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## Monday, 28 April 2014

### Sunday Afternoon Maths X Answers & Extensions

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

#### Three Digit Numbers

The sum of the products is:
$$\sum_{i=1}^{9}\sum_{j=1}^{9}\sum_{k=1}^{9}ijk$$ $$=(\sum_{i=1}^{9}i)(\sum_{j=1}^{9}j)(\sum_{k=1}^{9}k)$$ $$=45\times45\times45$$ $$=45^3$$
##### Extension
Find the sum of the products of the digits of all n digit numbers.

#### Princess Problem

Imagine a chessboard with 17 columns and a large number of rows. Let each column correspond to one of the rooms. In the first row, mark the room in which the princess is. Every day following, mark her location in the next row.
The princess will always move one square diagonally, so will always be on the same colour square she started on. Begin by checking the 17th room, then the 16th room, then continue down the the first room. If the princess started on a black square you will have found her, as she has no way of getting past you.
If you have not caught the princess, then she must have started on a white square. Checking rooms 17 down to one again will find her as this time it will start on a white square.
##### Extension
Is there a quicker way to find the princess?

#### Lights

If two lights opposite each other are on and the other two are off, then toggling two opposite lights will always lead to your release. It can be checked that the following combination will always lead to this:
• Toggle two opposite lights
• Toggle two opposite lights
• Toggle one light
• Toggle two opposite lights
• Toggle two opposite lights
##### Extension
Can the same be done with eight lights?

## Sunday, 27 April 2014

### Sunday Afternoon Maths X

Here's this week's collection, including problems from last week's Maths Jam. Answers & extensions here. Why not discuss the problems on Twitter using #SundayAfternoonMaths or on Reddit.

#### Three Digit Numbers

Brigette wrote down a list of all 3-digit numbers. For each of the numbers on her list she found the product of the digits. She then added up all of these products. Which of the following is equal to her total?
A 45     B 452     C 453     D 245     E 345

#### Princess Problem

A princess lives in a row of 17 rooms. Each day she moves to a room adjacent to the one she wakes up in (eg. If she sleeps in room 5 today, then she will sleep in room 4 or 6 tomorrow). If you are able to find the princess by only opening one door each night then you will become her prince. Can you find her in a finite number of moves?

#### Lights

You have been taken captive and are blindfolded. There is a table in front of you with four lights on it. Some are on, some are off: you don't know how many and which ones. You need to get either all the lights on or all the lights off to be released. To do this, you can ask your captor to toggle the light switches of some of the lights. You captor will then rotate the table (so you don't know where the lights you toggled now are). Find a sequence of moves which will always lead to your release.

## Monday, 21 April 2014

### Sunday Afternoon Maths IX Answers & Extensions

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

#### Equal Areas

Let A be the area of the square (and the triangle).
The length of a side of the square is √A, so the perimeter of the square is 4√A
Let l be the length of a side the triangle. Then 1/2l 2sin60 = A, so l 2 = 4A/√3. Therefore l = 2√A/31/4 and the perimeter of the triangle is 6√A/31/4.
Hence the ratio of the perimeters is 6√A/31/4:4√A which simplifies to 33/4:2
##### Extension
If an n sided regular polygon has the area A, what is the length of one of its sides?

#### Chessboard Squares

There are 64 1×1 squares, 49 2×2 squares, 36 3×3 squares, 25 4×4 squares, 16 5×5 squares, 9 6×6 squares, 4 7×7 squares and 1 8×8 square on a chessboard.
This can be shown by counting how many positions the top left corner of the square can sit on. For example, the top left corner of a 5×5 square can be in the first four rows and columns of the board (otherwise the square will go off the board) and 4×4=16.
64+49+36+25+16+9+4+1=204.
##### Extension
How many rectangles are there on a chessboard?

#### Downing Street

The probabilities can be summarised as follows:
 First person truthful First person lying Second person truthful 3/4×4/5=12/20 1/4×4/5=4/20 Second person lying 3/4×1/5=3/20 1/4×1/5=1/20
As they both agree, only both lying and both truthful are possible. Hence the chance of them lying is 1/13 and the chance of them telling the truth, and it indeed being the Minister of Maths, is 12/13
##### Extension
If the first person said it was the Minister of English and the second said it was the Minister of Maths, what is the probability that it was the Minister of Maths?

#### Multiple Sums

The multiples of 3 less than 1000 are 3,6,9,...,999; the multiples of 5 are 5,10,15,...,995. Multiples of 15 (15,30,...,990) will appear in both lists so we are trying to find (3+6+9+...+999)+(5+10+15+...+995)-(15+30+...+990). This is:
$$\sum_{i=1}^{333}3i+\sum_{j=1}^{199}5j-\sum_{k=1}^{66}15k$$ $$=3\sum_{i=1}^{333}i+5\sum_{j=1}^{199}j-15\sum_{k=1}^{66}k$$ $$=3\times\frac{333\times334}{2}+5\times\frac{199\times200}{2}-15\times\frac{66\times67}{2}$$ $$=166833+99500-33165$$ $$=233168$$
Massive thanks to MathJax for allowing me to typeset maths!
##### Extension
Find the sum of all the multiples of 3 or 5 below n.

## Sunday, 20 April 2014

### Sunday Afternoon Maths IX

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

#### Equal Areas

An equilateral triangle and a square have the same area. What is the ratio of the perimeter of the triangle to the perimeter of the square?

#### Chessboard Squares

It was once claimed that there are 204 squares on a chessboard. Can you justify this claim?

#### Downing Street

A knot of spectators in Downing Street was watching members of the Cabinet as they arrived for a critical meeting.
"Who's that?" I asked my neighbour, as a silk-hatted figure, carrying rolled umbrella, rang the bell at No. 10. "Is it the Minister of Maths?"
"Yes," he said.
"Quite right," said a second spectator. "The Minister of Maths it is. Looks grim, doesn't he?"
The first of the speakers tells the truth three times out of four. The second tells the truth four times out of five.
What is the probability that the gentleman in question was in fact the Minister of Maths?

#### Multiple Sum

If we list all the natural numbers below 10 that are multiples of 3 or 5, we get 3, 5, 6 and 9. The sum of these multiples is 23.
Find the sum of all the multiples of 3 or 5 below 1000.

## Monday, 14 April 2014

### Sunday Afternoon Maths VIII Answers & Extensions

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

#### Two

a+(a+A)-1 = a + A = 2.
So, a+2-1 = 2
So, a = 3/2
By the same method, b = 2-√2, c = -2 and d = 2-2k
##### Extension
For which values of k does this converge?

#### Reverse Bases Again

445 in base 10 is equal to 544 in base 7
##### Extension
Find another pair of bases A and B so that there exist digits d, e and f such that def in base A is equal to fed in base B?

A=8, B=9 and C=7.
##### Extension
Find bases A, B and C such that 123A+456B=789C

#### Complex Squares

Any complex number can be written in the form z=reiθ.
This gives that z2=r2e2iθ, which will have positive real and complex parts when 0+2πn < 2θ < π/2+2πn.
This will occur when 0 < θ < π/4 and π < θ < /4.
This can be represented on an Argand diagram:
A complex number z falls in these regions when |Re(z)|>|Im(z)| and sign(Re(z))=sign(Im(z))
##### Extension
For which complex numbers, z, are Re(z3) and Re(z3) both positive?

#### Rebounds

If mirrors were placed along the walls of the rectangle, the ball would appear to travel in a straight line across a grid of rectangles:
Viewed this way, the ball will still stop once it reaches a corner:
For an n by m (in above example: n = 4, m = 3) rectangle, this will occur once the ball has travelled through lcm(n, m) squares.
 Let a = lcm(n, m) n
 Let b = lcm(n, m) m
On its way to the corner, the ball will bounce b – 1 times off the top and bottom and a – 1 times off the sides. It can be seen that if a – 1 is even, then the ball will end in one of the corners on the right hand side. The complete results can be seen in the following Carroll diagram:
 a – 1 odd a – 1 even b – 1 odd Top left Top right b – 1 even Bottom left Bottom right
It can be shown that the ball will never finish in the top left (where it started) as this would require it to travel through the bottom right first. Therefore the following holds:
 a even a odd b even Top right b odd Bottom left Bottom right
##### Extension
For which sizes of rectangle will the path of the ball make the same pattern?

## Sunday, 13 April 2014

### Sunday Afternoon Maths VIII

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

#### Two

Find a such that a+(a+A)-1 = 2, where A = (a+A)-1.
ie. a + /(a+1/(a+1/(a+1/(a+1/(a+1/(a+...)))))) = 2.

Find b such that b+(b+B)0.5 = 2, where B = (b+B)0.5.
ie. b + √(b+√(b+√(b+√(b+√(b+√(b+...)))))) = 2.

Find c such that c+(c+C)2 = 2, where C = (c+C)2.

In terms of k, find d such that d+(d+D)k = 2, where D = (d+D)k.

#### Reverse Bases Again

Find three digits a, b and c such that abc in base 10 is equal to cba in base 9?

Let ab denote a in base b.
Find bases A, B and C less than 10 such that 12A+34B=56C

#### Complex Squares

For which complex numbers, z, are Re(z2) and Im(z2) both positive?

#### Rebounds

In a 4x3 rectangle, a ball is fired from the top left corner at 45°.
It bounces around a rectangle until it hits a corner. Which corner does it end in?
Which corner will it end in for rectangles of other sizes?

## Friday, 11 April 2014

### Countdown Probability, pt. 2

As well as letters games, the contestants on Countdown also take part in numbers games. Six numbers are chosen from the large numbers (25,50,75,100) and small numbers (1-10, two cards for each number) and a total between 101 and 999 (inclusive) is chosen by CECIL. The contestants then use the six numbers, with multiplication, addition, subtraction and division, to get as close to the target number as possible.
The best way to win the numbers game is to get the target exactly. This got me wondering: is there a combination of numbers which allows you to get every total between 101 and 999? And which combination of large and small numbers should be picked to give the highest chance of being able to get the target?
To work this out, I got my computer to go through every possible combination of numbers, trying every combination of operations. (I had to leave this running overnight as there are a lot of combinations!)

#### Getting Every Total

There are 2646 combinations of numbers which allow every total to be obtained. These include the following (click to see how each total can be made):
By contrast, the following combination only allows 13 totals between 101 and 999 to be reached:
• 1 1 2 2 3 3
The number of attainable targets for each set of numbers can be found here.

#### Probability of Being Able to Reach the Target

Some combinations of numbers are more likely than others. For example, 1 2 25 50 75 100 is four times as likely as 1 1 25 50 75 100, as (ignoring re-orderings) in the first combination, there are two choices for the 1 tile and 2 tile, but in the second combination there is only one choice for each 1 tile. Different ordering of tiles can be ignored as each combination with the same number of large tiles will have the same number of orderings.
By taking into account the relative probability of each combination, the following probabilities can be found:
 Number of large numbers Probability of being able to reach target 0 0.964463439 1 0.983830962 2 0.993277819 3 0.985770510 4 0.859709475
So, in order to maximise the probability of being able to reach the target, two large numbers should be chosen.
However, as this will mean that your opponent will also be able to reach the target, a better strategy might be to pick no large numbers or four large numbers and get closer to the target than your opponent, especially if you have practised pulling off answers like this.
Edit: Numbers corrected.

## Monday, 7 April 2014

### Sunday Afternoon Maths VII Answers & Extensions

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

#### Ninety Nine

Every item bought will cause the pence in the total cost to fall by 1. So to spend £65.76, Susanna must have bought 24 items.
##### Extension
What is the smallest amount Susanna could spend for which we could not tell how many items she bought?

#### Reverse Bases

If ab in base 10 is equal to ba in base 4, then 10a+b=4b+a.
So, 9a=3b.
a and b must both be less than 4, as they are digits used in base 4, so a=1 and b=3.
So 13 in base 10 is equal to 31 in base 4.
By the same method, we find that:
23 in base 10 is equal to 32 in base 7
46 in base 10 is equal to 64 in base 7.

12 in base 9 is equal to 21 in base 5.
24 in base 9 is equal to 42 in base 5.
##### Extension
For which pairs of bases A and B can you find two digits g and h such that gh in base A is equal to hg in base B?

## Sunday, 6 April 2014

### Countdown Probability

On Countdown, contestants have to make words from nine letters. The contestants take turns to choose how many vowels and consonants they would like. This got me wondering which was the best combination to pick in order to get a nine letter word.
Assuming the letters in countdown are still distributed like this, the probability of getting combinations of letters can be calculated. As the probability throughout the game is dependent on which letters have been picked, I have worked out the probability of getting a nine letter word on the first letters game.

#### The Probability of YODELLING

YODELLING has three vowels and six consonants. There are 6 (3!) ways in which the vowels could be ordered and 720 (6!) ways in which the consonants can be ordered, although each is repeated at there are two Ls, so there are 360 distinct ways to order the consonants. The probability of each of these is:
 21×13×13×6×3×5×4×8×1 67×66×65×74×73×72×71×70×69
So the probability of getting YODELLING is:
 6×360×21×13×13×6×3×5×4×8×1 = 0.000000575874154 67×66×65×74×73×72×71×70×69

#### The Probability of Any Nine Letter Word

I got my computer to find the probability of every nine letter word and found the following probabilities:
 Consonants Vowels Probability of nine letter word 0 9 0 1 8 0 2 7 0 3 6 0.000546 4 5 0.019724 5 4 0.076895 6 3 0.051417 7 2 0.005662 8 1 0.000033 9 0 0
So the best way to get a nine letter word in the first letters game is to pick five consonants and four vowels.

### Sunday Afternoon Maths VII

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

#### Ninety Nine

In a ‘ninety nine’ shop, all items cost a number of pounds and 99 pence. Susanna spent £65.76. How many items did she buy?

#### Reverse Bases

Find two digits a and b such that ab in base 10 is equal to ba in base 4?
Find two digits c and d such that cd in base 10 is equal to dc in base 7?
Find two digits e and f such that ef in base 9 is equal to fe in base 5?