mscroggs.co.uk Blog & PuzzlesNew posts from mscroggs.co.uk.
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A non-converging LaTeX document<div class='paragraph'>Today, I wrote <a href="https://github.com/mscroggs/annoying-latex-file/blob/master/annoying.tex" target="new">this LaTeΧ document</a>.
It only uses default LaTeΧ functionality, and uses no packages.</div>
<div class='paragraph'>If you compile it (twice so that cross references are not <b>??</b>s), you will get a four page document (pages numbered i to iv), that repeatedly says:
"This document ends on page v."</div>
<a class='zoom' href='javascript:showlimage("annoying-latex-compiled1.png")'><img src='http://www.mscroggs.co.uk/img/320/annoying-latex-compiled1.jpg'></a>
<div class='caption'>This document ends on page iv and says: "This document ends on page v."</div>
<div class='paragraph'>If you then recompile it, you will get a five page document (pages numbered i to v), that repeatedly says:
"This document ends on page iv."</div>
<a class='zoom' href='javascript:showlimage("annoying-latex-compiled2.png")'><img src='http://www.mscroggs.co.uk/img/320/annoying-latex-compiled2.jpg'></a>
<div class='caption'>This document ends on page v and says: "This document ends on page iv."</div>
<div class='paragraph'>Repeatedly compiling it will alternately give these two results, and so the process of compiling this LaTeX document does not converge.</div>
http://www.mscroggs.co.uk/blog/68
http://www.mscroggs.co.uk/blog/6815 Sep 2019 12:00:00 GMTTMiP 2019 treasure punt<div class='paragraph'>This week, I've been in Cambridge for <a href="http://talkingmathsinpublic.uk/" target="new">Talking Maths in Public (TMiP)</a>. TMiP is a conference for anyone involved in—or interested in getting involved
in—any sort of maths outreach, enrichment, or public engagement activity. It was really good, and I highly recommend coming to TMiP 2021.</div>
<div class='paragraph'>The Saturday morning at TMiP was filled with a choice of activities, including a treasure punt (a treasure hunt on a punt) written by me. This post contains the puzzle from the treasure punt for
anyone who was there and would like to revisit it, or anyone who wasn't there and would like to give it a try. In case you're not current in Cambridge on a punt, the clues that you were meant to
spot during the punt are given behing spoiler tags (hover/click to reveal).</div>
<h3>Instructions</h3>
<div class='paragraph'>Each boat was given a copy of the instructions, and a box that was locked using a combination lock.</div>
<a class='zoom' href='javascript:showlimage("punt19_1.png")'><img src='http://www.mscroggs.co.uk/img/320/punt19_1.jpg'></a>
<a class='zoom' href='javascript:showlimage("punt19_2.png")'><img src='http://www.mscroggs.co.uk/img/320/punt19_2.jpg'></a>
<div class='caption'>The instructions, download a pdf <a href="https://drive.google.com/file/d/1c5kKdtNNMYAXzBCntGEl70A0blx1dz5d/view?usp=sharing" target="new">here</a>.</div>
<div class='paragraph'></div>
<a class='zoom' href='javascript:showlimage("punt19_boxes.jpg")'><img src='http://www.mscroggs.co.uk/img/320/punt19_boxes.jpg'></a>
<div class='caption'>Locked boxes.</div>
<div class='paragraph'>If you want to make your own treasure punt or similar activity, you can find the LaTeX code used to create the instructions and the Python code I used to check that the puzzle
has a unique solution on <a href="https://github.com/mscroggs/tmip2019-treasure-punt" target="new">GitHub</a>. It's licensed with a <a href="https://creativecommons.org/licenses/by/4.0/" target="new">CC BY 4.0</a>
licence, so you can resuse an edit it in any way you like, as long as you attribute the bits I made that you keep.</div>
<h3>The puzzle</h3>
<div class='paragraph'>Four mathematicians—<a href="https://twitter.com/SparksMaths" target="new">Ben</a>, <a href="https://twitter.com/stecks" target="new">Katie</a>,
<a href="https://twitter.com/k_houston_math" target="new">Kevin</a>, and <a href="https://twitter.com/SamDurbin1" target="new">Sam</a>—each have one of the four clues needed to unlock a great treasure.
On a sunny/cloudy/rainy/snowy (delete as appropriate) day, they meet up in Cambridge to go punting, share their clues, work out the code for the lock,
and share out the treasure. One or more of the mathematicians, however, has decided to lie about their clue so they can steal all the treasure for themselves.
At least one mathematician is telling the truth.
(If the mathematicians say multiple sentences about their clue, then they are either all true or all false.)</div>
<div class='paragraph'>They meet at <a href="http://www.punting-in-cambridge.co.uk/" target="new">Cambridge Chauffeur Punts</a>, and head North under Silver Street Bridge.
Ben points out a plaque on the bridge with two years written on it:</div>
<div class='paragraph'>"My clue," he says, "tells me that the sum of the digits of the code is equal to the sum of the digits of the earlier year on that plaque (the year is <span class='spoiler'>1702</span>). My clue also tells me that at least one of the digits of the code is 7."</div>
<div class='paragraph'>The mathematicians next punt under the Mathematical Bridge, gasping in awe at its tangential trusses, then punt along the river under King's College Bridge and past King's College.
Katie points to a sign on the King's College lawn near the river:</div>
<div class='paragraph'>"See that sign whose initials are PNM?" says Katie. "My clue states that first digit of the code is equal to the number of vowels on that sign (The sign says "<span class='spoiler'>Private: No Mooring</span>").
My clue also tells me that at least one of the digits of the code is 1."</div>
<div class='paragraph'>They then reach Clare Bridge. Kevin points out the spheres on Clare Bridge:</div>
<div class='paragraph'>"My clue," he says, "states that the total number of spheres on both sides of this bridge is a factor of the code (there are <span class='spoiler'>14</span> spheres). My clue also tells me that at least one of the digits of the code is 2."
(Kevin has not noticed that one of the spheres had a wedge missing, so counts that as a whole sphere.)</div>
<div class='paragraph'>They continue past Clare College. Just before they reach Garret Hostel Bridge, Sam points out the Jerwood Library and a sign showing the year it was built (it was built in <span class='spoiler'>1998</span>):</div>
<div class='paragraph'>"My clue," she says, "says that the largest prime factor of that year appears in the code (in the same way that you might say the number 18 appears in 1018 or 2189).
My clue also says that the smallest prime factor of that year appears in the code. My clue also told me that at least one of the digits of the code is 0."</div>
<div class='paragraph'>They then punt under Garret Hostel Bridge, turn around between it and Trinity College Bridge, and head back towards Cambridge Chauffeur Punts.
<em>Zut alors</em>, the lies confuse them and they can't unlock the treasure. Can you work out who is lying and claim the treasure for yourself?</div>
<h3>The solution</h3>
<div class='paragraph'>The solution to the treasure punt is given below. Once you're ready to see it, click "Show solution".</div>
<div class='ans'><a id='RevB1-1' href='javascript:showRB("1")'><h4>Show solution</h4></a><div id='RevB2-1' style='display:none'><a href='javascript:hideRB("1")'><h4>Hide solution</h4></a>
<div class='paragraph'>Once all the information has been collected the clues are:</div>
<ul>
<li>Ben: The sum of the digits is 10. Code contains a 7.</li>
<li>Katie: First digit is 7. Code contains a 1.</li>
<li>Kevin: 14 is a factor of the code. Code contains a 2.</li>
<li>Sam: Code contains a 37, 2 and 0.</li>
</ul>
<div class='paragraph'>Sam's clue states that 37, 2 and 0 all appear in the code. If this is true, then the code contains a 7, and so Ben is also telling the truth.
But the sum of the digits is $3+7+2+0=12$; this is not 10, so Ben is also lying. This cannot be right, so Sam must be a liar.</div>
<div class='paragraph'>Sam is lying, and so there is no 2 in the code. Kevin is therefore also lying.</div>
<div class='paragraph'>If Ben is lying, then there is no 7 in the code. This means that the first digit cannot be a 7, and so Katie must also be lying. In this case, all four mathematicians are lying:
this is not allowed, as the questions states that at least one mathematician is telling the truth. Therefore, Ben is telling the truth.</div>
<div class='paragraph'>Ben is telling the truth, so the code contains a 7 and its digits add to 10. Sam is lying, and so there are no 0s or 2s in the code. The code must therefore be made of one 7 and three 1s.</div>
<div class='paragraph'>The code contains a 1, and so Katie is telling the truth. This means that the first digit it 7, and so the code is <b>7111</b>.</div>
</div></div>
http://www.mscroggs.co.uk/blog/67
http://www.mscroggs.co.uk/blog/6701 Sep 2019 12:00:00 GMTBig Internet Math-Off stickers 2019<div class='paragraph'>This year's <a href="https://aperiodical.com/category/the-big-internet-math-off/the-big-internet-math-off-2019/" target="new">Big Internet Math-Off</a>is now underway with 15 completely new contestants (plus one returning contender). As I'm not the returning contestant, I haven't been spendingmy time preparing <a href="http://www.mscroggs.co.uk/blog/55">my</a> <a href="http://www.mscroggs.co.uk/blog/57">pitches</a>. Instead, I've spent my time making an<a href="http://mathoffstickerbook.com" target="new">unofficial Big Internet Math-Off sticker book</a>.</div><div class='paragraph'>To complete the sticker book, you will need to collect 162 different stickers. Every day, you will be given a pack of 5 stickers; there arealso some bonus packs available if you can find them <small>(Hint: keep reading)</small>.</div><h3>How many stickers will I need?</h3><div class='paragraph'>Using the same method as I did for <a href="http://www.mscroggs.co.uk/blog/50">last year's World Cup sticker book</a>,you can work out that the expected number of stickers needed to finish the sticker book:</div><div class='paragraph'>If you have already stuck \(n\) stickers into your album, then the probability that the next sticker you get is new is</div>$$\frac{162-n}{162}.$$<div class='paragraph'>The probability that the second sticker you get is the next new sticker is</div>$$\mathbb{P}(\text{next sticker is not new})\times\mathbb{P}(\text{sticker after next is new})$$$$=\frac{n}{162}\times\frac{162-n}{162}.$$<div class='paragraph'>Following the same method, we can see that the probability that the \(i\)th sticker you buy is the next new sticker is</div>$$\left(\frac{n}{162}\right)^{i-1}\times\frac{162-n}{162}.$$<div class='paragraph'>Using this, we can calculate the expected number of stickers you will need to buy until you find a new one:</div>$$\sum_{i=1}^{\infty}i \left(\frac{162-n}{162}\right) \left(\frac{n}{162}\right)^{i-1} = \frac{162}{162-n}$$<div class='paragraph'>Therefore, to get all 162 stickers, you should expect to buy</div>$$\sum_{n=0}^{161}\frac{162}{162-n} = 918 \text{ stickers}.$$<div class='paragraph'>Using just your daily packs, it will take you until the end of the year to collect this many stickers.Of course, you'll only need to collect this many if you don't swap your duplicate stickers.</div><h3>How many stickers will I need if I swap?</h3><div class='paragraph'>To work out the expected number of stickers stickers you'd need if you swap, let's first think about two people who want to completetheir stickerbooks together. If there are \(a\) stickers that both collectors need and \(b\) stickers that one collector has and the other oneneeds, then let \(E_{a,b}\) be the expected number of stickers they need to finish their sticker books.The next sticker they get could be one of three things:</div><ul><li>A sticker they both need (with probability \(\frac{a}{162}\));</li><li>A sticker one of them needs (with probability \(\frac{b}{162}\));</li><li>A sticker they both have (with probability \(\frac{162-a-b}{162}\)).</li></ul><div class='paragraph'>Therefore, the expected number of stickers they need to complete their sticker books is</div>$$E_{a,b}=1+\frac{a}{162}E_{a-1,b+1}+\frac{b}{162}E_{a,b-1}+\frac{162-a-b}{162}E_{a,b}.$$<div class='paragraph'>This can be rearranged to give</div>$$E_{a,b}=\frac{162}{a+b}+\frac{a}{a+b}E_{a-1,b+1}+\frac{b}{a+b}E_{a,b-1}$$<div class='paragraph'>We know that $E_{0,0}=0$ (as if \(a=0\) and \(b=0\), both collectors have already finished their sticker books). Using this and theformula above, we can work out that</div>$$E_{0,1}=162+E_{0,0}=162$$$$E_{1,0}=162+E_{0,1}=324$$$$E_{0,2}=\frac{162}2+E_{0,1}=243$$$$E_{1,1}=\frac{162}2+\frac12E_{0,2}+\frac12E_{1,0}=364.5$$<div class='paragraph'>... and so on until we find that \(E_{162,0}=1269\), and so our collectors should expect to collect 634 stickers each to complete theirsticker books.</div><div class='paragraph'>For three people, we can work out that if there are \(a\) stickers that all three need, \(b\) stickers that two need, and \(c\) stickersthat one needs, then</div>$$E_{a,b,c} = \frac{162}{a+b+c}+\frac{a}{a+b+c}E_{a-1,b+1,c}+\frac{b}{a+b+c}E_{a,b-1,c+1}+\frac{c}{a+b+c}E_{a,b,c-1}.$$<div class='paragraph'>In the same way as for two people, we find that \(E_{162,0,0}=1572\), and so our collectors should expect to collect 524 stickers eachto complete their sticker books.</div><div class='paragraph'>Doing the same thing for four people gives an expected 463 stickers required each.</div><div class='paragraph'>After four people, however, the Python code I wrote to do these calculations takes too long to run, so instead I approximated the numbersby simulating 500 groups of \(n\) people collecting stickers, and taking the average number of stickers they needed. The results are shown inthe graph below.</div><a class='zoom' href='javascript:showlimage("mathoffstickerplot.png")'><img src='http://www.mscroggs.co.uk/img/320/mathoffstickerplot.jpg'></a><div class='paragraph'>The red dots are the expected values we calculated exactly, and the blue crosses are the simulated values.It looks like you'll need to collect at least 250 stickers to finish the album: in order to get this many before the end of the Math-Off,you'll need to find 20 bonus packs...</div><div class='paragraph'>Of course, these are just the mean values and you could get lucky and need fewer stickers. The next graph shows box plots with thequartiles of the data from the simulations.</div><a class='zoom' href='javascript:showlimage("mathoffstickerboxplot.png")'><img src='http://www.mscroggs.co.uk/img/320/mathoffstickerboxplot.jpg'></a><div class='paragraph'>So if you're lucky, you could complete the album with fewer stickers or fewer friends.</div><div class='paragraph'>As a thank you for reading to the end of this blog post, here's <a href="http://mathoffstickerbook.com/code/mscroggs" target="new">a link</a> thatwill give you two bonus packs and help you on your way to the 250 expected stickers...</div>
http://www.mscroggs.co.uk/blog/66
http://www.mscroggs.co.uk/blog/6603 Jul 2019 12:00:00 GMTProving a conjecture<div class='paragraph'>Last night at <a href="https://twitter.com/LONMathsJam" target="new">MathsJam</a>, <a href="http://www.peterkagey.com/" target="new">Peter Kagey</a> showed me
a conjecture about <a href="https://oeis.org/A308092" target="new">OEIS sequence A308092</a>.</div>
<blockquote><center><a href="https://oeis.org/A308092" target="new">A308092</a></center>
<div class='paragraph'>The sum of the first \(n\) terms of the sequence is the concatenation of the first \(n\) bits of the sequence read as binary, with \(a(1) = 1\).</div>
<center>1, 2, 3, 7, 14, 28, 56, 112, 224, 448, 896, 1791, 3583, 7166, ...</center></blockquote>
<div class='paragraph'>To understand this definition, let's look at the first few terms of this sequence written in binary:</div>
<div class='paragraph'><center>1, 10, 11, 111, 1110, 11100, 111000, 1110000, 11100000, 111000000, ...</center></div>
<div class='paragraph'>By "the concatenation of the first \(n\) bits of the sequence", it means the first \(n\) binary digits of the whole sequence written in order:
1, then 11, then 110, then 1101, then 11011, then 110111, and so on. So the definition means:</div>
<ul>
<li>The first term is 1, as given in the definition (\(a(1)=1\)).</li>
<li>The sum of the first 2 terms is the first 2 digits: \(1+10=11\).</li>
<li>The sum of the first 3 terms is the first 3 digits: \(1+10+11=110\).</li>
<li>The sum of the first 4 terms is the first 4 digits: \(1+10+11+111=1101\).</li>
<li>The sum of the first 5 terms is the first 5 digits: \(1+10+11+111+1110=11011\).</li>
</ul>
<div class='paragraph'>As we know that the sum of the first \(n-1\) terms is the first \(n-1\) digits, we can calculate the third term of this sequence onwards using:
"\(a(n)\) is the concatenation of the first \(n\) bits of the sequence subtract concatenation of the first \(n-1\) bits of the sequence":</div>
<ul>
<li>The third term is \(110 - 11 = 11\).</li>
<li>The fourth term is \(1101 - 110 = 111\).</li>
<li>The fourth term is \(11011 - 1101 = 1110\).</li>
<li>The fifth term is \(110111 - 11011 = 11100\).</li>
</ul>
<h3>The conjecture</h3>
<div class='paragraph'>Peter's conjecture is that the number of 1s in each term is greater than or equal to the number of 1s in the previous term.</div>
<div class='paragraph'>I'm going to prove this conjecture. If you'd like to have a try first, stop reading now and come back when you're ready for spoilers.
(If you'd like a hint, read the next section then pause again.)</div>
<h3>Adding a digit</h3>
<div class='paragraph'>The third term of the sequence onwards can be calculated by subtracting the first \(n-1\) digits from the first \(n\) digits.
If the first \(n-1\) digits form a binary number \(x\), then the first \(n\) digits will be \(2x+d\), where \(d\) is the \(n\)th digit
(because moving all the digits to the left one place in binary is multiplying by two).</div>
<div class='paragraph'>Therefore the different is \(2x+d-x=x+d\), and so we can work out the \(n\)th term of the sequence by adding the \(n\)th digit in the
sequence to the first \(n-1\) digits. (Hat tip to <a href="https://twitter.com/MarHarStar" target="new">Martin Harris</a>, who spotted this first.)</div>
<h3>Carrying</h3>
<div class='paragraph'>Adding 1 to a binary number the ends in 1 will cause 1 to carry over to the left. This carrying will continue until the 1 is carried into
a position containing 0, and after this all the digits to the left of this 0 will remain unchanged.</div>
<div class='paragraph'>Therefore adding a digit to
the first \(n-1\) digits can only change the digits from the rightmost 0 onwards.</div>
<h3>Endings</h3>
<div class='paragraph'>We can therefore disregard all the digits before the rightmost 0, and look at how the \(n\)th term compares to the \((n-1)\)th term.
There are 5 ways in which the first \(n\) digits could end:</div>
<ul>
<li>\(00\)</li>
<li>\(010\)</li>
<li>\(01...10\) (where \(1...1\) is a string of 2 or more ones)</li>
<li>\(01\)</li>
<li>\(01...1\) (where \(1...1\) is again a string of 2 or more ones)</li>
</ul>
<div class='paragraph'>We now look at each of these in turn and show that the \(n\)th term will contain at least as many ones at the \((n-1)\)th term.</div>
<h4>Case 1: \(00\)</h4>
<div class='paragraph'>If the first \(n\) digits of the sequence are \(x00\) (a binary number \(x\) followed by two zeros), then the \((n-1)\)th term of the
sequence is \(x+0=x\), and the \(n\)th term of the sequence is \(x0+0=x0\). Both \(x\) and \(x0\) contain the same number of ones.</div>
<h4>Case 2: \(010\)</h4>
<div class='paragraph'>If the first \(n\) digits of the sequence are \(x010\), then the \((n-1)\)th term of the sequence is \(x0+1=x1\),
and the \(n\)th term of the sequence is \(x01+0=x01\). Both \(x1\) and \(x01\) contain the same number of ones.</div>
<h4>Case 3: \(01...10\)</h4>
<div class='paragraph'>If the first \(n\) digits of the sequence are \(x01...10\), then the \((n-1)\)th term of the sequence is \(x01...1+1=x10...0\),
and the \(n\)th term of the sequence is \(x01...10+1=x01...1\). \(x01...1\) contains more ones than \(x10...0\).</div>
<h4>Case 4: \(01\)</h4>
<div class='paragraph'>If the first \(n\) digits of the sequence are \(x01\), then the \((n-1)\)th term of the sequence is \(x+0=x\),
and the \(n\)th term of the sequence is \(x0+1=x1\). \(x1\) contains one more one than \(x\).</div>
<h4>Case 5: \(01...1\)</h4>
<div class='paragraph'>If the first \(n\) digits of the sequence are \(x01...1\), then the \((n-1)\)th term of the sequence is \(x01...1+1=x10...0\),
and the \(n\)th term of the sequence is \(x01...1+1=x10...0\). Both these contain the same number of ones.</div>
<br />
<div class='paragraph'>In all five cases, the \(n\)th term contains more ones or an equal number of ones to the \((n-1)\)th term, and so the conjecture is true.</div>
http://www.mscroggs.co.uk/blog/65
http://www.mscroggs.co.uk/blog/6519 Jun 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 GMTHarriss and other spirals<div class='paragraph'>In the latest issue of <a href="http://chalkdustmagazine.com/read/read-issue-09-now/" target="new"><i>Chalkdust</i></a>,I wrote <a href="http://chalkdustmagazine.com/regulars/on-the-cover/on-the-cover-harriss-spiral/" target="new">an article</a>with <a href="https://maxwelldemon.com/" target="new">Edmund Harriss</a> about the Harriss spiral that appears on the cover of the magazine.To draw a Harriss spiral, start with a rectangle whose side lengths are in the plastic ratio; that is the ratio \(1:\rho\)where \(\rho\) is the real solution of the equation \(x^3=x+1\), approximately 1.3247179.</div><a class='zoom' href='javascript:showlimage("plastic.png")'><img src='http://www.mscroggs.co.uk/img/320/plastic.jpg'></a><div class='caption'>A plastic rectangle</div><div class='paragraph'>This rectangle can be split into a square and two rectangles similar to the original rectangle. These smaller rectangles can then be split up in the same manner.</div><a class='zoom' href='javascript:showlimage("plastic2.png")'><img src='http://www.mscroggs.co.uk/img/320/plastic2.jpg'></a><a class='zoom' href='javascript:showlimage("plastic3.png")'><img src='http://www.mscroggs.co.uk/img/320/plastic3.jpg'></a><div class='caption'>Splitting a plastic rectangle into a square and two plastic rectangles.</div><div class='paragraph'>Drawing two curves in each square gives the Harriss spiral.</div><a class='zoom' href='javascript:showlimage("plastic4.png")'><img src='http://www.mscroggs.co.uk/img/320/plastic4.jpg'></a><a class='zoom' href='javascript:showlimage("plastic5.png")'><img src='http://www.mscroggs.co.uk/img/320/plastic5.jpg'></a><div class='caption'>A Harriss spiral</div><div class='paragraph'>This spiral was inspired by the golden spiral, which is drawn in a rectangle whose side lengths are in the golden ratio of \(1:\phi\),where \(\phi\) is the positive solution of the equation \(x^2=x+1\) (approximately 1.6180339). This rectangle can be split into a square and onesimilar rectangle. Drawing one arc in each square gives a golden spiral.</div><a class='zoom' href='javascript:showlimage("golden2.png")'><img src='http://www.mscroggs.co.uk/img/320/golden2.jpg'></a><a class='zoom' href='javascript:showlimage("golden3.png")'><img src='http://www.mscroggs.co.uk/img/320/golden3.jpg'></a><div class='caption'>A golden spiral</div><h3>Continuing the pattern</h3><div class='paragraph'>The golden and Harriss spirals are both drawn in rectangles that can be split into a square and one or two similar rectangles.</div><a class='zoom' href='javascript:showlimage("golden1.png")'><img src='http://www.mscroggs.co.uk/img/320/golden1.jpg'></a><a class='zoom' href='javascript:showlimage("plastic2.png")'><img src='http://www.mscroggs.co.uk/img/320/plastic2.jpg'></a><div class='caption'>The rectangles in which golden and Harriss spirals can be drawn.</div><div class='paragraph'>Continuing the pattern of these arrangements suggests the following rectangle, split into a square and three similar rectangles:</div><a class='zoom' href='javascript:showlimage("third0.png")'><img src='http://www.mscroggs.co.uk/img/320/third0.jpg'></a><div class='paragraph'>Let the side of the square be 1 unit, and let each rectangle have sides in the ratio \(1:x\). We can then calculate that the lengths ofthe sides of each rectangle are as shown in the following diagram.</div><a class='zoom' href='javascript:showlimage("third.png")'><img src='http://www.mscroggs.co.uk/img/320/third.jpg'></a><div class='paragraph'>The side lengths of the large rectangle are \(\frac{1}{x^3}+\frac{1}{x^2}+\frac2x+1\) and \(\frac1{x^2}+\frac1x+1\). We want these to alsobe in the ratio \(1:x\). Therefore the following equation must hold:</div>$$\frac{1}{x^3}+\frac{1}{x^2}+\frac2x+1=x\left(\frac1{x^2}+\frac1x+1\right)$$<div class='paragraph'>Rearranging this gives:</div>$$x^4-x^2-x-1=0$$$$(x+1)(x^3-x^2-1)=0$$<div class='paragraph'>This has one positive real solution:</div>$$x=\frac13\left(1+\sqrt[3]{\tfrac12(29-3\sqrt{93})}+\sqrt[3]{\tfrac12(29+3\sqrt{93})}\right).$$<div class='paragraph'>This is equal to 1.4655712... Drawing three arcs in each square allows us to make a spiral from a rectangle with sides in this ratio:</div><a class='zoom' href='javascript:showlimage("myspiral2.png")'><img src='http://www.mscroggs.co.uk/img/320/myspiral2.jpg'></a><a class='zoom' href='javascript:showlimage("myspiral3.png")'><img src='http://www.mscroggs.co.uk/img/320/myspiral3.jpg'></a><div class='caption'>A spiral which may or may not have a name yet.</div><h3>Continuing the pattern</h3><div class='paragraph'>Adding a fourth rectangle leads to the following rectangle.</div><a class='zoom' href='javascript:showlimage("fourth.png")'><img src='http://www.mscroggs.co.uk/img/320/fourth.jpg'></a><div class='paragraph'>The side lengths of the largest rectangle are \(1+\frac2x+\frac3{x^2}+\frac1{x^3}+\frac1{x^4}\) and \(1+\frac2x+\frac1{x^2}+\frac1{x^3}\).Looking for the largest rectangle to also be in the ratio \(1:x\) leads to the equation:</div>$$1+\frac2x+\frac3{x^2}+\frac1{x^3}+\frac1{x^4} = x\left(1+\frac2x+\frac1{x^2}+\frac1{x^3}\right)$$$$x^5+x^4-x^3-2x^2-x-1 = 0$$<div class='paragraph'>This has one real solution, 1.3910491... Although for this rectangle, it's not obvious which arcs to draw to make aspiral (or maybe not possible to do it at all). But at least you get a pretty fractal:</div><a class='zoom' href='javascript:showlimage("my2spiral1.png")'><img src='http://www.mscroggs.co.uk/img/320/my2spiral1.jpg'></a><h3>Continuing the pattern</h3><div class='paragraph'>We could, of course, continue the pattern by repeatedly adding more rectangles. If we do this, we get the following polynomialsand solutions:</div><table><thead><tr><td>Number of rectangles</td><td>Polynomial</td><td>Solution</td></tr></thead><tr><td>1</td><td>\(x^2 - x - 1=0\)</td><td>1.618033988749895</td></tr><tr><td>2</td><td>\(x^3 - x - 1=0\)</td><td>1.324717957244746</td></tr><tr><td>3</td><td>\(x^4 - x^2 - x - 1=0\)</td><td>1.465571231876768</td></tr><tr><td>4</td><td>\(x^5 + x^4 - x^3 - 2x^2 - x - 1=0\)</td><td>1.391049107172349</td></tr><tr><td>5</td><td>\(x^6 + x^5 - 2x^3 - 3x^2 - x - 1=0\)</td><td>1.426608021669601</td></tr><tr><td>6</td><td>\(x^7 + 2x^6 - 2x^4 - 3x^3 - 4x^2 - x - 1=0\)</td><td>1.4082770325090774</td></tr><tr><td>7</td><td>\(x^8 + 2x^7 + 2x^6 - 2x^5 - 5x^4 - 4x^3 - 5x^2 - x - 1=0\)</td><td>1.4172584399350432</td></tr><tr><td>8</td><td>\(x^9 + 3x^8 + 2x^7 - 5x^5 - 9x^4 - 5x^3 - 6x^2 - x - 1=0\)</td><td>1.412713760332943</td></tr><tr><td>9</td><td>\(x^{10} + 3x^9 + 5x^8 - 5x^6 - 9x^5 - 14x^4 - 6x^3 - 7x^2 - x - 1=0\)</td><td>1.414969877544769</td></tr></table><div class='paragraph'>The numbers in this table appear to be heading towards around 1.414, or \(\sqrt2\).This shouldn't come as too much of a surprise because \(1:\sqrt2\) is the ratio of the sides of A\(n\) paper (for \(n=0,1,2,...\)).A0 paper can be split up like this:</div><a class='zoom' href='javascript:showlimage("a4split.png")'><img src='http://www.mscroggs.co.uk/img/320/a4split.jpg'></a><div class='caption'>Splitting up a piece of A0 paper</div><div class='paragraph'>This is a way of splitting up a \(1:\sqrt{2}\) rectangle into an infinite number of similar rectangles, arranged following the pattern,so it makes sense that the ratios converge to this.</div><h3>Other patterns</h3><div class='paragraph'>In this post, we've only looked at splitting up rectangles into squares and similar rectangles following a particular pattern. Thinking aboutother arrangements leads to the following question:</div><div class='puart in-blog'><div class='paragraph'>Given two real numbers \(a\) and \(b\), when is it possible to split an \(a:b\) rectangle into squares and \(a:b\) rectangles?</div></div><div class='paragraph'>If I get anywhere with this question, I'll post it here. Feel free to post your ideas in the comments below.</div>
http://www.mscroggs.co.uk/blog/64
http://www.mscroggs.co.uk/blog/6409 Apr 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='namepurple'>Kip Urples</span>, <span class='namegreen'>Meg Reeny</span>, <span class='namered'>Fred Metcalfe</span>, <span class='nameorange'>Jo Ranger</span>, and <span class='nameblue'>Bob Luey</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><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'>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><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>
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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>
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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>$$ABCCDE\times 4 = EDCCBA$$
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http://www.mscroggs.co.uk/puzzles/LXIV06 May 2018 12:00:00 GMT