Library
Mathematics

Divisibility and Group Structure Modulo 20

We explore the elegant digit-sum test for divisibility by 9 and dive into the finite group G = {1, 3, 7, 9, 11, 13, 17, 19} under multiplication modulo 20, mapping out its structure, inverses, and subgroups.

Scene 1 of 4
The Magic of 9s
Divisibility by 9Number = 13,306,617Sum of digits = 1+3+3+0+6+6+1+7 = 27
10n=(9+1)n≡1n≡1(mod9)10^n = (9+1)^n \equiv 1^n \equiv 1 \pmod 9
Have you ever wondered why adding the digits of a number tells us if it's divisible by 9? Because 10 is 9 plus 1, every power of 10 is one more than a multiple of 9, leaving behind just the digit sum.
Step-by-step solver
1

(i) Divisibility

Sum the digits of the number: 1+3+3+0+6+6+1+7 = 27. Since 2+7=9, and 9 is divisible by 9, the original number is divisible by 9.

9∣13,306,6179 \mid 13,306,617
2

(ii)(a) Cayley Table

Construct the 8x8 table by calculating the product of each pair of elements (a*b) mod 20.

a×20ba \times_{20} b
3

(ii)(b) Inverse of 7

Identify the element x in G such that 7*x = 1 (mod 20). From the table, 7*3 = 21, which is 1 mod 20.

7−1=37^{-1} = 3
4

(ii)(c) Order of 13

Calculate powers of 13: 13^1=13, 13^2=169=9, 13^3=117=17, 13^4=221=1. The smallest exponent is 4.

ord(13)=4ord(13) = 4
5

(ii)(d) Subgroup

The powers of an element of order 4 form a cyclic subgroup. Using 13, we get {1, 9, 13, 17}.

H={1,9,13,17}H = \{1, 9, 13, 17\}

Original question

Further Maths question (see attached image). 2(i) Using a suitable algorithm and without performing any division, determine whether 13 306 617 is divisible by 9. 2(ii) The group G = {1, 3, 7, 9, 11, 13, 17, 19} has multiplication modulo 20 as its operation. (a) Complete the Cayley table for G. (b) State the inverse of the element 7. (c) Determine the order of the element 13. (d) Write down a subgroup of G of order 4.

Follow-up chat

Ask me anything about this lesson — I'll answer using what we just covered.