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.

Geometry question

Here's a nice geometry question a student recently asked me about:

The surface area of a cube is 150 sq. cm. What is the length of its diagonal in cm?

The one measurement that's always useful when we're looking at cubes is the length of the side. How do we figure that out here?

Well, there are six sides to a cube, they're all the same size and they're all squares. So each of them has an area of 150÷6 = 25 square centimetres. If the area of a square is 25, the length of the side is √25 = 5cm.

Now comes the tricky bit. When I'm doing 3-d geometry problems, I find it helps to think about a rectangular room with four rectangular walls. In this case, we're trying to find the distance from one corner on the floor to the opposite corner at the top of the wall, and I imagine a bit of string running from one to the other.

Straight away, I can see a right-angled triangle - one side runs diagonally across the floor and the other up the wall, and the string makes the hypotenuse.

We know how big the wall is - it's 5cm, like all of the other sides of the cube. However, we don't know how far it is across the floor. To find that, we need to consult with everyone's favourite bean-eating ancient Greek mathematician, Pythagoras.

The line across the floor is also the hypotenuse of a right-angled triangle - again, its shorter sides are both 5cm long. Pythagoras says the hypotenuse squared is equal to one of the short sides squared plus the other short side squared: H² = 5² + 5² = 50. So the line across the floor is √50 cm long - a smidge over 7cm.

Now we have to work out the longer diagonal. The line on the floor is now one of the shorter sides, along with the wall. So D² = (√50)² + 5² = 50 + 25 = 75. The main diagonal is √75 cm - around 8.66cm.

Bonus question: why is this the same as 5√3?

In which year will I be twice your age?

This question, of course, depends on who you are, and who "I" is*, but we can work out a general case using algebra. First, though, I'll use myself and my friend John as:


An example
If we know the year of birth of any particular person we can work out their age by subtracting their birth year from the current year.

• TeaKay was born in 1982 and it's now 2009, so my age is      2009 - 1982 = 27      years old**.
• John was born in 1955, and it's still 2009, so his age is      2009 - 1955 = 54      years old.

Now, you can see already that John will become twice my age this year (2009), but how could we work it out if it wasn't so easy?

How would we work out how old each person is in any given year? Well, we'd use the same method- subtract their birth year from the year you're looking at. In general, we could say that:

• TeaKay is/was/will be      y - 1982      years old in the year y.
• John is/was/will be      y - 1955      years old in the year y.

To answer the question we want to find what year a must be for John's age to be twice TeaKay's age. So we can write:     John's age = 2 x TeaKay's age

Algebraically, that would look like this:

     y - 1955     =     2 x (y - 1982)

Now all we have to do is solve to find a value for a:

     y - 1955     =     2y - 3964               Expand the brackets

     y + 2009    =     2y                          Use the inverse to get the numbers on the same side of the '=' sign

            2009   =       y                          Use the inverse to get the letters on the same side of the '=' sign

So now we have proven that John will be twice as old as TeaKay in 2009.


A general case
We can go a step further and develop a formula for finding out when anyone will be twice as old as anyone else:

Call your people Person A and Person B. Person A is always the oldest of the two. Now call a the year that Person A was born in, and b the year that Person B was born in. In the year y:

• Person A will be      y - a      years old.
• Person B will be      y - b      years old.

So Person A will be twice as old as Person B when:

     y - a      =      2(y - b)

So now we can solve to find a value for y, given any years a and b:

     y - a      =      2y - 2b

  y - a + 2b =      2y

      -a + 2b =       y

And neaten up:

                y = 2b - a

So, in English, this formula says "to find the year in which Person A will be twice as old as Person B, double Person B's birth year and then subtract Person A's birth year."

Which is why mathematicians like algebra so much!

*Difficult to parse, I know, but grammatically sound in the sense that I mean it...
** Of course this assumes that your birthday has already happened- I won't actually be 27 until October 10th, but lets simplify in order to get the idea across.