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Runge's Phenomenon<div class='note'>This is a post I wrote for round 2 of The Aperiodical's <a href="https://aperiodical.com/2018/06/announcing-the-big-internet-math-off/" target="new">Big Internet Math-Off 2018</a>. As I went out in round 1 of the Big Math-Off, you got to read about <a href="https://aperiodical.com/2018/07/the-big-internet-math-off-round-2-matt-parker-v-jo-morgan/" target="new">the real projective plane</a> instead of this.</div><div class='paragraph'>Polynomials are very nice functions: they're easy to integrate and differentiate, it's quick to calculate their value at points, and they're generally friendly to deal with. Because of this, it can often be useful to find a polynomial that closely approximates a more complicated function.</div><div class='paragraph'>Imagine a function defined for \(x\) between -1 and 1. Pick \(n-1\) points that lie on the function. There is a unique degree \(n\) polynomial (a polynomial whose highest power of \(x\) is \(x^n\)) that passes through these points. This polynomial is called an <em>interpolating polynomial</em>, and it sounds like it ought to be a pretty good approximation of the function.</div><div class='paragraph'>So let's try taking points on a function at equally spaces values of \(x\), and try to approximate the function:</div>$$f(x)=\frac1{1+25x^2}$$<a class='zoom' href='javascript:showlimage("runge-uniform.gif")'><img src='http://www.mscroggs.co.uk/img/320/runge-uniform.jpg'></a><div class='caption'>Polynomial interpolations of \(\displaystyle f(x)=\frac1{1+25x^2}\) using equally spaces points</div><div class='paragraph'>I'm sure you'll agree that these approximations are pretty terrible, and they get worse as more points are added. The high error towards 1 and -1 is called Runge's phenomenon, and was discovered in 1901 by Carl David Tolmé Runge.</div><div class='paragraph'>All hope of finding a good polynomial approximation is not lost, however: by choosing the points more carefully, it's possible to avoid Runge's phenomenon. Chebyshev points (named after Pafnuty Chebyshev) are defined by taking the \(x\) co-ordinate of equally spaced points on a circle.</div><a class='zoom' href='javascript:showlimage("chebdef.png")'><img src='http://www.mscroggs.co.uk/img/320/chebdef.jpg'></a><div class='caption'>Eight Chebyshev points</div><div class='paragraph'>The following GIF shows interpolating polynomials of the same function as before using Chebyshev points.</div><a class='zoom' href='javascript:showlimage("runge-chebyshev.gif")'><img src='http://www.mscroggs.co.uk/img/320/runge-chebyshev.jpg'></a><div class='paragraph'>Nice, we've found a polynomial that closely approximates the function... But I guess you're now wondering how well the Chebyshev interpolation will approximate other functions. To find out, let's try it out on the votes over time of <a href="http://aperiodical.com/2018/07/the-big-internet-math-off-round-1-matt-parker-v-matthew-scroggs/" target="new">my first round Big Internet Math-Off match.</a></div><a class='zoom' href='javascript:showlimage("data.png")'><img src='http://www.mscroggs.co.uk/img/320/data.jpg'></a><div class='caption'>Scroggs vs Parker, 6-8 July 2018</div><div class='paragraph'>The graphs below show the results of the match over time interpolated using 16 uniform points (left) and 16 Chebyshev points (right). You can see that the uniform interpolation is all over the place, but the Chebyshev interpolation is very close the the actual results.</div><a class='zoom' href='javascript:showlimage("uniform.png")'><img src='http://www.mscroggs.co.uk/img/320/uniform.jpg'></a><a class='zoom' href='javascript:showlimage("chebyshev.png")'><img src='http://www.mscroggs.co.uk/img/320/chebyshev.jpg'></a><div class='caption'>Scroggs vs Parker, 6-8 July 2018, approximated using uniform points (left) and Chebyshev points (right)</div><div class='paragraph'>But maybe you still want to see how good Chebyshev interpolation is for a function of your choice... To help you find out, I've written <a href="https://twitter.com/RungeBot" target="new">@RungeBot</a>, a Twitter bot that can compare interpolations with equispaced and Chebyshev points. Just tweet it a function, and it'll show you how bad Runge's phenomenon is for that function, and how much better Chebysheb points are.</div><div class='paragraph'>For example, if you were to tweet <a href="https://twitter.com/mscroggs/status/1013052921288720384" target="new">@RungeBot f(x)=abs(x)</a>, then RungeBot would reply: <a href="https://twitter.com/RungeBot/status/1013061582933430272" target="new">Here's your function interpolated using 17 equally spaced points (blue) and 17 Chebyshev points (red). For your function, Runge's phenomenon is terrible.</a></div><a class='zoom' href='javascript:showlimage("runge-abs.jpg")'><img src='http://www.mscroggs.co.uk/img/320/runge-abs.jpg'></a><div class='paragraph'>A list of constants and functions that RungeBot understands can be found <a href="https://github.com/mscroggs/Equation/blob/master/doc/source/Functions.rst" target="new">here</a>.</div>
http://www.mscroggs.co.uk/blog/57
http://www.mscroggs.co.uk/blog/5713 Sep 2018 12:00:00 GMTWorld Cup stickers 2018, pt. 3<div class='paragraph'>So you've <a href="http://www.mscroggs.co.uk/blog/50">calculated how much you should expect the World Cup sticker book to cost</a>and <a href="http://www.mscroggs.co.uk/blog/54">recorded how much it actually cost</a>. You might be wondering what else you can do with your sticker book.If so, look no further: this post contains 5 mathematical things involvolving your sticker book and stickers.</div><h3>Test the birthday paradox</h3><a class='zoom' href='javascript:showlimage("birthdaysbrazil.jpg")'><img src='http://www.mscroggs.co.uk/img/320/birthdaysbrazil.jpg'></a><div class='caption'>Stickers 354 and 369: Alisson and Roberto Firmino</div><div class='paragraph'>In a group of 23 people, there is a more than 50% chance that two of them will share a birthday. This is often called the birthday paradox, as the number 23 is surprisingly small.</div><div class='paragraph'>Back in 2014 when <a href="https://www.theguardian.com/science/alexs-adventures-in-numberland/2014/jun/10/world-cup-birthday-paradox-footballers-born-on-the-same-day" target="new">Alex Bellos</a> suggested testing the birthday paradox on World Cup squads, as there are 23 players in a World Cup squad. I recently discovered that even further back in 2012, <a href="https://www.youtube.com/watch?v=a2ey9a70yY0" target="new">James Grime made a video</a> about the birthday paradox in football games, using the players on both teams plus the referee to make 23 people.</div><div class='paragraph'>In this year's sticker book, each player's date of birth is given above their name, so you can use your sticker book to test it out yourself.</div><h3>Kaliningrad</h3><a class='zoom' href='javascript:showlimage("kaliningrad.jpg")'><img src='http://www.mscroggs.co.uk/img/320/kaliningrad.jpg'></a><div class='caption'>Sticker 022: Kaliningrad</div><div class='paragraph'>One of the cities in which games are taking place in this World Cup is Kaliningrad. Before 1945, Kaliningrad was called Königsberg. In Königsburg, there were seven bridges connecting four islands. The arrangement of these bridges is shown below.</div><a class='zoom' href='javascript:showlimage("7bridges.png")'><img src='http://www.mscroggs.co.uk/img/320/7bridges.jpg'></a><div class='paragraph'>The people of Königsburg would try to walk around the city in a route that crossed each bridge exactly one. If you've not tried this puzzle before, try to find such a route now before reading on...</div><div class='paragraph'>In 1736, mathematician Leonhard Euler proved that it is in fact impossible to find such a route. He realised that for such a route to exist, you need to be able to pair up the bridges on each island so that you can enter the island on one of each pair and leave on the other. The islands in Königsburg all have an odd number of bridges, so there cannot be a route crossing each bridge only once.</div><div class='paragraph'>In Kaliningrad, however, there are eight bridges: two of the original bridges were destroyed during World War II, and three more have been built. Because of this, it's now possible to walk around the city crossing each bridge exactly once.</div><a class='zoom' href='javascript:showlimage("kaliningrad-path.png")'><img src='http://www.mscroggs.co.uk/img/320/kaliningrad-path.jpg'></a><div class='attr'><a href="https://www.openstreetmap.org/" target="new">OpenStreetMap</a></div><div class='caption'>A route around Kaliningrad crossing each bridge exactly once.</div><div class='paragraph'>I wrote more about this puzzle, and using similar ideas to find the shortest possible route to complete a level of Pac-Man, in <a href="http://www.mscroggs.co.uk/blog/18">this blog post.</a></div><h3>Sorting algorithms</h3><div class='paragraph'>If you didn't convince many of your friends to join you in collecting stickers, you'll have lots of swaps. You can use these to practice performing your favourite sorting algorithms.</div><h4>Bubble sort</h4><div class='paragraph'>In the bubble sort, you work from left to right comparing pairs of stickers. If the stickers are in the wrong order, you swap them. After a few passes along the line of stickers, they will be in order.</div><a class='zoom' href='javascript:showlimage("bubble-sort-stickers.gif")'><img src='http://www.mscroggs.co.uk/img/320/bubble-sort-stickers.jpg'></a><div class='caption'>Bubble sort</div><h4>Insertion sort</h4><div class='paragraph'>In the insertion sort, you take the next sticker in the line and insert it into its correct position in the list.</div><a class='zoom' href='javascript:showlimage("insertion-sort-stickers.gif")'><img src='http://www.mscroggs.co.uk/img/320/insertion-sort-stickers.jpg'></a><div class='caption'>Insertion sort</div><h4>Quick sort</h4><div class='paragraph'>In the quick sort, you pick the middle sticker of the group and put the other stickers on the correct side of it. You then repeat the process with the smaller groups of stickers you have just formed.</div><a class='zoom' href='javascript:showlimage("quick-sort-stickers.gif")'><img src='http://www.mscroggs.co.uk/img/320/quick-sort-stickers.jpg'></a><div class='caption'>Quick sort</div><div class='paragraph'></div><h3>The football</h3><a class='zoom' href='javascript:showlimage("correct-football.jpg")'><img src='http://www.mscroggs.co.uk/img/320/correct-football.jpg'></a><div class='caption'>Sticker 007: The official ball</div><div class='paragraph'>Sticker 007 shows the official tournament ball. If you look closely (click to enlarge), you can see that the ball is made of a mixture of pentagons and hexagons. The ball is not made of only hexagons, as <a href="https://www.youtube.com/watch?v=btPqKAGyajM" target="new">road signs in the UK show</a>.</div><div class='paragraph'>Stand up mathematician Matt Parker started <a href="https://petition.parliament.uk/petitions/202305" target="new">a petition</a> to get the symbol on the signs changed, but the idea was rejected.</div><div class='paragraph'>If you have a swap of sticker 007, why not stick it to a letter to your MP about the incorrect signs as an example of what an actual football looks like.</div><h3>Psychic pets</h3><div class='paragraph'>Speaking of Matt Parker, during this World Cup, he's looking for <a href="https://www.youtube.com/watch?v=tQiiaFE1e-Y" target="new">psychic pets</a> that are able to predict World Cup results. Why not use your swaps to label two pieces of food that your pet can choose between to predict the results of the remaining matches?</div><a class='zoom' href='javascript:showlimage("timber-football1.jpg")'><img src='http://www.mscroggs.co.uk/img/320/timber-football1.jpg'></a><a class='zoom' href='javascript:showlimage("timber-football2.jpg")'><img src='http://www.mscroggs.co.uk/img/320/timber-football2.jpg'></a><div class='caption'>Timber using my swaps to wrongly predict the first match</div>
http://www.mscroggs.co.uk/blog/56
http://www.mscroggs.co.uk/blog/5607 Jul 2018 12:00:00 GMTMathsteroids<div class='note'>This is a post I wrote for round 1 of The Aperiodical's <a href="https://aperiodical.com/2018/06/announcing-the-big-internet-math-off/" target="new">Big Internet Math-Off 2018</a>, where <a href="https://aperiodical.com/2018/07/the-big-internet-math-off-round-1-matt-parker-v-matthew-scroggs/" target="new">Mathsteroids lost to MENACE</a>.</div><div class='paragraph'>A map projection is a way of representing the surface of a sphere, such as the Earth, on a flat surface. There is no way to represent all the features of a sphere on a flat surface, so if you want a map that shows a certain feature of the world, then you map will have to lose some other feature of the world in return.</div><div class='paragraph'>To show you what different map projections do to a sphere, I have created a version of the game <em>Asteroids</em> on a sphere. I call it <em>Mathsteroids</em></a>. You can play it <a href="http://www.mscroggs.co.uk/mathsteroids/sphereisometric">here</a>, or follow the links below to play on specific projections.</div><h3>Mercator projection</h3><div class='paragraph'>The most commonly used map projection is the Mercator projection, invented by Gerardus Mercator in 1569. The Mercator projection preserves angles: if two straight lines meet at an angle \(\theta\) on a sphere, then they will also meet at an angle \(\theta\) on the map. Keeping the angles the same, however, will cause the straight lines to no longer appear straight on the map, and the size of the same object in two different place to be very different.</div><div class='paragraph'>The angle preserving property means that lines on a constant bearing (eg 030° from North) will appear straight on the map. These constant bearing lines are not actually straight lines on the sphere, but when your ship is already being buffeted about by the wind, the waves, and the whims of drunken sailors, a reasonably straight line is the best you can expect.</div><div class='paragraph'>The picture below shows what three ships travelling in straight lines on the sphere look like on a Mercator projection.</div><a class='zoom' href='javascript:showlimage("sphere--Mercator.png")'><img src='http://www.mscroggs.co.uk/img/320/sphere--Mercator.jpg'></a><div class='paragraph'>To fully experience the Mercator projection, you can play <em>Mathsteroids</em> in Mercator projection mode <a href="http://www.mscroggs.co.uk/mathsteroids/sphereMercator">here</a>. Try flying North to see your spaceship become huge and distorted.</div><h3>Gall–Peters projection</h3><div class='paragraph'>The Gall–Peters projection (named after James Gall and Arno Peters) is an area-preserving projection: objects on the map will have the same area as objects on the sphere, although the shape of the object on the projection can be very different to its shape on the sphere.</div><div class='paragraph'>The picture below shows what three spaceships travelling in straight lines on a sphere look like on the Gall–Peters projection.</div><a class='zoom' href='javascript:showlimage("sphere--Gall.png")'><img src='http://www.mscroggs.co.uk/img/320/sphere--Gall.jpg'></a><div class='paragraph'>You can play <em>Mathsteroids</em> in Gall–Peters projection mode <a href="http://www.mscroggs.co.uk/mathsteroids/sphereGall">here</a>. I find this one much harder to play than the Mercator projection, as the direction you're travelling changes in a strange way as you move.</div><h3>Azimuthal projection</h3><a class='zoom' href='javascript:showlimage("UNemblem.png")'><img src='http://www.mscroggs.co.uk/img/320/UNemblem.jpg'></a><div class='caption'>The emblem of the UN</div><div class='paragraph'>An azimuthal projection makes a map on which the directions from the centre point to other points on the map are the same as the directions on the sphere. A map like this could be useful if, for example, you're a radio operator and need to quickly see which direction you should point your aerial to communicate with other points on the map.</div><div class='paragraph'>The azimuthal projection I've used in <em>Mathsteroids</em> also preserves distances: the distance from the centre to the another points on the map is proportional to the actual distance on the sphere. This projection is used as the emblem of the UN.</div><div class='paragraph'>The picture below shows three straight lines on this projection. You can play <em>Mathsteroids</em> in azimuthal mode <a href="http://www.mscroggs.co.uk/mathsteroids/sphereazim">here</a>.</div><a class='zoom' href='javascript:showlimage("sphere--azim.png")'><img src='http://www.mscroggs.co.uk/img/320/sphere--azim.jpg'></a><h3>Craig retroazimuthal projection</h3><div class='paragraph'>A retroazimuthal projection makes a map on which the directions to the centre point from other points on the map are the same as the directions on the sphere. If you're thinking that this is the same as the azimuthal projection, then you're too used to doing geometry on flat surfaces: on a sphere, the sum of the angles in a triangle depends on the size of the triangle, so the directions from A to B and from B to A aren't as closely related as you would expect.</div><div class='paragraph'>The Craig retroazimuthal projection was invented by James Ireland Craig in 1909. He used Mecca as his centre point to make a map that shows muslims across the world which direction they should face to pray.</div><div class='paragraph'>The picture below shows what three spaceships travelling in a straight lines on a sphere looks like on this projection.</div><a class='zoom' href='javascript:showlimage("sphere--Craig.png")'><img src='http://www.mscroggs.co.uk/img/320/sphere--Craig.jpg'></a><div class='paragraph'>You can play <em>Mathsteroids</em> in Craig retroazimuthal mode <a href="http://www.mscroggs.co.uk/mathsteroids/sphereCraig">here</a> to explore the projection yourself. This is perhaps the hardest of all to play, as (a) two different parts of the sphere overlap on the map, and (b) the map is actually infinitely tall, so quite a bit of it is off the edge of the visible game area.</div><h3>Stereographic projection</h3><div class='paragraph'>The final projection I'd like to show you is the stereographic projection.</div><div class='paragraph'>Imagine that a sphere is sitting on a 2D plane. Take a point on the sphere. Imagine a straight line through this point and the point at the top of the sphere. The point where this line meets the 2D plane is stereographic projection of the point on the sphere.</div><a class='zoom' href='javascript:showlimage("stereographic3D.png")'><img src='http://www.mscroggs.co.uk/img/320/stereographic3D.jpg'></a><div class='caption'>The stereographic projection</div><div class='paragraph'>This projection (backwards) can be used to represent the every complex number as a point on a sphere: this is called the Riemann sphere.</div><div class='paragraph'>To make <em>Mathseteroids</em> playable after this projection, I split the sphere into 2 hemisphere and projected each seperately to give two circles. You can play <em>Mathsteroids</em> in stereographic projection mode <a href="http://www.mscroggs.co.uk/mathsteroids/spherestereographic">here</a>. Three spaceships travelling in straight lines on this projection are shown below.</div><a class='zoom' href='javascript:showlimage("sphere--stereographic.png")'><img src='http://www.mscroggs.co.uk/img/320/sphere--stereographic.jpg'></a><div class='paragraph'>... and if you still don't like map projections, you can still enjoy playing <em>Mathsteroids</em> on an old fashioned <a href="http://www.mscroggs.co.uk/mathsteroids/torusflat">torus</a>. Or on a <a href="http://www.mscroggs.co.uk/mathsteroids/Kleinflat">Klein bottle</a> or the <a href="http://www.mscroggs.co.uk/mathsteroids/real-ppflat">real projective plane</a>. Don't forget to take a short break from playing to <s><a href="https://aperiodical.com/2018/07/the-big-internet-math-off-round-1-matt-parker-v-matthew-scroggs/" target="new">head over to The Aperiodical and vote</a></s> (voting now closed).</div>
http://www.mscroggs.co.uk/blog/55
http://www.mscroggs.co.uk/blog/5506 Jul 2018 12:00:00 GMTWorld Cup stickers 2018, pt. 2<div class='paragraph'>This year, like every World Cup year, I've been collecting stickers to fill the official Panini World Cup sticker album.
Back in March, I <a href="http://www.mscroggs.co.uk/blog/50">calculated that I should expect it to cost £268.99 to fill this year's album</a> (if I order the last 50 stickers).
As of 6pm yesterday, I need 47 stickers to complete the album (and have placed an order on the Panini website for these).</div>
<h3>So... How much did it cost?</h3>
<div class='paragraph'>In total, I have bought 1781 stickers (including the 47 I ordered) at a cost of £275.93. The plot below shows
the money spent against the number of stickers stuck in, compared with the what I predicted in March.</div>
<a class='zoom' href='javascript:showlimage("panini2018plot1.png")'><img src='http://www.mscroggs.co.uk/img/320/panini2018plot1.jpg'></a>
<div class='paragraph'>To create this plot, I've been keeping track of exactly which stickers were in each pack I bought. Using this data, we can
look for a few more things. If you want to play with the data yourself, there's a link at the bottom to download it.</div>
<h3>Swaps</h3>
<div class='paragraph'>The bar chart below shows the number of copies of each sticker I got (excluding the 47 that I ordered). Unsurprisingly, it looks a lot like
random noise.</div>
<a class='zoom' href='javascript:showlimage("panini2018plot3.png")'><img src='http://www.mscroggs.co.uk/img/320/panini2018plot3.jpg'></a>
<div class='paragraph'>The sticker I got most copies of was sticker 545, showing Panana player Armando Cooper.</div>
<a class='zoom' href='javascript:showlimage("sticker545.jpg")'><img src='http://www.mscroggs.co.uk/img/320/sticker545.jpg'></a>
<div class='caption'>Armando Cooper on sticker 545</div>
<div class='paragraph'>I got swaps of 513 different stickers, meaning I'm only 169 stickers short of filling a second album.</div>
<h3>First pack of all swaps</h3>
<div class='paragraph'>Everyone who has every done a sticker book will remember the awful feeling you get when you first get a pack of all swaps.
For me, the first time this happened was the 50th pack. The plot below shows when the first pack of all swaps occurred in 500,000 simulations.</div>
<a class='zoom' href='javascript:showlimage("panini2018plot2.png")'><img src='http://www.mscroggs.co.uk/img/320/panini2018plot2.jpg'></a>
<div class='paragraph'>Looks like I was really quite unlucky to get a pack of all swaps so soon.</div>
<h3>Duplicates in a pack</h3>
<div class='paragraph'>In all the 345 packs that I bought, there wasn't a single pack that contained two copies of the same sticker.
In fact, I don't remember <em>ever</em> getting two of the same sticker in a pack. For a while I've been wondering if this is because Panini
ensure that packs don't contain duplicates, or if it's simply very unlikely that they do.</div>
<div class='paragraph'>If it was down to unlikeliness, the probability of having no duplicates in one pack would be:</div>
\begin{align}
\mathbb{P}(\text{no duplicates in a pack}) &= 1 \times\frac{681}{682}\times\frac{680}{682}\times\frac{679}{682}\times\frac{678}{682}\\
&= 0.985
\end{align}
<div class='paragraph'>and the probability of none of my 345 containing a duplicate would be:</div>
\begin{align}
\mathbb{P}(\text{no duplicates in 345 packs})
&= 0.985^{345}\\
&= 0.00628
\end{align}
<div class='paragraph'>This is very very small, so it's safe to conclude that Panini do indeed ensure that packs do not contain duplicates.</div>
<h3>The data</h3>
<div class='paragraph'>If you'd like to have a play with the data yourself, you can download it <a href="https://drive.google.com/open?id=1b0d-mVH5i135_MFz2HaUbq2nMfhA87-Y" target="new">here</a>. <a href="http://www.mscroggs.co.uk/contact">Let me know</a> if
you do anything with it...</div>
http://www.mscroggs.co.uk/blog/54
http://www.mscroggs.co.uk/blog/5416 Jun 2018 12:00:00 GMTSunday Afternoon Maths LXVI<h3>Cryptic crossnumber #2</h3><div class='paragraph'>In this puzzle, the clues are written like clues from a cryptic crossword, but the answers are all numbers. You can download a printable pdf of this puzzle <a href="https://drive.google.com/open?id=18fgs3nnakOyqcqyYQ7FVRfMl-UU_luNk" target="new">here</a>.</div><table class='crossnumber'><tr><td style='background-color:black'></td><td style='background-color:black'></td><td>1</td><td>2</td><td> </td></tr><tr><td style='background-color:black'></td><td style='background-color:black'></td><td>3</td><td> </td><td style='background-color:black'></td></tr><tr><td style='background-color:black'></td><td style='background-color:black'></td><td> </td><td style='background-color:black'></td><td style='background-color:black'></td></tr><tr><td style='background-color:black'></td><td>4</td><td> </td><td style='background-color:black'></td><td style='background-color:black'></td></tr><tr><td>5</td><td> </td><td> </td><td style='background-color:black'></td><td style='background-color:black'></td></tr></table><table class='invisible'><tr><td width='49%' style='vertical-align:top'><h3>Across</h3><table class='invisible'><tr><td width='15px' style='text-align:left;vertical-align:top'><b>1</b></td><td style='text-align:left;vertical-align:top'>Found your far dented horn mixed to make square.</td><td width='15px' style='text-align:right;vertical-align:top'><b>(3)</b></td></tr><tr><td width='15px' style='text-align:left;vertical-align:top'><b>3</b></td><td style='text-align:left;vertical-align:top'>Eno back in Bowie's evening prime.</td><td width='15px' style='text-align:right;vertical-align:top'><b>(2)</b></td></tr><tr><td width='15px' style='text-align:left;vertical-align:top'><b>4</b></td><td style='text-align:left;vertical-align:top'>Prime legs.</td><td width='15px' style='text-align:right;vertical-align:top'><b>(2)</b></td></tr><tr><td width='15px' style='text-align:left;vertical-align:top'><b>5</b></td><td style='text-align:left;vertical-align:top'>Palindrome ends cubone, starts ninetales, inside poison ekans.</td><td width='15px' style='text-align:right;vertical-align:top'><b>(3)</b></td></tr></table></td><td></td><td width='49%' style='vertical-align:top'><h3>Down</h3><table class='invisible'><tr><td width='15px' style='text-align:left;vertical-align:top'><b>1</b></td><td style='text-align:left;vertical-align:top'>Odd confused elven elves hounded deerhound antenna.</td><td width='15px' style='text-align:right;vertical-align:top'><b>(5)</b></td></tr><tr><td width='15px' style='text-align:left;vertical-align:top'><b>2</b></td><td style='text-align:left;vertical-align:top'>Prime try of confused Sven with Beckham's second.</td><td width='15px' style='text-align:right;vertical-align:top'><b>(2)</b></td></tr><tr><td width='15px' style='text-align:left;vertical-align:top'><b>4</b></td><td style='text-align:left;vertical-align:top'>Prime even and Ian fed the being.</td><td width='15px' style='text-align:right;vertical-align:top'><b>(2)</b></td></tr></table></td></tr></table>
http://www.mscroggs.co.uk/puzzles/LXVI
http://www.mscroggs.co.uk/puzzles/LXVI20 May 2018 12:00:00 GMTSunday Afternoon Maths LXV<h3>Cryptic crossnumber #1</h3><div class='paragraph'>In this puzzle, the clues are written like clues from a cryptic crossword, but the answers are all numbers. You can download a printable pdf of this puzzle <a href="https://drive.google.com/open?id=1cS0TN2_qxYX5Jo-ijSio1ywRW33zPz04" target="new">here</a>.</div><table class='crossnumber'><tr><td>1</td><td> </td><td>2</td><td style='background-color:black'></td><td style='background-color:black'></td></tr><tr><td> </td><td style='background-color:black'></td><td> </td><td style='background-color:black'></td><td style='background-color:black'></td></tr><tr><td>3</td><td> </td><td> </td><td> </td><td>4</td></tr><tr><td style='background-color:black'></td><td style='background-color:black'></td><td> </td><td style='background-color:black'></td><td> </td></tr><tr><td style='background-color:black'></td><td style='background-color:black'></td><td>5</td><td> </td><td> </td></tr></table><table class='invisible'><tr><td width='49%' style='vertical-align:top'><h3>Across</h3><table class='invisible'><tr><td width='15px' style='text-align:left;vertical-align:top'><b>1</b></td><td style='text-align:left;vertical-align:top'>Triangular one then square.</td><td width='15px' style='text-align:right;vertical-align:top'><b>(3)</b></td></tr><tr><td width='15px' style='text-align:left;vertical-align:top'><b>3</b></td><td style='text-align:left;vertical-align:top'>Audible German no between tutus, for one square.</td><td width='15px' style='text-align:right;vertical-align:top'><b>(5)</b></td></tr><tr><td width='15px' style='text-align:left;vertical-align:top'><b>5</b></td><td style='text-align:left;vertical-align:top'>Irreducible ending Morpheus halloumi fix, then Trinity, then mixed up Neo.</td><td width='15px' style='text-align:right;vertical-align:top'><b>(3)</b></td></tr></table></td><td></td><td width='49%' style='vertical-align:top'><h3>Down</h3><table class='invisible'><tr><td width='15px' style='text-align:left;vertical-align:top'><b>1</b></td><td style='text-align:left;vertical-align:top'>Inside Fort Worth following unlucky multiple of eleven.</td><td width='15px' style='text-align:right;vertical-align:top'><b>(3)</b></td></tr><tr><td width='15px' style='text-align:left;vertical-align:top'><b>2</b></td><td style='text-align:left;vertical-align:top'>Palindrome two between two clickety-clicks.</td><td width='15px' style='text-align:right;vertical-align:top'><b>(5)</b></td></tr><tr><td width='15px' style='text-align:left;vertical-align:top'><b>4</b></td><td style='text-align:left;vertical-align:top'>Confused Etna honored thundery din became prime.</td><td width='15px' style='text-align:right;vertical-align:top'><b>(3)</b></td></tr></table></td></tr></table><h3>Breaking Chocolate</h3><div class='paragraph'>You are given a bar of chocolate made up of 15 small blocks arranged in a 3×5 grid.</div>
<a class='zoom' href='javascript:showlimage("choc.png")'><img src='http://www.mscroggs.co.uk/img/320/choc.jpg'></a>
<div class='paragraph'>You want to snap the chocolate bar into 15 individual pieces. What is the fewest number of snaps that you need to break the bar? (One snap consists of picking up one piece of chocolate and snapping it into two pieces.)</div><h3>Square and cube endings</h3><div class='paragraph'>How many positive two-digit numbers are there whose square and cube both end in the same digit?</div>
http://www.mscroggs.co.uk/puzzles/LXV
http://www.mscroggs.co.uk/puzzles/LXV13 May 2018 12:00:00 GMTSunday Afternoon Maths LXIV<h3>Equal lengths</h3><div class='paragraph'>The picture below shows two copies of the same rectangle with red and blue lines. The blue line visits the midpoint of the opposite side. The lengths shown in red and blue are of equal length.</div><a class='zoom' href='javascript:showlimage("rectangles506.png")'><img src='http://www.mscroggs.co.uk/img/320/rectangles506.jpg'></a><div class='paragraph'>What is the ratio of the sides of the rectangle?</div><h3>Digitless factor</h3><div class='paragraph'>Ted thinks of a three-digit number. He removes one of its digits to make a two-digit number.</div><div class='paragraph'>Ted notices that his three-digit number is exactly 37 times his two-digit number. What was Ted's three-digit number?</div><h3>Backwards fours</h3><div class='paragraph'>If A, B, C, D and E are all unique digits, what values would work with the following equation?</div>$$ABCCDE\times 4 = EDCCBA$$
http://www.mscroggs.co.uk/puzzles/LXIV
http://www.mscroggs.co.uk/puzzles/LXIV06 May 2018 12:00:00 GMTA bad Puzzle for Today<div class='paragraph'>Every morning just before 7am, one of the Today Programme's presenters reads out a puzzle. Yesterday, it was this puzzle:</div>
<blockquote>
In a given month, the probability of a certain daily paper either running a story about inappropriate behaviour at a party conference
or running one about somebody's pet being retrieved from a domestic appliance is exactly half the probability of the same paper containing
a photo of a Tory MP jogging. The probability of no such photo appearing is the same as that of there being a story about inappropriate
behaviour at a party conference. The probability of the paper running a story about somebody's pet being retrieved from a domestic appliance
is a quarter that of its containing a photo of a Tory MP jogging. What are the probabilities the paper will
(a) run the conference story, (b) run the pet story, (c) contain the jogging photo?
<div class='attr'><a href="http://www.bbc.co.uk/programmes/articles/nPJsJnqcGxtzyWZvTkpLJx/puzzle-for-today" target="new">Puzzle for Today, 1 May 2018</a></div>
</blockquote>
<div class='paragraph'>I'm <a href="http://www.robeastaway.com/blog/puzzle-maths" target="new">not the only one</a> to notice that some of Radio 4's daily puzzles are not great.
I think this puzzle is a great example of a terrible puzzle. You can already see the first problem with it: it's long and words and very hard to follow on the radio.
But maybe this isn't so important, as you can
<a href="http://www.bbc.co.uk/programmes/articles/nPJsJnqcGxtzyWZvTkpLJx/puzzle-for-today" target="new">read it here</a> after it's been read out.</div>
<div class='paragraph'>Once you've done this, you can re-write the puzzle as follows:
there are three news stories (\(A\), \(B\) and \(C\)) that the newspaper might publish in a month. We are given the following information:</div>
$$\mathbb{P}(A\text{ or }B)=\tfrac12\mathbb{P}(C)$$
$$1-\mathbb{P}(C)=\mathbb{P}(A)$$
$$\mathbb{P}(B)=\tfrac14\mathbb{P}(C)$$
<div class='paragraph'>To solve this puzzle, we need use the formula \(\mathbb{P}(A\text{ or }B)=\mathbb{P}(A)+\mathbb{P}(B)-\mathbb{P}(A\text{ and }B)\).
These Venn diagrams justify this formula:</div>
<a class='zoom' href='javascript:showlimage("venn-probab.png")'><img src='http://www.mscroggs.co.uk/img/320/venn-probab.jpg'></a>
<div class='caption'>Venn diagrams showing that \(\mathbb{P}(A\text{ or }B)=\mathbb{P}(A)+\mathbb{P}(B)-\mathbb{P}(A\text{ and }B)\).</div>
<div class='paragraph'>Using the information we were given in the question, we get:</div>
\begin{align}
\tfrac12\mathbb{P}(C)&=\mathbb{P}(A\text{ or }B)\\
&=1-\mathbb{P}(C)+\tfrac14\mathbb{P}(C)-\mathbb{P}(A\text{ and }B)\\
\mathbb{P}(C)&=\tfrac45(1-\mathbb{P}(A\text{ and }B)).
\end{align}
<div class='paragraph'>At this point we have reached the second problem with this puzzle: there's no answer unless we make some extra assumptions, and the question doesn't make it clear what we can assume.
But let's give the puzzle the benefit of the doubt and try some assumptions.</div>
<h3>Assumption 1: The events are mutually exclusive</h3>
<div class='paragraph'>If we assume that the events \(A\) and \(B\) are mutually exclusive—or, in other words, only one of these two articles can be published,
perhaps due to a lack of space—then we can use the fact that</div>
$$\mathbb{P}(A\text{ and }B)=0.$$
<div class='paragraph'>This means that
\(\mathbb{P}(C)=\tfrac45\),
\(\mathbb{P}(A)=\tfrac15\), and
\(\mathbb{P}(B)=\tfrac15\). There's a problem with this answer, though: the three probabilities add up to more than 1.</div>
<div class='paragraph'>This wouldn't be a problem, except we assumed that only one of the articles \(A\) and \(B\) could be published.
The probabilities adding up to more than 1 means that either \(A\) and \(C\) are not mutually exclusive or \(A\) and \(B\) are not mutually exclusive,
so \(C\) could be published alongside \(A\) or \(B\). There seems to be nothing special about the three news stories to mean that only some combinations
could be published together, so at this point I figured that this assumption was wrong and moved on.</div>
<div class='paragraph'>Today, however, the answer was <a href="http://www.bbc.co.uk/programmes/articles/nPJsJnqcGxtzyWZvTkpLJx/puzzle-for-today" target="new">posted</a>, and
this answer was given (without an working out). So we have a third problem with this puzzle: the answer that was given is wrong, or at the very best
is based on questionable assumptions.</div>
<h3>Assumption 2: The events are independent</h3>
<div class='paragraph'>If we assume that the events are independent—so one article being published doesn't affect whether or not another is published—then
we may use the fact that</div>
$$\mathbb{P}(A\text{ and }B)=\mathbb{P}(A)\mathbb{P}(B).$$
<div class='paragraph'>If we let \(c=\mathbb{P}(C)\), then we get:</div>
\begin{align}
\tfrac12c&=\mathbb{P}(A)+\mathbb{P}(B)-\mathbb{P}(A\text{ and }B)\\
&=\mathbb{P}(A)+\mathbb{P}(B)-\mathbb{P}(A)\mathbb{P}(B)\\
&=1-c+\tfrac14c-\tfrac14(1-c)c\\
\tfrac14c^2-\tfrac32c+1&=0.
\end{align}
<div class='paragraph'>You can use your favourite formula to solve this to find that \(c=3-\sqrt5\), and therefore
\(\mathbb{P}(A)=\sqrt5-2\) and
\(\mathbb{P}(B)=\tfrac34-\tfrac{\sqrt5}4\).</div>
<div class='paragraph'>In this case, our assumption appears to be more reasonable—as over the course of a month the stories published by a paper probably don't have
much of an effect on each other—but we have the fourth, and probably biggest problem with the puzzle: the question and answer are not interesting or surprising, and
the method is a bit tedious.</div>
http://www.mscroggs.co.uk/blog/53
http://www.mscroggs.co.uk/blog/5302 May 2018 12:00:00 GMTSunday Afternoon Maths LXIII<h3>Is it equilateral?</h3><div class='paragraph'>In the diagram below, \(ABDC\) is a square. Angles \(ACE\) and \(BDE\) are both 75°.</div><a class='zoom' href='javascript:showlimage("square-cd07.png")'><img src='http://www.mscroggs.co.uk/img/320/square-cd07.jpg'></a><div class='paragraph'>Is triangle \(ABE\) equilateral? Why/why not?</div><h3>Cube multiples</h3><div class='paragraph'>Six different (strictly) positive integers are written on the faces of a cube. The sum of the numbers on any two adjacent faces is a multiple of 6.</div><div class='paragraph'>What is the smallest possible sum of the six numbers?</div>
http://www.mscroggs.co.uk/puzzles/LXIII
http://www.mscroggs.co.uk/puzzles/LXIII22 Apr 2018 12:00:00 GMTAdvent calendar 2017<h3>Advent 2017 logic puzzle</h3><div class='paragraph'>2016's Advent calendar ended with a logic puzzle: It's nearly Christmas and something terrible has happened: Santa and his two elves have been cursed! The curse has led Santa to forget which present three childrenâ€”Alex, Ben and Carolâ€”want and where they live.</div><div class='paragraph'>The elves can still remember everything about Alex, Ben and Carol, but the curse is causing them to lie. One of the elves will lie on even numbered days and tell the truth on odd numbered days; the other elf will lie on odd numbered days and tell the truth on even numbered days. As is common in elf culture, each elf wears the same coloured clothes every day.</div><div class='paragraph'>Each child lives in a different place and wants a different present. (But a present may be equal to a home.) The homes and presents are each represented by a number from 1 to 9.</div><div class='paragraph'>Here are the clues:</div><div class="advent ad2017"><div class="door solved white"><span class="advent_solved_num">21</span><br>White shirt says: "Yesterday's elf lied: Carol wants <b>4</b>, <b>9</b> or <b>6</b>."</div><div class="door solved orange"><span class="advent_solved_num">10</span><br>Orange hat says: "<b>249</b> is my favourite number."</div><div class="door solved red"><span class="advent_solved_num">5</span><br>Red shoes says: "Alex lives at <b>1</b>, <b>9</b> or <b>6</b>."</div><div class="door solved blue"><span class="advent_solved_num">16</span><br>Blue shoes says: "I'm the same elf as yesterday. Ben wants <b>5</b>, <b>7</b> or <b>0</b>."</div><div class="door solved red"><span class="advent_solved_num">23</span><br>Red shoes says: "Carol wants a factor of <b>120</b>. I am yesterday's elf."</div><div class="door solved blue"><span class="advent_solved_num">4</span><br>Blue shoes says: "<b>495</b> is my favourite number."</div><div class="door solved blue"><span class="advent_solved_num">15</span><br>Blue shoes says: "Carol lives at <b>9</b>, <b>6</b> or <b>8</b>."</div><div class="door solved purple"><span class="advent_solved_num">22</span><br>Purple trousers says: "Carol wants a factor of <b>294</b>."</div><div class="door solved white"><span class="advent_solved_num">11</span><br>White shirt says: "<b>497</b> is my favourite number."</div><div class="door solved pink"><span class="advent_solved_num">6</span><br>Pink shirt says: "Ben does not live at the last digit of <b>106</b>."</div><div class="door solved blue"><span class="advent_solved_num">9</span><br>Blue shoes says: "Ben lives at <b>5</b>, <b>1</b> or <b>2</b>."</div><div class="door solved orange"><span class="advent_solved_num">20</span><br>Orange hat says: "Carol wants the first digit of <b>233</b>."</div><div class="door solved red"><span class="advent_solved_num">1</span><br>Red shoes says: "Alex wants <b>1</b>, <b>2</b> or <b>3</b>."</div><div class="door solved green"><span class="advent_solved_num">24</span><br>Green hat says: "The product of the six final presents and homes is <b>960</b>."</div><div class="door solved grey"><span class="advent_solved_num">17</span><br>Grey trousers says: "Alex wants the first digit of <b>194</b>."</div><div class="door solved pink"><span class="advent_solved_num">14</span><br>Pink shirt says: "One child lives at the first digit of <b>819</b>."</div><div class="door solved white"><span class="advent_solved_num">3</span><br>White shirt says: "Alex lives at <b>2</b>, <b>1</b> or <b>6</b>."</div><div class="door solved green"><span class="advent_solved_num">18</span><br>Green hat says: "Ben wants <b>1</b>, <b>5</b> or <b>4</b>."</div><div class="door solved green"><span class="advent_solved_num">7</span><br>Green hat says: "Ben lives at <b>3</b>, <b>4</b> or <b>3</b>."</div><div class="door solved grey"><span class="advent_solved_num">12</span><br>Grey trousers says: "Alex lives at <b>3</b>, <b>1</b> or <b>5</b>."</div><div class="door solved purple"><span class="advent_solved_num">19</span><br>Purple trousers says: "Carol lives at <b>2</b>, <b>6</b> or <b>8</b>."</div><div class="door solved red"><span class="advent_solved_num">8</span><br>Red shoes says: "The digits of <b>529</b> are the toys the children want."</div><div class="door solved green"><span class="advent_solved_num">13</span><br>Green hat says: "One child lives at the first digit of <b>755</b>."</div><div class="door solved red"><span class="advent_solved_num">2</span><br>Red shoes says: "Alex wants <b>1</b>, <b>4</b> or <b>2</b>."</div></div><br /><h3>24 December</h3><div class='paragraph'>Today's number is the smallest number with exactly 28 factors (including 1 and the number itself as factors).</div><h3>23 December</h3><div class='paragraph'>In the song <a href="https://en.wikipedia.org/wiki/Twelve_Days_of_Christmas_(song)" target="new"><i>The Twelve Days of Christmas</i></a>, how many presents have been given after 8 days?</div><h3>22 December</h3><div class='paragraph'>22 is two times an odd number. Today's number is the mean of all the answers on days (including today) that are two times an odd number.</div><div class='note'>Clarification: You are taking the mean for answers on days that are two times an odd numbers; ie. the days are two times odd, not the answers.</div><h3>21 December</h3><div class='paragraph'>The factors of 6 (excluding 6 itself) are 1, 2 and 3. \(1+2+3=6\), so 6 is a <i>perfect number</i>.</div><div class='paragraph'>Today's number is the only three digit perfect number.</div><h3>20 December</h3><div class='paragraph'>What is the largest number that cannot be written in the form \(10a+27b\), where \(a\) and \(b\) are nonnegative integers (ie \(a\) and \(b\) can be 0, 1, 2, 3, ...)?</div><h3>19 December</h3><div class='paragraph'>Put the digits 1 to 9 (using each digit exactly once) in the boxes so that the products are correct. Today's number is the smallest number that can be made using the digits in the red boxes.</div><table class='grid'><tr><td class='bsq red'></td><td>×</td><td class='bsq'></td><td>×</td><td class='bsq'></td><td>= 90</td></tr><tr><td>×</td><td class='g'> </td><td>×</td><td class='g'> </td><td>×</td><td></td></tr><tr><td class='bsq'></td><td>×</td><td class='bsq'></td><td>×</td><td class='bsq'></td><td>= 84</td></tr><tr><td>×</td><td class='g'> </td><td>×</td><td class='g'> </td><td>×</td><td></td></tr><tr><td class='bsq red'></td><td>×</td><td class='bsq red'></td><td>×</td><td class='bsq'></td><td>= 48</td></tr><tr><td>=<br />64</td><td></td><td>=<br />90</td><td></td><td>=<br />63</td><td class='nb nr'></td></tr></table><h3>18 December</h3><div class='paragraph'>Today's number is the maximum number of pieces that a (circular) pancake can be cut into with 17 straight cuts.</div><h3>17 December</h3><div class='paragraph'>Arrange the digits 1-9 in a 3×3 square so that every row makes a three-digit square number, the first column makes a multiple of 7 and the second column makes a multiple of 4.The number in the third column is today's number.</div><table class='biggrid'><tr><td class='bsq'></td><td class='bsq'></td><td class='bsq'></td><td align=left>square</td></tr><tr><td class='bsq'></td><td class='bsq'></td><td class='bsq'></td><td>square</td></tr><tr><td class='bsq'></td><td class='bsq'></td><td class='bsq'></td><td align=left>square</td></tr><tr><td>multiple of 7</td><td>multiple of 4</td><td><b>today's number</b></td></tr></table><h3>16 December</h3><div class='paragraph'>There are <a href="http://www.mscroggs.co.uk/puzzles/25">204 squares (of any size) in an 8×8 grid of squares</a>. Today's number is the number of rectangles (of any size) in a 2×19 grid of squares</div><h3>15 December</h3><div class='paragraph'>The string <i>ABBAABBBBB</i> is 10 characters long, contains only <i>A</i> and <i>B</i>, and contains at least three <i>A</i>s.</div><div class='paragraph'>Today's number is the number of different 10 character strings of <i>A</i>s and <i>B</i>s that have at least three <i>A</i>s.</div><h3>14 December</h3><div class='paragraph'>There are <a href="http://www.mscroggs.co.uk/puzzles/25">204 squares (of any size) in an 8×8 grid of squares</a>. Today's number is the number of squares in a 13×13 grid of squares</div><h3>13 December</h3><div class='paragraph'>A book has 754 pages (numbered 1 to 754: page 1 is on the left of the first double page spread, and page 754 is on the right of the final double page spread). What do the page numbers of the middle two-page spread add up to?</div><h3>12 December</h3><div class='paragraph'>Put the digits 1 to 9 (using each digit exactly once) in the boxes so that the sums are correct. Today's number is the product of the numbers in the red boxes.</div><table class='grid'><tr><td class='bsq'></td><td>+</td><td class='bsq red'></td><td>+</td><td class='bsq red'></td><td>= 17</td></tr><tr><td>+</td><td class='g'> </td><td>+</td><td class='g'> </td><td>+</td><td></td></tr><tr><td class='bsq'></td><td>+</td><td class='bsq'></td><td>+</td><td class='bsq'></td><td>= 7</td></tr><tr><td>+</td><td class='g'> </td><td>+</td><td class='g'> </td><td>+</td><td></td></tr><tr><td class='bsq'></td><td>+</td><td class='bsq red'></td><td>+</td><td class='bsq'></td><td>= 21</td></tr><tr><td>=<br />11</td><td></td><td>=<br />20</td><td></td><td>=<br />14</td><td class='nb nr'></td></tr></table><h3>11 December</h3><div class='paragraph'>Two more than today's number is the reverse of two times today's number.</div><h3>10 December</h3><div class='paragraph'>How many zeros does 1000! (ie 1000 × 999 × 998 × ... × 1) end with?</div><h3>9 December</h3><div class='paragraph'>Write the numbers 1 to 15 in a row. Underneath, write the same list without the first and last numbers. Continue this until you have just one number left.</div><table class='invisible'><tr><td> 1 </td><td> 2 </td><td> 3 </td><td> 4 </td><td> 5 </td><td> 6 </td><td> 7 </td><td> 8 </td><td> 9 </td><td> 10 </td><td> 11 </td><td> 12 </td><td> 13 </td><td> 14 </td><td> 15 </td></tr><tr><td> </td><td> 2 </td><td> 3 </td><td> 4 </td><td> 5 </td><td> 6 </td><td> 7 </td><td> 8 </td><td> 9 </td><td> 10 </td><td> 11 </td><td> 12 </td><td> 13 </td><td> 14 </td><td> </td></tr><tr><td> </td><td> </td><td> 3 </td><td> 4 </td><td> 5 </td><td> 6 </td><td> 7 </td><td> 8 </td><td> 9 </td><td> 10 </td><td> 11 </td><td> 12 </td><td> 13 </td><td> </td><td> </td></tr><tr><td colspan=15>etc.</td></tr></table><div class='paragraph'>Today's number is the sum of all the numbers you have written.</div><h3>8 December</h3><div class='paragraph'>The odd numbers are written in a pyramid.</div><table class='invisible'><tr><td></td><td></td><td>1</td><td></td><td></td></tr><tr><td></td><td>3</td><td></td><td>5</td><td></td></tr><tr><td>7</td><td></td><td>9</td><td></td><td>11</td></tr><tr><td colspan=5>etc.</td></tr></table><div class='paragraph'>What is the mean of the numbers in the 23rd row?</div><h3>7 December</h3><div class='paragraph'>The odd numbers are written in a pyramid.</div><table class='invisible'><tr><td></td><td></td><td>1</td><td></td><td></td></tr><tr><td></td><td>3</td><td></td><td>5</td><td></td></tr><tr><td>7</td><td></td><td>9</td><td></td><td>11</td></tr><tr><td colspan=5>etc.</td></tr></table><div class='paragraph'>What is the sum of the numbers in the seventh row?</div><h3>6 December</h3><div class='paragraph'>\(p(x)\) is a quadratic with real coefficients. For all real numbers \(x\),</div>$$x^2+4x+14\leq p(x)\leq 2x^2+8x+18$$<div class='paragraph'>\(p(2)=34\). What is \(p(6)\)?</div><h3>5 December</h3><div class='paragraph'>You start at A and are allow to walk left, right, up or down along the grid. The grid continues forever in every direction. After you have walked thirteen units, how many different locations could you be in?</div><a class='zoom' href='javascript:showlimage("grid2017.png")'><img src='http://www.mscroggs.co.uk/img/320/grid2017.jpg'></a><h3>4 December</h3><div class='paragraph'>Pick a three digit number whose digits are all different.</div><div class='paragraph'>Sort the digits into ascending and descending order to form two new numbers. Find the difference between these numbers.</div><div class='paragraph'>Repeat this process until the number stops changing. The final result is today's number.</div><h3>3 December</h3><div class='paragraph'>Put the digits 1 to 9 (using each digit exactly once) in the boxes so that the sums are correct. The sums should be read left to right and top to bottom ignoring the usual order of operations. For example, 4+3×2 is 14, not 10. Today's number is the product of the numbers in the red boxes.</div><table class='grid'><tr><td class='bsq'></td><td>+</td><td class='bsq'></td><td>+</td><td class='bsq red'></td><td>= 17</td></tr><tr><td>+</td><td class='g'> </td><td>÷</td><td class='g'> </td><td>×</td><td></td></tr><tr><td class='bsq'></td><td>÷</td><td class='bsq'></td><td>-</td><td class='bsq red'></td><td>= 1</td></tr><tr><td>-</td><td class='g'> </td><td>×</td><td class='g'> </td><td>÷</td><td></td></tr><tr><td class='bsq'></td><td>÷</td><td class='bsq red'></td><td>-</td><td class='bsq'></td><td>= 0</td></tr><tr><td>=<br />4</td><td></td><td>=<br />12</td><td></td><td>=<br />27</td><td class='nb nr'></td></tr></table><h3>2 December</h3><div class='paragraph'>There are three cards; one number is written on each card. You are told that the sums of pairs of cards are 99, 83 and 102.What is the sum of all three cards?</div><h3>1 December</h3><div class='paragraph'>Today's number is the smallest three digit number such that the sum of its digits is equal to the product of its digits.</div>
http://www.mscroggs.co.uk/puzzles/advent2017
http://www.mscroggs.co.uk/puzzles/advent201725 Dec 2017 12:00:00 GMT