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2017-12-18
Just like last year, TD and I spent some time in November this year designing a puzzle Christmas card for Chalkdust.
![](javascript:showlimage("card2017.jpg"))
The card looks boring at first glance, but contains 10 puzzles. Converting the answers to base 3, writing them in the boxes on the front, then colouring the 1s black and 2s orange will reveal a Christmassy picture.
If you want to try the card yourself, you can download this pdf. Alternatively, you can find the puzzles below and type the answers in the boxes. The answers will be automatically converted to base 3 and coloured...
| # | Answer (base 10) | Answer (base 3) | ||||||||
| 1 | 0 | 0 | 0 | 0 | 0 | 0 | 0 | |||
| 2 | 0 | 0 | 0 | 0 | 0 | 0 | 0 | |||
| 3 | 0 | 0 | 0 | 0 | 0 | 0 | 0 | |||
| 4 | 0 | 0 | 0 | 0 | 0 | 0 | 0 | |||
| 5 | 0 | 0 | 0 | 0 | 0 | 0 | 0 | |||
| 6 | 0 | 0 | 0 | 0 | 0 | 0 | 0 | |||
| 7 | 0 | 0 | 0 | 0 | 0 | 0 | 0 | |||
| 8 | 0 | 0 | 0 | 0 | 0 | 0 | 0 | |||
| 9 | 0 | 0 | 0 | 0 | 0 | 0 | 0 | |||
| 10 | 0 | 0 | 0 | 0 | 0 | 0 | 0 | |||
- In a book with 116 pages, what do the page numbers of the middle two pages add up to?
- What is the largest number that cannot be written in the form \(14n+29m\), where \(n\) and \(m\) are non-negative integers?
- How many factors does the number \(2^6\times3^{12}\times5^2\) have?
- How many squares (of any size) are there in a \(15\times14\) grid of squares?
- You take a number and make a second number by removing the units digit. The sum of these two numbers is 1103. What was your first number?
- What is the only three-digit number that is equal to a square number multiplied by the reverse of the same square number? (The reverse cannot start with 0.)
- What is the largest three-digit number that is equal to a number multiplied by the reverse of the same number? (The reverse cannot start with 0.)
- What is the mean of the answers to questions 6, 7 and 8?
- How many numbers are there between 0 and 100,000 that do not contain the digits 0, 1, 2, 3, 4, 5, or 6?
- What is the lowest common multiple of 52 and 1066?
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2019-01-04
@Jose: There is a mistake in your answer: 243 (0100000) is the number of numbers between 10,000 and 100,000 that do not contain the digits 0, 1, 2, 3, 4, 5, or 6.Matthew
Thanks for the puzzle!
Is it possible that the question 9 is no correct?
I get a penguin with perfect simetrie except at answer 9 : 0100000 that breaks the simetry.
Is it correct or a mistake in my answer?
Thx
Is it possible that the question 9 is no correct?
I get a penguin with perfect simetrie except at answer 9 : 0100000 that breaks the simetry.
Is it correct or a mistake in my answer?
Thx
Jose
@C: look up something called Frobenius numbers. This problem's equivalent to finding the Frobenius number for 14 and 29. ×1
Lewis
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2017-11-28
This year, the front page of mscroggs.co.uk will once again feature an advent calendar, just like last year and the year before. Behind each door, there will be a puzzle with a three digit solution. The solution to each day's puzzle forms part of a logic puzzle:
It's nearly Christmas and something terrible has happened: Santa and his two elves have been cursed! The curse has led Santa to forget which present three children—Alex, Ben and Carol—want and where they live.
The elves can still remember everything about Alex, Ben and Carol, but the curse is causing them to lie. One of the elves will lie on even numbered days and tell the truth on odd numbered days; the other elf will lie on odd numbered days and tell the truth on even numbered days. As is common in elf culture, each elf wears the same coloured clothes every day.
Each child lives in a different place and wants a different present. (But a present may be equal to a home.) The homes and presents are each represented by a number from 1 to 9.
Santa has called on you to help him work out the details he has forgotten. Behind each day (except Christmas Day), there is a puzzle with a three-digit answer. Each of these answers forms part of a fact that one of the elves tells you. You must work out which combination of clothes each elf wears, which one lies on each day, then put all the clues together to work out which presents need delivering to Alex, Ben and Carol, and where to deliver them.
Ten randomly selected people who solve all the puzzles and submit their answers to the logic puzzle using the form behind the door on the 25th will win prizes! A selection of the prizes are shown below, and will be added to throughout December.
![](javascript:showlimage("prizes2017.jpg"))
The ten winners will also will one of these winners' medals:
![](javascript:showlimage("medal-2017.jpg"))
As you solve the puzzles, your answers will be stored. This year, there is a new feature allowing you to synchronise your answers between multiple computers: simply enter your email address below the calendar, and you will be emailed a magic link to visit on your other devices.
Behind the door on Christmas Day, there will be a form allowing you to submit your answers. The winner will be randomly chosen from all those who submit the correct answer before the end of 2017. Each day's puzzle (and the entry form on Christmas Day) will be available from 5:00am GMT. But as the winners will be selected randomly, there's no need to get up at 5am on Christmas Day to enter!
To win a prize, you must submit your entry before the end of 2017. Only one entry will be accepted per person. If you have any questions, ask them in the comments below or on Twitter.
So once December is here, get solving! Good luck and have a very merry Christmas!
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Me again
Just for info (clarification?): I read question on 22nd as
22 is two times an odd number. Today's number is the mean of all the answers, on days (including today), that are two times an odd number."
Note my added commas. I was averaging the answers, not the dates. Certainly ambiguous as far as I am concerned.
Only fixed it by 'cheating'. Trying best guessses of averages until I got the correct one.
Just for info (clarification?): I read question on 22nd as
22 is two times an odd number. Today's number is the mean of all the answers, on days (including today), that are two times an odd number."
Note my added commas. I was averaging the answers, not the dates. Certainly ambiguous as far as I am concerned.
Only fixed it by 'cheating'. Trying best guessses of averages until I got the correct one.
neal (@zbvif)
Wow. Just discovered I meisread 15th Dec puzzle.
I can tell you that the number of combinations of n As and Bs which contain at at least one uninterrupted sequence of 3 As is 2^n - F3(n+3) where F3 is the fibonaccia variant adding 3 numbers (1,1,2,4,7,13,24 etc.).
Only took me about 8 hours (with some small help form OEIS for the 2 As problem)
I can tell you that the number of combinations of n As and Bs which contain at at least one uninterrupted sequence of 3 As is 2^n - F3(n+3) where F3 is the fibonaccia variant adding 3 numbers (1,1,2,4,7,13,24 etc.).
Only took me about 8 hours (with some small help form OEIS for the 2 As problem)
Neal (@zbvif)
With todays puzzle does the pancake have any thickness i.e can we slice the pancake into 2 circular pancakes each with half the thickness or are we to assume its 2D
Alex
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2017-11-14
A few weeks ago, I took the copy of MENACE that I built to Manchester Science Festival, where it played around 300 games against the public while learning to play Noughts and Crosses. The group of us operating MENACE for the weekend included Matt Parker, who made two videos about it. Special thanks go to Matt, plus
Katie Steckles,
Alison Clarke,
Andrew Taylor,
Ashley Frankland,
David Williams,
Paul Taylor,
Sam Headleand,
Trent Burton, and
Zoe Griffiths for helping to operate MENACE for the weekend.
As my original post about MENACE explains in more detail, MENACE is a machine built from 304 matchboxes that learns to play Noughts and Crosses. Each box displays a possible position that the machine can face and contains coloured beads that correspond to the moves it could make. At the end of each game, beads are added or removed depending on the outcome to teach MENACE to play better.
Saturday
On Saturday, MENACE was set up with 8 beads of each colour in the first move box; 3 of each colour in the second move boxes; 2 of each colour in third move boxes; and 1 of each colour in the fourth move boxes. I had only included one copy of moves that are the same due to symmetry.
The plot below shows the number of beads in MENACE's first box as the day progressed.
![](javascript:showlimage("sat-same-axes.png"))
Sunday
Originally, we were planning to let MENACE learn over the course of both days, but it learned more quickly than we had expected on Saturday, so we reset is on Sunday, but set it up slightly differently. On Sunday, MENACE was set up with 4 beads of each colour in the first move box; 3 of each colour in the second move boxes; 2 of each colour in third move boxes; and 1 of each colour in the fourth move boxes. This time, we left all the beads in the boxes and didn't remove any due to symmetry.
The plot below shows the number of beads in MENACE's first box as the day progressed.
![](javascript:showlimage("sun-same-axes.png"))
The data
You can download the full set of data that we collected over the weekend here. This includes the first two moves and outcomes of all the games over the two days, plus the number of beads in each box at the end of each day. If you do something interesting (or non-interesting) with the data, let me know!
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WRT the comment 2017-11-17, and exactly one year later, I had the same thing happen whilst running MENACE in a 'Resign' loop for a few hours, unattended. When I returned, the orange overlay had appeared, making the screen quite difficult to read on an iPad.
g0mrb
On the JavaScript version, MENACE2 (a second version of MENACE which learns in the same way, to play against the original) keeps setting the 6th move as NaN, meaning it cannot function. Is there a fix for this?
Lambert
what would happen if you loaded the boxes slightly differently. if you started with one bead corresponding to each move in each box. if the bead caused the machine to lose you remove only that bead. if the game draws you leave the bead in play if the bead causes a win you put an extra bead in each of the boxes that led to the win. if the box becomes empty you remove the bead that lead to that result from the box before
Ian
Hi, I was playing with MENACE, and after a while the page redrew with a Dragon Curves design over the top. MENACE was still working alright but it was difficult to see what I was doing due to the overlay. I did a screen capture of it if you want to see it.
Russ
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2017-06-03
As a child, I was a huge fan of Captain Scarlet and the Mysterons, Gerry Anderson's puppet-starring sci-fi series.
As an adult, I am still a huge fan of Captain Scarlet and the Mysterons.
Set in 2068, the series follows Captain Scarlet and the other members of Spectrum as they attempt to protect Earth from the Mysterons.
One of my favourite episodes of the series is the third: Big Ben Strikes Again.
In this episode, the Mysterons threaten to destroy London. They do this by hijacking a vehicle carrying a nuclear device, and driving it to
a car park. In the car park, the driver of the vehicle wakes up and turns the radio on. Then something weird happens: Big Ben strikes thirteen!
![](javascript:showlimage("radio.jpg"))
The driver turning on the radio. Good to know that BBC Radio 4 will still broadcast at 92-95FM in 2068.
Following this, the driver is knocked out again and wakes up in a side road somewhere. After hearing his story, Captain Blue works out that
the car park must be 1500 yards away from Big Ben. Using this information, Captains Blue and Scarlet manage to track
down the nuclear device and save the day.
![](javascript:showlimage("scarlet_map.jpg"))
A map of London with a circle of radius 1500 yards drawn on it.
After rewatching the episode recently, I realised that it would be possible to recreate this scene and hear Big Ben striking thirteen.
Where does Big Ben strike thirteen?
At the end of the episode, Captain Blue explains to Captain Scarlet that the effect was due to light travelling faster than sound: as the
driver had the radio on, he could hear Ben's bongs both from the tower and through the radio. As radio waves travel faster than
sound, the bongs over the radio can be heard earlier than the sound waves travelling through the air.
Further from the tower, the gap between when the two bongs are heard
is longer; and at just the right distance, the second bong on the radio will be heard at the same time as the first bong from the tower.
This leads to the appearance of thirteen bongs: the first bong is just from the radio, the next eleven are both radio and from the tower, and
the final bong is only from the tower.
Big Ben's bongs are approximately 4.2s apart, sound travels at 343m/s, and light travels at 3×108m/s (this is so fast that it
could be assumed that the radio waves arrive instantly without changing the answer). Using these, we perform the following calculation:
$$\text{time difference} = \text{time for sound to arrive}-\text{time for light to arrive}$$
$$=\frac{\text{distance}}{\text{speed of sound}}-\frac{\text{distance}}{\text{speed of light}}$$
$$=\text{distance}\times\left(\frac1{\text{speed of sound}}-\frac1{\text{speed of light}}\right)$$
$$\text{distance}=\text{time difference}\div\left(\frac1{\text{speed of sound}}-\frac1{\text{speed of light}}\right)$$
$$=4.2\div\left(\frac1{343}-\frac1{3\times10^8}\right)$$
$$=1440\text{m}\text{ or }1574\text{ yards}$$
This is close to Captain Blue's calculation of 1500 yards (and to be fair to the Captain, he had to calculate it in his head in a few seconds).
Plotting a circle of this radius centred at Big Ben gives the points where it may be possible to hear 13 bongs.
![](javascript:showlimage("scarlet_my_map.jpg"))
Again, the makers of Captain Scarlet got this right: their circle shown earlier is a very similar size to this one.
To demonstrate that this does work (and with a little help from TD and her camera),
I made the following video yesterday near Vauxhall station. I recommend using earphones to watch it as the later bongs are quite faint.
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Very late to this but how excellent! My Gerry Andersen experiences were at the start of his output...I even watched Stingray in black and white.×4 ×3 ×3 ×3 ×3
(anonymous)
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2017-03-27
Tomorrow, the new 12-sided one pound coin is released.
![](javascript:showlimage("one-pound.png"))
Although I'm excited about meeting this new coin, I am also a little sad,
as its release ends the era in which all British coins are shapes of constant
width.
Shapes of constant width
A shape of constant width is a shape that is the same width in every direction,
so these shapes can roll without changing height. The most obvious such shape
is a circle. But there are others, including the shape of the seven-sided 50p
coin.
![](javascript:showlimage("50p2008.jpg"))
As shown below, each side of a 50p is part of a circle centred around the opposite corner.
As a 50p rolls, its height is always the distance between one of the corners and
the side opposite, or in other words the radius of this circle. As these circles
are all the same size, the 50p is a shape of constant width.
![](javascript:showlimage("50p2008circle.jpg"))
Shapes of constant width can be created from any regular polygon with an
odd number of sides, by replacing the sides by parts of circles centred at the
opposite corner. The first few are shown below.
![](javascript:showlimage("Reuleaux_polygons.png"))
It's also possible to create shapes of constant width from irregular polygons with an odd number,
but it's not possible to create them from polygons with an even number of sides.
Therefore, the new 12-sided pound coin will be the first non-constant width British coin since
the (also 12-sided) threepenny bit was phased out in 1971.
Back in 2014, I wrote to my MP in an attempt to find
out why the new coin was not of a constant width. He forwarded my letter to
the Treasury, but I never heard back from them.
Pizza cutting
When cutting a pizza into equal shaped pieces, the usual approach is to
cut along a few diameters to make triangles. There are other ways to fairly
share pizza, including the following (that has appeared here before as an answer to this puzzle):
![](javascript:showlimage("pizza-answer.png"))
The slices in this solution are closely related to a triangle of constant
width. Solutions can be made using other shapes of constant width,
including the following, made using a constant width pentagon and heptagon (50p):
![](javascript:showlimage("pizza-answer-pent.png"))
![](javascript:showlimage("pizza-answer-hept.png"))
There are many more ways to cut a pizza into equal pieces. You can find them in Infinite families of monohedral disk tilings by Joel Haddley and Stephen Worsley [1].
You can't use the shape of a new pound coin to cut a pizza though.
Edit: Speaking of new £1 coins, I made this stupid video with Adam "Frownsend" Townsend about them earlier today:
References
[1] Infinite families of monohedral disk tilings by Joel Haddley and Stephen Worsley. December 2015. [link]
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