When do 11 and 3 more equal 2?

This is modulo arithmetic, also called clock arithmetic (even when the modulus is not 12) as the numbers start back at zero when the modulus is reached - the numbers go round and round in a cycle like the numbers (or times) on a clock.

When doing modulo arithmetic, only the reminder when the number is divided by the modulus is considered. For example 15 modulo 6 is 3 as 15 ÷ 6 = (2←ignored) remainder 3; 15 modulo 3 is 0 as 15 ÷ 3 = (5←ignored) remainder 0.

This is an important area of maths.

One application is finding the highest common factor of two numbers (which is Euclid's Algorithm or method):

1. find the first number modulo the second
2. replace the first number by the second number
3. replace the second number by the number found in step 1
4. if the (new) second number is not zero repeat from step 1
5. the highest common factor is the (final) first number.

Usually the method specifies that the first number should be the larger of the two, but if it isn't, the first time through the method will swap the two over so that it is now the larger of the two.

eg find the hcf of 84 and 30:

84 MOD 30 = 24

30 MOD 24 = 6

24 MOD 6 = 0

→ hcf(84, 30) = 6.

Another use is in the RSA public key encryption.

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In the case of the question, the modulus is 12:

(11 + 2) MOD 12 = 13 MOD 12 = 1

(9 + 5) MOD 12 = 14 MOD 12 = 2