This blog has moved to www.mscroggs.co.uk.

## Monday, 22 September 2014

### Sunday Afternoon Maths XXIX Answers & Extensions

This post contains the answers to this week's Sunday Afternoon Maths and some extension problems based around the originals.

#### Three Squares

Draw three more squares and add these lines (I have coloured the angles to make equal angles clearer):
Triangles $$ACE$$, $$LDK$$ and $$IKE$$ are congruent, so angle $$KDL$$ is equal to $$\beta$$.
The congruence of these triangles tells us that angles $$DKL$$ and $$EKI$$ add up to a right angle, so angle $$EKD$$ is also a right angle.
The congruence of the triangles also tells us that $$KD$$ and $$KE$$ are the same length and so angle $$EDK$$ is the angle in an isosceles right-angled triangle. $$\alpha$$ is also the angle in an isosceles right-angled triangle, so these two angles are equal.
Therefore $$\alpha+\beta+\gamma=90^\circ$$.
##### Extension
The diagram shows three squares with diagonals drawn on and three angles labelled.
What is the value of $$\alpha+\beta+\gamma$$?

#### Equal Opportunity

Let $$p_1$$, $$p_2$$, ..., $$p_6$$ be the probabilities of getting 1 to 6 on one die and $$q_1$$, ..., $$q_6$$ on the other. The probability of getting a total of 2 is $$p_1q_1$$ and the probabilty of getting a total of 12 is $$p_6q_6$$. Therefore $$p_1q_1=p_6q_6$$.
If $$p_1\geq p_6$$ then $$q_1\leq q_6$$ (and vice-versa) as otherwise the above equality could not hole. Therefore:
$$(p_1-p_6)(q_1-q_6)\leq 0$$ $$p_1q_1-p_6q_1-p_1q_6+p_6q_6\leq 0$$ $$p_1q_1+q_6p_6\leq p_1q_6+p_6q_1$$
The probability of rolling a total of 7 is $$p_1q_6+p_2q_5+...+p_6q_1$$. This is larger than $$p_1q_6+p_6q_1$$, which is larger than (or equal to) $$p_1q_1+q_6p_6$$, which is larger than $$p_1q_1$$.
Therefore the probability of rolling a 7 is larger than the probability of rolling a two, so it is not possible.
##### Extension
Can two $$n$$-sided dice be weighted so that the probability of each of the numbers 2, 3, …, 2$$n$$ is the same?
Can a $$n$$-sided die and a $$m$$-sided die be weighted so that the probability of each of the numbers 2, 3, …, $$n+m$$ is the same?

#### Double Derivative

(i) $$\frac{dy}{dx}=1$$, so $$\frac{d}{dy}\left(\frac{dy}{dx}\right)=0$$
(ii) Differentiating $$y=x^2$$ with respect to $$x$$ $$\frac{dy}{dx}=2x$$. Let $$g=\frac{dy}{dx}$$. By the chain rule:
$$\frac{dg}{dy}=\frac{dg}{dx}\frac{dx}{dy}$$ $$=2\frac{1}{2x}$$ $$=\frac{1}{x}$$
So $$\frac{d}{dy}\left(\frac{dy}{dx}\right)=\frac{1}{x}$$
(iii) By the same method, $$\frac{d}{dy}\left(\frac{dy}{dx}\right)=\frac{2}{x}$$
(iv) $$\frac{d}{dy}\left(\frac{dy}{dx}\right)=\frac{n-1}{x}$$
(v) $$\frac{d}{dy}\left(\frac{dy}{dx}\right)=1$$
(vi) $$\frac{d}{dy}\left(\frac{dy}{dx}\right)=-\tan(x)$$
##### Extension
What is
$$\frac{d}{dy}\left(\frac{dy}{dx}\right)$$
when $$y=f(x)$$?

## Sunday, 21 September 2014

### Sunday Afternoon Maths XXIX

Here's this week's collection. Answers & extensions tomorrow. Why not discuss the problems on Twitter using #SundayAfternoonMaths or on Reddit. This Tuesday is Maths Jam, make sure you're there!

#### Three Squares

The diagram shows three squares with diagonals drawn on and three angles labelled.
What is the value of $$\alpha+\beta+\gamma$$?

#### Equal Opportunity

Can two (six-sided) dice be weighted so that the probability of each of the numbers 2, 3, …, 12 is the same?

#### Double Derivative

What is
$$\frac{d}{dy}\left(\frac{dy}{dx}\right)$$
when:
(i) $$y=x$$
(ii) $$y=x^2$$
(iii) $$y=x^3$$
(iv) $$y=x^n$$
(v) $$y=e^x$$
(vi) $$y=\sin(x)$$?

## Monday, 15 September 2014

### Sunday Afternoon Maths XXVIII Answers & Extensions

This post contains the answers to this week's Sunday Afternoon Maths and some extension problems based around the originals.

#### 2009

$$2009=7\times 7\times 41$$, so there are four possible sets of dimensions of the cuboid:
$$1\times 1\times 2009$$ $$1\times 7\times 287$$ $$1\times 41\times 49$$ $$7\times 7\times 41$$
In the first three cuboids, there is a face with an area of 2009 units ($$2009\times 1$$, $$7\times 287$$ and $$41\times 49$$ respectively) and so 2009 stickers will not be enough. Therefore the cuboid has dimensions $$7\times 7\times 41$$ and a surface area of 1246, leaving 764 stickers left over
##### Extension
For which numbers $$n$$ can a cuboid be made with $$n$$ unit cube such that $$n$$ unit square stickers can cover the faces of the cuboid.

#### 3$$n$$+1

(i) Let $$a,b\in S$$. Then $$\exists \alpha,\beta\in \mathbb{N}$$ such that $$a=3\alpha+1$$ and $$b=3\beta+1$$.
(This says that if $$a$$ and $$b$$ are in $$S$$ then they can be written as a multiple of three plus one.)
$$a\times b=(3\alpha+1)\times (3\beta+1)$$ $$=9\alpha\beta+3\alpha+3\beta+1$$ $$=3(3\alpha\beta+\alpha+\beta)+1$$
This is a multiple of three plus one, so $$a\times b\in S$$.
(ii) No, as $$36\times 22=4\times 253$$ and 36,22,4 and 253 are all irreducible.
##### Extension
Try the task again with $$S=\{a+b\sqrt{2}:a,b\in\mathbb{Z}\}$$.

The problem can be expressed using the following probability tree:
The probability that the card turned over from C is an Ace of Spades is:
$$\frac{1\times 2\times 2+1\times 50\times 1+50\times 1\times 2+50\times 51\times 1}{51\times 52\times 53}$$ $$=\frac{52}{51\times 53}$$

## Sunday, 14 September 2014

### Sunday Afternoon Maths XXVIII

Here's this week's collection. Answers & extensions tomorrow. Why not discuss the problems on Twitter using #SundayAfternoonMaths or on Reddit.

#### 2009

2009 unit cubes are glued together to form a cuboid. A pack, containing 2009 stickers, is opened, and there are enough stickers to place 1 sticker on each exposed face of each unit cube.
How many stickers from the pack are left?

#### 3$$n$$+1

Let $$S=\{3n+1:n\in\mathbb{N}\}$$ be the set of numbers one more than a multiple of three.
(i) Show that $$S$$ is closed under multiplication.
ie. Show that if $$a,b\in S$$ then $$a\times b\in S$$.
Let $$p\in S$$ be irreducible if $$p\not=1$$ and the only factors of $$p$$ in $$S$$ are $$1$$ and $$p$$. (This is equivalent to the most commonly given definition of prime.)
(ii) Can each number in $$S$$ be uniquely factorised into irreducibles?

I have three packs of playing cards with identical backs. Call the packs A, B and C.
I draw a random card from pack A and shuffle it into pack B.
I now turn up the top card of pack A, revealing the Queen of Hearts.
Next, I draw a card at random from pack B and shuffle it into pack C. Then, I turn up the top card of pack B, revealing another Queen of Hearts.
I now draw a random card from pack C and place it at the bottom of pack A.
What is the probability that the card at the top of pack C is the Ace of Spades?

## Monday, 8 September 2014

### Sunday Afternoon Maths XXVII Answers & Extensions

This post contains the answers to this week's Sunday Afternoon Maths and some extension problems based around the originals.

#### Triangles Between Squares

Let $$T_a$$ represent the $$a$$th triangle number. This means that $$T_a=\frac{1}{2}a(a+1)$$.
Suppose that for some integer $$n$$, $$n^2 \leq T_a <(n+1)^2$$. This means that:
$$n^2 \leq T_a$$ $$n^2 \leq \frac{1}{2}a(a+1)$$ $$2n^2 \leq a^2+a$$
But for every positive integer $$a \leq a^2$$, so:
$$2n^2 \leq 2a^2$$ $$n^2 \leq a^2$$
$$n$$ and $$a$$ are both positive integers, so:
$$n \leq a$$
Now consider $$T_{a+2}$$:
$$T_{a+2}=\frac{1}{2}(a+2)(a+3)$$ $$=\frac{1}{2}(a^2+5a+6)$$ $$=\frac{1}{2}(a^2+a)+\frac{1}{2}(4a+6)$$ $$=\frac{1}{2}a(a+1)+2a+3$$ $$=T_a+2a+3$$
We know that $$a \geq n$$ and $$T_a \geq n^2$$, so:
$$T_a+2a+3 \geq n^2+2n+3$$ $$>n^2+2n+1 = (n+1)^2$$
And so $$T_{a+2}$$ is not between $$n^2$$ and $$(n+1)^2$$. So if a triangle number $$T_a$$ is between $$n^2$$ and $$(n+1)^2$$ then the next but one triangle number $$T_{a+2}$$ cannot also be between $$n^2$$ and $$(n+1)^2$$. So there cannot be more than two triangle numbers between $$n^2$$ and $$(n+1)^2$$.
##### Extension
Given an integer $$n$$, how many triangle numbers are there between $$n^2$$ and $$(n+1)^2$$?

#### Sine

Cosine can be drawn the same way as sine but starting B at the top of the circle.
Tangent can be drawn by giving the following instructions:
A. Stand on the spot.
B. Walk around A in a circle, holding this string to keep you the same distance away.
C. Make a straight line with A and B, staying on the line tangent to the circle through B's starting point.
D. Walk in a straight line perpendicular to C's line.
E. Stay in line with C and D.
##### Extension
Could people be used to draw graphs of secant, cosecant and cotangent?

## Sunday, 7 September 2014

### Sunday Afternoon Maths XXVII

Here's this week's collection. Answers & extensions tomorrow. Why not discuss the problems on Twitter using #SundayAfternoonMaths or on Reddit.

#### Triangles Between Squares

Prove that there are never more than two triangle numbers between two consecutive square numbers.

#### Sine

A sine curve can be created with five people by giving the following instructions to the five people:
A. Stand on the spot.
B. Walk around A in a circle, holding this string to keep you the same distance away.
C. Stay in line with B, staying on this line.
D. Walk in a straight line perpendicular to C's line.
E. Stay in line with C and D. E will trace the path of a sine curve as shown here:
What instructions could you give to five people to trace a cos(ine) curve?
What instructions could you give to five people to trace a tan(gent) curve?

## Thursday, 4 September 2014

### Electromagnetic Field Talk

Last weekend, I attended Electromagnetic Field, a camp for hackers, geeks, makers and the interested. On the Sunday, I gave a talk on four mathematical ideas/tasks which I have encountered over the past few years: Flexagons, Folding Tube Maps, Braiding and Sine Curves. I'd love to see photos, hear stories, etc from anyone who tries these activities: either comment on here or tweet @mscroggs.

#### Flexagons

It's probably best to start by showing you what a flexagon is...
What you saw there is called a trihexaflexagon. Tri- because it has three faces; -hexa- because it is a hexagon; and -flexagon because it can be flexed to reveal the other faces.
The story goes that, in 1939, Arthur H. Stone, who was an Englishman studying mathematics at Harvard, was trimming the edges off his American paper to fit in his English folder. He was fiddling with the offcuts and found that if he folded the paper under itself in a loop, he could make a hexagon; and when this hexagon was folded up as we saw, it would open out to reveal a different face.
The way it flexes can be show on a diagram: In the circles, the colour on either side of the flexagon is shown and the lines show flexes which can be made.
When Stone showed his flexagon to other students at Harvard, they were equally amazed by it, and they formed what they called 'The Flexagon Committee'. Members of the committee included Richard Feynman, who was then still a graduate student. The committee could meet regularly and soon discovered other flexagons, the first of which was the hexahexaflexagon: Again shaped like a hexagon, but this time with six faces.
A hexahexaflexagon is created by taking a longer strip of paper and rolling it around itself like this. The shorter strip at the end is then folded and glued in the same way the trihexaflexagon was. Once made, the hexahexaflexagon can be flexed. From some positions, the flexagon can be flexed in different ways to reveal different faces. Due to this, finding some of the faces can be quite difficult. The committee went on to find other flexagons which could be made, again made by first folding into a shorter strip, then folding up like the trihexaflexagon.
The committee later found that hexaflexagons with any number of faces could be made by starting with a certain shaped strip, rolling it up then folding it like a trihexaflexagon.
An excellent article by Martin Gardner on flexagons can be found in this book.
Trihexaflexagon templates (click to enlarge then print):

#### Folding Tube Maps

Our second story starts with me sitting on the tube reading Alex's Adventures in Numberland by Alex Bellos on the tube. In his book, Alex describes how to fold a tetrahedron, or triangle-based pyramid, from two business cards. With no business cards to hand, I picked up two tube maps and followed the steps: first, I folded it corner to corner; then I folded the overlaps over.I made another one of these, but the second a mirror image of the first, slotted them together and I had my tetrahedron.
Then I made a tube map cube by making six squares like so and slotting them together.
While making these shapes, I discovered an advantage of tube maps over business cards: Due to the pages, folded tube maps have slots to tuck the tabs into, so the solids are pretty sturdy.
Making these shapes got me wondering: what other Platonic solids could I make?
In 2D, we have regular shapes: shapes with all the sides of the same length and all angles equal. Platonic solids are sort of the 3D equivalent of this: they are 3D shapes where every face is the same regular shape and at each vertex the same number of faces meet.
For example, our tetrahedron is a Platonic solid because every side is a regular triangle, and three triangles meet at every vertex. Our cube is a Platonic solid because every side is a square (which is a regular shape) and three squares meet at every vertex.
In order to fold all the Platonic solids, we must first find out how many there are.
To do this, we're going to start with a triangle, as it is the 2D shape with the smallest number of sides, and make Platonic solids.
If we try to put two triangles at each vertex, then they'll squash flat; so that's no good. We've seen that three triangles at each vertex makes a tetrahedron. If we put four triangles at each vertex then we get an octahedron.
Five triangles at each vertex gives us an icosahedron.
Each angles in an equilateral triangle is 60°. So if we put six triangles at each vertex the angles add up to 360°, a full turn. This means that the triangles will lie flat, giving us a nice pattern for a kitchen floor, but not a solid. Any more than 6 triangles will add up to more than 360 and also not give a solid. So we have found all the Platonic solids whose faces are triangles.
Next, four sided faces. Three squares at each vertex gives us a cube. Four squares at each vertex will add up to 4 times 90°... 360° again, so another kitchen floor and as before we have all the Platonic solids whose faces are squares.
Now moving up again to five sided faces. Three pentagons at each vertex will gives us a dodecahedron, which looks like this.
This is the best I could do.
(After the talk, I was shown a few better ways to fold pentagons. Watch this space for my attempts...) Now if we try four pentagons around a vertex: the internal angle in a pentagon is 108°. 4 times 108° is 432°. This is more than a full turn, so we don't get a solid.
Moving up again, if we take three hexagons we get another tessellation. Shapes with more than six sides will all have larger angles than this so three make more than a full turn. Therefore, we have found and folded all the Platonic solids.
In 2012, I posted this on my blog and got the following comment:
I'm pretty sure this was a joke, but one hour, 48 tube maps and a lot of glue later:
Alex's Adventures in Numberland by Alex Bellos introduced business card folding and takes it further, finishing with a business card Menger sponge.

#### Braiding

A few months ago, my mother showed my a way to make braids using a cardboard octagon with a slot cut on each side and a hole in the middle.
To make a braid, seven strands of wool are tied together, fed through the hole, then one tucked into each slot.
Now, we jump over two strands, pick the third strand and move it to the vacant slot. So first, we jump over the orange and green and move the red strand.
Then we jump the light blue and yellow and move the dark blue.
And so on..
After a while, the braid looks like this:
Once I'd made a few braids, I began to wonder which other numbers of threads could be used to make braids like this. To investigate this I found it useful to represent braids by drawing connections to show where a thread is moved. This shows the first move:
Then the second move:
And so on until you get:
After the octagon, I tried braiding on a hexagon, moving the second thread each time. Here's what happened:
I only moved the yellow and green threads and nothing interesting happened. When I drew this out as before, it demonstrated what had gone wrong: three slots are missed so three threads are never moved.
So we need to find out when slots are missed and when all the slots are hit. To do this, let's call the number of slots $$a$$, and let $$b$$ be the number thread we pick each time. For example, in the first braid that worked $$a$$ was 8 and $$b$$ was 3. This question is very similar to the Wool Circles Problem from Sunday Afternoon Maths XXI. You might like to have a go at it before you read on.
First we'll label the slots. Label the slot which starts empty 0, then number anti-clockwise. This numbering puts all the multiples of $$a$$ at the bottom slot.
Now let's look at which slots we visit. We start at0, then visit $$b$$, then $$2b$$, then $$3b$$ and so on. We visit all the multiples of $$b$$.
Therefore we will reach the bottom slot again and finish our loop when we reach a common multiple of $$a$$ and $$b$$. The first time this happens will be at the lowest common multiple, or:
$$\mbox{lcm}(a,b)$$
On our way to this slot, we visited one slot for every $$a$$ we passed, so the number of slots we have visited is
$$\frac{\mbox{lcm}(a,b)}{a}$$
and we will visit every slot if
$$\frac{\mbox{lcm}(a,b)}{a}=b$$
or, equivalently if
$$\mbox{lcm}(a,b)=ab.$$
This is true when, $$a$$ and $$b$$ have no common factors, or in other words are coprime; which can be written
$$\mbox{hcf}(a,b)=1.$$
So we've found that if $$a$$ is the number of slots and $$b$$ is the jump then the braid will not work unless $$a$$ and $$b$$ are coprime.
For example, if $$a$$ is 6 and $$b$$ is 2 then 2 is a common factor so the braid fails. And, if $$a$$ is 8 and $$b$$ is 3 then there are no common factors and the braid works. And, if $$a$$ is 12 and $$b$$ is 5 then there are no common factors and the braid works.
But, if $$a$$ is 5 and $$b$$ is 2 then there are no common factors but the braid fails.
The rule I've explained is still correct, and explains why some braids fail. But if $$a$$ and $$b$$ are coprime, we need more rules to decide whether or not the braid works.
And that's as far as I've got, so I'm going to finish braiding with two open questions: Why does the 5 and 2 braid fail? And for which numbers $$a$$ and $$b$$ does the braid work?

#### Sine Curves

For the last part of the talk, I did a practical demonstration of how to draw a sine curve using five people.
I told the first person to stand on the spot and the second person to stand one step away, hold a length of string and walk.
The third person was instructed to stay in line with the second person, while staying on a vertical line.
The fourth person was told to walk in a straight line at a constant speed.
And the fifth person had to stay in line with both the third and fourth people. This led them to trace a sine curve.
To explain why this is a sine curve, consider the following triangle:
As our first two people are one step apart, the hypotenuse of this triangle is 1. And so the opposite (vertical) side is equal to the sine of the angle.
I like to finish with a challenge, and this task leads nicely into two challenge questions:
1. How could you draw a cosine curve with five people?
2. How could you draw a tan(gent) curve with five people?
(These questions will be one of this week's Sunday Afternoon Maths puzzles, so I'll post the answer on Monday.)
People Maths: Hidden Depths is full of this kind of dynamic task involving moving people.

## Monday, 25 August 2014

### Sunday Afternoon Maths XXVI Answers & Extensions

This post contains the answers to this week's Sunday Afternoon Maths and some extension problems based around the originals.

#### Twenty-One

Virgil should go second. Whatever Scott adds, Virgil should then add to make four. For example, if Scott says 3, Virgil should say 1.
Using this strategy, Virgil will say 4, 8, 12, 16 then 20, forcing Scott to go above 21.
##### Extension
(i) If instead of 21, 22 cannot be said/beaten, how should Virgil win? How about 23? Or 24? How about $$n$$?
(ii) If instead of adding 1 to 3, 1 to 4 can be added, how should Virgil win? How about 1 to 5? Or 2 to 5? How about $$m$$ to $$l$$?
(iii) Alan wants to join the game. Can Virgil win if there are three people? Can he win if there are $$k$$ people?

#### Odd and Even Outputs

 $$n$$ odd even $$m$$ odd odd odd e odd odd
$$g(n,m)=1$$

 $$n$$ odd even $$m$$ odd odd odd e odd even
$$g(n,m)=n\times m + n + m$$

 $$n$$ odd even $$m$$ odd odd odd e even odd
$$g(n,m)=n\times m +n+1$$

 $$n$$ odd even $$m$$ odd odd odd e even even
$$g(n,m)=m$$
 $$n$$ odd even $$m$$ odd odd even e odd odd
$$g(n,m)=n\times m+m+1$$

 $$n$$ odd even $$m$$ odd odd even e odd even
$$g(n,m)=n$$

 $$n$$ odd even $$m$$ odd odd even e even odd
$$g(n,m)=n+m+1$$

 $$n$$ odd even $$m$$ odd odd even e even even
$$g(n,m)=n\times m$$
 $$n$$ odd even $$m$$ odd even odd e odd odd
$$g(n,m)=n\times m+1$$

 $$n$$ odd even $$m$$ odd even odd e odd even
$$g(n,m)=n+m$$

 $$n$$ odd even $$m$$ odd even odd e even odd
$$g(n,m)=n+1$$

 $$n$$ odd even $$m$$ odd even odd e even even
$$g(n,m)=n\times m+n$$
 $$n$$ odd even $$m$$ odd even even e odd odd
$$g(n,m)=m+1$$

 $$n$$ odd even $$m$$ odd even even e odd even
$$g(n,m)=n\times m+n$$

 $$n$$ odd even $$m$$ odd even even e even odd
$$g(n,m)=n\times m+n+m+1$$

 $$n$$ odd even $$m$$ odd even even e even even
$$g(n,m)=2$$
##### Extension
Can you find functions $$h:\mathbb{N}\times\mathbb{N}\times\mathbb{N}\rightarrow\mathbb{N}$$ (call the inputs $$n$$, $$m$$ and $$l$$) to give the following outputs:
$$l$$ odd
 $$n$$ odd even $$m$$ odd even even e even even
$$l$$ even
 $$n$$ odd even $$m$$ odd even even e even even

$$l$$ odd
 $$n$$ odd even $$m$$ odd even even e even even
$$l$$ even
 $$n$$ odd even $$m$$ odd even even e even odd

etc

## Sunday, 24 August 2014

### Sunday Afternoon Maths XXVI

Here's this week's collection. Answers & extensions tomorrow. Why not discuss the problems on Twitter using #SundayAfternoonMaths or on Reddit.
It's no late to get tickets for EMF Camp, where I will be giving a talk on flexagons, folding tube maps and braiding.

#### Twenty-One

Scott and Virgil are playing a game. In the game the first player says 1, 2 or 3, then the next player can add 1, 2 or 3 to the number and so on. The player who is forced to say 21 or above loses. The first game went like so:
Scott: 3
Virgil: 4
Scott: 5
Virgil: 6
Scott: 9
Virgil: 12
Scott: 15
Virgil 17
Scott: 20
Virgil: 21
Virgil loses.
To give him a better chance of winning, Scott lets Virgil choose whether to go first or second in the next game. What should Virgil do?

#### Odd and Even Outputs

Let $$g:\mathbb{N}\times\mathbb{N}\rightarrow\mathbb{N}$$ be a function.
This means that $$g$$ takes two natural number inputs and gives one natural number output. For example if $$g$$ is defined by $$g(n,m)=n+m$$ then $$g(3,4)=7$$ and $$g(10,2)=12$$.
The function $$g(n,m)=n+m$$ will give an even output if $$n$$ and $$m$$ are both odd or both even and an odd output if one is odd and the other is even. This could be summarised in the following table:
 $$n$$ odd even $$m$$ odd even odd e odd even
Using only $$+$$ and $$\times$$, can you construct functions $$g(n,m)$$ which give the following output tables:
 $$n$$ odd even $$m$$ odd odd odd e odd odd
 $$n$$ odd even $$m$$ odd odd odd e odd even
 $$n$$ odd even $$m$$ odd odd odd e even odd
 $$n$$ odd even $$m$$ odd odd odd e even even
 $$n$$ odd even $$m$$ odd odd even e odd odd
 $$n$$ odd even $$m$$ odd odd even e odd even
 $$n$$ odd even $$m$$ odd odd even e even odd
 $$n$$ odd even $$m$$ odd odd even e even even
 $$n$$ odd even $$m$$ odd even odd e odd odd
 $$n$$ odd even $$m$$ odd even odd e odd even
 $$n$$ odd even $$m$$ odd even odd e even odd
 $$n$$ odd even $$m$$ odd even odd e even even
 $$n$$ odd even $$m$$ odd even even e odd odd
 $$n$$ odd even $$m$$ odd even even e odd even
 $$n$$ odd even $$m$$ odd even even e even odd
 $$n$$ odd even $$m$$ odd even even e even even

## Monday, 18 August 2014

### Sunday Afternoon Maths XXV Answers & Extensions

This post contains the answers to this week's Sunday Afternoon Maths and some extension problems based around the originals.

#### Whist

Represent each player by a compass point. Let B, D, A and S represent Messrs. Banker, Dentist, Apothecary and Scrivener respectively and b, d, a and s represent the four jobs.
Mr. Banker (B) partners the scrivener (s). Let B sit at West. This means s sits at East. As no other player can also be B or s, the table looks like this:
 North B D A S b d a s
 West B D A S b d a s
 East B D A S b d a s
 South B D A S b d a s
The dentist (d) sits on Mr. Scrivener's (S) right. East cannot be S, so North cannot be d. East cannot be d, so South cannot be S.
 North B D A S b d a s
 West B D A S b d a s
 East B D A S b d a s
 South B D A S b d a s
By elimination, only North can be S. This means that d must sit to the right of North (at West):
 North B D A S b d a s
 West B D A S b d a s
 East B D A S b d a s
 South B D A S b d a s
A and a are partners. This is only possible if A is South and a is North:
 North B D A S b d a s
 West B D A S b d a s
 East B D A S b d a s
 South B D A S b d a s
Therefore, Mr. Banker the dentist sits to the left of the banker.
##### Extension
If each person is partnered with their job namesake, how many possible combinations of names and jobs are possible?

#### Polya Strikes Out

1, 3, 7, 12, 19, ...
1, 7, 19, ...
1=1; 1+7=8; 1+7+19=27; ...
1, 8, 27, ...
The final sequence is the cube numbers. To show why, let $$n$$ be an integer and follow through the process.
Cross out every third number:
1, 2, 3, 4, 5, 6, ..., 3n, $$3n+1$$, $$3n+2$$, ...
1, 2, 4, 5, ..., $$3n+1$$, $$3n+2$$, ...
Find the cumulative sums:
$$1=1$$ $$1+2=1+2=3$$ $$1+2+4=1+2+3+4-3=7$$ $$1+2+4+5=1+2+3+4+5-3=12$$ $$...$$ $$1+2+4+5+...+(3n+1)=\sum_{i=1}^{3n+1}-\sum_{i=1}^{n}3i$$ $$=\frac{1}{2}(3n+1)(3n+2)-\frac{3}{2}n(n+1)$$ $$=3n^2+3n+1$$ $$1+2+4+5+...+(3n+2)=3n^2+3n+1+(3n+2)$$ $$=3n^2+6n+3$$ $$...$$
1, 3, 7, 12, ..., $$3n^2+3n+1$$, $$3n^2+6n+3$$, ...
Cross out every second number, starting with the second:
1, 3, 7, 12, ..., $$3n^2+3n+1$$, 3n2+6n+3, ...
1, 7, ..., $$3n^2+3n+1$$, ...
Find the cumulative sums. The $$m$$th sum is:
$$\sum_{n=0}^{m}3n^2+3n+1$$ $$=3\sum_{n=0}^{m}n^2+3\sum_{n=0}^{m}n+\sum_{n=0}^{m}1$$ $$=\frac{3}{6}m(m+1)(2m+1)+\frac{3}{2}m(m+1)+m+1$$ $$=\frac{1}{2}(m+1)(m(2m+1)+3m+2)$$ $$=\frac{1}{2}(m+1)(2m^2+m+3m+2)$$ $$=\frac{1}{2}(m+1)(2m^2+4m+2)$$ $$=(m+1)(m^2+2m+1)$$ $$=(m+1)(m+1)^2$$ $$=(m+1)^3$$
Hence the numbers obtained are the cube numbers.
##### Extension
What happens if you cross out every third number starting at the second? Or every fifth number starting at the fifth? Or every $$n$$th number starting at the $$m$$th?

## Sunday, 17 August 2014

### Sunday Afternoon Maths XXV

Here's this week's collection. Answers & extensions tomorrow. Why not discuss the problems on Twitter using #SundayAfternoonMaths or on Reddit. This Tuesday is Maths Jam, make sure you're there!
It's less than two weeks 'til EMF Camp, where there will be talks on maths and science and a whole load of other awesome stuff. Tickets here

#### Whist

Messrs. Banker, Dentist, Apothecary and Scrivener played whist last night. (whist is a four player card game where partners sit opposite each other.) Each of these gentlemen is the namesake of another's vocation.
Last night, the apothecary partnered Mr. Apothecary; Mr. Banker's partner was the scrivener; on Mr. Scrivener's right sat the dentist.
Who sat on the banker's left?

#### Polya Strikes Out

Write the numbers 1, 2, 3, ... in a row. Strike out every third number beginning with the third. Write down the cumulative sums of what remains:
1, 2, 3, 4, 5, 6, 7, ...
1, 2, 3, 4, 5, 6, 7, ...
1, 2, 4, 5, 7, ...
1=1; 1+2=3; 1+2+4=7; 1+2+4+5=12; 1+2+4+5+7=19; ...
1, 3, 7, 12, 19, ...
Now strike out every second number beginning with the second. Write down the cumulative sums of what remains. What is the final sequence? Why do you get this sequence?

## Monday, 11 August 2014

### Sunday Afternoon Maths XXIV Answers & Extensions

This post contains the answers to this week's Sunday Afternoon Maths and some extension problems based around the originals.

#### Square Cross

Label and orient the shape as follows:
The smaller diagonal square is made up of a 1×1 square and four 1×2 right-angled triangles. Therefore its area is 5.
The line FM, and therefore the line DN, has gradient 2. The line JM, and therefore the line ON, has gradient -½. ON passes through (1,2) and DN passes through (2,0). Therefore, DN has equation $$y=2x-4$$ and ON has equation $$y=\frac{5}{2}-\frac{1}{2}x$$. These lines intersect at $$(\frac{13}{5},\frac{6}{5})$$, these are the co-ordinates of N. By the same method, the co-ordinates of P are $$(-\frac{8}{5},-\frac{1}{5})$$.
By Pythagoras' Theorem, The diagonal of the larger square is $$\frac{7\sqrt{10}}{5}$$ and so the area of the larger square is 9.8.

#### Exact Change

In the UK, eight coins are needed: 1p, 1p, 2p, 5p, 10p, 20p, 20p, 50p.
In the US, ten coins are needed: 1¢, 1¢, 1¢, 1¢, 5¢, 10¢, 10¢, 25¢, 25¢, 25¢.