How many connections?

This post was inspired by @squiggle7's Challenge That 'let's get connected' Post.

In it, @squiggle7 wonders about the number of connections able to be made by people in a 'room', be that virtually or physically.

Let's keep this up to date and use connecting with like-minded individuals on twitter as an example:

Imagine there's only one person on twitter who shares a particular interest of yours- how many connections can be made? By 'connections' I'm meaning mutual followships (to keep it simple), so in this case it's obvious: only one connection can be made.

Spotting the pattern
How about two other like- minded people? That's three altogether. I'll call them A, B & C (because they're cool names, right?). Because we're looking at mutual followships only, we can assume that if A is linked with B, then B is linked with A, so we can have the following links:

• A with B
• A with C
• B with C
• 3 connections altogether.

O.k, now how about four people? A, B, C & D this time:

• A with B
• A with C
• A with D
• B with C
• B with D
• C with D
• 6 connections available to be made.

[I'm being systematic when listing these possibilities- can you see the system I'm using?]

And five people (A, B, C, D & E):

• A with B
• A with C
• A with D
• A with E
• B with C
• B with D
• B with E
• C with D
• C with E
• D with E
• 10 connections possible.

We could carry on, but it's fairly clear by now that it's going to get a lot more complicated a lot more quickly, so can we do anything to simplify things? Can we find a pattern to follow?

People  2 3 4 5
Connections  1 3 6 10

Look at the numbers in that table: as the number of people goes up, the number of connections increases too. But by how much? It's fairly easy to see that as the number of people increases by 1, the number of connections increases by 2, then 3, then 4... so we can be confident that the number of connections for 6 people would be 10 +5= 15, and the number of connections for 7 people would be 15+6= 21 and so on. If we wanted to find out how many connections would be possible for, say, 20 people, we could just carry on the sequence. But what about 100 people? 150 people? 2000? 6000? 132,947 people?

Using algebra
Time to think about it a different way... Imagine there's a large number of people all with the same interest who all connect with each other. How many connections will there be? How many people exactly doesn't matter, so we'll call it "n" (so it can stand for any number).

• Person A can make a connection with everybody (except him/herself), so that's n-1 connections.
• Everyone else can make the same number of connections, so that's n-1 connections, n times: n(n-1)
• But we're forgetting that once a person makes a connection, they're connected both ways, so we've actually counted each connection twice. That's easy to fix, though, we just need to halve what we've got so far: 

And that's that. In simple(ish) terms, this means that to find the number of connections that can be made with any number of people in this situation you have to multiply the number of people by one less than the number of people, and then halve the answer.

So, for 132,947 people, there are (132947 x 132946) / 2 = 8,837,385,931 connections that can be made. That's well over eight billion!

12 balls: The solution.

At least, I think it's a solution. Please have a good, long look at it, figure out what it's talking about and then tell me if I've got anything wrong.

If you haven't had a go yourself, don't look at the image, just go here for the original problem, and here for a clue. It wouldn't be any fun if you went straight for the solution, would it!


The solution
Warning! This is the solution and contains spoilers!

O.k, that's not very easy to read, is it? Might be a good idea to click on it and view it full-size.

Just in case you're not sure what's going on, here's a walk-through of just one of the possibilities to give you an idea:

1. First up, you need to name your balls. I've gone for the rather unimaginative (but not too confusing) 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11 & 12. You could go for A-L if you wanted to, or you could think of 12 different names, but then you might get too attached.
2. Weigh* any four balls against any other four. I've gone for 1, 2, 3 & 4 against 5, 6, 7 & 8. Three things can happen here:
   i) The two sets of balls balance;
   ii) The two sets don't balance, with the first set rising and the second falling;
   iii) The two sets don't balance, with the first set falling and the second rising.
   Let's pretend they're balanced:
3. If the two sets are balanced, you know that balls 1 - 8 are fine, and the odd one out is either 9, 10, 11 or 12. To find out which one, first you need to weigh** two of the suspects against one suspect and one fine ball. In my example, I've gone with 9 & 10 against 11 & 1. Again, three things can happen:
   i) The two sets balance;
   ii) The two sets don't balance, with the first set rising;
   iii) The two sets don't balance, with the first set rising.
   Let's pretend they're unbalanced, and that the first set (9 & 10) rise.
4. If 9 & 10 rise, then one of two things is true:
   i) Either 9 or 10 is too light;
   ii) 11 is too heavy.
   To work out which it is, weigh*** the balls in the first set against each other (e.g. 9 against 10). Once more, three things can happen, and the solution depends on which:
   i) The two balls are balanced - in which case, it must be that 11 was the odd one out, and it was too heavy;
   ii) The balls are unbalanced, with 9 rising - in which case, it must be that 9 is odd, and it's too light (as 10 has already been in a rising set, so can't be too heavy);
   iii)The balls are unbalanced, with 10 rising - in which case, it must be that 10 is odd, and it's too light (for similar reasons as in the previous case).

I hope that makes some kind of sense. If not, or if you'd like any parts of my flow chart explaining in greater detail, feel free to get in touch (leave a comment, or send me a message on Twitter (@TeaKayB), or something).

* This is the first weigh.

** This is the second weigh.

*** This is the third weigh.

12 balls: A clue

Carrying on from the 12 pool balls problem posted the other day, here's a clue for anyone who's not too sure where to get started.

The clue comes in the form of a photograph of my whiteboard after a colleague and I had spent part of a free period muddling the problem through. It's not complete and it's not explicit, but then it wouldn't be a clue if it was, would it?

If you want to have a go without seeing the clue first, then CLICK HERE and close your eyes for a few seconds!



The clue
Warning! This is a clue and contains spoilers!

Click to view full-size image!

I'll post the full solution in another few days... subscribe to make sure you catch it!

I've you've got a maths-related question of your own, you could ask it in the comments or use the form at my main blog.

The 12 pool balls problem

Not my pic... click for originator

O.k, here's one for ya...

You have twelve pool balls and visually they're all exactly identical. One of them, however, has a different mass/weight to the other eleven, but you don't know whether it's heavier or lighter.

You have at your disposal a set of weighing scales so you can weigh the balls against each other, but no weights to go with it. You can only use these scales three times*.

Your job is to find out two things:

1. Which ball is the odd one out?
2. Is it heavier or lighter?

The rules

1. You're only allowed to use the instrumentation and apparatus as described above. No magicking weights or digital scales out of thin air, or inventing science prep rooms nearby.
2. No cheating. There are solutions available online, but if you look them up you're only cheating yourself. If you look it up and post it here, then you degrade yourself to a point a little beneath most traffic wardens.
3. There is a solution.
4. Feel free to comment and ask for clues. My HOD and I have worked out what we think is a pretty nifty solution, so I might post a clue-post in a few days or so if anybody asks for one.
5. I will post my solution at a future date as yet undecided, then edit this post to link to it. If you're reading this after that's happened, please give it a shot on your own (or with friends/colleagues) first. It's more fun that way!
6. Feel free to post your own solutions on your own blogs or websites and then let me know so I can link to them.
7. Subscribe to MathsQS for updates and new posts!

* No, I don't know why. Maybe they're set to explode, or something. Think Die Hard II with the jugs problem.

Movies and maths

Another maths blog that I follow, Mathspig, has made a series of posts debunking ten popular movie clichés. I won't go into too much detail here, as I'll just be repeating the posts themselves, so here they are:

10 Movie Clichés Debunked with Maths (introductory post)

1. You can duck bullets
2. You can out run a fireball
3. You can out run machine gun fire
4. If you are involved in a car chase, hijacking, explosion, volcanic eruption or alien invasion, you will not go into shock
5. You can jump out of an airplane [sic] that is about to crash and catch the baddie who took the last parachute
6. If the pilot dies the control tower can talk you down
7. You can safely jump from a burning skyscraper / bridge / aircraft into water
8. You can survive any amount of beating in a fight
9. A single match or candle will light a room the size of a football stadium
10. You can out run a lava flow

Number 5 reminded me of this (if you get bored you might want to skip to around 6 minutes to get what I'm talking about...):

Hare Lift - Watch more Videos at Vodpod.

What is zero to the power of zero?

To answer this, we need to think about what powers are.

What are powers?

A power is a number written small, slightly above and to the right of something, like this: 10³. In this case, it means 10 times 10 times 10 - or "ten multiplied by itself three times". The same way, x⁴ = x times x times x times x.

Multiplying powers

What happens if you have to multiply 10² by 10³? You could do it by saying "10 squared is 100 and 10 cubed is 1,000, and if you times them together you get 100,000," just by counting the noughts. What you're really doing, even if you don't realise it, is using the multiplying powers rule.

10² x 10³ = (10 x 10) x (10 x 10 x 10) = 10 x 10 x 10 x 10 x 10 = 10⁵

When you multiply powers together, you add the numbers above* and leave the bottom number alone. x¹² times x⁹⁷ = x¹⁰⁹. Easy, eh?

Dividing powers

It is a well-established fact that maths was invented by the same pedantically-minded two-year-old that invented the German language. That's what makes it easy. If multiplying powers together makes you do an addition sum, it makes Logical Sense that if you divide powers, you have to do a take away. That's exactly what happens. Try it if you don't believe me:

10⁵ ÷ 10³ = 10².

Powers of 0

How can you multiply something by itself no times? Well, you can't, really. But with powers, you can. You can turn it into a division calculation:

10⁰ = 10¹ ÷ 10¹ = 10 ÷ 10 = 1.

This is true... as long as the number underneath - 10, in this case - is not exactly zero.

Why not?

The reason why not is, 0 to the power of anything** is 0. Zero multiplied by anything, even itself, is zero. So we have a bit of a conflict: zero to the power of anything is 0 and anything to the power of 0 is 1.

Now, when maths sees a conflict, it has several options. One is to decide arbitrarily that one rule is more important than the other. Another is to figure out which is more important. A third is to throw up its hands and say "It just doesn't work," which is precisely what it does in this case.

0⁰ is undefined, just the same as 0÷0. You can't do it, it breaks the calculator. There is no answer that makes sense. It's an excellent question, and a really disappointing answer.

* As long as the number below is the same
** Anything positive, strictly

The new Odeon Premiére Reward Card: is it worth it?

The Odeon cinema chain has recently brought out its own reward card, but there's a catch: you have to pay for it. They've made it even more confusing by introducing three different levels of payment, each accompanied by a different number of 'free' rewards points.

The question is, is it worth forking out for a card? And if it is, which level do I go for?

I used the power of maths to work out an answer to both questions, and posted it at my regular blog, here.

Averages (for a list of numbers)

There are three kinds of average you need to know about for now: the MEAN, the MEDIAN and the MODE. They aren't the only ones, but they're the important ones for GCSE.

Let's say you have a list of numbers: 20, 19, 18, 18, 17, 13, 11, 10 and 8.

The MODE is the MOST COMMON number - here, it's 18. You can remember it by making an elaborate MO! sound, but be careful not to sound like a cow. If you do French, you can remember that la mode means 'fashion' - the mode is the most fashionable number in the list.

The MEDIAN is the MIDDLE number. It's 17 for this list - you get there by counting in from both ends until you reach the middle. Sometimes you get two middle numbers (when the length of the list is even). If the middle numbers are the same, that's your median; otherwise, the median is the number halfway between them (so, if you had 17 and 13 as your middle numbers, the median would be 15). You can remember that median sounds like medium, which means middling; or, if you speak American, you'll know that the middle of a highway (what we call the central reservation) is called the median strip because it's in the middle.

The MEAN is the nasty one. It's the meanest thing they can ask, hence the name. What you have to do is add all of the numbers up and divide by how many there are. The numbers here add to 116, which we have to divide by 9 to get 14.

What do 'nth' and 'nth term' mean in maths?

When you're describing how to get somewhere you might tell someone to 'go up the 5th street on the left'. This just means go past the first four, and go up the next one. When choosing an item of clothing from a rail, you might say 'the 7th one along'. This just means count along 7 and pick that one.

If I wanted to simplify that description of how to find the 5th road, or the 7th item of clothing (or even the 18th bottle on the wall, the 48th marble in the bag or the 239th name on a list), I might say something like this:

"To find the nth object, just count along the objects until you get to n."

This might look more complicated at first, but think about what it means: In maths, we can use a letter to:

1. represent numbers we don't know yet, or
2. work as place holder for something that works with any number.

In the example above, it's the second case that works for us. We could swap any number for n, and the statement would still make sense:

"To find the 5th object, just count along the objects until you get to 5."
"To find the 8th object, just count along the objects until you get to 8."
"To find the 846th object, just count along the objects until you get to 846.

What's the point?
In the example above, it may seem a little pointless, but remember that this is an over-simplified example. A slightly more complicated example is:

Imagine that a shop sells sea shells. The price of the shells is three for £1. The shop keeper could say, then, that:

"The number of sea shells I give a customer is three times the number of pounds the customer gives me."

Using our notation, we can write down the number of shells that he'd have to give his customer with a bit less effort. If the customer gives £n, then he gets 3 x n shells in return. If you can remember that 'n' stands for the number of pounds the customer gives, then remembering (and writing) "3 x n" takes a lot less effort than remembering "multiply the number of pounds by 3".


What about 'nth term'?
The 'nth term' is to do with sequences. A sequence is basically a list of numbers that follows some kind of rule. If you'd like to know more about sequences, please leave a comment and ask!

Take the following sequence as an example:

5, 8, 11, 14, 17, 20, 23, 26, 29... and so on.

This sequence goes on forever, and writing the whole thing down would take us just as long. We could just say "the sequence starts at 5, then you add three on each time." Or you could write it in the following way:

3n + 2

Then, if you wanted to find the 1st number in the list, you'd just say that n = 1 (3 x 1 + 2 = 5: it works!). If you wanted to find the 2nd, then n = 2 (3 x 2 + 2 = 8: that works too!).
The beauty of writing it down in this way is that you could also skip straight to finding the 48th number in the list: n would be replaced by 48 (3 x 48 + 2 = 146).
Writing down a sequence in this way is called writing down the 'nth term rule'. That just means that if we know which number term* we want to find (for example, the 8th, the 17th or the 4,243rd), we just swap n for that number




* A 'term' is just a posh word for each number in the list. For example, the first 'term' in the sequence we're looking at here is 5. The 7th 'term' in this sequence is 23: it's just the 7th number along when you write the whole list out.

What is a quadrillion?

The word 'quadrillion' comes from the same family of number names as 'ten', 'hundred', 'million' and 'googol'. It's just a name for a particular number.

But which number?

Another way of saying 'one quadrillion' would be 'one thousand million million'. What's that in digits? Let's work it out:

We know that 'one thousand' is written like this: 1,000 (a one with three zeroes, or 10³),

and 'one million' is written like this: 1,000,000 (a one with six zeroes or 10⁶).

So 'one thousand million' is just the 'million' bit with the 'one' replaced with 'one thousand', like this: 1,000,000,000.

So 'one thousand million million' should just be the 'million' bit with the 'one' replaced with 'one thousand million', like this: 1,000,000,000,000,000.

So 'one quadrillion' is 1,000,000,000,000,000 (a one with fifteen zeroes, or 10¹⁵).


Facts* about the word 'quadrillion'

• The SI prefix for a quadrillion is 'peta-'. This means that a 'petabyte' is one quadrillion bytes, or a 'petametre' would be a quadrillion metres.
• 'Quadrillion' can also refer to the number 'one million million million million', or 1,000,000,000,000,000,000,000,000, or 10²⁴, in the 'long scale' system of number naming. This definition is, however, falling out of use in the English language, which is why I have demonstrated the first definition in more detail (which uses the 'short scale' number naming conventions).
• The SI prefix for the long scale definition of a quadrillion is 'yotta-'. In the short scale system, 'yotta-' is the prefix for a 'septillion'.

*Whether they're interesting or not is entirely down to personal preference and circumstances.