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Normal Numbers – and random number generators

Numberphile have a nice new video where Matt Parker discusses all different types of numbers – including “normal numbers”.  Normal numbers are defined as irrational numbers for which the probability of choosing any given 1 digit number is the same, the probability of choosing any given 2 digit number is the same etc.  For example in the normal number 0.12345678910111213141516… , if I choose any digit in the entire number at random P(1) = P(2) = P(3) = … P(9) = 1/10.  Equally if I choose any 2 digit number at random I have P(10) = P(11) = P(12) = P(99) = 1/100.

It is incredibly hard to find normal numbers, but there is a formula to find some of them.

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In base 10, we are restricted to choosing a value of c such that 10 and c are relatively prime (i.e share no common factors apart from 1).  So if we choose c = 3 this gives:

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We can now put this into Wolfram Alpha and see what number this gives us:

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So we can put the first few digits into an online calculator to find the distributions

0.000333333444444444444448148148148148148148148148148148148148148149382716049382716049382716049382716049382716049382716049382716049382716049382716049382716049382716049382716049382716049382716049827160493827160493827160479423863312 7572016460905349794238683127572016460905349794238683127572016460 9053497942386831275720164609053497942386831275720164609053497942

4: 61
1: 41
8: 40
3: 38
0: 36
2: 33
7: 33
9: 33
6: 32
5: 10

We can see that we are already seeing a reasonably similar distribution of single digits, though with 4 and 5 outliers.  As the number progressed we would expect these distributions to even up (otherwise it would not be a normal number).

One of the potential uses of normal numbers is in random number generators – if you can use a normal number and specify a digit (or number of digits) at random then this should give an equal chance of returning each number.

To finish off this,  let’s prove that the infinite series:

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does indeed converge to a number (if it diverged then it could not be used to represent a real number).  To do that we can use the ratio test (only worry about this bit if you have already studied the Calculus Option for HL!):

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We can see that in the last limit 3 to the power n+1 will grow faster than 3 to the power n, therefore as n increases the limit will approach 0.  Therefore by the ratio test the series converges to a real number.

Is pi normal?

Interestingly we don’t know if numbers like e, pi and ln(2) are normal or not.  We can analyse large numbers of digits of pi – and it looks like it will be normal, but as yet there is no proof.  Here are the distribution of the first 100,000 digits of pi:

1: 10137
6: 10028
3: 10026
5: 10026
7: 10025
0: 9999
8: 9978
4: 9971
2: 9908
9: 9902

Which we can see are all very close to the expected value of 10,000 (+/- around 1%).

So, next I copied the first 1 million digits of pi into a character frequency counter which gives the following:

5: 100359
3: 100230
4: 100230
9: 100106
2: 100026
8: 99985
0: 99959
7: 99800
1: 99758
6: 99548

This is even closer to the expected values of 100,000 with most with +/- 0.25 %.

Proving that pi is normal would be an important result in number theory – perhaps you could be the one to do it!

 

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Volume optimization of a cuboid

This is an extension of the Nrich task which is currently live – where students have to find the maximum volume of a cuboid formed by cutting squares of size x from each corner of a 20 x 20 piece of paper.  I’m going to use an n x 10 rectangle and see what the optimum x value is when n tends to infinity.

First we can find the volume of the cuboid:

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Next we want to find when the volume is a maximum, so differentiate and set this equal to 0.

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Next we use the quadratic formula to find the roots of the quadratic, and then see what happens as n tends to infinity (i.e we want to see what the optimum x values are for our cuboid when n approaches infinity).  We only take the negative solution of the + – quadratic solutions because this will be the only one that fits the initial problem.

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Next we try and simplify the square root by taking out a factor of 16, and then we complete the square for the term inside the square root (this will be useful next!)

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Next we make a u substitution.  Note that this means that as n approaches infinity, u approaches 0.

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Substituting this into the expression gives us:

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We then manipulate the surd further to get it in the following form:

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Now, the reason for all that manipulation becomes apparent – we can use the binomial expansion for the square root of 1 + u2 to get the following:

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Therefore we have shown that as the value of n approaches infinity, the value of x that gives the optimum volume approaches 2.5cm.

So, even though we start with a pretty simple optimization task, it quickly develops into some quite complicated mathematics.  We could obviously have plotted the term in n to see what its behavior was as n approaches infinity, but it’s nicer to prove it.  So, let’s check our result graphically.

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As we can see from the graph, with n plotted on the x axis and x plotted on the y axis we approach x = 2.5 as n approaches infinity – as required.

An m by n rectangle.

So, we can then extend this by considering an n by m rectangle, where m is fixed and then n tends to infinity.  As before the question is what is the value of x which gives the maximum volume as n tends to infinity?

We do the same method.  First we write the equation for the volume and put it into the quadratic formula.

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Next we complete the square, and make the u substitution:

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Next we simplify the surd, and then use the expansion for the square root of 1 + u2

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This then gives the following answer:

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So, we can see that for an n by m rectangle, as m is fixed and n tends to infinity, the value of x which gives the optimum volume tends to m/4.  For example when we had a 10 by n rectangle (i.e m = 10) we had x = 2.5.  When we have a 20 by n rectangle we would have x = 5 etc.

And we’ve finished!  See what other things you can explore with this problem.

 

Narcissistic Numbers

Narcissistic Numbers are defined as follows:

An n digit number is narcissistic if the sum of its digits to the nth power equal the original number.

For example with 2 digits, say I choose the number 36:

32 + 62 = 45. Therefore 36 is not a narcissistic number, as my answer is not 36.

For example with 3 digits, say I choose the number 124:

13 + 23 + 43 = 73. Therefore 124 is not a narcissistic number as my answer is not 124.

The question is how to find all the narcissistic numbers less than 1000, without checking 1000 different numbers? Let’s start with 1 digit numbers.

1 digit numbers

01 = 0
11 = 1
21 = 2 etc.

Therefore all numbers from 0-9 are narcissistic.

2 digit numbers

For 2 digit numbers in the form ab we need the following:

a2 + b2 = 10a + b.

Therefore

a2 – 10a + b2 – b = 0.

Next if we choose a = 1,2,3,4,5 we get the following simultaneous equations:

b2 – b -16 = 0

b2 – b -21 = 0

b2 – b -24 = 0

b2 – b -25 = 0

None of these factorise for integer solutions, therefore there are no 2 digit solutions from 11 to 59. Trying a = 6,7,8,9 we find that we get the same as the first four equations. This is because a and 10-a give equivalent solutions. In other words, when a = 1 we get the equation b2 – b -9 = 0 and when a = 9 we also get the equation b2 – b -9 = 0. This is because:

for:

a2 – 10a

if we substitute a = (10 – a) we get

(10 – a)2 – 10(10 – a) = a2 – 10a.

Therefore we prove that there are no 2 digit narcissistic numbers.

3 digit numbers

First we list the cube numbers:

13 = 1, 23 = 8, 33 = 27, 43 = 64, 53 = 125, 63 = 216, 73 = 343, 83 = 512, 93 = 729.

and then consider 3 digit numbers of the form 1bc first. We need:

13+ b3 + c3 = 100 + 10b + c.

If our first digit is 1, then b3 + c3 need to add up to give us a number in the one hundreds, therefore:

99 ≤  b3 + c3≤ 198.

We can then check the cube numbers and see that the only possible combinations for a and b are 0 5, 5 0, 1 5, 5 1, 2 5, 5 2, 3 5, 5 3, 4 4, 4 5, 5 4. We can check these (only have to use the calculator for half as the reversed numbers give equivalent answers) and find that for 153 we do indeed get a narcissistic number i.e:

13+ 53 + 33 = 153.

Next we consider 3 digit numbers of the form 2bc first. We need:

192 ≤  b3 + c3≤ 292

This gives the following possibilities for b and c: 6 0, 0 6, 6 1, 16, 2 6, 6 2, 6 3, 3 6, 6 4, 4 6.

None of these give narcissistic numbers.

Next we consider 3 digit numbers of the form 3bc first. We need:

273 ≤  b3 + c3≤373

This gives the following possibilities for b and c: 6 4, 4 6, 6 5, 5 6, 7 1, 1 7, 7 2, 2 7, 7 3, 3 7, 7 0, 0 7.

Checking these we find 2 more narcissistic numbers:

370 = 33+ 73 + 03

371= 33+ 73 + 13

Using the same method, we can find that the only possibilities for 4bc are: 5 6, 6 5, 7 1, 1 7, 7 2, 2 7, 7 3, 3 7, 7 4, 4 7, 7 0, 0 7.  Checking these gives us 1 more narcissistic number:

407= 43+ 03 + 73

We can carry on with this method checking up to the form 9ab (it gets easier as there are less combinations possible, but will find no more narcissistic numbers.  Therefore we have all the narcissistic numbers less than 1000:

0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 153, 370, 371, 407.

Is there a limit to how many narcissistic numbers there are?

Surprisingly there is a limit – there are exactly 88 narcissistic numbers in base 10. To see why we can consider the following:

In 3 digits the biggest number we can choose is 999. That would give 93+ 93 + 93 (or 3(9)3). This needs to give a number in the hundreds (102) otherwise it would be impossible to achieve a narcissistic number. Therefore with an n digit number the largest number we can make is n(9)n and if we can’t make a number in the 10n-1, then a narcissistic number is not possible. If we can prove that the inequality:

n(9)n < 10n-1

is true for some values of n, then there will be an upper bound to the narcissistic numbers we can make. We could simply plot this directly, but let’s see if we can convince ourselves it’s true for some n without using graphical software first. Let’s see if we can find an equality:

n(9)n = 10n-1

First we take log base 10 of both sides

log n(9)n = n-1

log(n) + nlog(9) = n-1

n(log9 -1) + logn +1 = 0

Next we make the substitution logn = u and therefore 10u = n. This gives:

10u(log9 -1) + u + 1 = 0.

Now we can clearly see that 10u will grow much larger than u + 1, so any root must be for u is small. Let’s see, when u = 1 we get a positive number (as log9 -1 is a negative number close to 0), but when u = 2 we get a negative number. Therefore we have a root between u = 1 and u = 2. Given that we made the substitution logn = u, that means we have found the inequality n(9)n < 10n-1 will hold for n somewhere between 101 and 10 2.

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Using Wolfram we can see that the equality is reached when u = 1.784, i.e when n = 101.784 or approx 60.8. Therefore we can see that when we have more than 60 digit numbers, it is no longer possible to make narcissistic numbers.


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.

Dichotomy Paradox

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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

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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.

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A geometric proof for the Arithmetic and Geometric Mean

There is more than one way to define the mean of a number.  The arithmetic mean is the mean we learn at secondary school – for 2 numbers a and b it is:

(a + b) /2.

The geometric mean on the other hand is defined as:

(x1.x2.x3…xn)1/n

So for example with the numbers 1,2,3 the geometric mean is (1 x 2 x 3)1/3.

With 2 numbers, a and b, the geometric mean is (ab)1/2.

We can then use the above diagram to prove that (a + b) /2 ≥ (ab)1/2 for all a and b. Indeed this inequality holds more generally and it can be proved that the Arithmetic mean ≥ Geometric mean.

Step (1) We draw a triangle as above, with the line MQ a diameter, and therefore angle MNQ a right angle (from the circle theorems).  Let MP be the length a, and let PQ be the length b.

Step (2) We can find the length of the green line OR, because this is the radius of the circle.  Given that the length a+b was the diameter, then (a+b) /2 is the radius.

Step (3) We then attempt to find an equation for the length of the purple line PN.

We find MN using Pythagoras:  (MN)2 = a2 +x2

We find NQ using Pythagoras:  (NQ)2 = b2 +x2

Therefore the length MQ can also be found by Pythagoras:

(MQ)2 = (MN) + (NQ)2

(MQ) = a2 +x2 + b2 +x2

But MQ = a + b.  Therefore:

(a + b) = a2 +x2 + b2 +x2

a2+ b2 + 2ab = a2 +x2 + b2 +x2

2ab = x2 +x2

ab = x2

x = (ab)1/2

Therefore our green line represents the arithmetic mean of 2 numbers (a+b) /2 and our purple line represents the geometric mean of 2 numbers (ab)1/2. The green line will always be greater than the purple line (except when a = b which gives equality) therefore we have a geometrical proof of our inequality.

There is a more rigorous proof of the general case using induction you may wish to explore as well.

This is an example of how an investigation into area optimisation could progress.  The problem is this:

A farmer has 40m of fencing.  What is the maximum area he can enclose?

Case 1:  The rectangle:

Reflection – the rectangle turns out to be a square, with sides 10m by 10m.  Therefore the area enclosed is 100 metres squared.

Case 2:  The circle:

Reflection:  The area enclosed is greater than that of the square – this time we have around 127 metres squared enclosed.

Case 3: The isosceles triangle:

Reflection – our isosceles triangle turns out to be an equilateral triangle, and it only encloses an area of around 77 metres squared.

Case 4, the n sided regular polygon

Reflection:  Given that we found the cases for a 3 sided and 4 sided shape gave us the regular shapes, it made sense to look for the n-sided regular polygon case.  If we try to plot the graph of the area against n we can see that for n ≥3 the graph has no maximum but gets gets closer to an asymptote.  By looking at the limit of this area (using Wolfram Alpha) as n gets large we can see that the limiting case is the circle. This makes sense as regular polygons become closer to circles the more sides they have.

Proof of the limit using L’Hospital’s Rule

  

Here we can prove that the limit is indeed 400/pi by using L’Hospital’s rule.  We have to use it twice and also use a trig identity for sin(2x) – but pleasingly it agrees with Wolfram Alpha.

So, a simple example of how an investigation can develop – from a simple case, getting progressively more complex and finishing with some HL Calculus Option mathematics.

Sequence Investigation

This is a nice investigation idea from Nrich.  The above screen capture is from their Picture Story puzzle.  We have successive cubes – a 1x1x1 cube, a 2x2x2 cube etc.

The cubes are then rearranged to give the following shape.  The puzzle is then to use this information to discover a mathematical relationship.  This was my first attempt at this:

13 = 12
23 = (1+2)2 – 12
33 = (1+2+3)2 – (1+2)2
43 = (1+2+3+4)2 – (1+2+3)2

n3 = (1+2+3+4+…+n)2 – (1+2+3+…+ (n-1))2

This is not an especially attractive relationship – but nevertheless we have discovered a mathematical relationship using the geometrical figures above. Next let’s see why the RHS is the same as the LHS.

RHS:

(1+2+3+4+…+n)2 – (1+2+3+…+ (n-1))2

= ([1+2+3+4+…+ (n-1)] + n)2 – (1+2+3+…+ (n-1))2

= (1+2+3+…+ (n-1))2 + n2 + 2n(1+2+3+4+…+ (n-1)) – (1+2+3+…+ (n-1))2

= n2 + 2n(1+2+3+4+…+ (n-1))

next we notice that 1+2+3+4+…+ (n-1) is the sum of an arithmetic sequence first term 1, common difference 1 so we have:

1+2+3+4+…+ (n-1) = (n-1)/2 (1 + (n-1) )

1+2+3+4+…+ (n-1) = (n-1)/2 + (n-1)2/2

1+2+3+4+…+ (n-1) = (n-1)/2 + (n2 – 2n + 1)/2

Therefore:

2n(1+2+3+4+…+ (n-1)) = 2n ( (n-1)/2 + (n2 – 2n + 1)/2 )

2n(1+2+3+4+…+ (n-1)) = n2 -n + n3 – 2n2 + n

Therefore

n2 + 2n(1+2+3+4+…+ (n-1)) = n2 + n2 -n + n3 – 2n2 + n
n2 + 2n(1+2+3+4+…+ (n-1)) = n3

and we have shown that the RHS does indeed simplify to the LHS – as we would expect.

An alternative relationship

13 = 12
13+23 = (1+2)2
13+23+33  = (1+2+3)2
13+23+33+…n3  = (1+2+3+…+n)2

This looks a bit nicer – and this is a well known relationship between cubes and squares.  Could we prove this using induction?  Well we can show it’s true for n =1.  Then we can assume true for n=k:

13+23+33+…k3  = (1+2+3+…+k)2

Then we want to show true for n = k+1

ie.

13+23+33+… k3  + (k+1)3= (1+2+3+…+k + (k+1))2

LHS:

13+23+33+… k3  + (k+1)3

= (1+2+3+…+k)2 + (k+1)3

RHS:

(1+2+3+…+k + (k+1))2

= ([1+2+3+…+k] + (k+1) )2

= [1+2+3+…+k]2 + (k+1)2 + 2(k+1)[1+2+3+…+k]

= [1+2+3+…+k]+ (k+1)2 + 2(k+1)(k/2 (1+k))   (sum of a geometric formula)

= [1+2+3+…+k]+ (k+1)2 + 2(k+1)(k/2 (1+k))

= [1+2+3+…+k] + k3+ 3k2 + 3k + 1

= (1+2+3+…+k)2 + (k+1)3

Therefore we have shown that the LHS = RHS and using our induction steps have shown it’s true for all n.  (Write this more formally for a real proof question in IB!)

So there we go – a couple of different mathematical relationships derived from a simple geometric pattern – and been able to prove the second one (the first one would proceed in a similar manner).  This sort of free-style pattern investigation where you see what maths you can find in a pattern could make an interesting maths IA topic.

 

puzzle

This is a great puzzle which the Guardian ran last week:

Fill in the equations below using any of the basic mathematical operations, +, –, x, ÷, and as many brackets as you like, so that they make arithmetical sense.

10 9 8 7 6 5 4 3 2 1 = 2017

There are many different ways of solving this – see if you can find the simplest way!

Scroll down to see some possible answers.

 

 

 

 

 

I had a play around with this and this is my effort:

(10+(9 x 8 x 7) -(6 + 5) ) x 4 + 3 + (2 x 1) = 2017

An even nicer answer was provided on the Guardian – which doesn’t even need brackets:

10 x 9 x 8 x 7 / 6 / 5  x 4  x  3 + 2 – 1 = 2017

Any other solutions?

The Watson Selection Task – a logical puzzle

The Watson Selection Task is a logical problem designed to show how bad we are at making logical decisions.  Watson first used it in 1968 – and found that only 10% of the population would get the correct answer.  Indeed around 65% of the population make the same error.  Here is the task:

The participants were given the following instructions:

Here is a rule: “every card that has a D on one side has a 3 on the other.” Your task is to select all those cards, but only those cards, which you would have to turn over in order to discover whether or not the rule has been violated.  Each card has a number on one side and a letter on the other.

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Give yourself a couple of minutes to work out what you think the answer is – and then highlight the space below where the answer is written in white text.

The correct answer is to pick the D card and the 7 card

 

 

 

 

This result is normally quite unexpected – but it highlights one of the logical fallacies that we often fall into:

A implies B does not mean that B implies A

All cats have 4 legs (Cats = A, legs = B, A implies B)
All 4 legged animals are cats (B implies A)

We can see that here we would make a logical error if we concluded that all 4 legged animals were cats.

In the logic puzzle we need to turn over only 2 cards, D and 7.  This is surprising because most people will also say that you need to turn over card with a 3.  First we need to be clear about what we are trying to do:  We want to find evidence that the rule we are given is false.

If we turn over the D and find a number other than 3, we have evidence that the rule is false – therefore we need to turn over D.

If we turn over the 7 and find a D on the other side, we have evidence that the rule is false – therefore we need to turn over the 7.

But what about the 3?  If we turn over the 3 and find a D then we have no evidence that the rule is false (which is what we are looking for).  If we turn over the 3 and find another letter then this also gives us no evidence that the rule is false.  After all our rule says that all Ds have 3s on the other side, but it doesn’t say that all 3s have Ds on the other side.

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Are mathematicians better at this puzzle than historians?

Given the importance of logical thought in mathematics, people have done studies to see if undergraduate students in maths perform better than humanities students on this task.  Here are the results:

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You can see that there is a significant difference between the groups.  Maths students correctly guessed the answer D7 29% of the time, but only 8% of history students did.  The maths university lecturers performed best – getting the answer right 43% of the time.

Making different mistakes

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You can also analyse the mistakes that students made- by only looking at the proportions of incorrect selections.  Here again are significant differences which show that the groups are thinking about the problem in different ways.  DK7 was chosen by around 1/5 of both maths students and lecturers, but by hardly any history students.

You can read about these results in much more depth in the following research paper Mathematicians and the Selection Task – where they also use Chi Squared testing for significance levels.

 

 

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This is a nice example of using some maths to solve a puzzle from the mindyourdecisions youtube channel (screencaptures from the video).

How to Avoid The Troll: A Puzzle

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In these situations it’s best to look at the extreme case first so you get some idea of the problem.  If you are feeling particularly pessimistic you could assume that the troll is always going to be there.  Therefore you would head to the top of the barrier each time.  This situation is represented below:

The Pessimistic Solution:

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Another basic strategy would be the optimistic strategy.  Basically head in a straight line hoping that the troll is not there.  If it’s not, then the journey is only 2km.  If it is then you have to make a lengthy detour.  This situation is shown below:

The Optimistic Solution:

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The expected value was worked out here by doing 0.5 x (2) + 0.5 x (2 + root 2) = 2.71.

The question is now, is there a better strategy than either of these?  An obvious possibility is heading for the point halfway along where the barrier might be.  This would make a triangle of base 1 and height 1/2.  This has a hypotenuse of root (5/4).  In the best case scenario we would then have a total distance of 2 x root (5/4).  In the worst case scenario we would have a total distance of root(5/4) + 1/2 + root 2.  We find the expected value by multiply both by 0.5 and adding.  This gives 2.63 (2 dp).  But can we do any better?  Yes – by using some algebra and then optimising to find a minimum.

The Optimisation Solution:

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To minimise this function, we need to differentiate and find when the gradient is equal to zero, or draw a graph and look for the minimum.  Now, hopefully you can remember how to differentiate polynomials, so here I’ve used Wolfram Alpha to solve it for us.  Wolfram Alpha is incredibly powerful -and also very easy to use.  Here is what I entered:

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and here is the output:

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So, when we head for a point exactly 1/(2 root 2) up the potential barrier, we minimise the distance travelled to around 2.62 miles.

So, there we go, we have saved 0.21 miles from our most pessimistic model, and 0.01 miles from our best guess model of heading for the midpoint.  Not a huge difference – but nevertheless we’ll save ourselves a few seconds!

This is a good example of how an exploration could progress – once you get to the end you could then look at changing the question slightly, perhaps the troll is only 1/3 of the distance across?  Maybe the troll appears only 1/3 of the time?  Could you even generalise the results for when the troll is y distance away or appears z percent of the time?

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