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One of the main benefits of flipping the classroom is allowing IB maths students to self-teach IB content. There are currently a good number of videos on youtube which allow students to self teach syllabus content, but no real opportunity to watch videos going through IB Higher Level past paper questions. So, I’ve started to put some of these together:

Playlist, Worked Exam Solutions:

The videos above are all around 10 minutes long and consist of talking through the solutions to 2-3 IB HL maths questions. The best way to use these videos is to pause the video at the start of the question, attempt it, then watch the video to check the answer and make notes on the method. Click on the top left hand corner to change the video being shown in the playlist.

The playlists below combine these worked solutions with the syllabus content videos, all grouped into the relevant syllabus strands:

Playlist 1, Algebra 1:

Sequences, Binomial, Logs, Induction, Permutations, Gaussian elimination:

Playlist 2, Complex numbers:

Converting from Cartesian to Polar, De Moivre’s Theorem, Roots of Unity:

Playlist 3: Functions:

Sketching graphs, Finding Inverses, Factor and Remainder Theorem, Sketching 1/f(x), sketching absolute f(x), translating f(x):

chineseremaindertheorem

The Chinese Remainder Theorem is a method to solve the following puzzle, posed by Sun Zi around the 4th Century AD.

What number has a remainder of 2 when divided by 3, a remainder of 3 when divided by 5 and a remainder of 2 when divided by 7?

There are a couple of methods to solve this.  Firstly it helps to understand the concept of modulus – for example 21 mod 6 means the remainder when 21 is divided by 6.  In this case the remainder is 3, so we can write 21 ≡ 3 (mod 6).  The ≡ sign means “equivalent to” and is often used in modulus questions.

Method 1:

1) We try to solve the first part of the question, What number has a remainder of 2 when divided by 3,

to do this we list the values of x ≡ 2 (mod 3).  x = 2,5,8,11,14,17……

2) We then look at the values in this list and see which ones also satisfy the second part of the question, What number has a remainder of 3 when divided by 5

From the list x = 2,5,8,11,14,17…… we can see that x = 8 has a remainder of 3 when divided by 5 (i.e 8  ≡ 3 (mod 5) )

3) We now start from 8 and count up in multiples of 15 (3 x 5 because we have mod 3 and mod 5 )

8, 23, 38, 53 …..

4) We look for a number on this list which satisfies the last part of the question, What number has a remainder of 2 when divided by 7?

With x = 8, 23, 38, 53 ….. we can see that x = 23 has a remainder of 2 when divided by 7 (i.e 23 ≡ 2 (mod 7) )

Therefore 23 satisfies all parts of the question.  When you divide it by 3 you get a remainder of 2, when you divide it by 5 you get a remainder of 3, and when you divide it by 7 you get a remainder of 2.  So, 23 is our answer.

The second method is quite a bit more complicated – but is a better method when dealing with large numbers.

Method 2

1) We rewrite the problem in terms of modulus.

x ≡ 2 (mod 3)

x ≡ 3 (mod 5)

x ≡ 2 (mod 7)

We assign a = 2, b = 3, c = 2.

2) We give the values m1 = 3 (because the first line is mod 3), m2 = 5 (because the second line is mod 5), m3 = 2 (because the third line is mod 7).

3) We calculate M = m1m2m3 = 3x5x7 = 105

4) We calculate M1 = M/m1 = 105/3 = 35
M2 = M/m2 = 105/5 = 21
M3 = M/m3 = 105/7 = 15

4) We then note that M1 ≡ 2 (mod 3), M2 ≡ 1 (mod 5), M3 ≡ 1 (mod 7)

5) We then look for the multiplicative inverse of M1. This is the number which when multiplied by M1 will give an answer of 1 (mod 3). This number is 2 because 2×2 = 4 ≡ 1 (mod 3). We assign A = 2

We then look for the multiplicative inverse of M2. This is the number which when multiplied by M2 will give an answer of 1 (mod 5). This number is 1 because 1×1 = 1 ≡ 1 (mod 3). We assign B = 1.

We then look for the multiplicative inverse of M3. This is the number which when multiplied by M3 will give an answer of 1 (mod 7). This number is 1 because 1×1 = 1 ≡ 1 (mod 7). We assign C = 1.

6) We now put all this together:

x = aM1A + bM2B + cM3C (mod M)

x = 2x35x2 + 3x21x1 + 2x15x1 = 233 ≡ 23 (mod 105)

That last method may seem a lot slower – but when working with large numbers is actually a lot quicker.  So there we go – that’s the method that Sun Zi noted more than 1500 years ago.  This topic whilst seemingly quite abstract is a good introduction to number theory – the branch of mathematics which deals with the properties of whole numbers.

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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!

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