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http://www.mscroggs.co.uk/
A surprising fact about quadrilaterals<div class='note'>This is a post I wrote for The Aperiodical's <a href="https://aperiodical.com/2020/03/announcing-the-big-lockdown-math-off/" target="new">Big Lock-Down Math-Off</a>. You can vote for (or against) me <a href="https://aperiodical.com/2020/05/the-big-lock-down-math-off-match-18" target="new">here</a> until 9am on Sunday...</div>
<div class='paragraph'>Recently, I came across a surprising fact: if you take any quadrilateral and join the midpoints of its sides, then you will form a parallelogram.</div>
<a class='zoom' href='javascript:showlimage("parallelograms.png")'><img src='http://www.mscroggs.co.uk/img/320/parallelograms.jpg'></a>
<div class='caption'>The blue quadrilaterals are all parallelograms.</div>
<div class='paragraph'>The first thing I thought when I read this was: "oooh, that's neat." The second thing I thought was: "why?" It's not too difficult to show why this is true; you might like to pause here and try to work out why yourself before reading on...</div>
<div class='paragraph'>To show why this is true, I started by letting \(\mathbf{a}\), \(\mathbf{b}\), \(\mathbf{c}\) and \(\mathbf{d}\) be the position vectors of the vertices of our quadrilateral. The position vectors of the midpoints of the edges are the averages of the position vectors of the two ends of the edge, as shown below.</div>
<a class='zoom' href='javascript:showlimage("quadproof1.png")'><img src='http://www.mscroggs.co.uk/img/320/quadproof1.jpg'></a>
<a class='zoom' href='javascript:showlimage("quadproof2.png")'><img src='http://www.mscroggs.co.uk/img/320/quadproof2.jpg'></a>
<div class='caption'>The position vectors of the corners and the midpoints of the edges.</div>
<div class='paragraph'>We want to show that the orange and blue vectors below are equal (as this is true of opposite sides of a parallelogram).</div>
<a class='zoom' href='javascript:showlimage("quadproof3.png")'><img src='http://www.mscroggs.co.uk/img/320/quadproof3.jpg'></a>
<div class='paragraph'>We can work these vectors out: the orange vector is$$\frac{\mathbf{d}+\mathbf{a}}2-\frac{\mathbf{a}+\mathbf{b}}2=\frac{\mathbf{d}-\mathbf{b}}2,$$
and the blue vector is$$\frac{\mathbf{c}+\mathbf{d}}2-\frac{\mathbf{b}+\mathbf{c}}2=\frac{\mathbf{d}-\mathbf{b}}2.$$</div>
<div class='paragraph'>In the same way, we can show that the other two vectors that make up the inner quadrilateral are equal, and so the inner quadrilateral is a parallelogram.</div>
<h3>Going backwards</h3>
<div class='paragraph'>Even though I now saw why the surprising fact was true, my wondering was not over. I started to think about going backwards.</div>
<div class='paragraph'>It's easy to see that if the outer quadrilateral is a square, then the inner quadrilateral will also be a square.</div>
<a class='zoom' href='javascript:showlimage("quad-sqsq.png")'><img src='http://www.mscroggs.co.uk/img/320/quad-sqsq.jpg'></a>
<div class='caption'>If the outer quadrilateral is a square, then the inner quadrilateral is also a square.</div>
<div class='paragraph'>It's less obvious if the reverse is true: if the inner quadrilateral is a square, must the outer quadrilateral also be a square? At first, I thought this felt likely to be true, but after a bit of playing around, I found that there are many non-square quadrilaterals whose inner quadrilaterals are squares. Here are a few:</div>
<a class='zoom' href='javascript:showlimage("square_shapes.png")'><img src='http://www.mscroggs.co.uk/img/320/square_shapes.jpg'></a>
<div class='caption'>A kite, a trapezium, a delta kite, an irregular quadrilateral and a cross-quadrilateral whose innner quadrilaterals are all a square.</div>
<div class='paragraph'>There are in fact infinitely many quadrilaterals whose inner quadrilateral is a square. You can explore them in this Geogebra applet by dragging around the blue point:</div>
<div class='paragraph'>As you drag the point around, you may notice that you can't get the outer quadrilateral to be a non-square rectangle (or even a non-square parallelogram). I'll leave you to figure out why not...</div>
http://www.mscroggs.co.uk/blog/84
http://www.mscroggs.co.uk/blog/8415 May 2020 12:00:00 GMTInteresting tautologies<div class='note'>This is a post I wrote for The Aperiodical's <a href="https://aperiodical.com/2020/03/announcing-the-big-lockdown-math-off/" target="new">Big Lock-Down Math-Off</a>. You can vote for (or against) me <a href="https://aperiodical.com/2020/05/the-big-lock-down-math-off-match-12/" target="new">here</a> until 9am on Tuesday...</div>
<div class='paragraph'>A few years ago, I made <a href="https://twitter.com/mathslogicbot" target="new">@mathslogicbot</a>, a Twitter bot that tweets logical tautologies.</div>
<div class='paragraph'>The statements that @mathslogicbot tweets are made up of variables (a to z) that can be either true or false, and the logical symbols
\(\lnot\) (not), \(\land\) (and), \(\lor\) (or), \(\rightarrow\) (implies), and \(\leftrightarrow\) (if and only if), as well as brackets.
A tautology is a statement that is always true, whatever values are assigned to the variables involved.</div>
<div class='paragraph'>To get an idea of how to interpret @mathslogicbot's statements, let's have a look at a few tautologies:</div>
<div class='paragraph'>\(( a \rightarrow a )\). This says "a implies a", or in other words "if a is true, then a is true". Hopefully everyone agrees that this is an always-true statement.</div>
<div class='paragraph'>\(( a \lor \lnot a )\). This says "a or not a": either a is true, or a is not true</div>
<div class='paragraph'>\((a\leftrightarrow a)\). This says "a if and only if a".</div>
<div class='paragraph'>\(\lnot ( a \land \lnot a )\). This says "not (a and not a)": a and not a cannot both be true.</div>
<div class='paragraph'>\(( \lnot a \lor \lnot \lnot a )\). I'll leave you to think about what this one means.</div>
<div class='paragraph'>(Of course, not all statements are tautologies. The statement \((b\land a)\), for example, is not a tautology as is can be true or false depending on the
values of \(a\) and \(b\).)</div>
<div class='paragraph'>While looking through @mathslogicbot's tweets, I noticed that a few of them are interesting, but most are downright rubbish.
This got me thinking: could I get rid of the bad tautologies like these, and make a list of just the "interesting" tautologies. To do this, we first need to
think of different ways tautologies can be bad.</div>
<div class='paragraph'>Looking at tautologies the @mathslogicbot has tweeted, I decided to exclude:</div>
<ul>
<li>tautologies like \((a\rightarrow\lnot\lnot\lnot\lnot a)\) that contain more than one \(\lnot\) in a row.</li>
<li>tautologies like \(((a\lor\lnot a)\lor b)\) that contain a shorter tautology. Instead, tautologies like \((\text{True}\lor b)\) should be considered.</li>
<li>tautologies like \(((a\land\lnot a)\rightarrow b)\) that contain a shorter contradiction (the opposite of a tautology).
Instead, tautologies like \((\text{False}\rightarrow b)\) should be considered.</li>
<li>tautologies like \((\text{True}\lor\lnot\text{True})\) or \(((b\land a)\lor\lnot(b\land a)\) that are another tautology (in this case \((a\lor\lnot a)\)) with a variable replaced
with something else.</li>
<li>tautologies containing substatements like \((a\land a)\), \((a\lor a)\) or \((\text{True}\land a)\) that are equivalent to just writing \(a\).</li>
<li>tautologies that contain a \(\rightarrow\) that could be replaced with a \(\leftrightarrow\), because it's more interesting if the implication goes both ways.</li>
<li>tautologies containing substatements like \((\lnot a\lor\lnot b)\) or \((\lnot a\land\lnot b)\) that could be replaced with similar terms (in these cases \((a\land b)\) and \((a\lor b)\)
respectively) without the \(\lnot\)s.</li>
<li>tautologies that are repeats of each other with the order changed. For example, only one of \((a\lor\lnot a)\) and \((\lnot a\lor a)\) should be included.</li>
</ul>
<div class='paragraph'>After removing tautologies like these, some of my favourite tautologies are:</div>
<ul>
<li>\(( \text{False} \rightarrow a )\)</li>
<li>\(( a \rightarrow ( b \rightarrow a ) )\)</li>
<li>\(( ( \lnot a \rightarrow a ) \leftrightarrow a )\)</li>
<li>\(( ( ( a \leftrightarrow b ) \land a ) \rightarrow b )\)</li>
<li>\(( ( ( a \rightarrow b ) \leftrightarrow a ) \rightarrow a )\)</li>
<li>\(( ( a \lor b ) \lor ( a \leftrightarrow b ) )\)</li>
<li>\(( \lnot ( ( a \land b ) \leftrightarrow a ) \rightarrow a )\)</li>
<li>\(( ( \lnot a \rightarrow b ) \leftrightarrow ( \lnot b \rightarrow a ) )\)</li>
</ul>
<div class='paragraph'>You can find a list of the first 500 "interesting" tautologues <a href="http://www.mscroggs.co.uk/blog/interesting-tautologies.txt">here</a>. Let me know <a href="https://twitter.com/mscroggs" target="new">on Twitter</a>
which is your favourite. Or let me know which ones you think are rubbish, and we can further refine the list...</div>
http://www.mscroggs.co.uk/blog/83
http://www.mscroggs.co.uk/blog/8303 May 2020 12:00:00 GMTLog-scaled axes<div class='paragraph'>Recently, you've probably seen a lot of graphs that look like this:</div>
<a class='zoom' href='javascript:showlimage("exp2.png")'><img src='http://www.mscroggs.co.uk/img/320/exp2.jpg'></a>
<div class='paragraph'>The graph above shows something that is growing exponentially: its equation is \(y=kr^x\), for some constants \(k\) and \(r\).
The value of the constant \(r\) is very important, as it tells you how quickly the value is going to grow. Using a graph of some data,
it is difficult to get an anywhere-near-accurate approximation of \(r\).</div>
<div class='paragraph'>The following plot shows three different exponentials. It's very difficult to say anything about them except that they grow very quickly above around \(x=15\).</div>
<a class='zoom' href='javascript:showlimage("exp_both.png")'><img src='http://www.mscroggs.co.uk/img/320/exp_both.jpg'></a>
<div class='caption'>\(y=2^x\), \(y=40\times 1.5^x\), and \(y=0.002\times3^x\)</div>
<div class='paragraph'>It would be nice if we could plot these in a way that their important properties—such as the value of the ratio \(r\)—were more clearly evident from the
graph. To do this, we start by taking the log of both sides of the equation:</div>
$$\log y=\log(kr^x)$$
<div class='paragraph'>Using the laws of logs, this simplifies to:</div>
$$\log y=\log k+x\log r$$
<div class='paragraph'>This is now the equation of a straight line, \(\hat{y}=m\hat{x}+c\), with \(\hat{y}=\log y\), \(\hat{x}=x\), \(m=\log r\) and \(c=\log k\). So if we plot
\(x\) against \(\log y\), we should get a straight line with gradient \(\log r\). If we plot the same three exponentials as above using a log-scaled \(y\)-axis, we get:</div>
<a class='zoom' href='javascript:showlimage("exp_both_log.png")'><img src='http://www.mscroggs.co.uk/img/320/exp_both_log.jpg'></a>
<div class='caption'>\(y=2^x\), \(y=40\times 1.5^x\), and \(y=0.002\times3^x\) with a log-scaled \(y\)-axis</div>
<div class='paragraph'>From this picture alone, it is very clear that the blue exponential has the largest value of \(r\), and we could quickly work out a decent approximation of this value
by calculating 10 (or the base of the log used if using a different log) to the power of the gradient.</div>
<h3>Log-log plots</h3>
<div class='paragraph'>Exponential growth isn't the only situation where scaling the axes is beneficial. <a href="http://www.mscroggs.co.uk/blog/series/my PhD thesis">In my research</a> in finite and boundary element methods,
it is common that the error of the solution \(e\) is given in terms of a grid parameter \(h\) by a polynomial of the form
\(e=ah^k\),
for some constants \(a\) and \(k\).</div>
<div class='paragraph'>We are often interested in the value of the power \(k\). If we plot \(e\) against \(h\), it's once again difficult to judge the value of \(k\) from the graph alone. The following
graph shows three polynomials.</div>
<a class='zoom' href='javascript:showlimage("e_h_both.png")'><img src='http://www.mscroggs.co.uk/img/320/e_h_both.jpg'></a>
<div class='caption'>\(y=x^2\), \(y=x^{1.5}\), and \(y=0.5x^3\)</div>
<div class='paragraph'>Once again is is difficult to judge any of the important properties of these. To improve this, we once again begin by taking the log of each side of the equation:</div>
$$\log e=\log (ah^k)$$
<div class='paragraph'>Applying the laws of logs this time gives:</div>
$$\log e=\log a+k\log h$$
<div class='paragraph'>This is now the equation of a straight line, \(\hat{y}=m\hat{x}+c\), with \(\hat{y}=\log e\), \(\hat{x}=\log h\), \(m=k\) and \(c=\log a\). So if we plot
\(\log x\) against \(\log y\), we should get a straight line with gradient \(k\).</div>
<div class='paragraph'>Doing this for the same three curves as above gives the following plot.</div>
<a class='zoom' href='javascript:showlimage("e_h_both_log.png")'><img src='http://www.mscroggs.co.uk/img/320/e_h_both_log.jpg'></a>
<div class='caption'>\(y=x^2\), \(y=x^{1.5}\), and \(y=0.5x^3\) with log-scaled \(x\)- and \(y\)-axes</div>
<div class='paragraph'>It is easy to see that the blue line has the highest value of \(k\) (as it has the highest gradient, and we could get a decent approximation of this value by finding the line's gradient.</div>
<br />
<div class='paragraph'>As well as making it easier to get good approximations of important parameters, making curves into straight lines in this way also makes it easier to plot the trend of real data.
Drawing accurate exponentials and polynomials is hard at the best of times; and real data will not exactly follow the curve, so drawing an exponential or quadratic of best fit will be an
even harder task. By scaling the axes first though, this task simplifies to drawing a straight line through the data; this is much easier.</div>
<div class='paragraph'>So next time you're struggling with an awkward curve, why not try turning it into a straight line first.</div>
http://www.mscroggs.co.uk/blog/82
http://www.mscroggs.co.uk/blog/8231 Mar 2020 12:00:00 GMTPhD thesis, chapter infinity<div class='paragraph'>Before we move on to some concluding remarks and notes about possible future work,
I must take this opportunity to thank my co-supervisors <a href="https://sites.google.com/site/timobetcke/" target="new">Timo Betcke</a> and <a href="https://iris.ucl.ac.uk/iris/browse/profile?upi=ENBUR31" target="new">Erik Burman</a>,
as without their help, support, and patience, this work would never have happened.</div>
<h3>Future work</h3>
<div class='paragraph'>There are of course many things related to the work in my thesis that could be worked on in the future by me or others.</div>
<div class='paragraph'>In the thesis, we presented the analysis of the weak imposition of Dirichlet, Neumann, mixed Dirichlet–Neumann, Robin, and Signorini boundary conditions on Laplace's equation;
and Dirichlet, and mixed Dirichlet–Neumann conditions on the Helmholtz equation. One area of future work would be to extend this analysis to other conditions, such as the imposition of
Robin conditions on the Helmholtz equation. It would also be of great interest to extend the method to other problems, such as Maxwell's equations. For Maxwell's equations, it looks like the
analysis will be significantly more difficult.</div>
<div class='paragraph'>In the problems in later chapters, in particular <a href="http://www.mscroggs.co.uk/blog/79">chapter 4</a>, the ill-conditioning of the matrices obtained from the method led to slow or even inaccurate solutions.
It would be interesting to look into alternative preconditioning methods for these problems as a way to improve the conditioning of these matrices. Developing these preconditioners appears to be
very important for Maxwell's equations: general the matrices involved when solving Maxwell's equations tend to be very badly ill-conditioned, and in the few experiments I ran to try out
the weak imposition of boundary conditions on Maxwell's equations, I was unable to get a good solution due to this.</div>
<h3>Your work</h3>
<div class='paragraph'>If you are a undergraduate or master's student and are interested in working on similar stuff to me, then you could look into
doing a PhD with <a href="https://sites.google.com/site/timobetcke/" target="new">Timo</a> and/or <a href="https://iris.ucl.ac.uk/iris/browse/profile?upi=ENBUR31" target="new">Erik</a> (my supervisors).
There are also many other people around working on similar stuff, including:
</div>
<ul>
<li><a href="http://www.homepages.ucl.ac.uk/~ucahism/" target="new">Iain Smears</a>, <a href="https://www.ucl.ac.uk/~ucahdhe/" target="new">David Hewett</a>,
and others in the <a href="https://www.ucl.ac.uk/maths/" target="new">Department of Mathematics at UCL</a>;</li>
<li><a href="http://www3.eng.cam.ac.uk/~gnw20/" target="new">Garth Wells</a> in the <a href="http://www.eng.cam.ac.uk/" target="new">Department of Engineering at Cambridge</a>;</li>
<li><a href="https://www.maths.ox.ac.uk/people/patrick.farrell" target="new">Patrick Farrell</a>, <a href="https://www.maths.ox.ac.uk/people/carolina.urzuatorres" target="new">Carolina Urzua-Torres</a>,
and others in the <a href="https://www.maths.ox.ac.uk/" target="new">Department of Mathematics at Oxford</a>;</li>
<li><a href="https://people.bath.ac.uk/eas25/" target="new">Euan Spence</a>, <a href="https://people.bath.ac.uk/masigg/" target="new">Ivan Graham</a>,
and others in the <a href="https://www.bath.ac.uk/departments/department-of-mathematical-sciences/" target="new">Department of Mathematics at Bath</a>;</li>
<li><a href="https://perso.ensta-paris.fr/~chaillat/index.php?page=home" target="new">Stéphanie Chaillat-Loseille</a>, and others at <a href="https://www.ensta-paristech.fr/" target="new">ENSTA in Paris</a>;</li>
<li><a href="https://www.simula.no/people/meg" target="new">Marie Rognes</a>, and others at <a href="https://www.simula.no/" target="new">Simula in Oslo</a>.</li>
</ul>
<div class='paragraph'>There are of course many, many more people working on this, and this list is in no way exhaustive. But hopefully this list can be a useful starting point for anyone interested in studying this area
of maths.</div>
http://www.mscroggs.co.uk/blog/81
http://www.mscroggs.co.uk/blog/8117 Feb 2020 12:00:00 GMTPhD thesis, chapter 5<div class='paragraph'>In the fifth and final chapter of my thesis, we look at how boundary conditions can be weakly imposed on the Helmholtz equation.</div><h3>Analysis</h3><div class='paragraph'>As in chapter 4, we must adapt the analysis of chapter 3 to apply to Helmholtz problems. The boundary operators for the Helmholtz equation satisfy less strongconditions than the operators for Laplace's equation (for Laplace's equation, the operators satisfy a condition called <i>coercivity</i>; for Helmholtz, the operatorssatisfy a weaker condition called <i>Gårding's inequality</i>), making proving results about Helmholtz problem harder.</div><div class='paragraph'>After some work, we are able to prove an <i>a priori</i> error bound (with \(a=\tfrac32\) for the spaces we use):</div>$$\left\|u-u_h\right\|\leqslant ch^{a}\left\|u\right\|$$<h3>Numerical results</h3><div class='paragraph'>As in the previous chapters, we use Bempp to show that computations with this method match the theory.</div><a class='zoom' href='javascript:showlimage("thesis5-numerics.png")'><img src='http://www.mscroggs.co.uk/img/320/thesis5-numerics.jpg'></a><div class='caption'>The error of our approximate solutions of a Dirichlet (left) and mixed Dirichlet–Neumann problems in the exterior of a sphere withmeshes with different values of \(h\). The dashed lines show order \(\tfrac32\) convergence.</div><h3>Wave scattering</h3><div class='paragraph'>Boundary element methods are often used to solve Helmholtz wave scattering problems. These are problems in which a sound wave is travelling though a medium (eg the air),then hits an object: you want to know what the sound wave that scatters off the object looks like.</div><div class='paragraph'>If there are multiple objects that the wave is scattering off, the boundary element method formulation can get quite complicated. When using weak imposition,the formulation is simpler: this one advantage of this method.</div><div class='paragraph'>The following diagram shows a sound wave scattering off a mixure of sound-hard and sound-soft spheres.Sound-hard objects reflect sound well, while sound-soft objects absorb it well.</div><a class='zoom' href='javascript:showlimage("thesis5-many.png")'><img src='http://www.mscroggs.co.uk/img/320/thesis5-many.jpg'></a><div class='caption'>A sound wave scattering off a mixture of sound-hard (white) and sound-soft (black) spheres.</div><div class='paragraph'>If you are trying to design something with particular properties—for example, a barrier that absorbs sound—you may want to solvelots of wave scattering problems on an object on some objects with various values taken for their reflective properties. This type of problem is often called an <i>inverse problem</i>.</div><div class='paragraph'>For this type of problem, weakly imposing boundary conditions has advantages: the discretisation of the Calderón projector can be reused for each problem, and only the termsdue to the weakly imposed boundary conditions need to be recalculated. This is an advantages as the boundary condition terms are much less expensive (ie they use much less time and memory)to calculate than the Calderón term that is reused.</div><br /><div class='paragraph'>This concludes chapter 5, the final chapter of my thesis.Why not celebrate reaching the end by cracking open the following figure before reading the concluding blog post.</div><a class='zoom' href='javascript:showlimage("champ.png")'><img src='http://www.mscroggs.co.uk/img/320/champ.jpg'></a><div class='caption'>An acoustic wave scattering off a sound-hard champagne bottle and a sound-soft cork.</div>
http://www.mscroggs.co.uk/blog/80
http://www.mscroggs.co.uk/blog/8016 Feb 2020 12:00:00 GMTAdvent calendar 2019<h3>25 December</h3><div class='paragraph'>It's nearly Christmas and something terrible has happened: while out on a test flight, Santa's sled was damaged and Santa, Rudolph and Blitzen fell to the ground over the Advent Isles.You need to find Santa and his reindeer before Christmas is ruined for everyone.</div><div class='paragraph'>You have gathered one inhabitant of the four largest Advent Isles—<span class='namered'>Rum</span>, <span class='namegreen'>Land</span>, <span class='nameblue'>Moon</span> and <span class='nameorange'>County</span>—and they are going to give you a series of clues about where Santa and his reindeer landed.However, one or more of the islanders you have gathered may have been involved in damaging Santa's sled and causing it to crash: any islander involved in this will lie to you to attempt to stopyou from finding Santa and his reindeer.Once you are ready to search for Santa, Rudolph and Blitzen, <a href="http://www.mscroggs.co.uk/adventmap">you can find the map by following this link</a>.</div><div class='paragraph'>Each of the clues will be about Santa's, Rudolph's or Blitzen's positions in Advent Standard Coordinates (ASC): ASC are given by six two-digit numbers with dots inbetween, for example<b>12.52.12.13.84.55</b>.For this example coordinate, the islanders will refer to(the first) <b>12</b> as the first coordinate,<b>52</b> as the second coordinate,(the second) <b>12</b> as the third coordinate,<b>13</b> as the fourth coordinate,<b>84</b> as the fifth coordinate, and<b>55</b> as the sixth coordinate.</div><div class='paragraph'>Here are the clues:</div><div class='advent ad2019'><div class='door solved red'><div class='advent_solved_num'>3</div>Rum says: "The product of all the digits in Blitzen's six coords is <b>432</b>."</div><div class='door solved blue'><div class='advent_solved_num'>21</div>Moon says: "Blitzen's fifth coord is <b>23</b>."</div><div class='door solved blue'><div class='advent_solved_num'>9</div>Moon says: "Blitzen's third coord is <b>23</b>."</div><div class='door solved green'><div class='advent_solved_num'>1</div>Land says: "Santa's third coord ends in <b>3</b>, <b>0</b> or <b>1</b>."</div><div class='door solved green'><div class='advent_solved_num'>2</div>Land says: "Santa's third coord ends in <b>2</b>, <b>0</b> or <b>3</b>."</div><div class='door solved red'><div class='advent_solved_num'>4</div>Rum says: "Santa's second coord ends in <b>3</b>, <b>4</b> or <b>1</b>."</div><div class='door solved red'><div class='advent_solved_num'>12</div>Rum says: "Rudolph's second and sixth coords are both <b>64</b>."</div><div class='door solved red'><div class='advent_solved_num'>10</div>Rum says: "All six of Rudolph's coords are factors of <b>256</b>."</div><div class='door solved blue'><div class='advent_solved_num'>18</div>Moon says: "Santa's fourth and fifth coords are both <b>79</b>."</div><div class='door solved orange'><div class='advent_solved_num'>24</div>County says: "Santa's third coord ends in <b>3</b>, <b>2</b> or <b>1</b>."</div><div class='door solved green'><div class='advent_solved_num'>22</div>Land says: "Santa's sixth coord is not <b>43</b>."</div><div class='door solved red'><div class='advent_solved_num'>7</div>Rum says: "Santa's sixth coord is <b>43</b>."</div><div class='door solved orange'><div class='advent_solved_num'>23</div>County says: "One of the digits of Santa's third coord is <b>7</b>."</div><div class='door solved final'><div class='advent_solved_num'>25</div>✔</div><div class='door solved green'><div class='advent_solved_num'>14</div>Land says: "Santa's third coord is <b>12</b>."</div><div class='door solved red'><div class='advent_solved_num'>5</div>Rum says: "Santa's first coord is <b>36</b>."</div><div class='door solved red'><div class='advent_solved_num'>15</div>Rum says: "Blitzen's first coord is <b>23</b>."</div><div class='door solved red'><div class='advent_solved_num'>17</div>Rum says: "The first digit of Santa's third coord is <b>1</b>."</div><div class='door solved orange'><div class='advent_solved_num'>8</div>County says: "Santa's third coord shares a factor (≠1) with <b>270</b>."</div><div class='door solved orange'><div class='advent_solved_num'>6</div>County says: "Santa's second coord is <b>21</b>."</div><div class='door solved green'><div class='advent_solved_num'>16</div>Land says: "Blitzen's second coord is <b>21</b>."</div><div class='door solved blue'><div class='advent_solved_num'>20</div>Moon says: "All six of Rudolph's coords are multiples of <b>8</b>."</div><div class='door solved blue'><div class='advent_solved_num'>11</div>Moon says: "The sum of Rudolph's six coords is <b>192</b>."</div><div class='door solved blue'><div class='advent_solved_num'>13</div>Moon says: "Santa's second coord is <b>21</b> or <b>11</b>."</div><div class='door solved blue'><div class='advent_solved_num'>19</div>Moon says: "Blitzen's fourth and sixth coords are both <b>11</b>."</div></div><div class='paragraph'>To find a point's ASC coordinates, split a map of the islands into a 9×9 grid, then number the rows and columns 1 to 9: the first two digits of ASC give the vertical then horizontal position of a square in this grid.The next two digits then give a smaller square when this square is then itself split into a 9×9 grid, and so on. An example is show below.</div><a class='zoom' href='javascript:showlimage("asc_example.png")'><img src='http://www.mscroggs.co.uk/img/320/asc_example.jpg'></a><div class='caption'>The ASC coordinates of this pair of flowers are <b>12.52.12.13.84.55</b> (click to enlarge).</div><div class='paragraph'>You can view the map <a href="http://www.mscroggs.co.uk/adventmap">here</a>.</div><h3>24 December</h3><div class='paragraph'>There are six 3-digit numbers with the property that the sum of their digits is equal to the product of their digits. Today's number is the largest of these numbers.</div><h3>23 December</h3><div class='paragraph'>Arrange the digits 1-9 in a 3×3 square so the 3-digits numbers formed in the rows and columns are the types of numbers given at the ends of the rows and columns.The number in the first column is today's number.</div><table class='biggrid'><tr><td class='bsq'></td><td class='bsq'></td><td class='bsq'></td><td class='rowend wide'>a multiple of 4</td></tr><tr><td class='bsq'></td><td class='bsq'></td><td class='bsq'></td><td class='rowend wide'>a cube</td></tr><tr><td class='bsq'></td><td class='bsq'></td><td class='bsq'></td><td class='rowend wide'>a multiple of 3</td></tr><tr><td><b>today's number</b></td><td>a cube</td><td>an odd number</td></tr></table><h3>22 December</h3><div class='paragraph'>In bases 3 to 9, the number 112 is: \(11011_3\), \(1300_4\), \(422_5\), \(304_6\), \(220_7\), \(160_8\), and \(134_9\).In bases 3, 4, 6, 8 and 9, these representations contain no digit 2.</div><div class='paragraph'>There are two 3-digit numbers that contain no 2 in their representations in all the bases between 3 and 9 (inclusive). Today's number is the smaller of these two numbers.</div><h3>21 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 smallest number you can make with the digits 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'></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 red'></td><td>= 8</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>=<br />12</td><td></td><td>=<br />5</td><td></td><td>=<br />28</td><td class='nb nr'></td></tr></table><h3>20 December</h3><div class='paragraph'>The integers from 2 to 14 (including 2 and 14) are written on 13 cards (one number per card). You and a friend take it in turns to take one of the numbers.</div><div class='paragraph'>When you have both taken five numbers, you notice that the product of the numbers you have collected is equal to the product of the numbers that your friend has collected.What is the product of the numbers on the three cards that neither of you has taken?</div><h3>19 December</h3><div class='paragraph'>The diagram below shows three squares and five circles.The four smaller circles are all the same size, and the red square's vertices are the centres of these circles.</div><a class='zoom' href='javascript:showlimage("advent2019circles.png")'><img src='http://www.mscroggs.co.uk/img/320/advent2019circles.jpg'></a><div class='paragraph'>The area of the blue square is 14 units. What is the area of the red square?</div><h3>18 December</h3><div class='paragraph'>The final round of game show starts with £1,000,000. You and your opponent take it in turn to take any value between £1 and £900.At the end of the round, whoever takes the final pound gets to take the money they have collected home, while the other player leaves with nothing.</div><div class='paragraph'>You get to take an amount first. How much money should you take to be certain that you will not go home with nothing?</div><h3>17 December</h3><div class='paragraph'>Eve picks a three digit number then reverses its digits to make a second number. The second number is larger than her original number.</div><div class='paragraph'>Eve adds her two numbers together; the result is 584. What was Eve's original number?</div><h3>16 December</h3><div class='paragraph'>Arrange the digits 1-9 in a 3×3 square so that:the median number in the first row is 6;the median number in the second row is 3;the mean of the numbers in the third row is 4;the mean of the numbers in the second column is 7;the range of the numbers in the third column is 2,The 3-digit number in the first column is today's number.</div><table class='biggrid'><tr><td class='bsq'></td><td class='bsq'></td><td class='bsq'></td><td class='rowend wide'>median 6</td></tr><tr><td class='bsq'></td><td class='bsq'></td><td class='bsq'></td><td class='rowend wide'>median 3</td></tr><tr><td class='bsq'></td><td class='bsq'></td><td class='bsq'></td><td class='rowend wide'>mean 4</td></tr><tr><td><b>today's number</b></td><td>mean 7</td><td>range 2</td></tr></table><h3>15 December</h3><div class='paragraph'>There are 5 ways to make 30 by multiplying positive integers (including the trivial way):</div><ul><li>30</li><li>2×15</li><li>3×10</li><li>5×6</li><li>2×3×5</li></ul><div class='paragraph'>Today's number is the number of ways of making 30030 by multiplying.</div><h3>14 December</h3><div class='paragraph'>During one day, a digital clock shows times from 00:00 to 23:59. How many times during the day do the four digits shown on the clock add up to 14?</div><h3>13 December</h3><div class='paragraph'>Each clue in this crossnumber (except 5A) gives a property of that answer that is true of no other answer. For example: 7A is a multiple of 13; but 1A, 3A, 5A, 1D, 2D, 4D, and 6D are all not multiples of 13.No number starts with 0.</div><table class='invisible'><tr><td style='margin-right:5px'><table class='crossnumber'><tr><td>1</td><td>2</td><td style='background-color:black'></td><td style='background-color:black'></td></tr><tr><td>3</td><td> </td><td>4</td><td style='background-color:black'></td></tr><tr><td style='background-color:black'></td><td>5</td><td> </td><td>6</td></tr><tr><td style='background-color:black'></td><td style='background-color:black'></td><td>7</td><td> </td></tr></table></td><td><table class='invisible'><tr><td colspan=3><b>Across</b></td></tr><tr><td style='text-align:left;vertical-align:top' width='15px'><b>1</b></td><td style='text-align:left;vertical-align:top'>The only multiple of 3.</td><td style='text-align:right;vertical-align:top' width='15px'><b>(2)</b></td></tr><tr><td style='text-align:left;vertical-align:top' width='15px'><b>3</b></td><td style='text-align:left;vertical-align:top'>The only number larger than 300.</td><td style='text-align:right;vertical-align:top' width='15px'><b>(3)</b></td></tr><tr><td style='text-align:left;vertical-align:top' width='15px'><b>5</b></td><td style='text-align:left;vertical-align:top'>Today's number.</td><td style='text-align:right;vertical-align:top' width='15px'><b>(3)</b></td></tr><tr><td style='text-align:left;vertical-align:top' width='15px'><b>7</b></td><td style='text-align:left;vertical-align:top'>The only multiple of 13.</td><td style='text-align:right;vertical-align:top' width='15px'><b>(2)</b></td></tr><tr><td colspan=3><b>Down</b></td></tr><tr><td style='text-align:left;vertical-align:top' width='15px'><b>1</b></td><td style='text-align:left;vertical-align:top'>The only square number.</td><td style='text-align:right;vertical-align:top' width='15px'><b>(2)</b></td></tr><tr><td style='text-align:left;vertical-align:top' width='15px'><b>2</b></td><td style='text-align:left;vertical-align:top'>The only number whose digits have product 4.</td><td style='text-align:right;vertical-align:top' width='15px'><b>(3)</b></td></tr><tr><td style='text-align:left;vertical-align:top' width='15px'><b>4</b></td><td style='text-align:left;vertical-align:top'>The only number whose digits add to 11.</td><td style='text-align:right;vertical-align:top' width='15px'><b>(3)</b></td></tr><tr><td style='text-align:left;vertical-align:top' width='15px'><b>6</b></td><td style='text-align:left;vertical-align:top'>The only number less than 20.</td><td style='text-align:right;vertical-align:top' width='15px'><b>(2)</b></td></tr></tbody></table></td></tr></table><h3>12 December</h3><div class='paragraph'>For a general election, the Advent isles are split into 650 constituencies. In each constituency, exactly 99 people vote: everyone votes for one of the two main parties: the <span class='namered'>Rum</span> party or the<span class='namegreen'>Land</span> party. The party that receives the most votes in each constituency gets an MAP (Member of Advent Parliament) elected to parliament to represent that constituency.</div><div class='paragraph'>In this year's election, exactly half of the 64350 total voters voted for the <span class='namered'>Rum</span> party. What is the largest number of MAPs that the <span class='namered'>Rum</span> party could have?</div><h3>11 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 red digits.</div><table class='grid'><tr><td class='bsq'></td><td>+</td><td class='bsq'></td><td>÷</td><td class='bsq red'></td><td>= 2</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>= 3</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>= 1</td></tr><tr><td>=<br />2</td><td></td><td>=<br />1</td><td></td><td>=<br />1</td><td class='nb nr'></td></tr></table><h3>10 December</h3><div class='paragraph'>For all values of \(x\), the function \(f(x)=ax+b\) satisfies</div>$$8x-8-x^2\leqslant f(x)\leqslant x^2.$$<div class='paragraph'>What is \(f(65)\)?</div><div class='note'>Edit: The left-hand quadratic originally said \(8-8x-x^2\). This was a typo and has now been corrected.</div><h3>9 December</h3><div class='paragraph'>Arrange the digits 1-9 in a 3×3 square so that:all the digits in the first row are odd;all the digits in the second row are even;all the digits in the third row are multiples of 3;all the digits in the second column are (strictly) greater than 6;all the digits in the third column are non-prime.The number in the first column is today's number.</div><table class='biggrid'><tr><td class='bsq'></td><td class='bsq'></td><td class='bsq'></td><td class='rowend wide'>all odd</td></tr><tr><td class='bsq'></td><td class='bsq'></td><td class='bsq'></td><td class='rowend wide'>all even</td></tr><tr><td class='bsq'></td><td class='bsq'></td><td class='bsq'></td><td class='rowend wide'>all multiples of 3</td></tr><tr><td><b>today's number</b></td><td>all >6</td><td>all non-prime</td></tr></table><h3>8 December</h3><div class='paragraph'>Carol uses the digits from 0 to 9 (inclusive) exactly once each to write five 2-digit even numbers, then finds their sum. What is the largest number she could have obtained?</div><h3>7 December</h3><div class='paragraph'>The sum of the coefficients in the expansion of \((x+1)^5\) is 32. Today's number is the sum of the coefficients in the expansion of \((2x+1)^5\).</div><h3>6 December</h3><div class='paragraph'>Noel's grandchildren were in born in November in consecutive years. Each year for Christmas, Noel gives each of his grandchildren their age in pounds.</div><div class='paragraph'>Last year, Noel gave his grandchildren a total of £208. How much will he give them in total this year?</div><h3>5 December</h3><div class='paragraph'>28 points are spaced equally around the circumference of a circle. There are 3276 ways to pick three of these points.The three picked points can be connected to form a triangle. Today's number is the number of these triangles that are isosceles.</div><h3>4 December</h3><div class='paragraph'>There are 5 ways to tile a 3×2 rectangle with 2×2 squares and 2×1 dominos.</div><div class='paragraph'>Today's number is the number of ways to tile a 9×2 rectangle with 2×2 squares and 2×1 dominos.</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 largest number you can make with the digits in the red boxes.</div><table class='grid'><tr><td class='bsq'></td><td>+</td><td class='bsq'></td><td>+</td><td class='bsq'></td><td>= 21</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'></td><td>+</td><td class='bsq'></td><td>= 10</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 red'></td><td>= 14</td></tr><tr><td>=<br />21</td><td></td><td>=<br />10</td><td></td><td>=<br />14</td><td class='nb nr'></td></tr></table><h3>2 December</h3><div class='paragraph'>You have 15 sticks of length 1cm, 2cm, ..., 15cm (one of each length). How many triangles can you make by picking three sticks and joining their ends?</div><div class='note'>Note: Three sticks (eg 1, 2 and 3) lying on top of each other does not count as a triangle.</div><div class='note'>Note: Rotations and reflections are counted as the same triangle.</div><h3>1 December</h3><div class='paragraph'>If you write out the numbers from 1 to 1000 (inclusive), how many times will you write the digit 1?</div>
http://www.mscroggs.co.uk/puzzles/advent2019
http://www.mscroggs.co.uk/puzzles/advent201931 Dec 2019 12:00:00 GMTSunday Afternoon Maths LXVII<h3>Coloured weights</h3><div class='paragraph'>You have six weights. Two of them are red, two are blue, two are green. One weight of each colour is heavier than the other; the three heavy weights all weigh the same, and the three lighter weights also weigh the same.</div><div class='paragraph'>Using a scale twice, can you split the weights into two sets by weight?</div><h3>Not Roman numerals</h3><div class='paragraph'>The letters \(I\), \(V\) and \(X\) each represent a different digit from 1 to 9. If</div>$$VI\times X=VVV,$$<div class='paragraph'>what are \(I\), \(V\) and \(X\)?</div>
http://www.mscroggs.co.uk/puzzles/LXVII
http://www.mscroggs.co.uk/puzzles/LXVII19 May 2019 12:00:00 GMTAdvent calendar 2018<h3>Advent 2018 Logic Puzzle</h3><div class='paragraph'>2018's Advent calendar ended with a logic puzzle: It's nearly Christmas and something terrible has happened: one of Santa's five helpers—<span class='nameorange'>Jo Ranger</span>, <span class='namepurple'>Kip Urples</span>, <span class='nameblue'>Bob Luey</span>, <span class='namered'>Fred Metcalfe</span>, and <span class='namegreen'>Meg Reeny</span>—has stolen all the presents during the North Pole's annual Sevenstival. You need to find the culprit before Christmas is ruined for everyone.</div><div class='paragraph'>Every year in late November, Santa is called away from the North Pole for a ten hour meeting in which a judgemental group of elders decide who has been good and who has been naughty. While Santa is away, it is traditional for his helpers celebrate Sevenstival.Sevenstival gets in name from the requirement that every helper must take part in exactly seven activities during the celebration; this year'savailable activities were billiards, curling, having lunch, solving maths puzzles, table tennis, skiing, chess, climbing and ice skating.</div><div class='paragraph'>Each activity must be completed in one solid block: it is forbidden to spend some time doing an activity, take a break to do something else then return to the first activity.This year's Sevenstival took place between 0:00 and 10:00 (North Pole standard time).</div><div class='paragraph'>During this year's Sevenstival, one of Santa's helpers seven activities included stealing all the presents from Santa's workshop.Santa's helpers have 24 pieces of information to give to you, but the culprit is going to lie about everything in an attempt to confuse you, so be careful who you trust.</div><div class='paragraph'>Here are the clues:</div><div class="advent ad2018"><div class="door solved green"><span class="advent_solved_num">1</span><br>Meg says: "Between <b>2:33</b> and curling, I played billiards with Jo."</div><div class="door solved purple"><span class="advent_solved_num">15</span><br>Kip says: "The curling match lasted <b>323</b> mins."</div><div class="door solved red"><span class="advent_solved_num">24</span><br>Fred says: "In total, Jo and Meg spent <b>1</b> hour and <b>57</b> mins having lunch."</div><div class="door solved green"><span class="advent_solved_num">8</span><br>Meg says: "A total of <b>691</b> mins were spent solving maths puzzles."</div><div class="door solved orange"><span class="advent_solved_num">17</span><br>Jo says: "I played table tennis with Fred and Meg for <b>2</b>+<b>8</b>+<b>5</b> mins."</div><div class="door solved green"><span class="advent_solved_num">23</span><br>Meg says: "<b>1:32</b> was during my 83 min ski"</div><div class="door solved green"><span class="advent_solved_num">7</span><br>Meg says: "The number of mins the curling game lasted is a factor of <b>969</b>."</div><div class="door solved orange"><span class="advent_solved_num">16</span><br>Jo says: "I started skiing with Bob, and finished before Bob at <b>8:45</b>."</div><div class="door solved orange"><span class="advent_solved_num">5</span><br>Jo says: "At <b>4:45</b>, Fred, Bob, Kip and I started a curling match."</div><div class="door solved red"><span class="advent_solved_num">14</span><br>Fred says: "I spent <b>135</b> mins playing chess with Meg."</div><div class="door solved green"><span class="advent_solved_num">20</span><br>Meg says: "Jo started skiing at <b>7:30</b>."</div><div class="door solved blue"><span class="advent_solved_num">4</span><br>Bob says: "I went for a <b>150</b> min ski."</div><div class="door solved purple"><span class="advent_solved_num">13</span><br>Kip says: "Jo started skiing at <b>6:08</b>."</div><div class="door solved red"><span class="advent_solved_num">22</span><br>Fred says: "Bob, Kip and I finished lunch at <b>3:30</b>."</div><div class="door solved blue"><span class="advent_solved_num">6</span><br>Bob says: "I played billiards with Kip from 0:00 until <b>1:21</b>."</div><div class="door solved red"><span class="advent_solved_num">12</span><br>Fred says: "Between 3:30 and 4:45, there were <b>3</b> people climbing."</div><div class="door solved red"><span class="advent_solved_num">21</span><br>Fred says: "In total, Bob, Meg and I spent <b>269</b> mins ice skating."</div><div class="door solved green"><span class="advent_solved_num">10</span><br>Meg says: "Between 0:00 and <b>1:10</b>, I was ice skating."</div><div class="door solved orange"><span class="advent_solved_num">19</span><br>Jo says: "At <b>1:12</b>, Fred and I were both in the middle of maths puzzles."</div><div class="door solved orange"><span class="advent_solved_num">3</span><br>Jo says: "Straight after curling, I had a <b>108</b> min game of chess with Kip."</div><div class="door solved red"><span class="advent_solved_num">9</span><br>Fred says: "At <b>2:52</b>, I started having lunch with Bob and Kip."</div><div class="door solved orange"><span class="advent_solved_num">18</span><br>Jo says: "I spent <b>153</b> mins solving maths puzzles."</div><div class="door solved red"><span class="advent_solved_num">2</span><br>Fred says: "I was solving maths puzzles for <b>172</b> mins."</div><div class="door solved green"><span class="advent_solved_num">11</span><br>Meg says: "I spent <b>108</b> mins solving maths puzzles with Bob."</div></div><br /><h3>24 December</h3><div class='paragraph'>1,0,2,0,1,1</div><div class='paragraph'>The sequence of six numbers above has two properties:</div><ol><li>Each number is either 0, 1 or 2.</li><li>Each pair of consecutive numbers adds to (strictly) less than 3.</li></ol><div class='paragraph'>Today's number is the number of sequences of six numbers with these two properties</div><h3>23 December</h3><div class='paragraph'>Today's number is the area of the largest area rectangle with perimeter 46 and whose sides are all integer length.</div><h3>22 December</h3><div class='paragraph'>In base 2, 1/24 is0.0000101010101010101010101010...</div><div class='paragraph'>In base 3, 1/24 is0.0010101010101010101010101010...</div><div class='paragraph'>In base 4, 1/24 is0.0022222222222222222222222222...</div><div class='paragraph'>In base 5, 1/24 is0.0101010101010101010101010101...</div><div class='paragraph'>In base 6, 1/24 is0.013.</div><div class='paragraph'>Therefore base 6 is the lowest base in which 1/24 has a finite number of digits.</div><div class='paragraph'>Today's number is the smallest base in which 1/10890 has a finite number of digits.</div><div class='note'>Note: 1/24 always represents 1 divided by twenty-four (ie the 24 is written in decimal).</div><h3>21 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 smallest number you can make using the digits in the red boxes.</div><table class='grid'><tr><td class='bsq'></td><td>+</td><td class='bsq'></td><td>÷</td><td class='bsq'></td><td>= 2</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>= 31</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 red'></td><td>= 42</td></tr><tr><td>=<br />37</td><td></td><td>=<br />13</td><td></td><td>=<br />-2</td><td class='nb nr'></td></tr></table><h3>20 December</h3><div class='paragraph'>Today's number is the sum of all the numbers less than 40 that are not factors of 40.</div><h3>19 December</h3><div class='paragraph'>Today's number is the number of 6-dimensional sides on a 8-dimensional hypercube.</div><h3>18 December</h3><div class='paragraph'>There are 6 terms in the expansion of \((x+y+z)^2\):</div>$$(x+y+z)^2=x^2+y^2+z^2+2xy+2yz+2xz$$<div class='paragraph'>Today's number is number of terms in the expansion of \((x+y+z)^{16}\).</div><h3>17 December</h3><div class='paragraph'>For \(x\) and \(y\) between 1 and 9 (including 1 and 9), I write a number at the co-ordinate \((x,y)\): if \(x\lt y\), I write \(x\); if not,I write \(y\).</div><div class='paragraph'>Today's number is the sum of the 81 numbers that I have written.</div><h3>16 December</h3><div class='paragraph'>Arrange the digits 1-9 in a 3×3 square so that the first row makes a triangle number, the second row's digits are all even, the third row's digits are all odd; the first column makes a square number, and the second column makes a cube number.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 class='rowend'>triangle</td></tr><tr><td class='bsq'></td><td class='bsq'></td><td class='bsq'></td><td class='rowend'>all digits even</td></tr><tr><td class='bsq'></td><td class='bsq'></td><td class='bsq'></td><td class='rowend'>all digits odd</td></tr><tr><td>square</td><td>cube</td><td><b>today's number</b></td></tr></table><h3>15 December</h3><div class='paragraph'>Today's number is smallest three digit palindrome whose digits are all non-zero, and that is not divisible by any of its digits.</div><h3>14 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>= 10</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>= 3</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'></td><td>×</td><td class='bsq red'></td><td>= 33</td></tr><tr><td>=<br />7</td><td></td><td>=<br />3</td><td></td><td>=<br />3</td><td class='nb nr'></td></tr></table><h3>13 December</h3><div class='paragraph'>There is a row of 1000 lockers numbered from 1 to 1000. Locker 1 is closed and locked and the rest are open.</div><div class='paragraph'>A queue of people each do the following (until all the lockers are closed):</div><ul><li>Close and lock the lowest numbered locker with an open door.</li><li>Walk along the rest of the queue of lockers and change the state (open them if they're closed and close them if they're open) of all the lockersthat are multiples of the locker they locked.</li></ul><div class='paragraph'>Today's number is the number of lockers that are locked at the end of the process.</div><div class='note'>Note: closed and locked are different states.</div><h3>12 December</h3><div class='paragraph'>There are 2600 different ways to pick three vertices of a regular 26-sided shape. Sometime the three vertices you pick form a right angled triangle.</div><a class='zoom' href='javascript:showlimage("advent2018-12.png")'><img src='http://www.mscroggs.co.uk/img/320/advent2018-12.jpg'></a><div class='caption'>These three vertices form a right angled triangle.</div><div class='paragraph'>Today's number is the number of different ways to pick three vertices of a regular 26-sided shape so that the three vertices make a right angled triangle.</div><br style='clear:both'> <h3>11 December</h3><div class='note'>This puzzle is inspired by <a href="http://www.mscroggs.co.uk/puzzles/143">a puzzle Woody showed me at MathsJam</a>.</div><div class='paragraph'>Today's number is the number \(n\) such that $$\frac{216!\times215!\times214!\times...\times1!}{n!}$$ is a square number.</div><h3>10 December</h3><div class='paragraph'>The equation \(x^2+1512x+414720=0\) has two integer solutions.</div><div class='paragraph'>Today's number is the number of (positive or negative) integers \(b\) such that \(x^2+bx+414720=0\) has two integer solutions.</div><h3>9 December</h3><div class='paragraph'>Today's number is the number of numbers between 10 and 1,000 that contain no 0, 1, 2 or 3.</div><h3>8 December</h3><div class='paragraph'>Arrange the digits 1-9 in a 3×3 square so: each digit the first row is the number of letters in the (English) name of the previous digit, each digit in the second row is one less than the previous digit, each digit in the third row is a multiple of the previous digit; the second column is an 3-digit even number, and the third column contains one even digit.The number in the first column is today's number.</div><table class='biggrid'><tr><td class='bsq'></td><td class='bsq'></td><td class='bsq'></td><td class='rowend wide'>each digit is the number of letters in the previous digit</td></tr><tr><td class='bsq'></td><td class='bsq'></td><td class='bsq'></td><td class='rowend wide'>each digit is one less than previous</td></tr><tr><td class='bsq'></td><td class='bsq'></td><td class='bsq'></td><td class='rowend wide'>each digit is multiple of previous</td></tr><tr><td><b>today's number</b></td><td>even</td><td>1 even digit</td></tr></table><div class='edit'>Edit: There was a mistake in this puzzle: the original had two solutions. If you entered the wrong solution, it will automatically change to the correct one.</div><h3>7 December</h3><div class='paragraph'>There is a row of 1000 closed lockers numbered from 1 to 1000 (inclusive). Near the lockers, there is a bucket containing the numbers 1 to 1000 (inclusive) written on scraps of paper.</div><div class='paragraph'>1000 people then each do the following:</div><ul><li>Pick a number from the bucket (and don't put it back).</li><li>Walk along the row of lockers and change the state(open them if they're closed and close them if they're open)of all the lockers that are multiples of the number they picked (including the number they picked).</li></ul><div class='paragraph'>Today's number is the number of lockers that will be closed at the end of this process.</div><h3>6 December</h3><div class='note'>This puzzle is inspired by a puzzle that <a href="https://puzzlecritic.wordpress.com/2016/08/02/the-vault-my-favourite-puzzle/" target="new">Daniel Griller</a> showed me.</div><div class='paragraph'>Write down the numbers from 12 to 22 (including 12 and 22). Under each number, write down its largest odd factor*.</div><div class='paragraph'>Today's number is the sum of all these odd factors.</div><div class='note'>* If a number is odd, then its largest odd factor is the number itself.</div><h3>5 December</h3><div class='paragraph'>I make a book by taking 111 sheets of paper, folding them all in half, then stapling them all together through the fold.I then number the pages from 1 to 444.</div><div class='paragraph'>Today's number is the sum of the two page numbers on the centre spread of my book.</div><h3>4 December</h3><div class='paragraph'>Today's number is the number of 0s that 611! (611×610×...×2×1) ends in.</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 red'></td><td>+</td><td class='bsq'></td><td>+</td><td class='bsq'></td><td>= 11</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>= 11</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>= 11</td></tr><tr><td>=<br />-11</td><td></td><td>=<br />11</td><td></td><td>=<br />11</td><td class='nb nr'></td></tr></table><h3>2 December</h3><div class='paragraph'>Today's number is the area of the largest dodecagon that it's possible to fit inside a circle with area \(\displaystyle\frac{172\pi}3\).</div><h3>1 December</h3><div class='paragraph'>There are 5 ways to write 4 as the sum of 1s and 2s:</div><ul><li>1+1+1+1</li><li>2+1+1</li><li>1+2+1</li><li>1+1+2</li><li>2+2</li></ul><div class='paragraph'>Today's number is the number of ways you can write 12 as the sum of 1s and 2s.</div>
http://www.mscroggs.co.uk/puzzles/advent2018
http://www.mscroggs.co.uk/puzzles/advent201831 Dec 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 GMT