tag:blogger.com,1999:blog-57303912019-01-18T02:55:11.134-08:00Blaine's Puzzle BlogWeekly discussion on the NPR puzzler, brain teasers, math problems and more.Blainehttp://www.blogger.com/profile/06379274325110866036noreply@blogger.comBlogger63125tag:blogger.com,1999:blog-5730391.post-74382792796960710362018-11-18T06:05:00.001-08:002018-11-25T06:32:42.940-08:00NPR Sunday Puzzle (Nov 18, 2018): Taking the Next Logical Step<a href="https://www.npr.org/2018/11/18/669007480/sunday-puzzle-name-a-category">NPR Sunday Puzzle (Nov 18, 2018): Taking the Next Logical Step</a>: <blockquote><div class="separator" style="clear: both; text-align: center;"><a href="https://4.bp.blogspot.com/-dA6M9LjrPEk/W_F2eTWaBFI/AAAAAAADE14/PYlsghr8Qlk5hWvuM-26CzDE1snydQSsQCLcBGAs/s1600/welcome%2Bmat.png" imageanchor="1" style="clear: right; float: right; margin-bottom: 1em; margin-left: 1em;"><img border="0" src="https://4.bp.blogspot.com/-dA6M9LjrPEk/W_F2eTWaBFI/AAAAAAADE14/PYlsghr8Qlk5hWvuM-26CzDE1snydQSsQCLcBGAs/s200/welcome%2Bmat.png" width="200" height="200" data-original-width="625" data-original-height="625" /></a></div><b>Q: </b>In my trip to Europe two weeks ago I visited a friend in Amsterdam who literally has a puzzle on his doormat. Before you walk into his apartment, there's an original puzzle for you to solve. I was able to do it. See if you can. What number comes next in this series: 1, 2, 4, 8, 16, 23, 28?</blockquote>April 2nd, '07?<br/><br/><b>Edit: </b>In the OEIS (Online Encyclopedia of Integer Sequences), this is sequence <a href="http://oeis.org/A004207">A004207</a><blockquote>38 (Add the sum of the digits in the prior number, e.g. 16 + 1 + 6 = 23, 23 + 2 + 3 = 28, etc.)</blockquote>Blainehttp://www.blogger.com/profile/06379274325110866036noreply@blogger.com120tag:blogger.com,1999:blog-5730391.post-34479487137592232017-10-22T06:19:00.001-07:002017-10-26T12:08:05.908-07:00NPR Sunday Puzzle (Oct 22, 2017): Two Week Challenge: Move Two Numbers<a href="http://www.npr.org/2017/10/22/559206531/sunday-puzzle-read-into-it">NPR Sunday Puzzle (Oct 22, 2017): Two Week Challenge: Move Two Numbers</a>: <br /><blockquote><div class="separator" style="clear: both; text-align: center;"><a href="https://1.bp.blogspot.com/--DLMfHRNs_E/Weya5f16peI/AAAAAAACw20/m5KqxiC1LLAnykmd9wsVwvjGyujY5kVwQCLcBGAs/s1600/number%2Bpile.png" imageanchor="1" style="clear: right; float: right; margin-bottom: 1em; margin-left: 1em;"><img border="0" src="https://1.bp.blogspot.com/--DLMfHRNs_E/Weya5f16peI/AAAAAAACw20/m5KqxiC1LLAnykmd9wsVwvjGyujY5kVwQCLcBGAs/s200/number%2Bpile.png" width="200" height="134" data-original-width="954" data-original-height="640" /></a></div><b>Q: </b>This is a two-week challenge. Write down the equation:<br/>65 – 43 = 21.<br/>You'll notice that this is not correct. 65 minus 43 equals 22, not 21. The object is to move exactly two of the digits to create a correct equation. There is no trick in the puzzle's wording. In the answer, the minus and equal signs do not move.</blockquote>Keep it up; you still have another week to solve this.<br/><br/><b>Edit: </b>My hint last week was "I used to be an advocate...". I guess you could say if I'm no longer a proponent, I must be an ex-ponent. :) This week's clue is a hint to raise the numbers up.<blockquote><b>A: </b>65 - 4³ = 1² --> 65 - 64 = 1</blockquote>Blainehttp://www.blogger.com/profile/06379274325110866036noreply@blogger.com75tag:blogger.com,1999:blog-5730391.post-67381746483548425442017-10-15T06:27:00.004-07:002017-10-15T06:27:56.529-07:00NPR Sunday Puzzle (Oct 15, 2017): Two Week Challenge: Move Two Numbers<a href="http://www.npr.org/2017/10/15/557863670/sunday-puzzle-double-down">NPR Sunday Puzzle (Oct 15, 2017): Two Week Challenge: Move Two Numbers</a>: <br /><blockquote><a href="https://2.bp.blogspot.com/-zoVlI0kRBwg/WeNhcAI4BhI/AAAAAAACwxA/dWPE30XDB9kKDyq5WeNXfWmhgsEXCRlOQCLcBGAs/s1600/Number%2Bpile.jpg" imageanchor="1" style="clear: right; float: right; margin-bottom: 1em; margin-left: 1em;"><img border="0" data-original-height="480" data-original-width="852" height="113" src="https://2.bp.blogspot.com/-zoVlI0kRBwg/WeNhcAI4BhI/AAAAAAACwxA/dWPE30XDB9kKDyq5WeNXfWmhgsEXCRlOQCLcBGAs/s200/Number%2Bpile.jpg" width="200" /></a><b>Q: </b>This is a two-week challenge. Write down the equation:<br/>65 – 43 = 21.<br/>You'll notice that this is not correct. 65 minus 43 equals 22, not 21. The object is to move exactly two of the digits to create a correct equation. There is no trick in the puzzle's wording. In the answer, the minus and equal signs do not move.</blockquote>I used to be an advocate of more challenging number/math puzzles. But this one is <i>not</i> worthy of a two-week challenge.Blainehttp://www.blogger.com/profile/06379274325110866036noreply@blogger.com144tag:blogger.com,1999:blog-5730391.post-58003194585176654292017-01-22T06:27:00.000-08:002017-01-26T12:18:04.348-08:00NPR Sunday Puzzle (Jan 22, 2017): Think of a Number...<a href="http://www.npr.org/2017/01/22/511046359/youve-got-to-comb-together-to-solve-this-one">NPR Sunday Puzzle (Jan 22, 2017): Think of a Number...</a>: <blockquote><div class="separator" style="clear: both; text-align: center;"><a href="https://3.bp.blogspot.com/-b3y9x3i3CvU/WITA6aWt13I/AAAAAAAChck/IJp0MjoWHHo8-mBm7MSUgyOM-4CZnx8mQCLcB/s1600/MathNumbers.png" imageanchor="1" style="clear: right; float: right; margin-bottom: 1em; margin-left: 1em;"><img border="0" src="https://3.bp.blogspot.com/-b3y9x3i3CvU/WITA6aWt13I/AAAAAAAChck/IJp0MjoWHHo8-mBm7MSUgyOM-4CZnx8mQCLcB/s200/MathNumbers.png" width="200" height="125" /></a></div><b>Q: </b>This week's challenge is unusual. The numbers 5,000, 8,000, and 9,000 share a property that only five integers altogether have. Identify the property and the two other integers that have it.</blockquote>The hard part isn't figuring out the pattern, it's figuring out how we are supposed to extend it to find integers four and five.<br/><br/><b>Edit: </b>The title contains each of the vowels (a,e,i,o,u) exactly once. The 4 and 5 in my hint refer to the number of digits in the two other answers.<blockquote><b>A: </b>When spelled out in English, the numbers contain the 5 vowels (a, e, i, o, u, but not y) exactly once.<br/>The other two numbers would be 6,010 (s<b>i</b>x th<b>ou</b>s<b>a</b>nd t<b>e</b>n) and 10,006 (t<b>e</b>n th<b>ou</b>s<b>a</b>nd s<b>i</b>x).<br/><br/><i>I discounted answers like 80,000 and 90,000 which also contain y and wouldn't preclude 26,000 as an answer.</i></blockquote>Blainehttp://www.blogger.com/profile/06379274325110866036noreply@blogger.com139tag:blogger.com,1999:blog-5730391.post-24807924427183870652016-10-16T05:24:00.001-07:002016-11-13T06:26:01.993-08:00NPR Sunday Puzzle (Oct 16, 2016): Five, Four, Three, Two, One, Blast-off!<a href="http://www.npr.org/2016/10/16/497861459/when-it-comes-to-the-puzzle-dont-mail-it-in-just-exchange-a-letter">NPR Sunday Puzzle (Oct 16, 2016): Five, Four, Three, Two, One, Blast-off!</a>: <blockquote><div class="separator" style="clear: both; text-align: center;"><a href="https://2.bp.blogspot.com/-KwVSLpeDX3k/WANwrpsIu-I/AAAAAAACdko/7iCu3O6WGm8Ag35pIlw1cfSyR4M3uwO8gCLcB/s1600/Blast%2BOff.png" imageanchor="1" style="clear: right; float: right; margin-bottom: 1em; margin-left: 1em;"><img border="0" src="https://2.bp.blogspot.com/-KwVSLpeDX3k/WANwrpsIu-I/AAAAAAACdko/7iCu3O6WGm8Ag35pIlw1cfSyR4M3uwO8gCLcB/s200/Blast%2BOff.png" width="181" height="200" /></a></div><b>Q: </b>This is a <b><i>two-week</i></b> challenge. Take the digits 5, 4, 3, 2 and 1, in that order. Using those digits and the four arithmetic signs — plus, minus, times and divided by — you can get 1 with the sequence 5 - 4 + 3 - 2 - 1. You can get 2 with the sequence (5 - 4 + 3 - 2) x 1.<br /><br />The question is ... how many numbers from 1 to 40 can you get using the digits 5, 4, 3, 2, and 1 in that order along with the four arithmetic signs?<br /><br />You can group digits with parentheses, as in the example. There are no tricks to this, though. It's a straightforward puzzle. How many numbers from 1 to 40 can you get — and, specifically, what number or numbers can you not get? Will Shortz will reveal his solution in <b><i>two</i></b> weeks.</blockquote>Enjoy the two-week math challenge. Feel free to post *how many* numbers you can create, but just not specifics on which ones and with what expressions.<br/><br/><b>Edit: </b>A bit of a trick here. You can create 39 numbers, but not 39. :)<blockquote><b>A: </b>Without resorting to tricks, you can write expressions for all but number 39.</blockquote>Blainehttp://www.blogger.com/profile/06379274325110866036noreply@blogger.com268tag:blogger.com,1999:blog-5730391.post-17156396997915279872016-05-29T07:04:00.000-07:002016-06-05T06:22:17.271-07:00NPR Sunday Puzzle (May 29, 2016): Game, Set and Match<a href="http://www.npr.org/2016/05/29/479889322/to-solve-this-puzzle-you-must-name-a-person-of-fame">NPR Sunday Puzzle (May 29, 2016): Game, Set and Match</a>: <blockquote><div class="separator" style="clear: both; text-align: center;"><a href="https://3.bp.blogspot.com/-1MNfxJCFIPM/V0r2_8bCctI/AAAAAAACWME/7JmwFqnr0gQus7KsJi0YpP7fubzYU04kwCLcB/s1600/nicolas-mahut-right-posing-for-a-photo-next-to-the-scoreboard-following-their-record-breaking-men-s-singles-match-at-the-all-england-lawn-tennis-champ-157156760.jpg" imageanchor="1" style="clear: right; float: right; margin-bottom: 1em; margin-left: 1em;"><img border="0" src="https://3.bp.blogspot.com/-1MNfxJCFIPM/V0r2_8bCctI/AAAAAAACWME/7JmwFqnr0gQus7KsJi0YpP7fubzYU04kwCLcB/s200/nicolas-mahut-right-posing-for-a-photo-next-to-the-scoreboard-following-their-record-breaking-men-s-singles-match-at-the-all-england-lawn-tennis-champ-157156760.jpg" /></a></div><b>Q: </b>What is the most consecutive points a tennis player can lose and still win a best-of-five-sets match? There's no trick. It's a straightforward question. The modern tennis tiebreaker rule does not come into play.</blockquote>So basically you've got to figure how many sets you can get behind and still be able to catch up?<br/><br/><b>Edit: </b>You'll be almost ready to lose the 19th consecutive set. Add 57 (hint: Heinz "ketchup") and you have 76 points.<blockquote><b>A: 76 points. </b><i>You start by going up 5 games and 40-love (3 points). Your opponent then needs 5 points to win that game, and 6 more games (24 points) to win the first set 7-5, 24 points to win the second set 6-0 and then another 23 points to be at 5 games and 40-love in the third set</i></blockquote>Blainehttp://www.blogger.com/profile/06379274325110866036noreply@blogger.com163tag:blogger.com,1999:blog-5730391.post-19329825247490731552015-11-01T06:53:00.000-08:002015-11-05T11:58:26.574-08:00NPR Sunday Puzzle (Nov 1, 2015): And Three Nines are Twenty<a href="http://www.npr.org/2015/11/01/453463299/you-get-to-change-some-4-letter-words-in-this-weeks-puzzle">NPR Sunday Puzzle (Nov 1, 2015): And Three Nines are Twenty</a>: <div class="separator" style="clear: both; text-align: center;"><a href="http://3.bp.blogspot.com/-LCRpocKqHH0/VjYnFYX11oI/AAAAAAACOZI/8mjGqL0cbWw/s1600/Refrigerator%2BMath.jpg" imageanchor="1" style="clear: right; float: right; margin-bottom: 1em; margin-left: 1em;"><img border="0" src="http://3.bp.blogspot.com/-LCRpocKqHH0/VjYnFYX11oI/AAAAAAACOZI/8mjGqL0cbWw/s200/Refrigerator%2BMath.jpg" /></a></div><blockquote><b>Q: </b>This is one of the "lost" puzzles of Sam Loyd, the great American puzzlemaker from the 19th and early 20th centuries. It's from an old magazine with a Sam Loyd puzzle column. The object is to arrange three 9s to make 20. There is no trick involved. Simply arrange three 9s, using any standard arithmetic signs and symbols, to total 20. How can it be done?</blockquote>The whole point is there are no tricks involved. So you don't need to flip numbers upside down or involve higher order math like square roots or factorials... at least my solution doesn't need those.<br/><br/><b>Edit: </b>My hint was <i>point</i> as in decimal point.<blockquote><b>A: </b>(9 + 9)/.9 = 20</blockquote>Blainehttp://www.blogger.com/profile/06379274325110866036noreply@blogger.com213tag:blogger.com,1999:blog-5730391.post-32275843514084379532015-03-29T09:19:00.000-07:002015-04-02T12:01:24.791-07:00NPR Sunday Puzzle (Mar 29, 2015): May I Have Your Number?<a href="http://www.npr.org/2015/03/29/395983602/for-this-puzzle-watch-your-words">NPR Sunday Puzzle (Mar 29, 2015): May I Have Your Number?</a>: <blockquote><div class="separator" style="clear: both; text-align: center;"><a href="http://1.bp.blogspot.com/-Gr9xRUM83SU/VRgk3Onsk6I/AAAAAAAB_ls/ktjpKzborYk/s1600/calculator.png" imageanchor="1" style="clear: right; float: right; margin-bottom: 1em; margin-left: 1em;"><img border="0" src="http://1.bp.blogspot.com/-Gr9xRUM83SU/VRgk3Onsk6I/AAAAAAAB_ls/ktjpKzborYk/s200/calculator.png" /></a></div><b>Q: </b>This week's challenge is a little tricky. Given a standard calculator with room for 10 digits, what is the largest whole number you can register on it?</blockquote>I must have misdialed while trying to phone a friend; I got Ed Asner instead.<br/><br/><b>Edit: </b>I was trying to contact the mathematician Edward Kasner who, along with 9-year-old nephew Milton came up with the name "googol" for the large number 10¹⁰⁰.<blockquote><div class="separator" style="clear: both; text-align: center;"><a href="http://1.bp.blogspot.com/-QhC9wX7V_fE/VR2RRsYbNoI/AAAAAAAB_mI/ETOon-hObag/s1600/Calculator%2BGoogol.jpg" imageanchor="1" style="clear: right; float: right; margin-bottom: 1em; margin-left: 1em;"><img border="0" src="http://1.bp.blogspot.com/-QhC9wX7V_fE/VR2RRsYbNoI/AAAAAAAB_mI/ETOon-hObag/s200/Calculator%2BGoogol.jpg" /></a></div><b>A: </b>If you type 706006 (or 709009) and turn the calculator upside down, it spells gOOgOL (or GOOGOL). That's a 1 followed by 100 zeroes and is bigger than any regular number you could enter using 10 digits.</blockquote>Blainehttp://www.blogger.com/profile/06379274325110866036noreply@blogger.com153tag:blogger.com,1999:blog-5730391.post-37928743186760356772014-10-19T08:17:00.000-07:002014-10-26T07:08:20.437-07:00NPR Sunday Puzzle (Oct 19, 2014): Time To Flex Those Math Muscles<a href="http://www.npr.org/2014/10/19/357183310/time-to-flex-those-math-muscles">NPR Sunday Puzzle (Oct 19, 2014): Time To Flex Those Math Muscles</a>: <blockquote><div class="separator" style="clear: both; text-align: center;"><a href="http://4.bp.blogspot.com/-oJuzokXEXXw/VEPUOGTt01I/AAAAAAAAMHM/5KElLyEffuY/s1600/blue-eye.jpg" imageanchor="1" style="clear: right; float: right; margin-bottom: 1em; margin-left: 1em;"><img border="0" src="http://4.bp.blogspot.com/-oJuzokXEXXw/VEPUOGTt01I/AAAAAAAAMHM/5KElLyEffuY/s200/blue-eye.jpg" /></a></div><b>Q: </b>The following challenge is based on a puzzle from a Martin Gardner book, that may not be well-known. Out of a regular grade school classroom, two students are chosen at random. Both happen to have blue eyes. If the odds are exactly 50-50 that two randomly chosen students in the class will have blue eyes: How many students are in the class?</blockquote>It's going to be hard to provide hints to the answer this week.<br/><br/><b>Edit: </b>My hint was "Be" which is the symbol for Berylium, element 4 on the periodic table. While 4 is a technically correct answer (3 blue-eyed children --> 3/4 x 2/3 = 1/2) the intended answer was for a more typical class size.<blockquote><b>A: </b>21 students (15 with blue eyes and 7 without).<br/><br/>The probability the first child has blue eyes is 15/21 or 5/7. Once that child is taken out of consideration, there are 14 children with blue eyes out of 20 so the probability is 7/10. Multiplying that together, we have 5/7 x 7/10 = 5/10 = 1/2</blockquote>Blainehttp://www.blogger.com/profile/06379274325110866036noreply@blogger.com153tag:blogger.com,1999:blog-5730391.post-77815390296665514232014-08-17T09:05:00.001-07:002014-08-22T23:02:03.721-07:00NPR Sunday Puzzle (Aug 17, 2014): Target Practice<a href="http://www.npr.org/2014/08/17/340944712/is-there-an-echo-in-here">NPR Sunday Puzzle (Aug 17, 2014): Target Practice</a>: <blockquote><div class="separator" style="clear: both; text-align: center;"><a href="http://2.bp.blogspot.com/-xT0r2ImLLR8/U_DSnZ1YqDI/AAAAAAAALrQ/O6sf1cqd1TM/s1600/LloydTarget.png" imageanchor="1" style="clear: right; float: right; margin-bottom: 1em; margin-left: 1em;"><img border="0" src="http://2.bp.blogspot.com/-xT0r2ImLLR8/U_DSnZ1YqDI/AAAAAAAALrQ/O6sf1cqd1TM/s200/LloydTarget.png" /></a></div><b>Q: </b>You have a target with six rings, bearing the numbers 16, 17, 23, 24, 39, and 40. How can you score exactly 100 points, by shooting at the target.</blockquote>Finally a math puzzle, but then it has to be one that isn't very challenging. If you are having trouble, just keep firing arrows; you'll hit it eventually.<br/><br/><b>Edit: </b>You'll need the full complement of 6 arrows.<blockquote><b>A: </b>16 + 16 + 17 + 17 + 17 + 17 = 100</blockquote>Blainehttp://www.blogger.com/profile/06379274325110866036noreply@blogger.com78tag:blogger.com,1999:blog-5730391.post-22568665886518177832013-08-22T12:45:00.000-07:002013-08-22T12:46:29.302-07:00NPR Sunday Puzzle (Aug 18, 2013): Roman Numeral XXXVIII<div class="separator" style="clear: both; text-align: center;"><a href="http://2.bp.blogspot.com/-dXHRTchgeM8/UhDYlpkDiEI/AAAAAAAALFk/lFjgF6bq5VI/s1600/RomanNumeralClock.jpg" imageanchor="1" style="clear: right; float: right; margin-bottom: 1em; margin-left: 1em;"><img border="0" height="132" src="http://2.bp.blogspot.com/-dXHRTchgeM8/UhDYlpkDiEI/AAAAAAAALFk/lFjgF6bq5VI/s200/RomanNumeralClock.jpg" width="200" /></a></div><a href="http://www.npr.org/2013/08/18/213006718/a-matter-of-succession">NPR Sunday Puzzle (Aug 18, 2013): Roman Numeral XXXVIII</a>: <blockquote><b>Q: </b>The Roman numeral for 38 is XXXVIII. What is special or unusual about this Roman numeral that sets it apart from every other Roman numeral that can be written?</blockquote>I'm sure I'll figure this out next month when I go to New Jersey.<br/><br/><b>Edit: </b>If you sort the months alphabetically, September comes last. Similarly, if you sort the state capitals, Trenton comes last<blockquote><b>A: </b>XXXVIII sorts last alphabetically.</blockquote>Blainehttp://www.blogger.com/profile/06379274325110866036noreply@blogger.com107tag:blogger.com,1999:blog-5730391.post-11483091506135281592013-03-07T12:00:00.001-08:002018-01-23T01:52:01.465-08:00NPR Sunday Puzzle (Mar 3, 2013): Dinner Party Musical Chairs<a href="http://1.bp.blogspot.com/-2lN5hvx4FTA/UTYywGpipwI/AAAAAAAAK8s/ud0eN9zqS-E/s1600/Round+Table.png" imageanchor="1" style="clear: right; float: right; margin-bottom: 1em; margin-left: 1em;"><img border="0" height="200" src="http://1.bp.blogspot.com/-2lN5hvx4FTA/UTYywGpipwI/AAAAAAAAK8s/ud0eN9zqS-E/s200/Round+Table.png" width="200" /></a><a href="http://www.npr.org/2013/03/03/173344811/perfectly-puzzling">NPR Sunday Puzzle (Mar 3, 2013): Dinner Party Musical Chairs</a>: <br /><blockquote><b>Q: </b>Eight people are seated at a circular table. Each person gets up and sits down again — either in the same chair or in the chair immediately to the left or right of the one they were in. How many different ways can the eight people be reseated?</blockquote>For this puzzle, I think we have to assume each seat position and person is unique. Also, I assume Will wants seating arrangements where each person has their own chair (no sharing). What I don't see, is why the table has to be circular. Couldn't it be square and we could still figure out how to move left or right?<br/><br/><b>Edit: </b>The first case that might get overlooked is everyone returning to their original seat. The next two cases are where all 8 people move clockwise or counter-clockwise one seat. There can't be any other cycles involving more than two people because that would require someone to move more than one seat, so the remaining cases involve neighboring "couples" swapping seats while others stay still. All that is required is to enumerate the ways to swap couples. <blockquote><b>A: </b>There are <a href="https://drive.google.com/file/d/1kMFV4PrPX6NnLAdIoITS22G5wnZzh0LA/view?usp=sharing">49 ways that 8 people could stand up and be reseated</a> (link to PDF containing diagrams). Incidentally, the <i>Online Encyclopedia of Integer Sequences</i> has the <a href="http://oeis.org/A048162">answers for various table sizes (A0048162 = 1, 2, 6, 9, 13, 20, 31, 49...)</a> which confirms the answer for 8 people is 49 ways.</blockquote> Blainehttp://www.blogger.com/profile/06379274325110866036noreply@blogger.com157tag:blogger.com,1999:blog-5730391.post-12827489075381978262013-01-10T12:00:00.000-08:002013-01-10T12:02:10.509-08:00NPR Sunday Puzzle (Jan 6, 2013): Sam Loyd Puzzle (16 boxes, 10 markers)<a href="http://www.npr.org/2013/01/06/168698679/scrambling-to-ring-in-the-new-year">NPR Sunday Puzzle (Jan 6, 2013): Sam Loyd Puzzle (16 boxes, 10 markers)</a>: <blockquote><a href="http://3.bp.blogspot.com/-NxSjOAkM8wQ/UOmsn0z4xJI/AAAAAAAAK44/-dM68u9I-cU/s1600/GridDiagonals.png" imageanchor="1" style="clear:right; float:right; margin-left:1em; margin-bottom:1em"><img border="0" height="200" width="200" src="http://3.bp.blogspot.com/-NxSjOAkM8wQ/UOmsn0z4xJI/AAAAAAAAK44/-dM68u9I-cU/s200/GridDiagonals.png" alt="16 square grid"/></a><b>Q: </b>This challenge appeared in a puzzle column in the Woman's Home Companion in January 1913, exactly 100 years ago. Draw a square that is four boxes by four boxes per side, containing altogether 16 small boxes and 18 lines (across, down and diagonal). There are 10 ways to have four boxes in a line — four horizontal rows, four vertical columns, plus the two long diagonals. There are also eight other shorter diagonals of two or three squares each. The object is to place markers in 10 of the boxes so that as many of the lines as possible have either two or four markers. What is the maximum number of lines that can have either two or four markers, and how do you do it?</blockquote>Sorry, I needed my full 8 hours of sleep so wasn't able to post earlier, but I'm awake now. No doubt people are going to find this a tricky puzzle, so I added a diagram to help out. Apart from reflections or rotations, I've found a single solution that maximizes the number of lines. Please don't mention how many lines are involved until after the deadline. For those that are mathematically inclined, there are <a href="http://www.wolframalpha.com/input/?i=16+choose+10">8008</a> ways to place 10 markers into 16 squares and, accounting for symmetry, it shouldn't be too hard to brute force the answer. :)<br/><br/><b>Edit: </b>My hints above were to the number of lines being 16. Being asleep for 8 hours leaves 16 waking hours. The group "No Doubt" has a song called "Sixteen" on their album "Tragic Kingdom". Also the reference to the number of possible arrangements (8008) hinted at 8 horizontal/vertical lines and 8 diagonal lines. Finally, in a post I mentioned the Beatles' song "Taxman" which was parodied by Weird Al as "Pac Man"... I think the answer looks like Pac Man eating a dot.<blockquote><a href="http://1.bp.blogspot.com/-21exv_Mwf8w/UO8doDFLg8I/AAAAAAAAK5Q/EhtYG7WZogs/s1600/SixteenLines.png" imageanchor="1" style="clear:right; float:right; margin-left:1em; margin-bottom:1em"><img border="0" height="200" width="200" src="http://1.bp.blogspot.com/-21exv_Mwf8w/UO8doDFLg8I/AAAAAAAAK5Q/EhtYG7WZogs/s200/SixteenLines.png" /></a><b>A: </b>16 lines in one of 4 symmetric arrangements: <br/>Solution A: 1, 3, 7, 8, 9, 10, 11, 12, 14, 15<br/>Solution B: 2, 3, 5, 6, 7, 8, 9, 10, 14, 16<br/>Solution C: 2, 3, 5, 6, 7, 8, 11, 12, 13, 15<br/>Solution D: 2, 4, 5, 6, 9, 10, 11, 12, 14, 15 </blockquote> Blainehttp://www.blogger.com/profile/06379274325110866036noreply@blogger.com102tag:blogger.com,1999:blog-5730391.post-1448223169663694272012-10-11T12:00:00.000-07:002012-10-11T11:59:30.415-07:00NPR Sunday Puzzle (Oct 7, 2012): Hexagon Diagonals - Count the Triangles<a href="https://vimeo.com/51197832" imageanchor="1" style="clear:right; float:right; margin-left:1em; margin-bottom:1em"><img border="0" height="200" width="172" src="http://1.bp.blogspot.com/-xiX5ldzgZVM/UHIIvQVn11I/AAAAAAAAKv8/7DDrd34U9q8/s200/HexagonDiagonals.png" alt="Hexagon Diagonals (less one)"/></a><a href="http://www.npr.org/2012/10/06/162444203/frog-stuck-in-your-c-r-o-a-t">NPR Sunday Puzzle (Oct 7, 2012): Hexagon Diagonals - Count the Triangles</a>: <br/><br/><blockquote><b>Q: </b>Draw a regular hexagon, and connect every pair of vertices except one. The pair you don't connect are not on opposite sides of the hexagon, but along a shorter diagonal. How many triangles of any size are in this figure?</blockquote>The diagram in the upper right should help. I've removed one diagonal. It looks like a cool cube, don't you think?<br/><br/><b>Edit: </b>The words "cool cube" were a double hint. First, the diagram I drew reminded me of the isometric cubes in the <a href="http://en.wikipedia.org/wiki/Q*bert">Q*Bert video game</a> which was released in 19<b>82</b>. Additionally, if you cube the answer (82^3) you get 551368. I think the result is cool because 55+13=68. As a final clue, in several of my comments, I used the word "lead" which happens to be 82 on the Periodic Table of Elements.<blockquote><b>A: </b><a href="https://vimeo.com/51197832">82 Triangles</a> - be sure to watch the <a href="https://vimeo.com/51197832">video</a> for an explanation of the answer.</blockquote>Blainehttp://www.blogger.com/profile/06379274325110866036noreply@blogger.com88tag: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-2279265264096044802012-06-03T23:11:00.001-07:002018-01-23T01:54:55.521-08:00GeekDad Puzzle of the Week: Waffle Cuts<a href="http://www.wired.com/geekdad/2012/06/geekdad-puzzle-of-the-week-solution-waffle-cuts/">GeekDad Puzzle of the Week: Waffle Cuts</a>: <a href="https://drive.google.com/file/d/1fgzZryjCBRPpHYVayoE5Au3AzN2KaqjO/view?usp=sharing" imageanchor="1" style="clear:right; float:right; margin-left:1em; margin-bottom:1em"><img border="0" height="98" width="98" src="http://3.bp.blogspot.com/-IF8wudlJO2Q/T8oZKaZFYYI/AAAAAAAAKls/WNIch4YVyUQ/s320/Waffle.png" alt="Waffle, no cuts" /></a><blockquote><b>Q: </b>If we only cut along the ridges of a circular waffle, and if each cut traverses the waffle in a straight line from edge to edge, how many different ways can the waffle be cut?<br/><i><b>Note: </b>rotations, horizontal flips, and vertical flips of a set of cuts should only be counted once.</i></blockquote>Given that there are 6 places to cut vertically and 6 places to cut horizontally, that's a total of 12 cut lines. If you allow for any combination of these 12 lines to be cut or not, you have a total of 2^12 = 4096 ways to divide the waffle. But of course, the puzzle asks for the number of <b><i>unique</i></b> ways to cut the waffle, not including any mirrored or rotationally symmetric sets of cuts.<br/><br/>After the official answer to the puzzle is posted, I'll post my solution here.<br/><br/><b>Edit: </b>The solution is posted, but just the number without any detail. Also, I disagree with their counting of the "no cuts at all" solution as one of the ways to "cut" the waffle. In any case, a full detailing of my solution along with an enumeration of all <b>665 (or 666)</b> ways to uniquely cut the waffle can be found in <a href="https://drive.google.com/file/d/1fgzZryjCBRPpHYVayoE5Au3AzN2KaqjO/view?usp=sharing">Blaine's Solution to the GeekDad Waffle Puzzle</a>.Blainehttp://www.blogger.com/profile/06379274325110866036noreply@blogger.com2tag:blogger.com,1999:blog-5730391.post-35659360413084035432012-05-26T21:48:00.000-07:002012-06-02T06:55:43.325-07:00GeekDad Puzzle of the Week Answer: That Darn Achilles<a href="http://www.wired.com/geekdad/2012/05/gpotw-answer-that-darn-achilles/">That Darn Achilles</a>: <blockquote><b>Q: </b>You, Paris, have the luxury of launching an arrow at faraway Achilles either from the ramparts above the Hesperian Gate at a height of exactly 8 meters. Or you can stand atop Priam’s palace. This gains you another 7 meters of launch height, but it costs you 15 meters of horizontal distance. If the arrow leaves your bow at a somewhat modest 70 meters per second, are you best taking your pot-shot at far-off Achilles from the ramparts or the palace? Which perch offers the farthest reach?</blockquote><a href="http://3.bp.blogspot.com/-MHgjb5Jfl3Y/T8MA-Pfkg2I/AAAAAAAAKlc/c4mjuQVeNOw/s1600/AchillesArrowTrajectories.png" imageanchor="1" style="clear: right; float: right; margin-bottom: 1em; margin-left: 1em;"><img border="0" height="58" src="http://3.bp.blogspot.com/-MHgjb5Jfl3Y/T8MA-Pfkg2I/AAAAAAAAKlc/c4mjuQVeNOw/s320/AchillesArrowTrajectories.png" width="160" /></a><br />I had to look up the formulas for determining the maximum range of a projectile when fired on uneven ground. The following page was invaluable.<br /><a href="http://en.wikipedia.org/wiki/Range_of_a_projectile">Wikipedia: Range of a Projectile</a><br /><br />Because several things weren't stated, I'm going to assume that we can use acceleration of gravity (g) at sea level, we can assume no wind resistance and also assume that the ground is level between the target and the firing point (except for the elevation change and horizontal offset provided by the ramparts (0,8) and the palace (-15, 15).<br /><br />While the ideal case (firing on even ground) results in an optimal angle of 45°, when you are firing from a height, then you want to angle down slightly to maximize distance. I won't bore you with going through the details on that page, but basically there's an equation for the horizontal and vertical positions at time t, given an initial angle (theta) and velocity v. You can then set the final height to be 0 and solve for t. Using that you can get an equation for distance given an angle and by taking the derivative and setting it to zero, you can get a formula for the optimal angle to get the longest distance.<br /><br />Rampart (x0,y0) = (0,8)<br />Palace (x0,y0) = (-15,15)<br />Velocity (v) = 70 m/s<br />Gravity (g) = 9.80665 m/s^2<br /><br />Optimal angle (θ) = cos-1 [ √(2*g*y0 + v^2) / (2*g*y0 + 2v^2) ]<br /><br />Rampart angle (θ) = 0.777519 Radians or 44.54853°<br />Palace angle (θ) = 0.7708234 Radians or 44.164927°<br />Distance (d) = [ v*cos θ [v*sin θ + √((v*sin θ)^2 + 2*g*y0) ] /g + x0<br /><br />Rampart distance = 507.5979m<br />Palace distance = 499.44231m<br /><br /><blockquote><b>A: </b>When firing from the ramparts (8m), the optimal angle is around 44.55° and will net you a distance of 507.6 meters. <br /><br />When firing from the palace (15m), the optimal angle is around 44.16° but because of the -15m offset you only reach 499.4 meters.</blockquote>Blainehttp://www.blogger.com/profile/06379274325110866036noreply@blogger.com0tag:blogger.com,1999:blog-5730391.post-30444567485932666532012-05-13T21:31:00.000-07:002012-06-02T06:55:43.322-07:00GeekDad Puzzle of the Week Answer: Coffee Conundrum II<a href="http://www.wired.com/geekdad/2012/05/puzzle-answer-coffee-conundrum-ii/">Coffee Conundrum II</a>: <blockquote><b>Q: </b>Each cup of coffee gives me a jolt and then the jolt decays across time according to the following equation (t in minutes): Jitters=10-[(t-10)^2]/10<br />So at minute 10 after consuming a cup of coffee (which, for the purposes of this puzzle happens instantly), I reach a maximum of 10 jitters. At a combined 20 jitters, I go catatonic. With what frequency can I instantly consume coffee without the combined jitters passing this important tipping point?</blockquote><a href="http://1.bp.blogspot.com/-jnrs85mlI4Y/T8L-o9oFR3I/AAAAAAAAKlU/PceccxElFZk/s1600/Coffee%2BJitters.png" imageanchor="1" style="clear:right; float:right; margin-left:1em; margin-bottom:1em"><img border="0" height="85" width="160" src="http://1.bp.blogspot.com/-jnrs85mlI4Y/T8L-o9oFR3I/AAAAAAAAKlU/PceccxElFZk/s320/Coffee%2BJitters.png" /></a>The diagram shows the exact tipping point situation. Focus on the red curve which is the second cup of coffee. Right at the middle of the parabola, the second cup is at its maximum (10 jitters). At this point, the effects of the first cup are diminishing and the effects of the third cup are increasing. Notice, because of symmetry, they have the exact same value when they cross. The tipping point will be when cup 1 is contributing 5 jitters, cup 2 is at its maximum of 10 jitters and cup 3 is also contributing 5 jitters (5+10+5 = 20). (Note you can extend this diagram out to 4 cups, 5 cups, etc. but you'll never have more than 3 cups contributing to the total jitters at once. If you did, you'd definitely be over the tipping point of 20.)<br /><br />As stated in the puzzle, the height of one parabolic curve is given by the formula:<br />Jitters = 10 - [ ( t - 10 ) ^ 2 ] / 10<br /><br />Solving this for a Jitters value of 5 we have:<br />5 = 10 - [ ( t - 10 ) ^ 2] / 10<br />-5 = -[ ( t - 10 ) ^ 2] / 10<br />5 = [ ( t - 10 ) ^ 2] / 10<br />50 = ( t - 10 )^2<br />t - 10 = √50<br />t = 10 ± √50<br />t = 10 ± 5√2<br /><br />That means that the coffee Jitters are at a height of 5 either 5√2 seconds before the peak or 5√2 seconds after the peak. This also happens to be the minimum frequency between cups (5√2 seconds or approximately 7.071068 seconds apart) to avoid jitters.<br /><br />If you can handle exactly 20 jitters without going catatonic, then you could handle drinking cups of coffee every 5√2 seconds (≈7.071068 seconds)<br /><br />If you must stay under 20 jitters, then you would have to pick a frequency just over every 5√2 seconds.<blockquote><b>A: </b>Frequency of 5√2 seconds (or approximately every 7.0710678118654752440084436210484903928483593768847403 seconds)</blockquote>Blainehttp://www.blogger.com/profile/06379274325110866036noreply@blogger.com0tag:blogger.com,1999:blog-5730391.post-54763412231259273192012-03-11T22:02:00.000-07:002012-06-02T06:55:43.327-07:00GeekDad Puzzle of the Week Answer: Bigger than Googol!<a href="http://www.wired.com/geekdad/2012/03/geekdad-puzzle-of-the-week-solution-bigger-then-googol/">Bigger than Googol!</a>: <blockquote><b>Q: </b>What is the first Fibonacci number bigger than a googol? </blockquote>This was relatively easy in Haskell:<br /><br />let fib = 0 : 1 : zipWith (+) fib (tail fib)<br />let answer = [ (n, fib!!n) | n <- [1..1000], (fib!!(n-1) <= 10^100) && (fib!!n > 10^100) ]<br />answer<br />(481,14913169640232740127827512057302148063648650711209401966150219926546779697987984279570098768737999681)<br /><br /><blockquote><b>A: </b>F(481) is the first Fibonacci number greater than googol. The numerical value is given above, but just for fun here's that number in words:<br />fourteen duotrigintillion, nine hundred thirteen untrigintillion, one hundred sixty-nine trigintillion, six hundred forty novemvigintillion, two hundred thirty-two octovigintillion, seven hundred forty septemvigintillion, one hundred twenty-seven sesvigintillion, eight hundred twenty-seven quinquavigintillion, five hundred twelve quattuorvigintillion, fifty-seven tresvigintillion, three hundred two duovigintillion, one hundred forty-eight unvigintillion, sixty-three vigintillion, six hundred forty-eight novemdecillion, six hundred fifty octodecillion, seven hundred eleven septendecillion, two hundred nine sexdecillion, four hundred one quindecillion, nine hundred sixty-six quattuordecillion, one hundred fifty tredecillion, two hundred nineteen duodecillion, nine hundred twenty-six undecillion, five hundred forty-six decillion, seven hundred seventy-nine nonillion, six hundred ninety-seven octillion, nine hundred eighty-seven septillion, nine hundred eighty-four sextillion, two hundred seventy-nine quintillion, five hundred seventy quadrillion, ninety-eight trillion, seven hundred sixty-eight billion, seven hundred thirty-seven million, nine hundred ninety-nine thousand, six hundred eighty-one</blockquote>Blainehttp://www.blogger.com/profile/06379274325110866036noreply@blogger.com0tag:blogger.com,1999:blog-5730391.post-21078221933943783682012-02-02T13:00:00.000-08:002012-02-02T19:56:42.638-08:00NPR Sunday Puzzle (Jan 29, 2012): An Equation for 2012<a href="http://www.npr.org/2012/01/29/146034893/this-puzzle-is-the-pits">NPR Sunday Puzzle (Jan 29, 2012): An Equation for 2012</a>: <blockquote><b>Q: </b>Write the digits from 1 to 9 in a line. If you put a plus sign after the 2, a times sign after the 4, and plus signs after the 6 and 8, the line shows 12 + 34 x 56 + 78 + 9, which equals 2003. That's nine years off from our current year 2012. This example uses four arithmetic symbols. The object is to use just three of the following arithmetic operations: addition, subtraction, multiplication and division, in a line from 1 to 9 to get 2012 exactly. The operations should be performed in order from left to right. There are no tricks to this puzzle. Can you do it?</blockquote>I was just about to retire for the evening, but I figured you might need some assistance in solving the puzzle, so your help is... Gee, how do I give you a hint to a math puzzle?<br/><br/><b>Edit: </b>The hints were "retire" (Social Security Administration = SSA = subtract, subtract, add) and "assistance" and "help" (411 = number of digits to group together, with 3 being assumed for the remaining digits).<blockquote><b>A: </b>1234 - 5 - 6 + 789 = 2012</blockquote>Blainehttp://www.blogger.com/profile/06379274325110866036noreply@blogger.com56tag:blogger.com,1999:blog-5730391.post-18556286189544043912011-11-17T12:00:00.000-08:002011-11-17T12:02:55.760-08:00NPR Sunday Puzzle (Nov 13, 2011): What Comes Next?<a href="http://www.npr.org/2011/11/13/142276105/a-four-letter-word-for-capital-city">NPR Sunday Puzzle (Nov 13, 2011): What Comes Next?</a>: <br /><blockquote><b>Q: </b>What number comes next in the following series: 2, 4, 6, 9, 11, 15, 20, 40, <i>51*, 55*,</i> 60 and 90?</blockquote>See, I thought I had the answer to this, but if so, there are a couple numbers missing.<br/><br/><b>*Update: </b>The consensus seems to be that Henry Hook and Will Shortz overlooked a couple terms in the sequence and it should be 2, 4, 6, 9, 11, 15, 20, 40, <b><i>51, 55, </i></b>60 and 90. Hopefully everyone is able to solve it now with the corrected wording. If anyone has direct access to Will's email, perhaps they could ask for a similar correction to the puzzle on the NPR website.<br/><br/>Will Shortz has confirmed (see <a href="http://puzzles.blainesville.com/2011/11/npr-sunday-puzzle-nov-13-2011-what.html?showComment=1321284406668#c3310607708672120974">his comment</a>) that he extended Henry Hook's original series (2, 4, 6, 9, 11, 15, 20) and in the process overlooked the numbers above. The NPR website has been updated as well. Thanks to everyone that helped clear this up.<br/><br/><b>Edit: </b>My hint was "See, I..." which sounds like CI which is 101 in Roman numerals<blockquote><b>A: </b>101 is next in <a href="http://oeis.org/A195526">the sequence</a>. When represented as Roman numerals, each number in the series requires exactly two letters (II, IV, VI, IX, XI, XV, XX, XL, LI, LV, LX, XC, CI...)</blockquote>Blainehttp://www.blogger.com/profile/06379274325110866036noreply@blogger.com121tag:blogger.com,1999:blog-5730391.post-54844063058221779692011-04-07T15:14:00.000-07:002011-04-07T15:15:45.522-07:00NPR Sunday Puzzle (Apr 3, 2011): Moby Dick scores an 82<a href="http://www.npr.org/2011/04/03/135076213/try-one-on-for-size">NPR Sunday Puzzle (Apr 3, 2011): Moby Dick scores an 82</a>: <blockquote><b>Q: </b>Assign every letter of the alphabet a numerical value: A=1, B=2, C=3 and so forth. Think of a classic work of literature that has eight letters in its title. When the letters are given a numerical value, they add up to 35. What's the title? Clue: The title has two words.</blockquote>Clue: 12,672<br /><br /><b>Edit: </b>First hint was 12 as in <a href="http://en.wikipedia.org/wiki/Adam-12">Adam-12</a>, second hint was 672 as in the birth year of <a href="http://www.religionfacts.com/christianity/people/bede.htm">Venerable Bede</a><blockquote><b>A: </b>ADAM BEDE, the first novel by <a href="http://en.wikipedia.org/wiki/Adam_Bede">George Eliot</a></blockquote>Blainehttp://www.blogger.com/profile/06379274325110866036noreply@blogger.com79