You are currently browsing the category archive for the ‘puzzles’ category.

Can you find a sequence of consecutive integers that add up to 1000?

This puzzle is based on the excellent book A First Step to Mathematical Olympiad Problems – which is full of problems that could be extended to become exploration ideas.

Step 1 – arithmetic formula

Our first step is to write out what we want:

a + (a+1) + (a+2) + … (a +n) = 1000

next we notice that the LHS is an arithmetic series with first term a, last term a+n and n+1 terms.  Therefore we can use the sum of an arithmetic sequence formula:

Sn = 0.5n(u1 + un)

Sn = 0.5(n+1)(a + a+n) = 1000

Sn = (n+1)(2a+n) = 2000

Step 2 – logic

However, we currently have 2 unknowns, n and a, and only 1 equation – so we can’t solve this straight away.  However we do know that both a and n are integers – and n can be taken as positive.

The next step is to see that one of the brackets (n+1)(2a+n) must be odd and the other even (if  n is odd then 2a + n is odd.  If n is even then n+1 is odd).   Therefore we can look at the odd factors of 2000:

Step 3 – prime factorisation

Using prime factorisation: 2000 = 24 x 5³

Therefore any odd factors must solely come from the prime factor combinations of 5 – i.e 5, 25 and 125.

Step 4 – trial and error

So we now know that either (n+1) or (2a+n) must be 5, 25, 125.  And therefore the other bracket must be 400, 80 or 16 (as 5 x 400 = 2000 etc).  Next we can equate the (n+1) bracket to one of these 6 values, find the value of n and hence find a.  For example:

If one bracket is 5 then the other bracket is 400.

So if (n+1) = 5 and (2a+n) = 400 then n = 4 and a = 198.

This means that the sequence: 198+199+200+201+202 = 1000.

If (n+1) = 400 and (2a+n) = 5 then n = 399 and a = -197.

This means the sequence: -197 + -196+ -195 … + 201 + 202 = 1000.

We follow this same method for brackets 25, 80 and 125,16.  This gives the following other sequences:

28+29+30+…+51+52 = 1000

-54+-53+-52+…+69+70 = 1000

-27+-26+-25+…+51+52 = 1000

55+56+57+…+69+70 = 1000

So with a mixture of mathematical formulae, prime factorisation, logic and trial and error we have our solutions.  A good example of how mathematics is often solved in reality!

Tetrahedral Numbers – Stacking Cannonballs

This is one of those deceptively simple topics which actually contains a lot of mathematics – and it involves how spheres can be stacked, and how they can be stacked most efficiently.  Starting off with the basics we can explore the sequence:

1, 4, 10, 20, 35, 56….

These are the total number of cannons in a stack as the stack gets higher.  From the diagram we can see that this sequence is in fact a sum of the triangular numbers:

S1 = 1

S2 1+3

S3 1+3+6

S4 1+3+6+10

So we can sum the first n triangular numbers to get the general term of the tetrahedral numbers. Now, the general term of the triangular numbers is 0.5n2 + 0.5n therefore we can think of tetrahedral numbers as the summation:

$\bf \sum_{k=1}^{n}0.5k+0.5k^2 = \sum_{k=1}^{n}0.5k+\sum_{k=1}^{n}0.5k^2$

But we have known results for the 2 summations on the right hand side:

$\bf \sum_{k=1}^{n}0.5k =\frac{n(n+1)}{4}$

and

$\bf \huge \sum_{k=1}^{n}0.5k^2 = \frac{n(n+1)(2n+1)}{12}$

and when we add these two together (with a bit of algebraic manipulation!) we get:

$\bf S_n= \frac{n(n+1)(n+2)}{6}$

This is the general formula for the total number of cannonballs in a stack n rows high. We can notice that this is also the same as the binomial coefficient:

$\bf S_n={n+2\choose3}$

Therefore we also can find the tetrahedral numbers in Pascals’ triangle (4th diagonal column above).

The classic maths puzzle (called the cannonball problem), which asks which tetrahedral number is also a square number was proved in 1878. It turns out there are only 3 possible answers. The first square number (1) is also a tetrahedral number, as is the second square number (4), as is the 140th square number (19,600).

We can also look at something called the generating function of the sequence. This is a polynomial whose coefficients give the sequence terms. In this case the generating function is:

$\bf \frac{x}{(x-1)^4} = x + 4x^2 + 10x^3 + 20x^4 ...$

Having looked at some of the basic ideas behind the maths of stacking spheres we can look at a much more complicated mathematical problem. This is called Kepler’s Conjecture – and was posed 400 years ago. Kepler was a 17th century mathematician who in 1611 conjectured that there was no way to pack spheres to make better use of the given space than the stack above. The spheres pictured above fill about 74% of the given space. This was thought to be intuitively true – but unproven. It was chosen by Hilbert in the 18th century as one of his famous 23 unsolved problems. Despite much mathematical efforts it was only finally proved in 1998.

If you like this post you might also like:

The Poincare Conjecture – the search for a solution to one of mathematics greatest problems.

Hailstone Numbers

This is a post inspired by the article on the same topic by the ever brilliant Plus Maths. Hailstone numbers are created by the following rules:

if n is even: divide by 2

if n is odd: times by 3 and add 1

We can then generate a sequence from any starting number.  For example, starting with 10:

10, 5, 16, 8, 4, 2, 1, 4, 2, 1…

we can see that this sequence loops into an infinitely repeating 4,2,1 sequence.  Trying another number, say 58:

58, 29, 88, 44, 22, 11, 34, 17, 52, 26, 13, 40, 20, 10, 5, 16, 8, 4, 2, 1, 4, 2, 1…

and we see the same loop of 4,2,1.

In fact we can use the generator in the Plus Maths article to check any numbers we can think of, and we still get the pattern 4,2,1 looping.  The question is, does every number end in this loop?  Well, we don’t know.  Every number mathematicians have checked do indeed lead to this loop, but that is not a proof.  Perhaps there is a counter-example, we just haven’t found it yet.

Hailstone numbers are called as such because they fall, reach one (the ground) before bouncing up again.  The proper mathematical name for this investigation is the Collatz conjecture. This was made in 1937 by a German mathematian, Lothar Collatz.

One way to investigate this conjecture is to look at the length of time it takes a number to reach the number 1.  Some numbers take longer than others.  If we could find a number that didn’t reach 1 even in an infinite length of time then the Collatz conjecture would be false.

The following graphic from wikipedia shows how different numbers (x axis) take a different number of iterations (y axis) to reach 1.  We can see that some numbers take much longer than others to reach one.  For example, the number 73 has the following pattern:

73, 220, 110, 55, 166, 83, 250, 125, 376, 188, 94, 47, 142, 71, 214, 107, 322, 161, 484, 242, 121, 364, 182, 91, 274, 137, 412, 206, 103, 310, 155, 466, 233, 700, 350, 175, 526, 263, 790, 395, 1186, 593, 1780, 890, 445, 1336, 668, 334, 167, 502, 251, 754, 377, 1132, 566, 283, 850, 425, 1276, 638, 319, 958, 479, 1438, 719, 2158, 1079, 3238, 1619, 4858, 2429, 7288, 3644, 1822, 911, 2734, 1367, 4102, 2051, 6154, 3077, 9232, 4616, 2308, 1154, 577, 1732, 866, 433, 1300, 650, 325, 976, 488, 244, 122, 61, 184, 92, 46, 23, 70, 35, 106, 53, 160, 80, 40, 20, 10, 5, 16, 8, 4, 2, 1…

so investigating what it is about certain numbers that leads to long chains is one possible approach to solving the conjecture. This conjecture has been checked by computers up to a staggering 5.8 x 1018 numbers. That would suggest that the conjecture could be true – but doesn’t prove it is. Despite looking deceptively simple, Paul Erdos – one of the great 20th century mathematicians stated in the 1980s that “mathematics is not yet ready for such problems” – and it has remained unsolved over the past few decades.  Maybe you could be the one to crack this problem!

If you liked this post you might also like:

Friendly Numbers, Solitary Numbers, Perfect Numbers – a look at some other number sequence problems.

Stellar Numbers Investigation

This is an old IB internal assessment question and so can not be used for the new IB exploration – however it does give a good example of the sort of pattern investigation that is possible.

The task starts off with the fairly straightforward problem of trying to find the nth term formula for the triangular numbers:

Method 1

There are a number of ways to do this, probably the easiest is to notice that the second differences are always constant (+1 each time).  Therefore we have a quadratic sequence in the form an2 + bn + c

We can now substitute the known values when n = 1, 2, 3 into this to find 3 equations:

a(1) + b(1) + c = 1

a(2)2 + b(2) + c = 3

a(3)2 + b(3) + c = 6

this gives us:

a + b + c = 1

4a + 2b + c = 3

9a + 3b + c = 6

We can then eliminate using simultaneous equations to find a, b, c.  In fact our job is made easier by knowing that if the second difference is a constant, then the a in our formula will be half that value.  Therefore as our second difference was 1, the value of a will be 1/2.  We then find that b = 1/2 and c = 0.  So the formula for the triangular numbers is:

0.5n2 + 0.5n

Method 2

We can also derive this formula by breaking down triangular numbers into the following series:

1

1+2

1+2+3

1+2+3+4

Therefore we have the sum of an arithmetic series, with first term 1, common difference 1 and last term n, and so we can use the sum of an arithmetic series formula:

Sn = 0.5n(a1 + an)

Sn = 0.5n(1 + n) = 0.5n2 + 0.5n

Once this is done, we are asked to find the nth term for the 6-stellar numbers (with 6 vertices) below:

which give the pattern 1, 13, 37, 73

Method 1

Once again we can use the method for quadratic sequences.  The second difference is 12, giving us an2 + bn + c with a = 12/2 = 6. Substituting values gives us:

1 = 6(1)2 + b(1) + c
13 = 6(2)2 + b(2) + c

This simplifies to:

1 = 6 + b + c
13 = 24 + 2b + c

Therefore we can eliminate to find that b = -6 and c = 1.

which gives 6n2 – 6n + 1

Method 2

A more interesting method makes use of the triangular numbers.  We can first note a recurrence relationship in the stellar numbers – each subsequent pattern contains all the previous patterns inside.  In fact we can state the relationship as:

S1

S2 = S1 + outside star edge

S3 = S2 + outside star edge

S4 = S3 + outside star edge

The outside star edge of S2 can be thought of as 6 copies of the 2nd triangular number

The outside star edge of S3 can be thought of as 6 copies of the 3rd triangular number, but where we subtract 6×1 (the first triangular number) because we double count one of the internal points six times. We also subtract 6 as we double count each vertex.

The outside star edge of S4 can be thought of as 6 copies of the 4th triangular number, but where we subtract 6 x 3 (the second triangular number) because we double count three of the internal points six times. We also subtract 6 as we double count each vertex.

The outside star edge of S5 can be thought of as 6 copies of the 5th triangular number, but where we subtract 6 x 6 (the third triangular number) because we double count six of the internal points six times. We also subtract 6 as we double count each vertex.

Therefore we can form a formula for the outside star:

6(0.5n2 + 0.5n) – 6(0.5(n-2)2 + 0.5(n-2)) – 6

which simplifies to:

12(n -1)

We can now put this into our recurrence relationship:

S1 = 1

S2 = 1 + 12(n -1)

S3 = 1 + 12((n-1) -1) + 12(n -1)

S4 = 1 + 12((n-2) -1) + 12((n-1) -1) + 12(n -1)

Note that when we substituted the nth term formula for S2 into S3 we had to shift the n value to become n-1 as we were now on the 3rd term rather than 2nd term.

So:

S1 = 1

S2 = 1 + 12(n -1)

S3 = 1 + 12(n-1) + 12(n-2)

S4 = 1 + 12(n-1) + 12(n-2) + 12(n-3)

So:

S1 = 1 + 0

S2 = 1 + 12

S3 = 1 + 12+ 24

S4 = 1 + 12 + 24 + 36

So using the formula for the sum of an arithmetic Sn = 0.5n(a1 + an) we have

Sn = 1 + 0.5(n-1)(12 + 12(n-1))

Sn = 6n2 – 6n + 1

Quite a bit more convoluted – but also more interesting, and also more clearly demonstrating how the sequence is generated.

Generalising for p-stellar numbers

We can then generalise to find stellar number formulae for different numbers of vertices.  For example the 5-stellar numbers pictured above have the formula 5n2 – 5n + 1.  In fact the p-stellar numbers will have the formula pn2 – pn + 1.

We can prove this by using the same recurrence relationship before:

S1

S2 = S1 + outside star edge

S3 = S2 + outside star edge

S4 = S3 + outside star edge

and by noting that the outside star edge is found in the same way as before for a p-stellar shape – only this time we subtract p for the number of vertices counted twice.  This gives:

p(0.5n2 + 0.5n) – p(0.5(n-2)2 + 0.5(n-2)) – p

which simplifies to

2p(n-1)

and so substituting this into our recurrence formula:

S1 = 1

S2 = 1 + 2p(n-2)

S3 = 1 + 2p(n-2) + 2p(n-1)

S4 = 1 + 2p(n-3) + 2p(n-2) + 2p(n-1)

We have the same pattern as before – an arithmetic series in terms of 2p, and using Sn = 0.5n(a1 + an) we have:

Sn= 1 + 0.5(n-1)(2p + 2p(n-1) )

Sn = pn2 – pn + 1

Therefore, although our second method was slower, it allowed us to spot the pattern in the progression – and this then led very quickly to a general formula for the p-stellar numbers.

If you like this you might also like:

The Goldbach Conjecture – The Goldbach Conjecture states that every even integer greater than 2 can be expressed as the sum of 2 primes.  No one has ever managed to prove this.

Maths Puzzles

These should all be accessible for top sets in KS4 and post 16.  See if you can manage to get all 3 correct.

Puzzle Number 1

Why is xx undefined when x = 0 ?

Puzzle Number 2

I multiply 3 consecutive integers together. My answer is 8 times the middle of the 3 integers I multiplied. What 3 numbers could I have chosen?

Puzzle Number 3

You play a game as follows:

1 point for a prime number
2 points for an even number
-3 points for a square number

(note if you choose (say) the number 2 you get +1 for being a prime and +2 for being an even number giving a total of 3 points)

You have the numbers 1-9 to choose from. You need to choose 4 numbers such that their score adds to zero. How many different ways can you find to win this game?

Answers below in white text (highlight to reveal)

1) xx is undefined because using 2 different indices rules will give us contradictory results. 0 to any power will always be 0, however any number to the power 0 will always be 1. With 2 contradictory answers we leave it as undefined!

2) The equation we want is (x)(x+1)(x+2) = 8(x+1). This simplifies to x^3 + 3x^2 -6x – 8 = 0. We can solve this using the factor theorem, polynomial division or by plotting a graph to get 2 possible solutions – x = 2 or x = -4.

3) The numbers will have the following values: 1 = -3, 2 = 3, 3 = 1, 4 = -1, 5 = 1, 6 = 2, 7 = 1, 8 = 2, 9 = -3. There are at least the following possible combinations:
1,2,3,4
1,2,5,4
1,2,7,4
9,2,3,4
9,2,5,4
9,2,7,4
6,8,4,9
6,8,4,1
Check to see I haven’t missed any!

If you like this post, you might also like:

A Maths Snooker Puzzle. A great little puzzle which tests logic skills.

Visualising Algebra Through Geometry.  How to use geometry to simplify puzzles

Analytic Continuation and the Riemann Zeta Function

Analytic Continuation is a very important mathematical technique which allows us to extend the domain of functions.  It is essential in higher level mathematics and physics and leads to some remarkable results. For example, by using analytic continuation we can prove that the sum of the natural numbers (1 + 2 + 3 + ….) is -1/12. Results don’t get more surprising than that!

Analytic continuation concerns functions of the form:

f(z) where z is a complex number and f(z) is (complex) differentiable.

Remember complex numbers are of the form a + bi and can be thought of as coordinate points in an x,y axis.  For the purposes of this post we will only look at real values of z (real numbers are still a subset of complex numbers).

The idea of analytic continuation is to take an original function with a restricted domain, then to find another function which is the same within that restricted domain, but also is valid outside that domain.  This sounds very complicated – but let’s look at a couple of examples:

$f(z) = \frac{(z+1)(z+2)}{(z+1)}$

This is a function which is defined for all values except for z = -1. When z = -1 we have zero on the denominator so the function doesn’t exist. However we can write a new function:

$g(z) = (z+2)$

Now, g(z) = f(z) for all z when z is not -1, but g(z) also exists when z = -1. Therefore we can regard g(z) as the analytic continuation of f(z), and we have extended the domain of f(z) from all values except -1, to all values of z.

A more interesting example is the following:

$f(z) = \sum_{n=0}^\infty z^{n}$

This is the infinite series:

$1 + z + z^{2}+ ...$

This function is analytic (complex differentiable) only when   -1 < z< 1.  (Don’t worry about how this is calculated – though it is related to the domain of convergence). Therefore this is our restricted domain.

But we can notice that the sum of a geometric sequence formula allows us to calculate f(z) in a different way:

$\sum_{n=0}^\infty z^{n} = \frac{1}{(1-z)}$

Here we have used the formula for summing a geometric, with the first term 1 and common ratio z.

Therefore we could write:

$g(z) = \frac{1}{(1-z)}$

f(z) = g(z) when -1 < z< 1 , but g(z) is complex differentiable for all values except for z = 1 (when the denominator is 0).  Therefore g(z) is the analytic continuation of f(z) from  -1 < z< 1 to all values of z except z = 1.

One example of analytic continuation that I’ve written about before is the Riemann Sphere. This extends by analytic continuation the complex plane into the complex plane plus infinity.

Another example is used in showing that the sum of natural numbers is -1/12. There are a few different methods to show this – some discussed previously here. I’m going to try and talk through another proof of this result. It’s a bit difficult, but try and understand the general method!

The proof revolves around the Riemann Zeta function, (Riemann is pictured above). This is defined as:

$\zeta(z)= \sum_{n=1}^{\infty}n^{-z}$

This can also be written as:

$\zeta(z)=\frac{1}{1^{z}} +\frac{1}{2^{z}} +\frac{1}{3^{z}}..$

So, if we want to find the sum of 1 + 2 + 3 … then we need to substitute z = -1 into the above summation. However this formula for the zeta function is only valid for the domain z > 1, so we first need to extend the function through analytic continuation.

Through analytic continuation (where we extend the domain from z > 1 to all complex numbers apart from -1) we can rewrite the zeta function as:

$\zeta(1-z)=2^{1-z}\pi^{-z}cos(0.5\pi z)\Gamma(z)\zeta(z)$

and substituting z = 2 into this formula, so that we end up with zeta(-1) we get:

$\zeta(-1)=2^{-1}\pi^{-2}cos(\pi)\Gamma(2)\zeta(2)$

Now,

$\zeta(2) = \frac{\pi^{2}}{6}$

$\Gamma(2) = 1$

$cos(\pi) = -1$

Therefore

$\zeta(-1)=-\frac{1}{12}$

We have proved that 1 + 2 + 3 … = -1/12 !

If you enjoyed this post you might also like:

The Riemann Hypothesis Explained. What is the Riemann Hypothesis – and how solving it can win you $1 million. Unbelievable: 1+2+3+4…. = -1/12 ? A result that at first glance looks ridiculous – and yet can be shown to be correct. How? Murder in the Maths Department A murder has been committed in the maths department! A body has been discovered surrounded by mathematical objects and only the hardworking maths teachers were in school, doing long division sums for fun at the weekend. One of them must be the murderer. (The wall of fame of successful detectives will be posted here) Your task, should you choose to accept it, is to find: 1) The murderer 2) The room 3) The murder weapon The murder suspects are: 1) Al Jabra – who was wearing a white, T-shirt with 2 stripes and ripped jeans on the day of the murder. 2) Polly Gon – who was wearing a knee-length green skirt, white blouse and gold watch. 3) Lisa Perbound- who was wearing a blue Adidas T-shirt with 3 stripes on the sleeves, Bermuda shorts and a baseball cap. 4) May Trix- who was wearing a black and white pin-stripe suit with shiny black shoes. 5) Ella Ment- who was wearing a blue knitted jumper with a picture of pi on the front, and brown cords. The possible rooms are: The possible rooms are: 1) The Canteen 2) The Tuck-shop 3) Room 20 4) Room 18 5) Room 17 6) Room 7 The possible murder weapons are: 1) A wooden metre ruler 2) A large metal stapler 3) A dusty trundle wheel 4) A sharp compass 5) A large maths textbook 6) An oversized calculator Clue Number 1 When you have solved this clue – click here, and enter the last word only of the decoded message (no capital letters). The Poincare Conjecture and Grigori Perelman In 2006 the Russian mathematician Grigori Perelman was awarded the mathematical equivalent of the mathematical Nobel prize (the Fields Medal). He declined it. In 2010 he was the first mathematician to be awarded$1 million – he turned it down.  What had Perelman done to achieve such (apparently unwanted) acclaim?  He had solved a puzzle that had frustrated mathematicians for over 100 years – the Poincare conjecture.

What is the Poincare Conjecture?

The Poincare Conjecture is that, “Every simply connected, closed 3-manifold is homeomorphic to the 3-sphere.”  At first glance that may look quite complicated – so looking at the definitions in turn:

Simply connected means a shape without holes.  The two shapes on the left above are simply connected, the two on the right are not.  In 3 dimensions, a sphere and cube are simply connected, but a donut shape (torus) is not.

3D manifold means a 3 dimensional surface.  Imagine the surface of a sphere – that is a 2 dimensional surface.  So a 3 dimensional surface on a sphere would require a 4 dimensional sphere.  A 4 dimensional sphere is one which has a fixed radius in 4 dimensions (unlike in 3 dimensions for a sphere and 2 dimensions for a circle).

Homeomorphic means it is mathematically equivalent in terms of the relationship between points.  Basically, if 2 shapes can be sqeezed or stretched to form another shape then they are homeomorphic.  In the above animation, the coffee mug and the donut (torus) are shown to be homeomorphic.

3-Sphere means a sphere in 4 dimensions (i.e with a 3 dimensional surface area).

So, with those terms defined we can simplify the Poincare conjecture.  In regular 3 dimensions, conventional 3 dimensional shapes without a hole in them (cubes, cuboids etc) can all be squashed and squeezed to create a sphere.  Poincare conjectured that the same would be true in higher dimensions – i.e 4 dimensional cubes (a tesseract, as shown above) could be squashed and squeezed to make a 4 dimensional sphere.

Grigori Perelman however was not interested in either the acclaim or the money on offer for solving one of the world’s most difficult mathematics problems.  In explaining why he turned down \$1 million he said that the prize, “was completely irrelevant for me. Everybody understood that if the proof is correct, then no other recognition is needed.”

If you liked this post you might also like:

Imagining the 4th Dimension. How mathematics can help us explore the notion that there may be more than 3 spatial dimensions.

Non Euclidean Geometry V – The Shape of the Universe – Using mathematics to understand one of the most important questions of all.

Zeno’s Paradox – Achilles and the Tortoise

This is a very famous paradox from the Greek philosopher Zeno – who argued that a runner (Achilles) who constantly halved the distance between himself and a tortoise  would never actually catch the tortoise.  The video above explains the concept.

There are two slightly different versions to this paradox.  The first version has the tortoise as stationary, and Achilles as constantly halving the distance, but never reaching the tortoise (technically this is called the dichotomy paradox).  The second version is where Achilles always manages to run to the point where the tortoise was previously, but by the time he reaches that point the tortoise has moved a little bit further away.

The first version we can think of as follows:

Say the tortoise is 2 metres away from Achilles.  Initially Achilles halves this distance by travelling 1 metre.  He halves this distance again by travelling a further 1/2 metre.  Halving again he is now 1/4 metres away.  This process is infinite, and so Zeno argued that in a finite length of time you would never actually reach the tortoise.  Mathematically we can express this idea as an infinite summation of the distances travelled each time:

1 + 1/2 + 1/4 + 1/8 …

Now, this is actually a geometric series – which has first term a = 1 and common ratio r = 1/2.  Therefore we can use the infinite summation formula for a geometric series (which was derived about 2000 years after Zeno!):

sum = a/(1-r)

sum = 1/(1-0.5)

sum = 2

This shows that the summation does in fact converge – and so Achilles would actually reach the tortoise that remained 2 metres away.  There is still however something of a sleight of hand being employed here however – given an infinite length of time we have shown that Achilles would reach the tortoise, but what about reaching the tortoise in a finite length of time?  Well, as the distances get ever smaller, the time required to traverse them also gets ever closer to zero, so we can say that as the distance converges to 2 metres, the time taken will also converge to a finite number.

There is an alternative method to showing that this is a convergent series:

S = 1+ 1/2 + 1/4 + 1/8 + 1/16 + …

0.5S = 1/2+ 1/4 + 1/8 + 1/16 + …

S – 0.5S = 1

0.5S = 1

S = 2

Here we notice that in doing S – 0.5S all the terms will cancel out except the first one.

Achilles and the Tortoise

The second version also makes use of geometric series.  If we say that the tortoise has been given a 10 m head start, and that whilst the tortoise runs at 1 m/s, Achilles runs at 10 m/s, we can try to calculate when Achilles would catch the tortoise.  So in the first instance, Achilles runs to where the tortoise was (10 metres away).  But because the tortoise runs at 1/10th the speed of Achilles, he is now a further 1m away.  So, in the second instance, Achilles now runs to where the tortoise now is (a further 1 metre).  But the tortoise has now moved 0.1 metres further away.  And so on to infinity.

This is represented by a geometric series:

10 + 1 + 0.1 + 0.01 …

Which has first time a = 10 and common ratio r = 0.1.  So using the same formula as before:

sum = a/(1-r)

sum = 10/(1-0.1)

sum = 11.11m

So, again we can show that because this geometric series converges to a finite value (11.11), then after a finite time Achilles will indeed catch the tortoise (11.11m away from where Achilles started from).

We often think of mathematics and philosophy as completely distinct subjects – one based on empirical measurement, the other on thought processes – but back in the day of the Greeks there was no such distinction.  The resolution of Zeno’s paradox by use of calculus and limits to infinity some 2000 years after it was first posed is a nice reminder of the power of mathematics in solving problems across a wide range of disciplines.

The Chess Board Problem

The chess board problem is nothing to do with Zeno (it was first recorded about 1000 years ago) but is nevertheless another interesting example of the power of geometric series.  It is explained in the video above.  If I put 1 grain of rice on the first square of a chess board, 2 grains of rice on the second square, 4 grains on the third square, how much rice in total will be on the chess board by the time I finish the 64th square?

The mathematical series will be:

1+ 2 + 4 + 8 + 16 +……

So a = 1 and r = 2

Sum = a(1-r64)/(1-r)

Sum = (1-264)/(1-2)

Sum = 264 -1

Sum = 18, 446,744, 073, 709, 551, 615

This is such a large number that, if stretched from end to end the rice would reach all the way to the star Alpha Centura and back 2 times.

IB Maths Revision

I’d strongly recommend starting your revision of topics from Y12 – certainly if you want to target a top grade in Y13.  My favourite revision site is Revision Village – which has a huge amount of great resources – questions graded by level, full video solutions, practice tests, and even exam predictions.  Standard Level students and Higher Level students have their own revision areas.  Have a look!

The Telephone Numbers – Graph Theory

The telephone numbers are the following sequence:

1, 1, 2, 4, 10, 26, 76, 232, 764, 2620, 9496…

(where we start from n=0).

This pattern describes the total number of ways which a telephone exchange with n telephones can place a connection between pairs of people.

To illustrate this idea, the graph below is for n=4.  This is when we have 10 telephones:

Each red line represents a connection.  So the first diagram is for when we have no connections (this is counted in our sequence).  The next five diagrams all show a single connection between a pair of phones.  The last three diagrams show how we could have 2 pairs of telephones connected at the same time.  Therefore the 4th telephone number is 10.   These numbers get very large, very quickly.

Finding a recursive formula

The formula is given by the recursive relationship:

T(n) = T(n-1) + (n-1)T(n-2)

This means that to find (say) the 5th telephone number we do the following:

T(5) = T(5-1) + (5-1)T(5-2)

T(5) = T(4) + (4)T(3)

T(5) = 10 + (4)4

T(5) = 26

This is a quick way to work out the next term, as long as we have already calculated the previous terms.

Finding an nth term formula

The telephone numbers can be calculated using the nth term formula:

This is going to be pretty hard to derive!  I suppose the first step would start by working out the total number of connections possible between n phones – and this will be the the same as the graphs below:

These clearly follow the same pattern as the triangular numbers which is 0.5(n² +n) when we start with n = 1.  We can also think of this as n choose 2 – because this gives us all the ways of linking 2 telephones from n possibilities.  Therefore n choose 2 also generates the triangular numbers.

But then you would have to work out all the permutations which were allowed – not easy!

Anyway, as an example of how to use the formula to calculate the telephone numbers, say we wanted to find the 5th number:

We have n = 5.  The summation will be from k = 0 and k = 2 (as 5/2 is not an integer).

Therefore T(5) = 5!/(20(5-0)!0!) + 5!/(21(5-2)!1!) + 5!/(22(5-4)!2!)

T(5) = 1 + 10 + 15 = 26.

Finding telephone numbers through calculus

Interestingly we can also find the telephone numbers by using the function:

y = e0.5x2+x

and the nth telephone number (starting from n = 1)  is given by the nth derivative when x = 0.

For example,

So when x = 0, the third derivative is 4.  Therefore the 3rd telephone number is 4.

The fifth derivative of the function is:

So, when x =0 the fifth derivative is 26.  Therefore the 5th telephone number is 26.

If you liked this post you might also like:

Fermat’s Theorem on the Sum of two Squares – A lesser known theorem from Fermat – but an excellent introduction to the idea of proof.

Unbelievable: 1+2+3+4…. = -1/12 ? A result that at first glance looks ridiculous – and yet can be shown to be correct.  How?

### Website Stats

• 5,938,318 views