog-5730391.post-40202875153361271962013-09-12T15:40:00.000-07:002013-09-12T15:37:44.836-07:00NPR Sunday Puzzle (Sep 8, 2013): Shortest Path to the Answer<a href="http://3.bp.blogspot.com/-EL4-W0SYLd0/UiyOYLahG0I/AAAAAAAALKM/TooKhnLhB3Q/s1600/CurveLines.gif" imageanchor="1" style="clear: right; float: right; margin-bottom: 1em; margin-left: 1em;"><img border="0" height="107" src="http://3.bp.blogspot.com/-EL4-W0SYLd0/UiyOYLahG0I/AAAAAAAALKM/TooKhnLhB3Q/s200/CurveLines.gif" width="160" /></a><br /><a href="http://www.npr.org/2013/09/08/220033016/close-but-no-cigar">NPR Sunday Puzzle (Sep 8, 2013): Shortest Path to the Answer</a>: <blockquote><b>Q: </b>Name a famous person in history with four letters in the first name and six letters in the last. Move the first letter of all this to the end. The result will be a two-word phrase that might be defined as "the opposite of a curve." Who's the famous person, and what's the phrase?</blockquote>What is this? Another reference to the puzzle from three weeks ago? At least we know the answer isn't EDIR ECTLIN.<br/><br/><b>Edit: </b>The puzzle 3 weeks ago was the one involving a Roman numeral. And this week the puzzle involves a Roman general under Julius Caesar.<blockquote><b>A: </b>MARC ANTONY --> ARC ANTONYM</blockquote>Blainehttp://www.blogger.com/profile/06379274325110866036noreply@blogger.com121tag:blogger.com,1999:blog-5730391.post-13798222698270895362012-12-27T12:03:00.000-08:002012-12-27T12:03:21.098-08:00NPR Sunday Puzzle (Dec 23, 2012): Actor, Artist, God<a href="http://4.bp.blogspot.com/-_5TZWLEhOyc/UNdDYijcU0I/AAAAAAAAK2I/jJY2NqJTwdo/s1600/Gods.jpg" imageanchor="1" style="clear:right; float:right; margin-left:1em; margin-bottom:1em"><img border="0" height="124" width="200" src="http://4.bp.blogspot.com/-_5TZWLEhOyc/UNdDYijcU0I/AAAAAAAAK2I/jJY2NqJTwdo/s200/Gods.jpg" style="border: 8px solid rgb(255,255,255);" alt="Gods, National Geographic" /></a><a href="http://www.npr.org/2012/12/23/167885296/unwrap-christmas-for-your-gift">NPR Sunday Puzzle (Dec 23, 2012): Actor, Artist, God</a>: <blockquote><b>A: </b>Take the last name of a famous actor. Drop the first letter, and you'll get the last name of a famous artist. Drop the first letter again, and you'll get the name of a god in classical mythology. What names are these?</blockquote>I thought the puzzle deadline would be a day earlier this week, but it isn't so this comment can be mostly ignored.<br/><br/><b>Edit: </b>The first hint is Wednesday (day earlier) which is named after the Norse god Woden/Odin. Additionally, if you ignore one letter in "ignored" it anagrams to the name of the actor.<blockquote><b>A: </b>GRODIN --> RODIN --> ODIN</blockquote>Blainehttp://www.blogger.com/profile/06379274325110866036noreply@blogger.com111tag:blogger.com,1999:blog-5730391.post-9273698474928263372012-12-21T02:27:00.001-08:002016-11-18T09:12:15.252-08:00Christmas Puzzle 2012 - Snowflake Maze<a href="http://home.wavecable.com/~blaineblog@astound.net/xmas/2012/XmasPuzzle2012.pdf" imageanchor="1" style="clear:right; float:right; margin-left:1em; margin-bottom:1em"><img border="0" height="200" width="173" src="http://3.bp.blogspot.com/-RBIaPrwDWxE/UNQ0VqHWG1I/AAAAAAAAK1w/kiqOyewZigQ/s200/snowflake-small.png" /></a>Our annual Christmas puzzle is available now. It's a fun maze in the shape of a snowflake. As in prior years, the reward for solving is a <a href="http://vimeo.com/blainefelicia/xmas2012">video Christmas card</a>, but you'll need to figure out the password by <a href="http://home.wavecable.com/~blaineblog@astound.net/xmas/2012/XmasPuzzle2012.pdf">solving the puzzle</a>.<br /><br /><i>Note: If you need some help, the full answer is posted <a href="http://home.wavecable.com/~blaineblog@astound.net/xmas/2012/">here</a>, but try solving it <b>without</b> help first... it's more fun that way.</i><br /><br />Feel free to add a comment below to let us know that you successfully figured it out (without giving away the answer to others). We are always looking for new ideas for next year's <a href="http://family.blainesville.com/p/christmas-puzzles.html">Christmas puzzle</a>, so submit those too.Blainehttp://www.blogger.com/profile/06379274325110866036noreply@blogger.com19tag:blogger.com,1999:blog-5730391.post-85944618239771743042012-07-28T13:05:00.000-07:002012-07-28T13:07:01.836-07:00GeekDad Puzzle of the Week: Contiguous Consonants<a href="http://2.bp.blogspot.com/-Wwx7oWBuP4Q/UA69McmXdwI/AAAAAAAAKqg/5F8xjN7zvG4/s1600/ContiguousConsonants.png" imageanchor="1" style="clear: right; float: right; margin-bottom: 1em; margin-left: 1em;"><img border="0" height="152" src="http://2.bp.blogspot.com/-Wwx7oWBuP4Q/UA69McmXdwI/AAAAAAAAKqg/5F8xjN7zvG4/s200/ContiguousConsonants.png" width="160" style="border: 8px solid rgb(17, 174, 195);" alt="breastroke, matchstick, corkscrew, postscript"/></a><a href="http://www.wired.com/geekdad/2012/07/puzzle-contiguous-consonants/">GeekDad Puzzle of the Week: Contiguous Consonants</a>: <br /><blockquote><b>Q: </b><i>[GeekDad had]</i> a lot of time to work out several phrases that incorporated words with multiple adjacent non-vowels or "contiguous consonants." For purposes of this puzzle, please consider the letter "y" strictly as a consonant. In parentheses, after each phrase is the number of words, and each word's count of contiguous consonants.<br /><ul><li>Ice-free, super tall buildings in Scranton (3w/6c)</li><li>Encoding long words in a fixed orbit (3w/6c)</li><li>Crazy fish-studier’s two wheeled transport (3w/5c)</li><li>Melodic equivalents to the "Queen of Diamonds" (Condon)(3w/6c)</li><li>Sufficiently valuable magic during the America’s Cup (3w/5c)</li><li>Where playing Beethoven on your iPhone was invented (3w/5c)</li><li>Artificial disk-flip game (2w/5c)</li><li>Rotational energy "battery," 10<sup>-10</sup> meters across (2w/5c)</li></ul></blockquote>For those that have struggled with the math problems, this might be more up your alley. The hardest one, in my opinion, is the 4th one; I'm not completely happy with my answer. Which ones do you find tricky? Remember don't give anything away since this is a contest with a prize. Feel free to read the <a href="http://www.wired.com/geekdad/2012/07/puzzle-contiguous-consonants/">full puzzle details on the GeekDad site</a> and submit your answers by Friday for a chance at the $50 prize.<br/><br/><b>Edit: </b>The deadline has passed and I've posted our answers in the comments. I'm still waiting to see the intended answers, especially for #4.Blainehttp://www.blogger.com/profile/06379274325110866036noreply@blogger.com4tag:blogger.com,1999:blog-5730391.post-82010752979273162702012-07-21T08:21:00.000-07:002012-07-21T08:21:54.307-07:00GeekDad Puzzle of the Week: Dog Siblings<a href="http://3.bp.blogspot.com/-mrMtOxMy9Co/UAaqp-gbvdI/AAAAAAAAKp4/rj6B2G7Gp70/s1600/BlackLabs.png" imageanchor="1" style="clear: right; float: right; margin-bottom: 1em; margin-left: 1em;"><img alt="Black Labs, mrpattersonsir@flickr" border="0" height="128" src="http://3.bp.blogspot.com/-mrMtOxMy9Co/UAaqp-gbvdI/AAAAAAAAKp4/rj6B2G7Gp70/s320/BlackLabs.png" style="border: 8px solid rgb(17, 174, 195);" width="160" /></a><a href="http://www.wired.com/geekdad/2012/07/geekdad-puzzle-of-the-week-dog-siblings/">GeekDad Puzzle of the Week: Dog Siblings</a>: <br /><blockquote><b>Q: </b>A guy with a black lab said that his dog, Selkie, has five brothers and sisters in town. "But I’ve never run into one of them," he said. "I wonder what are the chances of that?"<br /><br />Imagine that each of the six dogs goes out somewhere an average of once every three days. And imagine that between trails and parks and fields there are 200 places a dog can go, all (let's say...) with equal probability.<br /><br />If it's been exactly two years — 730 days — since Selkie's owner picked her up from the litter, what are the chances that during this time Selkie would NOT see a doggie sibling?<br />For extra credit, what are the chances over the same time that any sibling will meet any other sibling?</blockquote>I've got my answers which I will reveal after the deadline. In the meantime, feel free to solve it and submit your answer to GeekDad.<br /><br /><strong>Edit:</strong> The deadline was Friday, so here is how I went about solving the puzzle.<br /><br />The key to this puzzle is figure out the chances of dogs <strong><em>not</em></strong> meeting on one day. From there it is easy to figure out the chance of them <strong><em>not </em></strong>meeting for 730 days. And then if necessary, you can figure out the probability of the opposite case (meeting) by subtracting from 100%.<br /><br /><h4>Part 1 - Selkie doesn't meet a doggie sibling</h4>In order for a dog to be at a specific location, they must be out (with 1/3 probability) and at that specific spot (1/200 probability). That means there is a 1/600 chance of a specific dog being out at a specific location. Thinking of the negative probability, that means there is a 599/600 chance that a dog is *not* at a specific location.<br /><br />Selkie will *not* meet another dog on a specific day if,<br />1) Selkie is at home (2/3)<br />2) Selkie is out at any location (1/3) and dog 1 is not there (599/600) and dog 2 is not there (599/600) and dog 3 is not there...<br /><br />In other words, the chance that Selkie doesn't meet any other dog on a specific day is:<br />2/3 + 1/3 x (599/600)^5 ≈ 99.7231466062%<br /><br />And the chance that Selkie doesn't meet any dogs for 2 years (730 days) is:<br />[ 2/3 + 1/3 x (599/600)^5 ]^730 ≈ 13.2148023616%<br /><br /><blockquote><b>A: </b>(Part 1) The chance that Selkie does NOT see a doggie sibling is around 13.2148%</blockquote><h4><br />Part 2 - Chances that any sibling meets another sibling</h4>This is the much tougher question. First let's make a table of probabilities of having 0 dogs out, 1 dog out, 2 dogs out, etc.<br /><br />The chance that 'n' dogs are out on any one day is C(6,n) x (1/3)^n x (2/3)^(6-n).<br />And given 'n' dogs are out, the chance that they do NOT meet is 200/200 x 199/200 x 198/200 ... (using the number of dogs that are out.)<br /><br /><div class="separator" style="clear: both; text-align: center;"><a href="http://4.bp.blogspot.com/-OX8ZkMgUfss/UArHEMsjG6I/AAAAAAAAKqE/QcBuhY3jYuI/s1600/DoggieChart.png" imageanchor="1" style="margin-left: 1em; margin-right: 1em;"><img border="0" height="117" src="http://4.bp.blogspot.com/-OX8ZkMgUfss/UArHEMsjG6I/AAAAAAAAKqE/QcBuhY3jYuI/s400/DoggieChart.png" width="400" /></a></div><br />Thus, the chance that no dogs meet on a single day is around 99.1717389885%<br />The chance that no dogs meet for 730 days is (99.1717389885%)^730 ≈ 0.2307745729%<br />Subtracting from 100%, you get the probability that at least two siblings will meet (100% - 0.230774573%) = 99.769225427%<br /><br /><blockquote><b>A: </b>(Part 2) The chance that any of the siblings meet during those 2 years is around 99.7692%</blockquote>Blainehttp://www.blogger.com/profile/06379274325110866036noreply@blogger.com1tag:blogger.com,1999:blog-5730391.post-79851307582422409922012-07-18T04:51:00.001-07:002012-07-18T05:37:08.581-07:00GeekDad Puzzle of the Week: Math Trolls<a href="http://1.bp.blogspot.com/-neKn6nw7Ego/UAahahQGsRI/AAAAAAAAKpo/VAlgAtWF2Kc/s1600/TrollBridge.png" imageanchor="1" style="clear: right; float: right; margin-bottom: 1em; margin-left: 1em;"><img border="0" height="160" src="http://1.bp.blogspot.com/-neKn6nw7Ego/UAahahQGsRI/AAAAAAAAKpo/VAlgAtWF2Kc/s320/TrollBridge.png" width="128" alt="Troll Bridge Sign"/></a><a href="http://www.wired.com/geekdad/2012/07/geekdad-puzzle-of-the-week-solution-math-trolls/">GeekDad Puzzle of the Week: Math Trolls</a>: <br /><blockquote><b>Q: </b>Nora is taking a trip to visit her Grandmother in northernmost New York State this week, to bring her some freshly picked berries. On the way there, she has to cross a total of 30 bridges, and under each of the these bridges lives a troll. Each troll is aware of their bridge number, and either demands or gives berries based upon the rarest or most applicable description of their bridge. They demand or give berries according to the following schedule:<br /><ul><li>Trolls under odd numbered bridges demand half of your berries. </li><li>Trolls under even numbered bridges demand 20 berries. </li><li>Trolls under prime numbered bridges give you half again the number of berries you are carrying. </li><li>Trolls under perfect square numbered bridges demand a quarter of your berries. </li><li>Trolls under perfect cube numbered bridges give you the number of berries you are carrying, doubling your number of berries. </li></ul>If trolls round up in their demands (i.e., if you have 57 berries at the foot of a bridge best described as odd numbered, you will cross it with 28 berries), what is the minimum number of berries Nora must start with so that she ends up with 1,000 berries when she arrives at her Grandmother’s house?</blockquote>The solution has already been posted on the GeekDad website. If you want to solve it yourself, read no further. A detailed breakdown of my solution is given in the comments.Blainehttp://www.blogger.com/profile/06379274325110866036noreply@blogger.com4tag:blogger.com,1999:blog-5730391.post-9729275556528646272012-07-04T09:59:00.000-07:002012-07-08T07:39:54.230-07:00GeekDad Puzzle of the Week: Poaching Berries<a href="http://3.bp.blogspot.com/-nXfHof6byJQ/T_R09rWO28I/AAAAAAAAKow/t4xPgVxuvOg/s1600/RedRaspberries.jpg" imageanchor="1" style="clear: right; float: right; margin-bottom: 1em; margin-left: 1em;"><img alt="Red Raspberries, photofarmer@Flickr" border="0" height="107" src="http://3.bp.blogspot.com/-nXfHof6byJQ/T_R09rWO28I/AAAAAAAAKow/t4xPgVxuvOg/s320/RedRaspberries.jpg" style="border: 8px solid rgb(17, 174, 195);" width="160" /></a><a href="http://www.wired.com/geekdad/2012/07/geekdad-puzzle-week-poaching-berries/">GeekDad Puzzle of the Week: Poaching Berries</a>: <br /><blockquote><b>Q: </b>Leif and Kestrel are willing to give GeekDad a 20% cut of berries they poach for turning a blind eye. Imagine that Leif picks five berries per poach and Kestrel picks three berries per poach and that they attempt to poach once every day, with the exception of any day just after they’ve been caught. Now imagine that each time they poach berries, they have a 15% chance of getting caught. How many berries can GeekDad expect to eat each each week, averaged over time?</blockquote>I'll post my thoughts on the answer after the answer is revealed (generally early next week).<br /><br /><b>Edit: </b>The <a href="http://www.wired.com/geekdad/2012/07/geekdad-puzzle-of-the-week-answer-poaching-berries/">following answer</a> was what I submitted to end up winning the puzzle:<br /><br />Let's assume a few things:<br />1) Leif and Kestrel start in a state where they haven't ever been caught.<br />2) Each time they are caught, the berries that day don't count. Also, they must skip the next day until trying again.<br /><br />There are essentially 3 states that each child could be in. Since the probabilities for each child are the same we can group them together, though in reality the puzzle allows for one child to be caught and the other not. The results work out the same, so let's just simplify it to a pair of children trying to poach 8 berries a day.<br /><br />The children could be:<br />1) Caught the day before and therefore have no chance of getting berries that day (must sit out).<br />2) Caught today (15%, if not caught the day before)<br />3) Not caught today (85%, if not caught the day before)<br /><br />Each day, the chance of being caught the day before is just carried forward. So on day 2, the chance is 15% they are sitting out. That leaves 85% chance they are attempting poaching. Of that 85%, there's a 15% chance they are caught (0.85 x 0.15 = 12.75%) and an 85% chance they poach successfully (0.85 x 0.85 = 72.25%).<br /><br />If you repeat this, the next day (Day 3) they have a 12.75% chance of sitting out and a 87.25% chance of attempting poaching. Day 4, the chance of sitting out is 0.15 x 87.25% = 13.0875% and not sitting out is 86.9125%. Day 5, the chance of sitting out is 0.15 x 86.9125% = 13.036875% and not sitting out is 86.963125%. You can continue this progression and you will see that the chance they are sitting out approaches a value of about 13.043478%. The chance they are caught that day is equivalently about 13.043478%. That leaves a 73.913043% chance they are able to poach 8 berries with an expected return of 5.913043478 berries a day.<br /><br />That equates to approximately 41.39130435 berries a week. With a 20% "commission" after awhile you will be eating approximately 8.27826087 berries each week.<br /><br /><strong>Note:</strong> For a more accurate answer (rather than just a decimal approximation) we can solve this algebraically as follows:<br /><br />Let p be the chance that you ARE sitting out. <br />Let q be the chance you are NOT sitting out. <br /><br />Together these are mutually exclusive and therefore add up to 100% (or mathematically we say 1) <br />p + q = 1 <br />p = 1 - q <br /><br />The chance you are NOT sitting out, but CAUGHT is 0.15q <br />The chance you are NOT sitting out, and SUCCESSFULLY POACHED is 0.85q <br /><br />We know that eventually the two values p and 0.15q end up being the same, so equate them <br />p = 0.15q <br /><br />Substitute in 1-q: <br />1 - q = 0.15q <br /><br />Rearrange: <br />1 = 1.15q <br />q = 1/1.15 <br /><br />The chance we are caught that day is 0.15q<br />0.15q = 0.15(1/1.15) = 0.15/1.15 = 15/115 = 3/23 <br />And the chance we poach some berries is 0.85q <br />0.85q = 0.85(1/1.15) = 0.85/1.15 = 85/115 = 17/23 <br /><br />For the sake of completeness that means you have: <br />3/23 = chance child is sitting out <br />3/23 = chance child was caught today <br />17/23 = chance child was able to poach successfully <br /><br />Now multiply this last number by 8 berries attempted times 7 days and then times 1/5 (20% commission) to get the expected number of berries poached each week. <br />Berries per week = 17/23 x 8 x 7 x 1/5 <br /><br />If you reduce that to a fraction you end up with: <br />952/115 = 8 32/115 berries each week.<br /><br /><blockquote><b>A: </b>8 32/115 berries each week.<br/>8.2<u>7826086956521739130434</u>... (underlined portion repeats indefinitely)</blockquote>Blainehttp://www.blogger.com/profile/06379274325110866036noreply@blogger.com7tag:blogger.com,1999:blog-5730391.post-16696092587308840102012-06-16T14:57:00.002-07:002012-06-18T06:03:58.417-07:00GeekDad Puzzle of the Week: When Are the Odds Even?<a href="http://4.bp.blogspot.com/-FCyzq0gVmBg/T9z-QwDTq1I/AAAAAAAAKnw/BEHzwC8_6sY/s1600/Marbles.jpg" imageanchor="1" style="clear: right; float: right; margin-bottom: 1em; margin-left: 1em;"><img alt="Black and white marbles" border="0" height="120" src="http://4.bp.blogspot.com/-FCyzq0gVmBg/T9z-QwDTq1I/AAAAAAAAKnw/BEHzwC8_6sY/s320/Marbles.jpg" title="" width="120" style="border: 8px solid rgb(17,174,195);"/></a><a href="http://www.wired.com/geekdad/2012/06/geekdad-puzzle-of-the-week-when-are-the-odds-even/">GeekDad Puzzle of the Week: When Are the Odds Even?</a>:<br/><blockquote><b>Q: </b>If we have a bag containing equal numbers of black and white marbles, and we pull out <b>one</b> marble, the odds of it being black are even. If we have a bag containing 120 marbles, 85 of which are black, the odds of us pulling out <b>two</b> marbles and them both being black is also even — (85/120)x(84/119) = 0.5 or 50%.<br /><br />If the largest bag we have can hold 1,000,000 marbles, for how many sets of marbles (i.e., the 120 marbles described above are one set) can we pull <b>two</b> marbles and have a 50% chance of them being the same designated color? Are there any sets of marbles for which we can pull <b>three</b> marbles and have a 50% chance of them being the same designated color? If so, how many?</blockquote>After the solution is revealed, I'll post the details of my answer.<br/><br/><b>Edit: </b><a href="http://www.wired.com/geekdad/2012/06/geekdad-puzzle-of-the-week-solution-when-are-the-odds-even/">GeekDad Puzzle Solution:</a><br/>In the first case you are essentially looking for integer solutions to:<br/>a(a-1) = 2b(b-1)<br/><br/>There are EIGHT sets under 1 million that will result in even odds when 2 balls are drawn.<br/><br/>4 marbles (3 black) --> 4 x 3 = 2(3 x 2)<br/>21 marbles (15 black) --> 21 x 20 = 2(15 x 14)<br/>120 marbles (85 black) --> 120 x 119 = 2(85 x 84)<br/>697 marbles (493 black) --> 697 x 696 = 2(493 x 492)<br/>4,060 marbles (2,871 black) --> 4,060 x 4,059 = 2(2,871 x 2,870)<br/>23,661 marbles (16,731 black) --> 23,661 x 23,660 = 2(16,731 x 16,730)<br/>137,904 marbles (97,513 black) --> 137,904 x 137,903 = 2(97,513 x 97,512)<br/>803,761 marbles (568,345 black) --> 803,761 x 803,760 = 2(568,345 x 568,344)<br/><br/>Interestingly, the next number in each sequence can be computed as follows:<br/>a(n) = 6a(n-1) - a(n-2) - 2<br/><br/>So for example, the next numbers in the sequence would be:<br/>Total balls: 6 x 803,761 - 137,904 - 2 = 4,684,660 marbles<br/>Black balls: 6 x 568,345 - 97,513 - 2 = 3,312,555 black<br/><br/>Integer sequences: <a href="http://oeis.org/A011900">A011900</a> and <a href="http://oeis.org/A046090">A046090</a><br/><br/>In the second case you are looking for integer solutions to:</br>a(a-1)(a-2) = 2b(b-1)(b-2)<br/><br/>There is only ONE set under 1 million that will result in even odds when 3 balls are drawn.</br><br/>6 marbles (5 black) --> 6 x 5 x 4 = 2(5 x 4 x 3)<br/><br/>Blainehttp://www.blogger.com/profile/06379274325110866036noreply@blogger.com5tag:blogger.com,1999:blog-5730391.post-11633746466417020892011-06-16T12:00:00.000-07:002016-11-22T17:44:07.507-08:00NPR Sunday Puzzle (Jun 12, 2011): Sam Loyd's Hat Rack Puzzle<a href="http://www.npr.org/2011/06/12/137129959/its-lonely-at-the-top">NPR Sunday Puzzle (Jun 12, 2011): Sam Loyd's Hat Rack Puzzle</a>:<br /><br />This <b>Hat Rack Puzzle</b> by <b>Sam Loyd</b> was published 100 years ago in <i>Woman's Home Companion</i>: <blockquote><b>Q: </b>A hat room contains a wall with 49 pegs, arranged in a 7-by-7 square. The hat clerk has 20 hats that are to be hung on 20 different pegs. How many lines, containing four hats in a straight line, is it possible to produce? A line can go in any direction: horizontally, vertically or obliquely. To explain your answer, number the pegs in order, from 1 in the upper left corner to 49 in the lower right corner; list which pegs you put the 20 hats on, and give the total number of lines containing four hats in a row.</blockquote>Liane has left, but it also seems like the NPR website editors are gone. Last week they had "goose" as a two word phrase (instead of "roast goose") and this week they misspelled Sam Loyd (as Sam Lloyd). Anyway, back to the puzzle; not counting rotations and reflections, I have 3 ways to get the answer.<br /><br /><b>Edit: </b>If you re-read my post you'll see the phrase "are gone" at the end of the first sentence. This is a homophone of Argon with atomic number 18, a clue to there being 18 lines in the solution(s).<blockquote><b>A: </b>I found 3 main solutions (not counting reflections and rotations). Click each one to see a larger view with any rotated/reflected variants.<br /><a href="http://home.wavecable.com/~puzzleblog@astound.net/hatrack/HatRack1.gif"><img border="0" height="138" width="138" style="margin:0 7px 7px 0;cursor:pointer; cursor:hand;" src="http://home.wavecable.com/~puzzleblog@astound.net/hatrack/HatRack1thumb.gif" alt="Hat Rack Solution 1"/></a><a href="http://home.wavecable.com/~puzzleblog@astound.net/hatrack/HatRack2.gif"><img border="0" height="138" width="138" style="margin:0 7px 7px 0;cursor:pointer; cursor:hand;" src="http://home.wavecable.com/~puzzleblog@astound.net/hatrack/HatRack2thumb.gif" alt="Hat Rack Solution 2"/></a><a href="http://home.wavecable.com/~puzzleblog@astound.net/hatrack/HatRack3.gif"><img border="0" height="138" width="138" style="margin:0 7px 7px 0;cursor:pointer; cursor:hand;" src="http://home.wavecable.com/~puzzleblog@astound.net/hatrack/HatRack3thumb.gif" alt="Hat Rack Solution 3"/></a><br /></blockquote>Blainehttp://www.blogger.com/profile/06379274325110866036noreply@blogger.com76tag:blogger.com,1999:blog-5730391.post-27995107864781637402011-01-18T17:40:00.001-08:002016-11-22T17:46:35.867-08:00Write the Alphabet Backwards, Quicker than Forwards!<img style="float:right; margin:0 10px 10px 0;cursor:pointer; cursor:hand;" src="http://home.wavecable.com/~puzzleblog@astound.net/tebahpla.png" border="0" alt="The alphabet, backwards"/>I was reminded recently of a way to impress your friends and perhaps win a bet too. It involves writing the alphabet backwards faster than they can forwards. <br /><br />The key is to learn the backwards alphabet as "words" rather than individual letters. If you break it into chunks of 4 letters (with 2 left over) you have:<blockquote>ZYXW VUTS RQPO NMLK JIHG FEDC BA<br />Phonetically think of this as the phrase:<br />"Zixwa Vuts Irqpo Nimlick Jig Fedic Bah"</blockquote>Practice saying this as you write each set of letters one after the other. With a little practice you'll be able to write this very quickly.<br /><br />Now you are ready to challenge your friends to a race. You can even bet them that you'll write the alphabet backwards faster than they write it forwards. The reason it works is you won't need to stop, think, sing that alphabet song, go back a few letters, etc. You simply write down your 7 "words" as quickly as possible and you are sure to beat them.<br /><br />Enjoy!Blainehttp://www.blogger.com/profile/06379274325110866036noreply@blogger.com1tag:blogger.com,1999:blog-5730391.post-25749193830896020292010-12-21T23:19:00.001-08:002016-11-18T09:12:29.965-08:00Hack the Video Password, by Solving our Christmas Puzzle<a href="http://home.wavecable.com/~blaineblog@astound.net/xmas"><img style="float:right; margin:0 10px 10px 0;cursor:pointer; cursor:hand;" src="http://home.astound.net/~familyblog/Spot8Diffs_thumb.png" alt="Christmas Puzzle 2010"/></a>This year we've hidden a <a href="http://vimeo.com/17956757">secret video message</a> as the solution to our puzzle. Only those that can spot the 8 differences in our annual <a href="http://home.wavecable.com/~blaineblog@astound.net/xmas">Christmas Puzzle</a> will be able to figure out the password. Are you up to the challenge?<br /><br /><i>Note: If you need some help there are hints on the puzzle page and even the complete solution, but try solving it <b>without</b> help first... it's more fun that way.</i><br /><br />When you solve it, please don't give away the answer but feel free to add a comment to let us know that you successfully figured it out. And we are always looking for new ideas for next year's <a href="http://family.blainesville.com/p/christmas-puzzles.html">Christmas puzzle</a>, so submit those too.Blainehttp://www.blogger.com/profile/06379274325110866036noreply@blogger.com1tag:blogger.com,1999:blog-5730391.post-83381706221217135702009-05-01T00:34:00.001-07:002016-11-22T17:46:47.925-08:00Friday Fun: Rapidly Rotating Electronic Lock<img style="float:right; margin:0 10px 10px 0;cursor:pointer; cursor:hand;" src="http://home.wavecable.com/~puzzleblog@astound.net/uploaded_images/circularlock.gif" border="0" alt="Circular Electronic Lock"/>It's Friday and you are looking forward to the weekend, but an evil genius has locked you in a room. The door to the room is protected by a special electronic lock with four identical buttons equally spaced along the rim of a circular dial.<br /><br />Each button toggles an internal switch within the mechanism. You can attempt to open the lock by simultaneously pressing any set of the 4 buttons which will toggle the corresponding switches. If you are lucky enough to thereby align the switches so they are all on or all off, the lock will open. Otherwise the dial begins a spinning cycle that lasts for 1 full minute. When it comes to rest you have no way of knowing which button(s) you pressed previously.<br /><br />Your captor is returning in 15 minutes. Is there any possible method you can think of that will GUARANTEE that you can open the lock in less than 15 tries? If it is not possible, then let me know why that is the case... so I don't waste my time.Blainehttp://www.blogger.com/profile/06379274325110866036noreply@blogger.com16tag:blogger.com,1999:blog-5730391.post-69036262656348112472009-03-27T17:55:00.000-07:002009-03-27T17:55:00.281-07:00Friday Fun: What's the next number in the sequence?Can you figure out the next few terms in the following sequence?<blockquote><b>Q: </b>1, 3, 7, 12, 18, 26, 35, 45, 56, 69...</blockquote>I'll post the answer some time next week.Blainehttp://www.blogger.com/profile/06379274325110866036noreply@blogger.com10tag:blogger.com,1999:blog-5730391.post-2430597812616536992009-02-11T07:27:00.000-08:002010-01-10T08:01:31.872-08:00Word Game: If I give you BVI, the answer is obvious...Here's a fun game you can play with friends. I'll give you a sequence of letters. Your goal is to find a common English word that has those letters exactly in that order with no additional letters in between. As the title implies, if I gave you "BVI" you could reply with "oBVIous". Got it?<br /><br />To get you started, here are a few combinations to try:<br />HTG, WKW, UMF, PTC, GGP, AUE and HMM...<br /><br />Good luck, and feel free to post any letter combinations of your own that might be fun to try.Blainehttp://www.blogger.com/profile/06379274325110866036noreply@blogger.com16tag:blogger.com,1999:blog-5730391.post-70881824563055856642009-01-02T05:17:00.001-08:002016-11-22T17:46:50.791-08:00New Year's Resolution: Exercise Your Brain<a href="http://home.wavecable.com/~puzzleblog@astound.net/CrossNumberPuzzle090102.pdf"><img style="display:block; margin:0px auto 10px; text-align:center;cursor:pointer; cursor:hand;" src="http://home.wavecable.com/~puzzleblog@astound.net/uploaded_images/CrossNumber090102.gif" border="0" alt="Cross-Number Puzzle" /></a>As long as everyone is making New Year's resolutions, I hope you've made one to get more <b><i>mental</i></b> exercise. To start you off, here's a challenging "Cross Number Puzzle." The grid above is filled in like a traditional cross<i>word</i> puzzle, except every answer is a three-digit number (100-999) rather than a word. <i><b>Warning: </b>some of the clues may have you going in circles but there <i>is</i> a unique solution.</i>.<br /><br />Click here for a <a href="http://home.wavecable.com/~puzzleblog@astound.net/CrossNumberPuzzle090102.pdf">printable version of the puzzle</a>. And don't worry, you can get around to your other resolutions, like not procrastinating, and going to the gym later. Go sit on the sofa and work on a puzzle instead!<br /><br />Across:<br />1. 3 Down plus 5 Across<br />3. One-seventh of 8 Across<br />5. Half of 14 Across<br />6. A prime number<br />8. Seven times 3 Across<br />10. Twice 7 Down<br />12. A perfect square<br />14. 9 Down reversed<br />15. The sum of its own digits, times thirty-seven<br />16. A perfect square<br /><br />Down:<br />1. 13 Down plus 10 Down<br />2. Average of 9 Down and 14 Across<br />3. 1 Across minus 5 Across<br />4. A multiple of three<br />7. 16 Across minus 1 Across<br />9. 1 Across plus 5 Across<br />10. 13 Down plus three hundred<br />11. 12 Down minus 1 Down<br />12. Anagram of 4 Down<br />13. 1 Down minus 10 DownBlainehttp://www.blogger.com/profile/06379274325110866036noreply@blogger.com9tag:blogger.com,1999:blog-5730391.post-26955426459192380302008-12-26T19:15:00.003-08:002016-11-22T17:46:51.095-08:00Can you solve our Colorful Christmas Crossword (2008)?<a href="http://home.wavecable.com/~blaineblog@astound.net/xmas/2008"><img style="float:right; margin:0 10px 10px 0;cursor:pointer; cursor:hand;" src="http://home.wavecable.com/~puzzleblog@astound.net/uploaded_images/PuzzleThumb.jpg" alt="Christmas Puzzle 2008"/></a>This year's Christmas puzzle is a fun, themed crossword. And there is a <b><i>secret holiday surprise</i></b> if you are able to solve it. Take a look at our <a href="http://home.wavecable.com/~blaineblog@astound.net/xmas/2008">Christmas Puzzle for 2008</a>.<br /><br />Note: if you need some hints on solving the puzzle, post your questions here. But please don't give away any of the secrets, especially the final solution website address.Blainehttp://www.blogger.com/profile/06379274325110866036noreply@blogger.com3tag:blogger.com,1999:blog-5730391.post-3383304916843228142008-09-26T17:36:00.001-07:002016-11-22T17:46:53.909-08:00Friday Fun - Mini-Sudoku Puzzle (nine squares!)<a href="http://home.wavecable.com/~puzzleblog@astound.net/uploaded_images/mini-sudoku.gif"><img style="float:right; margin:0 10px 10px 0;cursor:pointer; cursor:hand;" src="http://home.wavecable.com/~puzzleblog@astound.net/uploaded_images/mini-sudoku-small.gif" border="0" alt="Mini-Sudoku (nine squares!)"/></a>For all of those that are tired of having to fill in a full Sudoku grid, here's a <a href="http://home.wavecable.com/~puzzleblog@astound.net/uploaded_images/mini-sudoku.gif">Mini-Sudoku Puzzle</a>. The goal is to fill the nine squares with just the digits 1 to 9. The only hints provided are the "L-block" hints at each corner. Each value tells you the sum of the five squares that make up the two adjacent edges.<br /><br /><i><b>Note: </b>This is <b>not</b> a magic square. You cannot make any assumptions about the totals of the rows, columns or diagonals.</i><br /><br />See how quickly you can come up with the unique solution. I'll probably post the answer next Friday. In the meantime, please don't reveal the answer so others can enjoy the puzzle too. Post comments on whether you find this puzzle easy, hard, fun or frustrating. I'd be interested in your solving techniques and times, too.Blainehttp://www.blogger.com/profile/06379274325110866036noreply@blogger.com3tag:blogger.com,1999:blog-5730391.post-36897290636630199502008-08-01T06:29:00.001-07:002016-11-22T17:46:55.646-08:00Friday Fun - Cycling on the Bridge<img style="float:right; margin:0 10px 10px 0;cursor:pointer; cursor:hand;" src="http://home.wavecable.com/~puzzleblog@astound.net/uploaded_images/BridgeBikes.jpg" border="0" alt="Cycling on the Bridge"/>Two bicyclists start cycling from opposite ends of a bridge. One cyclist is faster than the other and they meet at a point 2,000 feet from the nearest end. When each cyclist reaches the opposite end of the bridge, he takes a 15 minute rest break and then starts on his on return trip. The cyclists again meet 720 feet from the other end. Assuming each is cycling at a constant speed, how long is the bridge?<br /><br /><b>Note: </b><i>There is no mention of the actual speed of each cyclist, or the time that each takes but this problem is solvable. In fact, there is an elegant solution that could be understood by an elementary school student, with basic rules of addition and subtraction. It can also be solved the "hard" way. I'll post the elegant solution next week.</i><br /><br /><b>Edit: </b>I've provided an answer in the comments.Blainehttp://www.blogger.com/profile/06379274325110866036noreply@blogger.com10tag:blogger.com,1999:blog-5730391.post-21972264002359543352008-07-25T08:00:00.001-07:002016-11-22T17:46:55.983-08:00Friday Fun - How Long is the Ring Road around Iceland? - Answer<a href="http://home.wavecable.com/~puzzleblog@astound.net/uploaded_images/Iceland.jpg"><img style="float:left; margin:0 10px 10px 0;cursor:pointer; cursor:hand;" src="http://home.wavecable.com/~puzzleblog@astound.net/uploaded_images/icelandthumb.jpg" border="0" alt="Iceland Ring Road"/></a>We should be flying back home from Iceland about this time. Hopefully everyone has had fun with the puzzles while we have been gone. If you haven't had a chance to solve the puzzle about the Iceland Ring Road yet, take a look at last Friday's post and don't read any further. But if you want the answer, read on...<blockquote><b>A: </b>Let A be the speed of the first couple and B be the speed of the second couple. After an equivalent amount of time T, one couple has traveled AT miles and the other travels BT miles. For the return, the first couple now travels BT miles in 9 hours, while the other couple travels AT miles in 16 hours.<br /><br />A = BT/9<br />B = AT/16<br /><br />9A = BT<br />16B = AT<br /><br />T = 9A/B<br />T = 16B/A<br />9A^2 = 16B^2<br />Take the square root of both sides (which is okay because both are positive)<br />3A = 4B<br /><br />This tells us the ratio of their speeds is 4 to 3. In other words, over the same time, the faster couple will travel 4/7 of the ring road, the slower couple will travel 3/7. The difference is 120 miles. And if 1/7 is 120 miles, the whole road is 840 miles.</blockquote>Blainehttp://www.blogger.com/profile/06379274325110866036noreply@blogger.com2tag:blogger.com,1999:blog-5730391.post-42249675535062125832008-07-18T08:00:00.001-07:002016-11-22T17:46:56.352-08:00Friday Fun - How Long is the Ring Road around Iceland?<a href="http://home.wavecable.com/~puzzleblog@astound.net/uploaded_images/Iceland.jpg"><img style="float:left; margin:0 10px 10px 0;cursor:pointer; cursor:hand;" src="http://home.wavecable.com/~puzzleblog@astound.net/uploaded_images/icelandthumb.jpg" border="0" alt="Iceland Ring Road"/></a>My wife and I are taking a leisurely drive around Iceland on the Ring Road... at this point we should be a little more than half way on the East side of Iceland in <a href="http://www.cjwareing.net/images/07-09-11-egilstaddir-fjords-breddalsvik/index.html">Egilsstaðir</a>. However, I thought it might be fun to give you a little topical puzzle in honor of our trip.<blockquote><b>Q: </b>Two couples leave Reykjavik at exactly the same time traveling opposite directions on the Ring Road around Iceland. When they meet later, one couple has traveled 120 miles farther than the other. After a night's rest in a hotel and some refueling, the couples continue their respective drives. The first couple arrives back at Reykjavik 9 hours later, the second couple takes 16 hours. Assuming that each couple maintains the same constant speed each time they drive, how long is the Ring Road around Iceland?</blockquote>I'll post the answer next Friday.Blainehttp://www.blogger.com/profile/06379274325110866036noreply@blogger.com6tag:blogger.com,1999:blog-5730391.post-34239584386028020092008-07-11T00:54:00.001-07:002016-11-22T17:46:58.108-08:00How old is Mark?<img style="float:right; margin:0 10px 10px 0;cursor:pointer; cursor:hand;" src="http://home.wavecable.com/~puzzleblog@astound.net/uploaded_images/howold.gif" border="0" alt="How old is Mark?"/>For everyone that struggled with the pencil puzzle, here's another algebra puzzle to "stretch your neurons". Pay attention...<br /><br />The combined ages of Mark and Ann are forty-four years, and Mark is twice as old as Ann was when Mark was half as old as Ann will be when Ann is three times as old as Mark was when Mark was three times as old as Ann.<br /><br />How old is Mark?Blainehttp://www.blogger.com/profile/06379274325110866036noreply@blogger.com7