Modular arithmetic · Modular arithmetic

On a 12-hour clock, 5 hours after 9 o’clock is 2 o’clock, not 14. That’s modular arithmetic: numbers wrap round after a fixed number, called the modulus. In modulo 7, only 0, 1, 2, 3, 4, 5 and 6 are used. Any other number is replaced by its remainder when divided by 7.

\begin{gathered} a \equiv r \pmod n \\ \text{means } a \div n \text{ leaves remainder } r \end{gathered}

So $19 \equiv 5 \pmod 7$ , because $19 = 2 \times 7 + 5$ .

A modulo-5 clockOnly the remainder matters: where you stop

Try it

Modular arithmetic: a clockChange the numbers

5 + 4 = 9ordinary answerremainder 29 = 1 × 7 + 22so 5 + 4 ≡ 2 (mod 7)

modulus n = 7a = 5b = 4

Start at 0, walk 5 steps, then 4 more. The clock only has 0 to 6, so after 6 you go back to 0. You stop at 2, which is the remainder when 9 is divided by 7.

Add by walking round the clock, and multiply by jumping. Wherever you stop is the remainder of the ordinary answer.

Calculating

Work out the ordinary answer, then divide by the modulus and keep the remainder. With big numbers, replace each one by its remainder first. The answer is the same, and the numbers are smaller.

More: calculating modulo n

WAEC 2022 · Paper 1 · Q5Evaluate $23 \times 54 \pmod 7$ .

Solving an equation

Try each possible value $0, 1, 2, \ldots$ up to $n - 1$ in turn. There are only a few.

More: solving an equation

WAEC 2020 · Paper 1 · Q4Find the least value of $x$ which satisfies the equation $4x \equiv 7 \pmod 9$ .

Tables on a set

WAEC often gives a set, such as $\{2, 3, 5, 7\}$ , and asks for the multiplication (or addition) table modulo $n$ . Each entry is the ordinary product, reduced modulo $n$ . Then you use the table to solve equations: find the row, look along it for the answer, and read off the column.

Fill in a table yourself, then use it to solve equations.

A table modulo 7Tap the cells to fill the table

⊗	2	4	5	6
2
4
5
6

0 of 16cells filled

Each entry is the ordinary product reduced modulo 7: multiply, divide by 7, keep the remainder. Tap a cell to fill it.

Worked example · WAEC 2019

WAEC 2019 · Paper 2 · Q1

Draw a table for multiplication $\otimes$ in modulo 8 on the set $T = \{2, 3, 5, 7\}$ .

Use the table to find the solution set of: (i) $3 \otimes n = 5$ ; (ii) $n \otimes n = 1$ .

Fill in the table
Multiply, then take the remainder on dividing by 8. For example, ${5 \times 7 = 35 = 4 \times 8 + 3}$ , so ${5 \otimes 7 = 3}$ .

$\otimes$ 2 3 5 7
2 4 6 2 6
3 6 1 7 5
5 2 7 1 3
7 6 5 3 1
Think first. What is $5 \otimes 7$ in modulo 8?
(b)(i) 3 ⊗ n = 5
In the row of 3, the 5 is in the column of 7: $n = 7$ .
Think first. Go along the row of 3. Where is 5?
(ii) n ⊗ n = 1
The diagonal entries are 4, 1, 1, 1. So ${n \otimes n = 1}$ for ${n = 3, 5, 7}$ , and the solution set is ${\{3, 5, 7\}}$ .
Think first. Look down the diagonal.

$\otimes$	2	3	5	7
2	4	6	2	6
3	6	1	7	5
5	2	7	1	3
7	6	5	3	1

Your turn

WAEC 2014 · Paper 2 · Q8 (a)

(a)
Copy and complete the table for multiplication $\otimes$ modulo 11 on the set $\{1, 5, 9, 10\}$ .

$\otimes$ 1 5 9 10

1 1 5 9 10

5 5

9 9

10 10

Use the table to: (i) evaluate $(9 \otimes 5) \otimes (10 \otimes 10)$ ; (ii) find the truth set of $10 \otimes m = 2$ ; (iii) find the truth set of $n \otimes n = 4$ .
Model answer

$\otimes$ 1 5 9 10

1 1 5 9 10

5 5 3 1 6

9 9 1 4 2

10 10 6 2 1

Multiply, then take the remainder on dividing by 11: for example $9 \otimes 10 = 90 = 8 \times 11 + 2$ , so the entry is 2. From the table, (i) $1 \otimes 1 = 1$ ; (ii) $10 \otimes 9 = 2$ , so $\{9\}$ ; (iii) $9 \otimes 9 = 4$ , so $\{9\}$ .

$\otimes$	1	5	9	10
1	1	5	9	10
5	5
9	9
10	10

$\otimes$	1	5	9	10
1	1	5	9	10
5	5	3	1	6
9	9	1	4	2
10	10	6	2	1

Worked solution (try it first)

(a)

Multiply, then take the remainder on dividing by 11.
For example $9 \otimes 10 = 90 = 8 \times 11 + 2$ , so the entry is 2.
$\otimes$ 1 5 9 10

1 1 5 9 10

5 5 3 1 6

9 9 1 4 2

10 10 6 2 1

$\otimes$	1	5	9	10
1	1	5	9	10
5	5	3	1	6
9	9	1	4	2
10	10	6	2	1

Report a problem with this question