IB Maths and GCSE Maths Resources from British International School Phuket. Theory of Knowledge (ToK). Maths explorations and investigations. Real life maths. Maths careers. Maths videos. Maths puzzles and Maths lesson resources.

British International School Phuket

Welcome to the British International School Phuket’s maths website. I am currently working at BISP and so I am running my site as the school’s maths resources website for both our students and students around the world.

We are a British international school located on the tropical island of Phuket in Southern Thailand. We offer a number of scholarships each year, catering for a number of national and international standard sports stars as well as for academic excellence. You can find out more about our school here.

BISP has a very proud tradition in mathematical excellence. Our students have achieved the top in Thailand awards for the Cambridge Maths IGCSEs 3 years in a row. Pictured above are our world-class maths students for this year, Minjin Kang and Natchongrat (Oy) Terdkiatkhachorn.

Please explore the site – there is a huge amount of content! Some of the most popular includes:

A large “Flipping the classroom” videos section for IB students. This covers the entire IB HL, SL and Studies syllabus.

A new School Code Challenge activity which allows students to practice their code breaking skills – each code hides the password needed to access the next level.

Over 200 ideas to help with students’ Maths Explorations – many with links to additional information to research.

You might think that winning at rock, paper, scissors was purely a matter of chance – after all mathematically each outcome has the same probability. We can express the likelihood of winning in terms of a game theory grid:

It is clear that in theory you would expect to win, draw and lose with probability 1/3. However you can actually exploit human psychology to give yourself a significant edge at this game. Below is a report of a Chinese study on the psychology of game players:

Zhijian and co carried out their experiments with 360 students recruited from Zhejiang University and divided into 60 groups of six players. In each group, the players played 300 rounds of Rock-Paper-Scissors against each other with their actions carefully recorded.

As an incentive, the winners were paid in local currency in proportion to the number of their victories. To test how this incentive influenced the strategy, Zhijian and co varied the payout for different groups. If a loss is worth nothing and a tie worth 1, the winning payout varied from 1.1 to 100.

The results reveal a surprising pattern of behavior. On average, the players in all the groups chose each action about a third of the time, which is exactly as expected if their choices were random.

But a closer inspection of their behavior reveals something else. Zhijian and co say that players who win tend to stick with the same action while those who lose switch to the next action in a clockwise direction (where R → P → S is clockwise).

So, for example if person A chooses Rock and person B chooses Paper, then person B wins. Human nature therefore seems to mean that person B is more likely to stick to a winning strategy and choose Paper again, whilst person A is more likely to copy that previous winning behaviour and also choose Paper. A draw.

So you can exploit this by always moving anticlockwise i.e R → S → P. To look at our example again, person A chooses Rock and person B chooses Paper, then person B wins. This time person A follows his previous pattern and still chooses Paper, but person B exploits this new knowledge to choose Rock. Player B wins.

You can play against a Wolfram Alpha AI player here. This program will track your win percentage, and will also adapt its behavior to exploit any non-random behavior that you exhibit. Even though you may not be conscious of your biases, they probably will still be there – and the designers of this simulator are confident that the program will be beating you after about 50 games. Have a go!

There are some additional tips for winning at rock paper scissors – if you are in a single game competition then choose paper. This is because men are most likely to choose rock, and scissors are the least popular choice. Also you should try some reverse psychology and announce what you will throw. Most opponents will not believe you and modify their throw as a result.

Rock, Paper, Scissors, Lizard, Spock

You can of course make the game as complicated as you wish – the version above was popularised (though not invented by) The Big Bang Theory. The grid below shows the possible outcomes for this game:

And of course, why stop there? Below is a 15 throw version of the game

If you’ve honed your strategy then maybe you could compete in the a professional rock, paper, scissors tournament – here you can watch the final of a $50,000 Las Vegas competition.

If you liked this post you might also like:

Game Theory and Tic Tac Toe – Tic Tac Toe has already been solved using Game Theory – this topic also brings in an introduction to Group Theory.

This post builds on some of the ideas in the previous post on elliptical curves. The excellent Numberphile video above expands on some of the ideas below.

On a (slightly simplified) level elliptical curves they can be regarded as curves of the form:

y² = x³ +ax + b

So for example the curve below is an elliptical curve. This curve also has an added point at infinity though we don’t need to worry about that here. Elliptical curve cryptography is based on the difficulty in solving arithmetic problems on these curves. If you remember from the last post, we have a special way of defining the addition of 2 points.

Let’s say take 2 points A and B on y² = x³ -4x + 1. In the example we have A = (2,1) and B = (-2,1). We now want to find an answer for A + B which also is on the elliptical curve. If we add them as we might vectors we get (0,2) – but unfortunately this is not on the curve. So, we define the addition A + B through the following geometric steps.

We join up the points A and B. This line intersects the curve in one more place, C.

We then reflect the point C in the x axis. We then define this new point C’ = A + B. In this case this means that (2,1) + (-2,1) = (1/4, -1/8).

Trying another example, y² = x³ -5x + 4 (below), with A = (1,0) and B = (0,2) we have C = (3,-4) and C’ = (3,4). Therefore (1,0) + (0,2) = (3,4).

We also need to have a definition when A and B define the same point on the curve. This will give us the definition of 2A. In this case we take the tangent to the curve at that point, and then as before find the intersection of this line and the curve, before reflecting the point. This probably is easier to understand with another graph:

Here we used the graph y² = x³ -5x + 4 again. This time point A = B = (-1.2, 2.88) and we have drawn the tangent to the curve at this point, which gives point D, which is then reflected in the x axis to give D’. D’ = (2.41, -2.43). Therefore we can say 2A = D’, or 2(-1.2, 2.88) = (2.41, -2.43).

Now addition of points is defined we can see how elliptical curve cryptography works. The basic idea is that given 2 points on the curve, say A and B, it takes a huge amount of computing power to work out the value a such that aA = B. For example, say I use the curve y² = x³ -25x to encrypt, and the 2 points on the curve are A = (-4,6) and B = (1681/144 , -62279/1728). Someone who wanted to break my encryption would need to find the value a such that a(-4,6) = (1681/144 , -62279/1728). The actual answer is a =2 which we can show graphically. As we want to show that 2(-4,6) = (1681/144 , -62279/1728) , we can use the previous method of finding the tangent at the point (-4,6):

We can then check with Geogebra which shows that B’ is indeed (1681/144 , -62279/1728). When a is chosen so that it is very large, this calculation becomes very difficult to attack using brute force methods – which would require checking 2(4,-6), 3(4,-6), 4(4,-6)… until the solution (1681/144 , -62279/1728) was found.

NSA and hacking data

Elliptical curve cryptography has some advantages over RSA cryptography – which is based on the difficulty of factorising large primes – as less digits are required to create a problem of equal difficulty. Therefore data can be encoded more efficiently (and thus more rapidly) than using RSA encryption. Currently the digital currency Bitcoin uses elliptical curve cryptography, and it is likely that its use will become more widespread as more and more data is digitalised. However, it’s worth noting that as yet no-one has proved that it has to be difficult to crack elliptical curves – there may be a novel approach which is able to solve the problem in a much shorter time. Indeed many mathematicians and computer scientists are working in this field.

Government digital spy agencies like the NSA and GCHQ are also very interested in such encryption techniques. If there was a method of solving this problem quickly then overnight large amounts of encrypted data would be accessible – and for example Bitcoin currency exchange would no longer be secure. It also recently transpired that the NSA has built “backdoor” entries into some elliptical curve cryptography algorithms which have allowed them to access data that the people sending it thought was secure. Mathematics is at the heart of this new digital arms race.

If you enjoyed this post you might also like:

RSA Encryption – the encryption system which secures the internet.

Circular inversion – learn about some other geometrical transformations used in university level mathematics.

Elliptical curves are a very important new area of mathematics which have been greatly explored over the past few decades. They have shown tremendous potential as a tool for solving complicated number problems and also for use in cryptography.

Andrew Wiles, who solved one of the most famous maths problems of the last 400 years, Fermat’s Last Theorem, using elliptical curves. In the last few decades there has also been a lot of research into using elliptical curves instead of RSA encryption to keep data transfer safe online. So, what are elliptical curves? On a simple level they can be regarded as curves of the form:

y² = x³ +ax + b

So for example the following is an elliptical curve:

If we’re being a bit more accurate, we also need 4a³ + 27b² ≠ 0. This stops the graph having “singular points” which cause problems with the calculations. We also have a “point at infinity” which can be thought of as an extra point added on to the usual x,y plane representing infinity. This also helps with calculations – though we don’t need to go into this in any more detail here!

What makes elliptical curves so useful is that we can create a group structure using them. Groups are very important mathematical structures – again because of their usefulness in being applied to problem solving. A group needs to have certain properties. For example, we need to be able to combine 2 members of the group to create a 3rd member which is also in the group. This is how it is done with elliptical curves:

Take 2 points A and B on y² = x³ -4x + 1. In the example we have A = (2,1) and B = (-2,1). We now want to find an answer for A + B which also is on the elliptical curve. If we add them as we might vectors we get (0,2) – but unfortunately this is not on the curve. So, we define the addition A + B through the following geometric steps.

We join up the points A and B. This line intersects the curve in one more place, C.

We then reflect the point C in the x axis. We then define this new point C’ = A + B. In this case this means that (2,1) + (-2,1) = (1/4, -1/8).

Trying another example, y² = x³ -5x + 4 (above), with A = (1,0) and B = (0,2) we have C = (3,-4) and C’ = (3,4). Therefore (1,0) + (0,2) = (3,4).

We can also have elliptical curves as a collection of discrete points. We start with the curve y² = x³ +x+1, above, and just look at the positive integer solutions mod 7. For example, when x = 1,

y² = 1³ +1 + 1

y² = 3

So this has no integer solution.

Next, when x = 2 we have:

y² = 2³ +2 +1 = 11.

However when we are working mod 7 we look at the remainder when 11 is divided by 7 (which is 4). So:

y² = 4 (mod 7)

y = 2

When x = 3 we have:

y² = 3³ +3 +1 = 31

y² = 3 (mod 7)

which has no integer solutions.

In fact, all the following coordinate points satisfy the equation (mod 7):

(2,2), (0,1), (0,6), (2,5).

Even though all this might seem very abstract, these methods of calculating points on elliptical curves form the basis of elliptical cryptography. The basic idea is that it takes computers a very long time to make these sorts of calculations – and so they can be used very effectively to encrypt data.

If you enjoyed this post you might also like:

RSA Encryption – the encryption system which secures the internet.

Circular inversion – learn about some other geometrical transformations used in university level mathematics.

The one time pad is a very secure way of encoding messages – so secure in fact that it is unbreakable as long as the agents using it don’t make any mistakes. It was a very popular method of sending coded messages during the Second World War – and persisted well into the Cold War, when secret agents on the ground in a foreign country could encode their reports without fear that they would be broken by the enemy.

The picture above is from a one time pad code sheet which was used in the Second World War. What would happen is that the secret agent would be issued with a code book with lots of pages such as the page shown above. The person who was to receive the coded message would have an identical copy of this book kept securely in the home country. Now, the secret agent would infiltrate a foreign country and start to gather data. To encode it he would choose a page from the codebook, encode his message using the method will look at in a minute, and then the receiver would once he knew which page of the codebook had been used, would be able to decipher very quickly.

The one time pad is uncrackable as long as the codebook uses truly randomly generated numbers (not easy!) and as long as each page from the codebook is only used once – and is never discovered by the enemy. Sometimes this code was cracked because agents became lazy and reused the same pages, other times because the agent was captured and their codebook discovered. Nevertheless, in the age before digital communications it was a very secure method of encoding data.

How does the one time pad work?

It’s all based on Modulo arithmetic (mod 26). This sounds complicated, but basically we just work out the remainder when we divide by 26. For example 4 is still 4 mod 26, because 4 divided by 26 is 0 remainder 4. 30 is also 4 mod 26 because 30 divided by 26 is 1 remainder 4. For negatives we just keep adding 26 until we get a positive number. eg. -10 = 16 mod 26.

Now, say I want to encode the message ” Attack”. I assign each letter in the alphabet a value between 0 and 25 (A = 0, B = 1 etc).

So Attack = 0 19 0 2 10

Next I use my one time pad. Let’s use the one pictured at the top of the page. The first 5 random numbers are 54048, so I add these in turn working in mod 26 if required.

0 + 5 = 5

19 + 4 = 23

0 + 0 = 0

2 + 4 = 6

10 + 8 = 18

Now I convert these new number back into the alphabet using A = 0, B = 1 etc as before. This gives my code word as:

FXAGS

I can then send this back home using my chosen method (e.g. in a letter perhaps hidden as the first word of every sentence etc). The person receiving the code simply has to work in reverse using the same code pad. He converts FXAGS to numbers:

5 23 0 6 18

and he notes the numbers in his code pad are 54048 so he takes away to reveal the message:

5 – 5 = 0

23 – 4 = 19

0 – 0 = 0

6 – 4 = 2

18 – 8 = 10

This reveals the hidden message, Attack!

The beauty of this code is that it can even be done without a code pad – indeed deep cover secret agents were often too scared of being caught with a code book that they used ordinary books. This works in the same way, as long as both parties have agreed in advance on what copy book to use. Say I choose to use John Le Carre’s spy classic The Honourable Schoolboy, page 1 paragraph 1. This can sit innocuously on my bookshelf and yet I can create an equivalent to a code pad by taking each word in turn to make a number. In this case the first word is Afterwards – which would give me the code pad numbers 0 5 19 4 17 22 0 17 3 18. I could then use this to encode in the same manner as above.

We’re now running a huge school code breaking competition. There are a total of 7 competitions to enter – each one with a number of codes to crack. Each time a code is cracked this gives the password to access the next clue. Students who break all codes will be entered onto the leaderboard of fame. Levels of codes range from upper KS2 to post 16 – so there should be something for everyone.

Choose your level, and good luck!

Level 1: Easy, Age 11-16. This is intended to be a gentle introduction to code breaking. Hints provided throughout.

Level 2: Easy, Age 11-16. This builds on some of the skills from the Level 1 code and includes a mixture of common codes, from Morse Codes to substitution ciphers.

Level 3: Murder mystery, Medium, Age 11-16. A murder has been committed in the Maths Department! Solve the clues to uncover the murderer, the weapon and the room. Level 4: Murder mystery, Medium, Age 11-16. Another murder has been committed! Who can solve it first?

Level 5: Medium, Age 11-16. This level combines knowledge of a variety of codes and methods of solving them. Caesar shifts, transposition codes and more.

Level 6: Hard, Age 15-18. This is a real tough challenge – not for the faint hearted! It is primarily aimed at post 16 students and you can expect to look at binary codes, RSA codes and modulo arithmetic.

Level Extra: Easy, Age 11-16. A bonus code breaking challenge in case you solve all the others!

The last World Cup was a relatively rare one for England, with no heroic defeat on penalties, as normally seems to happen. England are in fact the worst country of any of the major footballing nations at taking penalties, having won only 1 out of 6 shoot-outs at the Euros and World Cup. In fact of the 31 penalties taken in shoot-outs England have missed 10 – which is a miss rate of over 30%. Germany by comparison have won 5 out of 7 – and have a miss rate of only 15%.

With the stakes in penalty shoot-outs so high there have been a number of studies to look at optimum strategies for players.

Shoot left when ahead

One study published in Psychological Science looked at all the penalties taken in penalty shoot-outs in the World Cup since 1982. What they found was pretty incredible – goalkeepers have a subconscious bias for diving to the right when their team is behind.

As is clear from the graphic, this is not a small bias towards the right, but a very strong one. When their team is behind the goalkeeper apparently favours his (likely) strong side 71% of the time. The strikers’ shot meanwhile continues to be placed either left or right with roughly the same likelihood as in the other situations. So, this built in bias makes the goalkeeper much less likely to help his team recover from a losing position in a shoot-out.

Shoot high

Analysis by Prozone looking at the data from the World Cups and European Championships between 1998 and 2010 compiled the following graphics:

The first graphic above shows the part of the goal that scoring penalties were aimed at. With most strikers aiming bottom left and bottom right it’s no surprise to see that these were the most successful areas.

The second graphic which shows where penalties were saved shows a more complete picture – goalkeepers made nearly all their saves low down. A striker who has the skill and control to lift the ball high makes it very unlikely that the goalkeeper will save his shot.

The last graphic also shows the risk involved in shooting high. This data shows where all the missed penalties (which were off-target) were being aimed. Unsurprisingly strikers who were aiming down the middle of the goal managed to hit the target! Interestingly strikers aiming for the right corner (as the goalkeeper stands) were far more likely to drag their shot off target than those aiming for the left side. Perhaps this is to do with them being predominantly right footed and the angle of their shooting arc?

Win the toss and go first

The Prozone data also showed the importance of winning the coin toss – 75% of the teams who went first went on to win. Equally, missing the first penalty is disastrous to a team’s chances – they went on to lose 81% of the time. The statistics also show a huge psychological role as well. Players who needed to score to keep their teams in the competition only scored a miserable 14% of the time. It would be interesting to see how these statistics are replicated over a larger data set.

Don’t dive

A different study which looked at 286 penalties from both domestic leagues and international competitions found that goalkeepers are actually best advised to stay in the centre of the goal rather than diving to one side. This had quite a significant affect on their ability to save the penalties – increasing the likelihood from around 13% to 33%. So, why don’t more goalkeepers stay still? Well, again this might come down to psychology – a diving save looks more dramatic and showcases the goalkeepers skill more than standing stationary in the centre.

Test yourself

You can test your penalty taking skills with this online game from the Open University – choose which players are best suited to the pressure, decide what advice they need and aim your shot in the best position.

The usual geometry taught in school is that of Euclidean geometry – in which angles in a triangle add up 180 degrees. This is based on the idea that the underlying space on which the triangle is drawn is flat. However, if the underlying space in curved then this will no longer be correct. On surfaces of constant curvature triangles will have angles greater than 180 degrees, and on surfaces of constant negative curvature triangles will have angles less than 180 degrees. Hyperbolic geometry is based on a geometry in which the underlying space is negatively curved.

Pseudosphere

The pseudosphere pictured above is an example of a surface of constant negative curvature. It was given the name pseudosphere as a usual sphere is the opposite – a surface of constant positive curvature. So the question arises, what would the geometry on a surface such as a pseudosphere look like? If we imagine 2 dimensional beings who live on the surfaces of a pseudosphere, what geometry would they believe their reality to be based on?

Poincare Disc

The Poincare disc is one way we can begin to understand what hyperbolic geometry would be like for its inhabitants. The Poincare disc is a map which translates points in the hyperbolic plane to points in the Euclidean disc. For example, we translate points from the spherical Earth onto a flat atlas – and this helps us understand our spherical reality. In the same way, we can arrive at a model of the hyperbolic plane that helps us to understand it, without actually having to picture a pseudosphere everytime we want to do some maths.

The Poincare disc maps the entire hyperbolic plane onto a finite disc. The shortest line between 2 points (which we would call a straight line in Euclidean geometry) is represented in the disc model by both circular arcs and diameters of the circle. These circular arcs are orthogonal (at 90 degrees) to the boundary circle edge. So our model shows that if we wanted to travel between points E and F on a pseudosphere, that the shortest route would be the curved path of the arc shown. Indeed, being able to calculate the shortest distance between 2 points is an integral part of understanding the underlying geometry.

Distance on the Poincare Disc

The easiest case to look at is the distance from point A on the centre, to a point x on the diameter. Remember that diameters are also straight lines in the Poincare model. What we want to calculate is what this distance from A to x actually represents in the hyperbolic plane. The formula to calculate this is:

Hyperbolic distance = 2tanh^{-1}x

Where tanh^{-1}x is the inverse of the hyperbolic tangent function. This can be calculated with a calculator, or we can use the alternative definition for tanh^{-1}x

2tanh^{-1}x = ln(1+x) – ln(1-x)

So, if our point F is 0.5 units away from A, then this relates to a hyperbolic distance of 2tanh^{-1}0.5 = 1.0986.

As we get further to the boundary of the unit circle this represents ever larger distances in the hyperbolic plane. For example if our point F is 0.9 units away from A, then this relates to a hyperbolic distance of 2tanh^{-1}0.9 = 2.9444

And if our point is 0.999999 units away from A we get a hyperbolic distance of 2tanh^{-1}0.999999 = 14.5087

In fact, as we get closer and closer to the edge of the disc (say x = 1) then the hyperbolic distance gets closer and closer to infinity. In this way the Poincare Disc, even though it has a radius of 1, is able to represent the entire (infinite) hyperbolic plane. We can see this behavior of 2tanh^{-1}x by considering the graph of tanh^{-1}x:

y = tanh^{-1}x has 2 vertical asymptotes at x =1 and x = -1, so as we approach these values, the graph tends to infinity.

So, remarkably the Poincare disc is able to represent an infinite amount of information within a finite space. This disc model was hugely helpful for mathematicians as they started to investigate hyperbolic geometry.

Mathematical art

Escher was a Dutch artist who worked in the 20th century. He used mathematical concepts within his pieces. The above piece is one of his most famous, which is art based on the Poincare disc. It consists of tessellations of the same shape – but because distance is not the same as in Euclidean geometry, those fish which are closer to the boundary edge appear smaller. It is a fantastic illustration of the strange nature of hyperbolic geometry and a pictorial representation of infinity.

Computer aided design gets ever more important in jobs – and with graphing software we can create art using maths functions. For example the above graph was created by a user, Kara Blanchard on Desmos. You can see the original graph here, by clicking on each part of the function you can see which functions describe which parts of Stewie. There are a total of 83 functions involved in this picture. For example, the partial ellipse:

when x is bounded between 3.24 and 0.9, and y is bounded as less than 1.5 generates Stewie’s left cheek. This is what he looks like without it:

By clicking on the various functions you can discover which ones are required to complete the full drawing. Other artwork designed by users includes:

See if you can create some designs of your own! This could make an interesting maths investigation for anyone thinking about a career in computer design or art – as it is a field which will grow in importance in the coming years.

A recent post by the excellent Maths Careers website looked at how we can model volcanic eruptions mathematically. This is an important branch of mathematics – which looks to assign risk to events and these methods are very important to statisticians and insurers. Given that large-scale volcanic eruptions have the potential to end modern civilisation, it’s also useful to know how likely the next large eruption is.

The Guardian has recently run a piece on the dangers that large volcanoes pose to humans. Iceland’s Eyjafjallajökull volcano which erupted in 2010 caused over 100,000 flights to be grounded and cost the global economy over $1 billion – and yet this was only a very minor eruption historically speaking. For example, the Tombora eruption in Indonesia (1815) was so big that the explosion could be heard over 2000km away, and the 200 million tones of sulpher that were emitted spread across the globe, lowering global temperatures by 2 degrees Celsius. This led to widespread famine as crops failed – and tens of thousands of deaths.

Super volcanoes

Even this destruction is insignificant when compared to the potential damage caused by a super volcano. These volcanoes, like that underneath Yellowstone Park in America, have the potential to wipe-out millions in the initial explosion and and to send enough sulpher and ash into the air to cause a “volcanic winter” of significantly lower global temperatures. The graphic above shows that the ash from a Yellowstone eruption could cover the ground of about half the USA. The resultant widespread disruption to global food supplies and travel would be devastating.

So, how can we predict the probability of a volcanic eruption? The easiest model to use, if we already have an estimated probability of eruption is the Poisson distribution:

This formula calculates the probability that X equals a given value of k. λ is the mean of the distribution. If X represents the number of volcanic eruptions we have Pr(X ≥1) = 1 – Pr(x = 0). This gives us a formula for working out the probability of an eruption as 1 -e^{-λ}. For example, the Yellowstone super volcano erupts around every 600,000 years. Therefore if λ is the number of eruptions every year, we have λ = 1/600,000 ≈ 0.00000167 and 1 -e ^{-λ} also ≈ 0.00000167. This gets more interesting if we then look at the probability over a range of years. We can do this by modifying the formula for probability as 1 -e^{-tλ} where t is the number of years for our range.

So the probability of a Yellowstone eruption in the next 1000 years is 1 -e^{-0.00167} ≈ 0.00166, and the probability in the next 10,000 years is 1 -e^{-0.0167} ≈ 0.0164. So we have approximately a 2% chance of this eruption in the next 10,000 years.

A far smaller volcano, like Katla in Iceland has erupted 16 times in the past 1100 years – giving a average eruption every ≈ 70 years. This gives λ = 1/70 ≈ 0.014. So we can expect this to erupt in the next 10 years with probability 1 -e^{-0.14} ≈ 0.0139. And in the next 30 years with probability 1 -e^{-0.42} ≈ 0.34.

The models for volcanic eruptions can get a lot more complicated – especially as we often don’t know the accurate data to give us an estimate for the λ. λ can be estimated using a technique called Maximum Likelihood Estimation – which you can read about here.

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:

S_{n} = 0.5n(u_{1} + u_{n})

S_{n} = 0.5(n+1)(a + a+n) = 1000

S_{n} = (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 = 2^{4} 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!