tag:blogger.com,1999:blog-91827054998982524962017-12-16T08:16:01.635-05:00Bill the Lizard"The time has come," the Walrus said,
"To talk of many things..."Bill the Lizardhttp://www.blogger.com/profile/09810099093752485841noreply@blogger.comBlogger222125tag:blogger.com,1999:blog-9182705499898252496.post-8877672242551042922017-12-16T08:16:00.000-05:002017-12-16T08:16:01.729-05:00Three Water Bottles<br />You have three water bottles with capacities of 8 quarts, 5 quarts, and 3 quarts. The largest bottle is filled with water, and the other two are empty. If there are no graduation marks on any of the bottles, how can you split the water evenly so that two of the bottles contain exactly 4 quarts each? You can only use these three bottles. Click below for the answer.<br /><br /><div class="separator" style="clear: both; text-align: center;"><img border="0" src="https://2.bp.blogspot.com/-_Zo_dWIyEWA/WiwGXD1MiCI/AAAAAAAADpM/CaPRbTD1Q7YAQXdpPZUatqAxGyYsltRqwCLcBGAs/s400/water.png" width="400" height="366" data-original-width="327" data-original-height="299" /></div><br /><div id="three_water_bottles" style="background-color: honeydew; display: none; padding: 5px;">There may be other ways to solve this problem, but here's one sequence that works.<br /><ol><li>Fill the 5 quart bottle, leaving 3 quarts in the 8 quart bottle.</li><li>Pour 3 quarts from the 5 quart bottle into the 3 quart bottle, leaving 2 quarts in the 5 quart bottle.</li><li>Empty the 3 quart bottle into the 8 quart bottle , leaving 6 quarts in the 8 quart bottle.</li><li>Pour the 2 quarts from the 5 quart bottle into the 3 quart bottle.</li><li>Fill the 5 quart bottle from the 8 quart bottle , leaving 1 quart in the 8 quart bottle.</li><li>Pour 1 quart from the 5 quart bottle into the 3 quart bottle (filling it), leaving 4 quarts in the 5 quart bottle.</li><li>Pour the 3 quarts from the 3 quart bottle into the 8 quart bottle, leaving 4 quarts in the 8 quart bottle.</li></ol>That's a lot to follow, so here are the steps in tabular form.<br /><br /><table><tr><th>8 qt.</th><th>5 qt.</th><th>3 qt.</th></tr><tr><td align="center">8</td><td align="center">0</td><td align="center">0</td></tr><tr><td align="center">3</td><td align="center">5</td><td align="center">0</td></tr><tr><td align="center">3</td><td align="center">2</td><td align="center">3</td></tr><tr><td align="center">6</td><td align="center">2</td><td align="center">0</td></tr><tr><td align="center">6</td><td align="center">0</td><td align="center">2</td></tr><tr><td align="center">1</td><td align="center">5</td><td align="center">2</td></tr><tr><td align="center">1</td><td align="center">4</td><td align="center">3</td></tr><tr><td align="center">4</td><td align="center">4</td><td align="center">0</td></tr></table><br /></div><br /><button onclick="showHideDiv('three_water_bottles');" title="Click to Show/Hide Answer">Show/Hide Answer</button>Bill the Lizardhttp://www.blogger.com/profile/09810099093752485841noreply@blogger.com0tag:blogger.com,1999:blog-9182705499898252496.post-41491941725018321442017-12-09T08:28:00.001-05:002017-12-10T21:30:03.934-05:00The Compulsive Gambler<br />You are approached by a compulsive gambler with the following proposal. You are to flip a fair coin four times. If heads and tails both appear twice each, he will pay you <span>$</span>11. If any other combination of heads and tails appears, you have to pay him only <span>$</span>10. Do you take the wager? Click below for the answer.<br /><br /><div id="the_compulsive_gambler" style="background-color: honeydew; display: none; padding: 5px;">This is not a good gamble. Below are all the possible outcomes for four successive coin flips.<br /><br /><div class="separator" style="clear: both; text-align: center;"><img border="0" src="https://3.bp.blogspot.com/-vA9V36nMCO0/WhGisQg07VI/AAAAAAAADok/2fVpC72YRXA14L7eVVNNqcF3UqdCUls2wCLcBGAs/s640/4_coin_flips.png" width="373" height="640" data-original-width="664" data-original-height="1139" /></div><br />Notice that exactly two heads and two tails only appear six out of sixteen times, so you can only expect to win this game about 37.5% of the time. At the offered stakes (<span>$</span>11 for a win, <span>$</span>10 for a loss) you'd be losing an average of around <span>$</span>2.12 every time you play.<br /><br />This problem appeared as an exercise in <a href="https://www.amazon.com/Introductory-Graph-Theory-Dover-Mathematics-ebook/dp/B008TVFB1U/">Introductory Graph Theory</a> by Gary Chartrand.<br />See my <a href="https://github.com/BillCruise/Probability/blob/master/scripts/four_flips.py">Probability GitHub repository</a> for a script that shows how to model this problem in Python.<br /></div><br /><button onclick="showHideDiv('the_compulsive_gambler');" title="Click to Show/Hide Answer">Show/Hide Answer</button>Bill the Lizardhttp://www.blogger.com/profile/09810099093752485841noreply@blogger.com2tag:blogger.com,1999:blog-9182705499898252496.post-51890013816252367872017-12-02T08:24:00.002-05:002017-12-02T08:24:12.539-05:00The Lily Pad<br />A lily pad starts out very small, but doubles in size every day. After 60 days it has completely covered a pond. After how many days had it covered one-quarter the area of the pond? Click below for the answer.<br /><br /><div class="separator" style="clear: both; text-align: center;"><img border="0" src="https://3.bp.blogspot.com/-dOep6ItLRLg/WhGdSRPpoUI/AAAAAAAADoU/-Zr4EsATW9QqMZ7VQYjxLxOHX91nCrsewCLcBGAs/s1600/Lily-Pad.jpg" data-original-width="420" data-original-height="315" /></div><br /><div id="the_lily_pad" style="background-color: honeydew; display: none; padding: 5px;">The instinctive answer might seem to be 15 days, but that would only be correct if the lily pad was growing linearly. Remember, the lily pad <i>doubles</i> in size every day, which is exponential. To solve this puzzle, just work backwards. If the lily pad completely covers the pond on day 60, then half the pond was covered on day 59, and one-quarter of the pond was covered on day <b>58</b>.<br /></div><br /><button onclick="showHideDiv('the_lily_pad');" title="Click to Show/Hide Answer">Show/Hide Answer</button><br />Bill the Lizardhttp://www.blogger.com/profile/09810099093752485841noreply@blogger.com0tag:blogger.com,1999:blog-9182705499898252496.post-35700811494153863952017-11-25T08:01:00.000-05:002017-11-25T08:01:53.684-05:00Three Coin Flips<br />The following game has a <span>$</span>10 entry fee. You are to flip a fair coin three times. The first time it comes up heads you are paid <span>$</span>5. The second time it comes up heads you're paid an additional <span>$</span>7. The third time it comes up heads you're paid <span>$</span>9 more, for a possible maximum prize of <span>$</span>21. Would you pay the <span>$</span>10 entry fee to play?<br /><br />If not, what would be a fair price for this game? Click below for the answer.<br /><br /><div id="three_coin_flips" style="background-color: honeydew; display: none; padding: 5px;">It's not a good idea to play this game at the offered entry fee. Here are the eight possible outcomes when flipping a coin three times, along with how much you would win after each flip. <br /><br /><div class="separator" style="clear: both; text-align: center;"><img border="0" src="https://3.bp.blogspot.com/-B_AAhrqI2zw/WhDBBQlholI/AAAAAAAADnw/hECTGTtnxtUF9IMiVYJPNN_TqmIjbbBXgCLcBGAs/s400/3_coin_flips.png" width="384" height="400" data-original-width="541" data-original-height="563" /></div><br />When you subtract out the <span>$</span>10 entry fee, you only win "big" (<span>$</span>11) one out of eight times. Three times you win only <span>$</span>2, three times you lose <span>$</span>5, and one out of the eight times you lose your entire <span>$</span>10 entry fee. If you average these, you can expect to lose <span>$</span>1 every time you play. So, if you lower the entry fee to <span>$</span>9 this would be a fair game.<br /><br />This problem appeared as an exercise in <a href="https://www.amazon.com/Introductory-Graph-Theory-Dover-Mathematics-ebook/dp/B008TVFB1U/">Introductory Graph Theory</a> by Gary Chartrand.<br />See my <a href="https://github.com/BillCruise/Probability/blob/master/scripts/three_flips.py">Probability GitHub repository</a> for a script that shows how to model this problem in Python.<br /></div><br /><button onclick="showHideDiv('three_coin_flips');" title="Click to Show/Hide Answer">Show/Hide Answer</button><br />Bill the Lizardhttp://www.blogger.com/profile/09810099093752485841noreply@blogger.com0tag:blogger.com,1999:blog-9182705499898252496.post-40238577305756009582017-11-18T08:54:00.000-05:002017-11-18T08:54:48.718-05:00One equals 0.999...<br />The following is a mathematical proof that 1 is equal to 0.999.... What's wrong with it? Click below for the answer.<br /><div class="separator" style="clear: both; text-align: center;">$x = 0.999...$<br />$10x = 9.999...$<br />$10x = 9 + 0.999...$<br />$10x = 9 + x$<br />$9x = 9$<br />$x = 1$<br /></div><br /><div id="one_equals_point_nine_nine_nine" style="background-color: honeydew; display: none; padding: 5px;">There's nothing wrong with it. 1 really is equal to <a href="https://en.wikipedia.org/wiki/0.999...">0.999...</a><br /></div><br /><button onclick="showHideDiv('one_equals_point_nine_nine_nine');" title="Click to Show/Hide Answer">Show/Hide Answer</button>Bill the Lizardhttp://www.blogger.com/profile/09810099093752485841noreply@blogger.com0tag:blogger.com,1999:blog-9182705499898252496.post-86621959475818267532017-11-11T08:16:00.000-05:002017-11-11T08:16:05.882-05:00Two equals one?<br />The following is a mathematical proof that two equals one. What's wrong with it? Click below for the answer.<br /><div class="separator" style="clear: both; text-align: center;">$a = b$<br />$aa = ab$<br />$aa - bb = ab - bb$<br />$(a + b)(a - b) = b(a - b)$<br />$a + b = b$<br />$a + a = a$<br />$2a = a$<br />$2 = 1$<br /></div><br /><div id="two_equals_one" style="background-color: honeydew; display: none; padding: 5px;">The problem is in the fourth step, where both sides of the equation are divided by $(a - b)$. Since $a = b$ is given at the start, $a - b$ is 0, and you can't divide by 0.<br /></div><br /><button onclick="showHideDiv('two_equals_one');" title="Click to Show/Hide Answer">Show/Hide Answer</button>Bill the Lizardhttp://www.blogger.com/profile/09810099093752485841noreply@blogger.com1tag:blogger.com,1999:blog-9182705499898252496.post-14271222612714640682017-11-04T08:12:00.002-04:002017-11-04T08:12:54.580-04:00State Names<br />There's only one letter in the English alphabet that is not used in the name of any of the 50 United States. Do you know which letter it is? Click below for the answer.<br /><br /><div id="state_names" style="background-color: honeydew; display: none; padding: 5px;">If you said 'J' or 'Z' you weren't far off. Each of those letters appear only once in the names of the 50 states (thanks to New Jersey and Arizona). The correct answer, though, is the letter <b>Q</b>, which does not appear in any state name.<br /><br />To find the solution, I used <a href="https://github.com/BillCruise/Probability/blob/master/scripts/state_names.py">a Python script</a> to load a list of state names and count the occurrence of each letter of the alphabet. This method takes a lot less time than consulting a map.<br /><br /><div class="separator" style="clear: both; text-align: center;"><a href="https://1.bp.blogspot.com/-50u-MnuwwdQ/WfTFBbz8okI/AAAAAAAADms/9MWrHPRMNQwJ-o1e3KCefoqatYlIkcW8QCLcBGAs/s1600/50-States.png" imageanchor="1" style="margin-left: 1em; margin-right: 1em;"><img border="0" src="https://1.bp.blogspot.com/-50u-MnuwwdQ/WfTFBbz8okI/AAAAAAAADms/9MWrHPRMNQwJ-o1e3KCefoqatYlIkcW8QCLcBGAs/s400/50-States.png" width="400" height="247" data-original-width="600" data-original-height="371" /></a></div><br /></div><br /><button onclick="showHideDiv('state_names');" title="Click to Show/Hide Answer">Show/Hide Answer</button><br />Bill the Lizardhttp://www.blogger.com/profile/09810099093752485841noreply@blogger.com0tag:blogger.com,1999:blog-9182705499898252496.post-33247298699123768392017-10-28T08:36:00.000-04:002017-10-28T08:58:22.900-04:00Tic-Tac-Toe<br />In a standard game of Tic-Tac-Toe, players take turns placing X's and O's on a 3x3 grid until one player makes three-in-a-row in any direction (horizontally, vertically, or diagonally). Because of these rules, you can only place a maximum of five of either symbol on the board during a game, often ending in a draw.<br /><br />Can you place <i>six</i> X's on a Tic-Tac-Toe board <i>without</i> making three-in-a-row in any direction? (Without placing any O's.) Click below for the solution.<br /><br /><div class="separator" style="clear: both; text-align: center;"><img border="0" src="https://2.bp.blogspot.com/-iysAo1_IoE0/WdjH-0ZmreI/AAAAAAAADlY/A6xWXQv6DloQPv3QlEg_r7o1CP-yZd39wCLcBGAs/s320/board_0.png" width="300" height="300" data-original-width="360" data-original-height="360" /></div><br /><div id="tic_tac_toe" style="background-color: honeydew; display: none; padding: 5px;">I found the solution below with a little trial and error. The trick to this puzzle is that you cannot solve it with an X in the center position.<br /><br /><div class="separator" style="clear: both; text-align: center;"><img border="0" src="https://4.bp.blogspot.com/-zQByHR1mzvM/WdjJb_U_PyI/AAAAAAAADlk/vVvPmQlEk18r1nXTQhn9n0CCmC7RcFgqwCLcBGAs/s320/board_6.png" width="300" height="300" data-original-width="360" data-original-height="360" /></div></div><br /><button onclick="showHideDiv('tic_tac_toe');" title="Click to Show/Hide Answer">Show/Hide Answer</button><br />Bill the Lizardhttp://www.blogger.com/profile/09810099093752485841noreply@blogger.com2tag:blogger.com,1999:blog-9182705499898252496.post-4172443500455156192017-10-21T07:34:00.001-04:002017-10-21T07:34:43.626-04:00Draw Two<br />Two numbers are drawn at random from the integers 1 through 10. What is the expected value of their sum? Does it change if the second draw is done with or without replacement? Click below for the answers.<br /><br /><div id="draw_two" style="background-color: honeydew; display: none; padding: 5px;">This puzzle is from <a href="https://twitter.com/MrHonner/status/917546796322377728">Patrick Honner</a>. It's easy to calculate the expected value with replacement. It's just two times the expected value of a random draw from 1...10, so 2 * 5.5, or 11. The interesting part is that when you draw two numbers <i>without</i> replacement the expected sum doesn't change. Why is that? <br /><br />To find out, take a look at what happens to the expected value of the second draw for each value of the first draw. (Expected value is just the average of the remaining numbers.)<br /><br /><table><tr><th>1st draw</th><th>E(2nd draw)</th><tr><td>$1$</td><td>$6$</td><tr><td>$2$</td><td>$5\frac{8}{9}$</td><tr><td>$3$</td><td>$5\frac{7}{9}$</td><tr><td>$4$</td><td>$5\frac{2}{3}$</td><tr><td>$5$</td><td>$5\frac{5}{9}$</td><tr><td>$6$</td><td>$5\frac{4}{9}$</td><tr><td>$7$</td><td>$5\frac{1}{3}$</td><tr><td>$8$</td><td>$5\frac{2}{9}$</td><tr><td>$9$</td><td>$5\frac{1}{9}$</td><tr><td>$10$</td><td>$5$</td> </table><br />As the value of the first draw increases, the expected value of the second draw decreases. If you take the sum of each column you get 55. Divide by 10 to get the average and you get 5.5, so when you add them together you arrive back at the solution of 11.<br /><br />See my <a href="https://github.com/BillCruise/Probability/blob/master/scripts/draw_two.py">Probability GitHub repository</a> for a script that shows how to model this problem in Python.<br /></div><br /><button onclick="showHideDiv('draw_two');" title="Click to Show/Hide Answer">Show/Hide Answer</button>Bill the Lizardhttp://www.blogger.com/profile/09810099093752485841noreply@blogger.com0tag:blogger.com,1999:blog-9182705499898252496.post-29614942147763367942017-10-14T07:55:00.000-04:002017-10-14T07:55:00.694-04:00Chicken McNuggets<br />You drove for hours last week to get your hands on <a href="http://www.snopes.com/2017/10/09/mcdonalds-szechuan-sauce-rick-and-morty/">McDonald's limited edition Szechuan sauce</a>, and now you need some chicken nuggets for you and all of your friends. You can buy McNuggets in boxes of 6, 9, and 20. What is the <i>largest</i> whole number of nuggets that it is <b>not possible</b> to obtain by purchasing some combination of boxes of 6, 9, and 20? Click below for the answer.<br /><br /><div class="separator" style="clear: both; text-align: center;"><img border="0" src="https://3.bp.blogspot.com/-aG4Y9Gj358c/WdvHmiO8tuI/AAAAAAAADl8/xdTzyiqe8S4yIKymoSIlW60SaltLH-W_wCLcBGAs/s320/Szechuan.jpg" width="259" height="320" data-original-width="375" data-original-height="464" /></div><br /><div id="chicken_mcnuggets" style="background-color: honeydew; display: none; padding: 5px;">There might be cleverer solutions to this problem, but we can do this fairly easily by listing combinations. Once we hit a streak of six numbers in a row that we <i>can</i> obtain, we know that the last number we <i>couldn't</i> obtain before the streak is the largest such number. Beyond that streak of six we can just add one or more boxes of 6 nuggets to one of those numbers to obtain any higher number. (There may be other combinations to obtain some of these numbers, but we only need one combination for each.)<br /><br /><table><tr><th>Number</th><th>Boxes</th></tr><tr><td>1</td><td>Not Possible</td></tr><tr><td>2</td><td>Not Possible</td></tr><tr><td>3</td><td>Not Possible</td></tr><tr><td>4</td><td>Not Possible</td></tr><tr><td>5</td><td>Not Possible</td></tr><tr><td>6</td><td>6</td></tr><tr><td>7</td><td>Not Possible</td></tr><tr><td>8</td><td>Not Possible</td></tr><tr><td>9</td><td>9</td></tr><tr><td>10</td><td>Not Possible</td></tr><tr><td>11</td><td>Not Possible</td></tr><tr><td>12</td><td>6 + 6</td></tr><tr><td>13</td><td>Not Possible</td></tr><tr><td>14</td><td>Not Possible</td></tr><tr><td>15</td><td>6 + 9</td></tr><tr><td>16</td><td>Not Possible</td></tr><tr><td>17</td><td>Not Possible</td></tr><tr><td>18</td><td>9 + 9</td></tr><tr><td>19</td><td>Not Possible</td></tr><tr><td>20</td><td>20</td></tr><tr><td>21</td><td>6 + 6 + 9</td></tr><tr><td>22</td><td>Not Possible</td></tr><tr><td>23</td><td>Not Possible</td></tr><tr><td>24</td><td>6 + 9 + 9</td></tr><tr><td>25</td><td>Not Possible</td></tr><tr><td>26</td><td>20 + 6</td></tr><tr><td>27</td><td>9 + 9 + 9</td></tr><tr><td>28</td><td>Not Possible</td></tr><tr><td>29</td><td>20 + 9</td></tr><tr><td>30</td><td>6 + 6 + 9 + 9</td></tr><tr><td>31</td><td>Not Possible</td></tr><tr><td>32</td><td>6 + 6 + 20</td></tr><tr><td>33</td><td>6 + 9 + 9 + 9</td></tr><tr><td>34</td><td>Not Possible</td></tr><tr><td>35</td><td>6 + 9 + 20</td></tr><tr><td>36</td><td>9 + 9 + 9 + 9</td></tr><tr><td>37</td><td>Not Possible</td></tr><tr><td>38</td><td>9 + 9 + 20</td></tr><tr><td>39</td><td>6 + 6 + 9 + 9 + 9</td></tr><tr><td>40</td><td>20 + 20</td></tr><tr><td>41</td><td>6 + 6 + 9 + 20</td></tr><tr><td>42</td><td>6 + 9 + 9 + 9 + 9</td></tr><tr><td>43</td><td>Not Possible</td></tr><tr><td>44</td><td>6 + 9 + 9 + 20</td></tr><tr><td>45</td><td>9 + 9 + 9 + 9 + 9</td></tr><tr><td>46</td><td>6 + 20 + 20</td></tr><tr><td>47</td><td>9 + 9 + 9 + 20</td></tr><tr><td>48</td><td>6 + 6 + 9 + 9 + 9 + 9</td></tr><tr><td>49</td><td>9 + 20 + 20</td></tr></table><br />That's six in a row, so we can get any higher number of nuggets just by adding boxes of 6 to those combinations. That means that <b>43</b> is the largest number of Chicken McNuggets that you cannot buy by combining boxes of 6, 9, and 20.<br /></div><br /><button onclick="showHideDiv('chicken_mcnuggets');" title="Click to Show/Hide Answer">Show/Hide Answer</button><br />Bill the Lizardhttp://www.blogger.com/profile/09810099093752485841noreply@blogger.com0tag:blogger.com,1999:blog-9182705499898252496.post-79094631238159027482017-10-07T07:35:00.000-04:002017-10-07T07:35:54.186-04:00Pennies<br />Would you rather have a ton of pennies, four miles of pennies lined up end-to-end, or a stack of pennies half a mile tall? Click below for a hint, or for the answer.<br /><br /><div class="separator" style="clear: both; text-align: center;"><img border="0" src="https://3.bp.blogspot.com/-N0XhKlsfaK0/WcexvY9xc3I/AAAAAAAADlA/F232PIh41bYG1LLA1o9Zbvhd3uEvKHPCgCLcBGAs/s1600/Pennies.jpg" data-original-width="512" data-original-height="288" /></div><br /><div id="pennies_hint" style="background-color: #ffffcc; display: none; padding: 5px;">One penny weighs 2.500 grams (according to the <a href="https://www.usmint.gov/learn/coin-and-medal-programs/coin-specifications">U.S. Mint</a>).<br />There are about 28.35 grams in an ounce<br />There are 16 ounces in a pound.<br />There are 2,000 pounds in a (U.S.) ton.<br /><br />One penny is 0.750 inches in diameter.<br />There are 12 inches in a foot.<br />There are 5,280 feet in a mile.<br /><br />One penny is 1.52 millimeters thick.<br />There are 25.4 millimeters in an inch.<br /></div><br /><button onclick="showHideDiv('pennies_hint');" title="Click to Show/Hide Hint">Show/Hide Hint</button><br /><br /><div id="pennies_answer" style="background-color: honeydew; display: none; padding: 5px;">Four miles of pennies lined up end-to-end would be <b>$3,379.20</b>, while one ton is <b>$3,628.80</b>, so between the first two options you would be better off to take the ton. However, a stack of pennies half a mile tall would be <b>$5,293.89</b>, so the stack is by far the best option.<br /></div><br /><button onclick="showHideDiv('pennies_answer');" title="Click to Show/Hide Answer">Show/Hide Answer</button>Bill the Lizardhttp://www.blogger.com/profile/09810099093752485841noreply@blogger.com0tag:blogger.com,1999:blog-9182705499898252496.post-11409406971016993382017-09-30T07:52:00.001-04:002017-09-30T07:52:26.622-04:00Western Leaders<br />Here's a freaky coincidence about World War II. If you add up the year of birth, age in 1944, year of taking power, and the number of years in office in 1944 for each of the five main leaders of the Western world during World War II, the sums are all the same.<br /><br /><table><tr><th></th><th>Churchill</th><th>Hitler</th><th>Mussolini</th><th>Roosevelt</th><th>Stalin</th></tr><tr><th>Year of birth</th><td align="right">1874</td><td align="right">1889</td><td align="right">1883</td><td align="right">1882</td><td align="right">1878</td></tr><tr><th>Age in 1944</th><td align="right">70</td><td align="right">55</td><td align="right">61</td><td align="right">62</td><td align="right">66</td></tr><tr><th>Took power</th><td align="right">1940</td><td align="right">1933</td><td align="right">1922</td><td align="right">1933</td><td align="right">1922</td></tr><tr><th>Years in office</th><td align="right">4</td><td align="right">11</td><td align="right">22</td><td align="right">11</td><td align="right">22</td></tr><tr><th>Sum</th><td align="right">3,888</td><td align="right">3,888</td><td align="right">3,888</td><td align="right">3,888</td><td align="right">3,888</td></tr></table><br />Can you explain this coincidence? Click below for the answer.<br /><br /><div id="western_leaders" style="background-color: honeydew; display: none; padding: 5px;">There's really no coincidence at all. If you take any person's year of birth and add their age in 1944, the sum will be... 1944. This also goes for the year a leader took office and the number of years they had held office in 1944, so the sum of all four values will always be 1944 + 1944 = 3888. You can make your own puzzle using any year as a reference point.<br /></div><br /><button onclick="showHideDiv('western_leaders');" title="Click to Show/Hide Answer">Show/Hide Answer</button>Bill the Lizardhttp://www.blogger.com/profile/09810099093752485841noreply@blogger.com0tag:blogger.com,1999:blog-9182705499898252496.post-19920450564357068592017-09-23T08:19:00.000-04:002017-09-23T08:19:54.196-04:00Concentric Shapes<br />See the image below of a square inscribed inside a circle inscribed inside a square. If the outer square has an area of 100 square inches, is there a quick way of figuring out the area of the inner square? Click below for the answer.<br /><br /><div class="separator" style="clear: both; text-align: center;"><img src="https://3.bp.blogspot.com/-1hy5k8jQnTw/Wb3Hx_YAs2I/AAAAAAAADkk/Djf9qmQPdVoA-mtcZ0x-jFnqTIwe3uCDwCLcBGAs/s400/Concentric-1.png" style="border: none" /></div><br /><div id="concentric_shapes" style="background-color: honeydew; display: none; padding: 5px;">Spin the inner square around so the corners touch the top, bottom, and sides of the outer square.<br /><br /><div class="separator" style="clear: both; text-align: center;"><img src="https://4.bp.blogspot.com/-84ncZyrET3s/Wb3H6TAQFWI/AAAAAAAADko/BhOJvSz9chcP0DranKup4mkPu05MuxdCgCLcBGAs/s400/Concentric-2.png" style="border: none" /></div><br />Now imagine taking each corner of the outer square and folding them in so they touch in the center of the image. It's easy to see that the area outside the inner square will perfectly cover the area inside, so the inner square has exactly half the area of the outer square, or 50 square inches.<br /></div><br /><button onclick="showHideDiv('concentric_shapes');" title="Click to Show/Hide Answer">Show/Hide Answer</button>Bill the Lizardhttp://www.blogger.com/profile/09810099093752485841noreply@blogger.com0tag:blogger.com,1999:blog-9182705499898252496.post-13708401374391763062017-09-16T09:00:00.003-04:002017-09-16T09:00:25.062-04:00Digit Frequency<br />If you write down all the numbers from 1 to 1000 (inclusive) which digit occurs most frequently? Which digit appears least frequently in the same range? Click below for the answers.<br /><br /><div id="digit_frequency" style="background-color: honeydew; display: none; padding: 5px;">You could write all the numbers down or write a program to count the digit frequencies, but this problem is much easier if you look for a pattern. The digits 0 through 9 all appear the same number of times in the ones place, but 1 through 9 appear a lot more often in the tens and hundreds place (because of numbers like 22, 222, etc., and the fact that we don't write leading 0s). This means that <b>0 is the least frequent digit</b>. It only appears 192 times in the specified range. What about the most frequent? The digits 1 through 9 appear with exactly the same frequency in the range 1 to 999 (300 times each), so that extra occurrence (in the number 1000) makes <b>1 the most frequent digit</b> from 1 to 1000.<br /></div><br /><button onclick="showHideDiv('digit_frequency');" title="Click to Show/Hide Answer">Show/Hide Answer</button>Bill the Lizardhttp://www.blogger.com/profile/09810099093752485841noreply@blogger.com0tag:blogger.com,1999:blog-9182705499898252496.post-77966285663368657282017-09-09T08:37:00.004-04:002017-09-09T08:37:55.340-04:00Counting Chickens<br />If one-and-a-half chickens lay one-and-a-half eggs in one-and-a-half days, how many eggs does one chicken lay in one day? Click below for the answer.<br /><br /><div id="counting_chickens" style="background-color: honeydew; display: none; padding: 5px;">For many people, the intuitive answer is one egg, but it pays double-check your math on this kind of problem. The daily rate of eggs per chicken per day is given by the formula<br /><br />daily rate = eggs / (chickens x days)<br /><br />Plugging in the numbers from the first part of the problem, we get<br /><br />daily rate = 1.5 / (1.5 x 1.5)<br />daily rate = 1.5 / 2.25<br />daily rate = 2/3<br /><br />So, one chicken lays two-thirds of an egg in one day.<br /></div><br /><button onclick="showHideDiv('counting_chickens');" title="Click to Show/Hide Answer">Show/Hide Answer</button>Bill the Lizardhttp://www.blogger.com/profile/09810099093752485841noreply@blogger.com0tag:blogger.com,1999:blog-9182705499898252496.post-57376924382986807272017-09-02T08:46:00.002-04:002017-09-02T08:46:30.505-04:00Number Sense<br />How good is your "number sense"? How many of the following can you answer without using a calculator or looking up a conversion factor?<br /><ol><li>Are there more inches in a mile, or Sundays in 1000 years?</li><li>Are there more seconds in a week, or feet in 100 miles?</li><li>Are there more millimeters in a mile, or seconds in a month?</li><li>Which is larger, multiplying all the numbers from 1 to 10, or multiplying just the even numbers from 1 to 16?</li><li>Which is longer, 666 days or 95 weeks?</li><li>Which is longer, 666 inches or 55 feet?</li><li>Which is longer, 666 hours or 28 days?</li><li>Are there more ounces in a ton or inches in a kilometer?</li><li>Which is hotter, $0^{\circ}C$ or $0^{\circ}F$?</li><li>Which is larger, $e^\pi$ or $\pi^e$?</li></ol><br />Click below for the answers.<br /><br /><div id="number_sense" style="background-color: honeydew; display: none; padding: 5px;"><ol><li>Inches in a mile. (63,360. There can be up to 52,178 Sundays in 1000 years.)</li><li>Seconds in a week. (604,800, compared to 528,000 feet in 100 miles.)</li><li>Seconds in a month. (Even if the month only has 28 days, that's 2,419,200 seconds, compared to only 1,609,340 millimeters in a mile.)</li><li>Just the even numbers from 1 to 16. (Multiplying all the numbers from 1 to 10 gives you 3,628,800. Multiplying the even numbers from 1 to 16 give you 10,321,920.)</li><li>666 days. (95 weeks is only 665 days.)</li><li>666 inches. (55 feet is 660 inches.)</li><li>28 days (which is 672 hours).</li><li>Inches in a kilometer. (39,370.1, compared to 35,840 ounces in a <a href="http://www.onlineconversion.com/faq_09.htm">long ton</a>, which is the heaviest ton.)</li><li>$0^{\circ}C$ is "hotter" since it is equal to $32^{\circ}F$</li><li>$e^\pi$ (23.14) is larger than $\pi^e$ (22.46).</li></ol></div><br /><button onclick="showHideDiv('number_sense');" title="Click to Show/Hide Answer">Show/Hide Answer</button>Bill the Lizardhttp://www.blogger.com/profile/09810099093752485841noreply@blogger.com2tag:blogger.com,1999:blog-9182705499898252496.post-77521852311842412922017-08-26T09:03:00.000-04:002017-08-26T09:03:33.743-04:00Replacing Marbles<br />We place 15 black marbles and 15 white marbles in an urn. We have 30 additional black marbles in a bag. Then we follow these rules.<br /><br />1. Remove two marbles from the urn.<br />2. If they are different colors, put the white marble back in the urn and the black marble in the bag.<br />3. If they are the same color, put both marbles in the bag, then put one black marble from the bag into the urn.<br /><br />Continue following these rules until only one marble is left in the urn. What color is that marble? Click below for the answer.<br /><br /><div id="replacing_marbles" style="background-color: honeydew; display: none; padding: 5px;">When I first heard this puzzle, I immediately thought of writing a Python script, since that's my favorite method for dealing with problems in probability. This is a logic problem in disguise, though. I realized that as I tried to figure out the best way to set up the problem in code. If you pay close attention to the rules for adding and removing marbles from the urn, and the initial conditions, you'll notice a couple of things.<br /><br />1. You start with an odd number of both black and white marbles in the urn.<br />2. The rules force you to always keep an odd number of white marbles in the urn (they can only be removed two at a time), but allow for both odd and even numbers of black marbles.<br /><br />From those observations it's easy to see that when you get down to one marble, it must be a white marble.<br /></div><br /><button onclick="showHideDiv('replacing_marbles');" title="Click to Show/Hide Answer">Show/Hide Answer</button>Bill the Lizardhttp://www.blogger.com/profile/09810099093752485841noreply@blogger.com0tag:blogger.com,1999:blog-9182705499898252496.post-42883588942914710272017-08-19T09:11:00.000-04:002017-08-19T09:11:03.014-04:00Factor Sums<br />Not counting itself, the number 6 has the factors 1, 2, and 3, which add to 6. The number 28 has the same property (its factors are 1, 2, 4, 7, and 14). Can you come up with a three-digit number that has this property? What about a four-digit number? Click below for the answers.<br /><br /><div id="factor_sums" style="background-color: honeydew; display: none; padding: 5px;">If you knew that a number that is the sum of its own proper divisors is called a <a href="https://en.wikipedia.org/wiki/Perfect_number">Perfect number</a>, this puzzle was pretty easy. You could just search for that name and find the solutions are <b>496</b> and <b>8,128</b>. Perfect numbers have been known at least as far back as <a href="https://en.wikipedia.org/wiki/Euclid">Euclid</a> (323–283 BCE), who included a formulation for then in his book of <a href="https://en.wikipedia.org/wiki/Euclid%27s_Elements">Elements</a>.<br /><br />The formulations states that $q(q + 1) / 2$ is a perfect number whenever $q$ is a prime of the form $2^p - 1$ for prime $p$ (now known as a <a href="https://en.wikipedia.org/wiki/Mersenne_prime">Mersenne prime</a>). So, if we know the first few Mersenne primes, we can calculate the first few perfect numbers.<br /><br />$3(3 + 1) / 2 = 6$<br />$7(7 + 1) / 2 = 28$<br />$31(31 + 1) / 2 = 496$<br />$127(127 + 1) / 2 = 8,128$<br />$8,191(8,191 + 1) / 2 = 33,550,336$<br /><br />The ancient Greek mathematicians would not have known that 8,191 was a prime, so Euclid would only have known the first four Perfect numbers. Now you can say you know something that Euclid didn't!<br /></div><br /><button onclick="showHideDiv('factor_sums');" title="Click to Show/Hide Answer">Show/Hide Answer</button>Bill the Lizardhttp://www.blogger.com/profile/09810099093752485841noreply@blogger.com0tag:blogger.com,1999:blog-9182705499898252496.post-78970575768431500102017-08-12T09:02:00.002-04:002017-08-12T09:02:33.275-04:00Minimum Percentage<br />75% of men from a certain group are tall, 75% have brown hair, and 75% have brown eyes. What is the minimum percentage that are tall, have brown hair, <i>and</i> have brown eyes? Click below to see the answer.<br /><br /><div id="minimum_percentage" style="background-color: honeydew; display: none; padding: 5px;">Instead of thinking in percentages to solve this problem, it's helpful to think back to the <a href="http://www.billthelizard.com/2017/06/the-pigeonhole-principle.html">Pigeonhole Principle</a>. Think of a group of 100 men, then 75 are tall, 75 have brown hair, and 75 have brown eyes. That's 225 individual attributes to assign to 100 men, so at least 25 of them (or <b>25%</b>) must have each of the three attributes.<br /></div><br /><button onclick="showHideDiv('minimum_percentage');" title="Click to Show/Hide Answer">Show/Hide Answer</button>Bill the Lizardhttp://www.blogger.com/profile/09810099093752485841noreply@blogger.com0tag:blogger.com,1999:blog-9182705499898252496.post-32450245227305312802017-08-05T08:33:00.000-04:002017-08-05T08:33:01.099-04:00A Two-Digit Number<br />Find a two-digit number that's equal to two times the result of multiplying its digits. Click below to see the answer.<br /><br /><div id="two_digit_number" style="background-color: honeydew; display: none; padding: 5px;">My first attempt at solving this puzzle was to set it up as an equation and try to solve it algebraically. Let's say the two digits are $x$ and $y$. Then the equation would be:<br /><br />$10x + y = 2xy$<br /><br />The left-hand side is the two-digit number ($x$ in the tens place, $y$ in the ones place) and the right-hand side is two times the result of multiplying its digits. If you try to isolate either $x$ or $y$, you'll see that it's not very easy to come up with a clean solution. That's because the equation above describes a hyperbola.<br /><br /><div class="separator" style="clear: both; text-align: center;"><img border="0" src="https://2.bp.blogspot.com/-ZmCgq6t5V0E/WX3hk42LFMI/AAAAAAAADi8/v_EYiDg41ogMywLfBhtVbtrrCqFLl9FxgCLcBGAs/s1600/hyperbola.gif" data-original-width="200" data-original-height="213" /></div><br />That's not exactly a dead end, but it isn't the kind of easy-to-understand (once you see it) solution I like in a logic puzzle. Luckily, there's an easier way. There aren't that many possibilities (we're only dealing with two digits), and we can eliminate a lot of them.<br /><br />For example, we know that neither digit is 0. Also, we know that $2xy$ is an even number, so $y$ must be even (because the result of adding it to an even number is even). We also know that the product of the digits must be less than 50, otherwise $2xy$ would have three digits. That gets us down to only 32 possibilities to test.<br /><br /><div class="separator" style="clear: both; text-align: center;"><img border="0" src="https://4.bp.blogspot.com/-N5YnHP9cx9o/WX3nb6U8oSI/AAAAAAAADjM/ESN3uWAJ46kND-kjxI4aGhaRp722WRauACLcBGAs/s1600/possibilities.png" data-original-width="321" data-original-height="201" /></div><br />Any other shortcuts that I can think of would only eliminate a few possibilities, but it's easy to just test the remaining ones (I went through them manually, but you could write a short script or use a spreadsheet), and find that the solution is<br /><br />$36 = 2 * 3 * 6$<br /></div><br /><button onclick="showHideDiv('two_digit_number');" title="Click to Show/Hide Answer">Show/Hide Answer</button><br /><br />Bill the Lizardhttp://www.blogger.com/profile/09810099093752485841noreply@blogger.com2tag:blogger.com,1999:blog-9182705499898252496.post-89933902456152111202017-07-29T09:03:00.003-04:002017-07-29T09:03:31.780-04:00Identical Twins<br />Alice and Eve are identical twin sisters. One always lies and the other always tells the truth, but we don't know which is which. We ask one of them "Is Alice the one that always lies?" and she replies "Yes." Did we speak to Alice or Eve? Click below to see the answer.<br /><br /><div id="identical_twins" style="background-color: honeydew; display: none; padding: 5px;">We spoke to Eve. A person who always lies or always tells the truth cannot admit that they are a liar, so Alice could not have answered "Yes" to that question. (Note that we still don't know which sister is the liar and which is the truth teller.)<br /></div><br /><button onclick="showHideDiv('identical_twins');" title="Click to Show/Hide Answer">Show/Hide Answer</button><br />Bill the Lizardhttp://www.blogger.com/profile/09810099093752485841noreply@blogger.com0tag:blogger.com,1999:blog-9182705499898252496.post-23704839958327485652017-07-22T08:35:00.002-04:002017-07-22T08:35:20.939-04:00Counting Socks<br />All my socks are red except two. All my socks are white except two. All my socks are blue except two. How many socks do I have? Click below for the answer.<br /><br /><div id="counting_socks" style="background-color: honeydew; display: none; padding: 5px;">Oddly, I only have three socks. Don't let the fact that socks normally come in matching pairs distract you. No other number satisfies all three conditions above.<br /></div><br /><button onclick="showHideDiv('counting_socks');" title="Click to Show/Hide Answer">Show/Hide Answer</button>Bill the Lizardhttp://www.blogger.com/profile/09810099093752485841noreply@blogger.com2tag:blogger.com,1999:blog-9182705499898252496.post-20033286181022604682017-07-15T08:51:00.000-04:002017-07-15T08:51:18.819-04:00Circumnavigation<br />From 1519 until 1522, Ferdinand Magellan's <i>Victoria </i> was the first ship to successfully circumnavigate the globe. (Magellan himself did not survive the entire voyage.) Can you tell me which part of the ship traveled the greatest distance? Click below for the answer.<br /><br /><div id="circumnavigation" style="background-color: honeydew; display: none; padding: 5px;">If you remembered the <a href="http://www.billthelizard.com/2017/01/rope-around-earth.html">Rope Around the Earth</a> puzzle I posted a few months ago, you probably got this one pretty quickly. Since the world is roughly spherical, the tip of the tallest mast of the ship would have traveled the greatest distance in sailing around the globe. Imagine if a boat sailed in a perfect circle around the equator. The part of the boat deepest under water (the keel) would create a smaller circle than the tip of the mast several feet above the water, so the tip of the mast travels the greatest distance during the voyage.<br /></div><br /><button onclick="showHideDiv('circumnavigation');" title="Click to Show/Hide Answer">Show/Hide Answer</button><br /><br /><div class="separator" style="clear: both; text-align: center;"><img border="0" src="https://2.bp.blogspot.com/-bBbSml44s2Q/WV-rP4V7fOI/AAAAAAAADic/uA9WE3etDxg0roaJzLCu5_HtEDJVa5rUgCLcBGAs/s320/Nao_Victoria.jpg" width="320" height="240" data-original-width="1280" data-original-height="960" /></div><div style="clear: both; text-align: center; font-size: 80%;">Replica of the Victoria, Photograph by <a href="https://commons.wikimedia.org/w/index.php?curid=371163">Gnsin - Own work, CC BY-SA 3.0</a></div>Bill the Lizardhttp://www.blogger.com/profile/09810099093752485841noreply@blogger.com0tag:blogger.com,1999:blog-9182705499898252496.post-85100362687461626332017-07-08T09:59:00.001-04:002017-07-08T09:59:04.355-04:0050 factorial<br />50! = 30414093201713378043612608166064768844377641568960512071337804000<br /><br />Without doing the full computation, can you tell whether the above statement is true or false? Click below for the answer.<br /><br /><div id="fifty_factorial" style="background-color: honeydew; display: none; padding: 5px;">You can probably guess that the statement is false, otherwise it wouldn't be much of a puzzle. The reasoning, though, is that the factorial for 50 must include the factors 10, 20, 30, 40, and 50, so it must end in at least five zeroes. The value above ends in only three zeroes, so it cannot be correct. (The correct value is 30414093201713378043612608166064768844377641568960512000000000000.)<br /></div><br /><button onclick="showHideDiv('fifty_factorial');" title="Click to Show/Hide Answer">Show/Hide Answer</button>Bill the Lizardhttp://www.blogger.com/profile/09810099093752485841noreply@blogger.com0tag:blogger.com,1999:blog-9182705499898252496.post-60670354582130822962017-07-01T08:09:00.002-04:002017-07-01T08:09:14.877-04:00The Missing Fish<br />Two fathers took their sons fishing. Each man and his son caught one fish, but when they all returned to camp they only had three fish. None of the fish were eaten, lost, or thrown back. How could this be? Click below to see the answer.<br /><br /><div id="missing_fish" style="background-color: honeydew; display: none; padding: 5px;">There were only three people on the fishing trip. One man was the father and grandfather of the other two.<br /></div><br /><button onclick="showHideDiv('missing_fish');" title="Click to Show/Hide Answer">Show/Hide Answer</button><br /><br /><div class="separator" style="clear: both; text-align: center;"><img border="0" src="https://1.bp.blogspot.com/-z897STj_WRQ/WUqNWeRKvdI/AAAAAAAADiA/ov3_yb28tTk6RpHrtCnCZUu9uHqnTRxYgCLcBGAs/s1600/Fish.JPG" data-original-width="480" data-original-height="283" /></div>Bill the Lizardhttp://www.blogger.com/profile/09810099093752485841noreply@blogger.com0