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Monday, 26 May 2014

World Cup Stickers

With the FIFA World Cup approaching, sticker fans across the world are filling up their official Panini sticker books. This got me wondering: how many stickers should I expect to need to buy to complete my album? And how much will this cost?

How many stickers?

There are 640 stickers required to fill the album. The last 100 stickers required can be ordered from the Panini website.
After $$n$$ stickers have been stuck into the album, the probability of the next sticker being the next new sticker is: $$\frac{640-n}{640}$$ The probability that the sticker after next is the next new sticker is: $$\frac{n}{640}\frac{640-n}{640}$$ The probability that the sticker after that is the next new sticker is: $$\left(\frac{n}{640}\right)^2\frac{640-n}{640}$$ Following this pattern, we find that the expected number of stickers bought to find a new sticker is: $$\sum_{i=1}^{\infty}i \left(\frac{640-n}{640}\right) \left(\frac{n}{640}\right)^i = \frac{640}{640-n}$$
Therefore, to get all 640 stickers, I should expect to buy: $$\sum_{n=0}^{639}n \frac{640}{640-n} = 4505 \mbox{ stickers.}$$ Or, if the last 100 stickers needed are ordered: $$\sum_{n=0}^{539}n \frac{640}{640-n} + 100 = 1285 \mbox{ stickers.}$$

How much?

The first 21 stickers come with the album for £1.99. Additional stickers can be bought in packs of 5 for 50p or multipacks of 30 for £2.75. To complete the album, 100 stickers can be bought for 25p each.
If I decided to complete my album without ordering the final stickers, I should expect to buy 4505 stickers. After the 21 which come with the album, I will need to buy 4484 stickers: just under 897 packs. These packs would cost £411.25 (149 multipacks and 3 single packs), giving a total cost of £413.24 for the completed album.
I'm not sure if I have a spare £413.24 lying around, so hopefully I can reduce the cost of the album by buying the last 100 stickers for £25. This would mean that once I've received the first 21 stickers with the album, I will need to buy 1164 stickers, or 233 packs. These packs would cost £107 (38 multipacks and 5 single packs), giving a total cost of £133.99 for the completed album, significantly less than if I decided not to buy the last stickers.

How many should I order?

The further reduce the number of stickers bought, I could get a friend to also order 100 stickers for me and so buy the last 200 stickers for 25p each. With enough friends the whole album could be filled this way, although as the stickers are more expensive than when bought in packs, this would not be the cheapest way.
If the last 219 or 250 stickers are bought for 25p each, then I should expect to spend £117.74 in total on the album. If I buy any other number of stickers at the end, the expected spend will be higher.

Fortunately, as you will be able to swap your duplicate stickers with your friends, the cost of a full album should turn out to be significantly lower than this. Although if saving money is your aim, then perhaps the Panini World Cup 2014 Sticker Book game would be a better alternative to a real sticker book.

Sunday Afternoon Maths XIV Answers & Extensions

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

Eight Peas

Extension
If you start with $$n$$ cups with one pea in each, how many moves is it possible to make?

Superdog

On hearing the eighth whistle, the dog will be running at 512m/s, faster than the speed of sound. Hence, no more of the master's whistles will reach the dog. However, the dog will catch up with the seven whistles she has already heard and hear them again. So fifteen whistles will be heard in total.
Extension
How much time will the dog have been running for when she hears the final whistle?

Sunday, 25 May 2014

Sunday Afternoon Maths XIV

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

Eight Peas

There are eight cups, with one pea in each cup. You are allowed to move a pea by picking up the pea in a pot with only one pea and jumping it to the left or the right over two peas into a pot with only one pea in it. For example, the following moves are allowed:
Starting with your eight cups, can you make four moves?

Superdog

A dog is running along a beach at 2m/s. The dog's owner blow a whistle every 10 seconds. Each time the dog hears a whistle, she doubles her speed. How many whistles will the dog hear?
(Hint: speed of sound = 343m/s)
(Hint 2: The answer is not eight!)

Monday, 19 May 2014

Sunday Afternoon Maths XIII Answers & Extensions

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

Mean, Mode, Median, Range

A 1, 3, 4, 6 & 6
B 1, 1, 1, 3 & 9
C 1, 2, 2, 4 & 6 or 2, 2, 2, 2 & 7
Extension
A and B have only one solution. C has two solutions. For which other numbers a, b, c and d does the problem
Find five one-digit positive integers which have a mean of a, mode of b, median of c and a range of d.
have more than one solution

Times Roamin'

One of the terms in the series will be $$(x-x)$$, so the product will be zero.

Sunday, 18 May 2014

Sunday Afternoon Maths XIII

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!

Mean, Mode, Median, Range

A Find five one-digit positive integers which have a mean of 4, mode of 6, median of 4 and a range of 5.
B Find five one-digit positive integers which have a mean of 3, mode of 1, median of 1 and a range of 8.
C Find five one-digit positive integers which have a mean of 3, mode of 2, median of 2 and a range of 5.

Times Roamin’

What is the product of this series?
$$(x-a)(x-b)(x-c)...(x-z)$$

Monday, 12 May 2014

Sunday Afternoon Maths XII Answers & Extensions

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

Circles

Let $$4x$$ be the side length of the square. This means that the radius of the red circle is $$2x$$ and the radius of a blue circle is $$x$$. Therefore the area of the red circle is $$4\pi x^2$$.
The area of one of the blue squares is $$\pi x^2$$ so the blue area is $$4\pi x^2$$. Therefore the two areas are the same.
Extension
Is the red or blue area larger?

Largest Triangle

As our shape is a triangle, the 4cm and 5cm sides must be adjacent. Call the angle between them be $$\theta$$.
The area of the triangle is $$\frac{1}{2}\times 4\times 5 \times \sin{\theta}$$ or $$10\sin{\theta}$$. This has a maximum value when $$\theta=90^\circ$$, so the largest triangle has and area of 10cm2 and looks like:
Extension
What is the largest area triangle with a perimeter of 12cm?

Unit Octagon

Name the regions as follows:
$$E$$ is a 1×1 square. Placed together, $$A$$, $$C$$, $$G$$ and $$I$$ also make a 1×1 square. $$B$$ is equal to $$H$$ and $$D$$ is equal to $$F$$.
Therefore $$B+E+F=A+C+D+G+H+I$$. Therefore The hatched region is $$C$$ larger than the shaded region. The area of $$C$$ (and therefore the difference) is $$\frac{1}{4}$$
Extension
What is the difference between the shaded and the hatched regions in this dodecagon?

Sunday, 11 May 2014

Sunday Afternoon Maths XII

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

Circles

Which is largest, the red or the blue area?

Largest Triangle

What is the largest area triangle which has one side of length 4cm and one of length 5cm?

Unit Octagon

The diagram shows a regular octagon with sides of length 1. The octagon is divided into regions by four diagonals. What is the difference between the area of the hatched region and the area of the region shaded grey?

Monday, 5 May 2014

Sunday Afternoon Maths XI Answers & Extensions

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

Light Work

Extension
If instead a window 3/4 of the size was required, how could this be done?

Semi Circle in a Triangle

Label the triangle as follows:
EC and BC are both tangents to the circle so FEA is a right angle and the lengths EC and BC are equal, so the length of EC is 5. Let r be the radius of the circle.
Using Pythagoras' Theorem in triangle FEA:
$$8^2+r^2=(12-r)^2$$ $$64+r^2=144-24r+r^2$$ $$24r=80$$ $$r=\frac{80}{24}=\frac{10}{3}$$
Extension
What is the radius of the circle whose diameter lies on the 5cm side and to which the 12cm and 13cm sides are tangents?
What is the radius of the circle whose diameter lies on the 13cm side and to which the 12cm and 5cm sides are tangents?
For each pair of semi circles, draw a straight line between the two points where the semi circles intersect. These lines all meet at a point.

Sunday, 4 May 2014

Sunday Afternoon Maths XI

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

Light Work

I don’t know if you are fond of puzzles, or not. If you are, try this. … A gentleman (a nobleman let us say, to make it more interesting) had a sitting-room with only one window in it–a square window, 3 feet high and 3 feet wide. Now he had weak eyes, and the window gave too much light, so (don’t you like ‘so’ in a story?) he sent for the builder, and told him to alter it, so as only to give half the light. Only, he was to keep it square–he was to keep it 3 feet high–and he was to keep it 3 feet wide. How did he do it? Remember, he wasn’t allowed to use curtains, or shutters, or coloured glass, or anything of that sort.

Semi Circle in a Triangle

This right-angled triangle above has sides of lengths 12cm, 5cm and 13cm. The diameter of the semicircle lies on the 12cm side and the 13cm side is a tangent to the circle. What is the radius of the semi circle?