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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 m_{1} = 3 (because the first line is mod 3), m_{2} = 5 (because the second line is mod 5), m_{3} = 2 (because the third line is mod 7).

3) We calculate M = m_{1}m_{2}m_{3} = 3x5x7 = 105

4) We calculate M_{1} = M/m_{1} = 105/3 = 35

M_{2} = M/m_{2} = 105/5 = 21

M_{3} = M/m_{3} = 105/7 = 15

4) We then note that M_{1} ≡ 2 (mod 3), M_{2} ≡ 1 (mod 5), M_{3} ≡ 1 (mod 7)

5) We then look for the multiplicative inverse of M_{1}. This is the number which when multiplied by M_{1} 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 M_{2}. This is the number which when multiplied by M_{2} 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 M_{3}. This is the number which when multiplied by M_{3} 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 = aM_{1}A + bM_{2}B + cM_{3}C (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.

**IB Revision**

If you’re already thinking about your coursework then it’s probably also time to start planning some revision, either for the end of Year 12 school exams or Year 13 final exams. There’s a really great website that I would strongly recommend students use – you choose your subject (HL/SL/Studies if your exam is in 2020 or Applications/Analysis if your exam is in 2021), and then have the following resources:

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The Practice Exams section takes you to ready made exams on each topic – again with worked solutions. This also has some harder exams for those students aiming for 6s and 7s and the Past IB Exams section takes you to full video worked solutions to every question on every past paper – and you can also get a prediction exam for the upcoming year.

I would really recommend everyone making use of this – there is a mixture of a lot of free content as well as premium content so have a look and see what you think.