Modular arithmetic: remainders and congruence

Imagine you look at your watch and it says 9 o'clock. Someone asks: what time will it be in 5 hours? You don't say "$14$ o'clock" — you say "$2$ o'clock." That's modular arithmetic in action. When you reach $12$, you don't keep counting upwards forever — you wrap around to $1$ and start again.

The same idea applies to days of the week. What day is it 10 days from Tuesday? You don't say "Tuesday plus 10 days." You count by sevens — 7 days from Tuesday is Tuesday again — so 10 days from Tuesday is Tuesday plus 3 more days = Friday.

This is the heart of modular arithmetic: a system where numbers wrap around after reaching a fixed value. WAEC tests this in nearly every paper, sometimes openly ("find $25 \bmod 4$") and sometimes disguised as clock, calendar, or remainder problems. This lesson opens the topic by defining the mod operation, showing how to find remainders quickly, and building the foundation for everything that follows.

Learning objectives

By the end of this lesson you should be able to:

• explain modular arithmetic as arithmetic of remainders
• find $a \bmod n$ quickly, including for negative $a$
• use the division algorithm $a = qn + r$
• recognise when two numbers are congruent modulo $n$
• connect modular arithmetic with clocks and calendars

The basic idea — remainders

When you divide one whole number by another, you get a quotient and a remainder. Modular arithmetic is about the remainder alone.

Example. Divide $17$ by $5$.

In modular notation, this says:

$$17 \bmod 5 = 2$$

We ignore the quotient ($3$) and keep only the remainder ($2$). The result tells us: if you wrap around every 5 steps, where do you land after 17 steps? Answer: $2$ steps past the start of the latest wrap.

The division algorithm

The foundation of modular arithmetic is the division algorithm, which says:

For any whole number $a$ and any positive whole number $n$, there are unique whole numbers $q$ (quotient) and $r$ (remainder) such that:

$a = q \times n + r, \quad \text{where } 0 \leq r < n$

The remainder $r$ is what we call $a \bmod n$. The constraint $0 \leq r < n$ is the key — the remainder is always non-negative and strictly less than the modulus.

Examples:

• $a = 23$, $n = 7$: $23 = 3 \times 7 + 2$, so $23 \bmod 7 = 2$.
• $a = 100$, $n = 9$: $100 = 11 \times 9 + 1$, so $100 \bmod 9 = 1$.
• $a = 50$, $n = 8$: $50 = 6 \times 8 + 2$, so $50 \bmod 8 = 2$.
• $a = 28$, $n = 7$: $28 = 4 \times 7 + 0$, so $28 \bmod 7 = 0$.

The last case is important: when $a$ is a multiple of $n$, the remainder is exactly $0$.

How to compute $a \bmod n$ quickly

For most WAEC questions, the modulus is small ($3$, $4$, $5$, $6$, $7$, $8$, $9$, $10$, $12$, $24$). For these, you can find the remainder in your head — or with one line of long division.

Method 1 — Long division

For $47 \bmod 6$:

• $47 \div 6$: $6$ goes into $47$ seven times ($7 \times 6 = 42$), with $5$ left over.
• So $47 = 7 \times 6 + 5$, and $47 \bmod 6 = 5$.

Method 2 — Subtract multiples of $n$

Keep subtracting $n$ from $a$ until what remains is between $0$ and $n - 1$.

For $47 \bmod 6$:

• $47 - 6 = 41$
• $41 - 6 = 35$
• $35 - 6 = 29$
• $29 - 6 = 23$
• $23 - 6 = 17$
• $17 - 6 = 11$
• $11 - 6 = 5$

Result: $5$. (Same answer as long division, but slower for big numbers.)

Method 3 — Jump by multiples

Find the largest multiple of $n$ that is $\leq a$. Subtract.

For $47 \bmod 6$:

• Multiples of $6$: $6, 12, 18, 24, 30, 36, 42, 48, \ldots$
• The largest $\leq 47$ is $42$.
• $47 - 42 = 5$.

Result: $5$. This is the fastest method once you know your multiplication tables well.

When $a$ is smaller than $n$

If $a < n$, then $a \bmod n = a$. There's no multiple of $n$ to subtract, so the remainder is $a$ itself.

• $3 \bmod 7 = 3$
• $5 \bmod 12 = 5$
• $0 \bmod 9 = 0$

When $a$ is negative — a special case

Negative numbers in modular arithmetic are uncommon at WAEC Core level, but worth understanding. The rule is: the remainder must still be in the range $[0, n-1]$.

For $-3 \bmod 7$:

• You want $r$ such that $-3 = q \times 7 + r$ with $0 \leq r < 7$.
• Try $q = -1$: $-3 = -1 \times 7 + 4$. So $r = 4$.
• So $-3 \bmod 7 = 4$.

Way to think about it: what number between 0 and 6 has the same wrap position as $-3$ in a 7-step cycle? Going $-3$ steps backwards is the same as going $4$ steps forwards (in a 7-cycle), so $-3 \bmod 7 = 4$.

This comes up occasionally in WAEC, particularly in clock problems ("$3$ hours before $1$ o'clock").

Equivalence — when two numbers have the same remainder

Two numbers $a$ and $b$ are said to be congruent modulo $n$ if they have the same remainder when divided by $n$. Written:

$$a \equiv b \pmod{n}$$

This is read: "$a$ is congruent to $b$ modulo $n$."

Examples:

• $17 \equiv 2 \pmod{5}$ (both leave remainder $2$ when divided by $5$).
• $24 \equiv 3 \pmod{7}$ (because $24 = 3 \times 7 + 3$).
• $20 \equiv 8 \pmod{6}$ (because $20 = 3 \times 6 + 2$ and $8 = 1 \times 6 + 2$).
• $50 \equiv 0 \pmod{10}$ (because $50$ is a multiple of $10$).

This congruence relation is the heart of modular arithmetic. We don't always reduce to the remainder; sometimes we say "these two numbers are equivalent in this modular system," which means we can use either one in calculations.

A visual picture — the modular clock

Think of modular arithmetic as a clock with $n$ positions, numbered $0$ to $n - 1$.

For mod 5, the clock has positions $0, 1, 2, 3, 4$ arranged in a circle.

Figure 1. Clocks and calendars are everyday modular arithmetic: mod 12 for hours, mod 7 for days. WASSCE Tutor diagram.

To find $a \bmod 5$, start at position $0$ and walk $a$ steps clockwise. Where you end up is the answer.

• $7 \bmod 5$: walk 7 steps. Pass 0, 1, 2, 3, 4, return to 0, walk to 1, then 2. End at position $2$. So $7 \bmod 5 = 2$.
• $11 \bmod 5$: walk 11 steps. After 5 steps you're back at 0, after 10 you're back at 0 again, then 1 step to position $1$. So $11 \bmod 5 = 1$.

The key insight: every time you complete a full circle, you can ignore those steps. So $11 = 2 \times 5 + 1$ means you go around twice and end one step further. The 2 full circles don't change your position — only the remainder ($1$) matters.

Real-life appearances of modular arithmetic

The 12-hour clock

A 12-hour clock is mod 12 arithmetic. Hours add modulo 12.

• If it's $10$ o'clock now, what time will it be in $5$ hours?
• $10 + 5 = 15$. But $15 \bmod 12 = 3$. Answer: $3$ o'clock.

The 24-hour clock

A military or international clock is mod 24.

• It's $22{:}00$. What time will it be in $7$ hours?
• $22 + 7 = 29$. $29 \bmod 24 = 5$. Answer: $05{:}00$ (5 a.m.).

Days of the week

Days cycle every 7 days, so mod 7 arithmetic applies.

• Today is Wednesday. What day will it be in $30$ days?
• Counting Wednesday as day 0, we want $30 \bmod 7$. $30 = 4 \times 7 + 2$, so $30 \bmod 7 = 2$. Two days after Wednesday is Friday.

Days of the month / week

Less obvious examples include: days of the year (mod 365 or 366), months of the year (mod 12), positions on a circular track (mod number of seats), and even something like the position of a player rotating around a snooker table.

The next lesson goes deeper into clock and day-of-the-week problems. For now, focus on the basic remainder operation.

Key properties — preview

We develop these in the next lesson, but here is a quick preview of why modular arithmetic is useful:

1. Reduce before computing. Instead of computing $147 + 198$ and then taking mod, you can take each mod first, add, then mod the result. This keeps the numbers small.

2. Multiples cancel. If $a \equiv 0 \pmod{n}$, then $a$ contributes nothing to a mod calculation involving multiples of $n$.

3. Speed. For large numbers, modular reduction can replace heavy arithmetic. WAEC questions like "Find the remainder when $2^{50}$ is divided by $7$" are solved this way.

WASSCE examination focus

Practice problems

1. Find $43 \bmod 6$.

2. Find $100 \bmod 8$.

3. Find $77 \bmod 11$.

4. Find $5 \bmod 12$.

5. Find $84 \bmod 12$.

6. Today is Sunday. What day will it be $25$ days from now?

7. It is $4$ p.m. on a 12-hour clock. What time will it be in $20$ hours?

8. Find $250 \bmod 9$.

9. Find $-5 \bmod 7$.

10. Find the smallest positive value of $x$ such that $x \equiv 23 \pmod{6}$.

Worked solutions

1. $43 \div 6 = 7$ remainder $1$ (because $7 \times 6 = 42$, and $43 - 42 = 1$). So $43 \bmod 6 = 1$.

2. $100 \div 8 = 12$ remainder $4$ (because $12 \times 8 = 96$, and $100 - 96 = 4$). So $100 \bmod 8 = 4$.

3. $77 \div 11 = 7$ remainder $0$ (because $7 \times 11 = 77$). So $77 \bmod 11 = 0$.

4. $5 < 12$, so $5 \bmod 12 = 5$.

5. $84 \div 12 = 7$ remainder $0$ (because $7 \times 12 = 84$). So $84 \bmod 12 = 0$.

6. $25 \bmod 7$: largest multiple of $7$ up to $25$ is $21$. So $25 - 21 = 4$. Four days after Sunday: Monday, Tuesday, Wednesday, Thursday.

7. $4 + 20 = 24$ on a 12-hour clock. $24 \bmod 12 = 0$, which on a clock face is 12 o'clock. (a.m. or p.m.? $20$ hours from $4$ p.m. is $12$ p.m. — i.e., noon the next day.)

8. $250 \div 9$: $9 \times 27 = 243$, so $250 - 243 = 7$. Answer: $250 \bmod 9 = 7$.

9. $-5 + 7 = 2$ (adding the modulus brings us into the positive range). So $-5 \bmod 7 = 2$. Check: $-5 = -1 \times 7 + 2$. ✓

10. $23 \bmod 6 = ?$ Largest multiple of $6$ up to $23$ is $18$. So $23 - 18 = 5$. Therefore $23 \equiv 5 \pmod 6$, and the smallest positive $x$ is $x = 5$.

What's next

In the next lesson we cover addition, subtraction and multiplication in modular arithmetic, clock and calendar problems, and solving simple congruences. The key principle: you can reduce before you compute. We will see how this turns long calculations into short ones, why $99 + 97 \bmod 7$ is easy without big multiplication, and how WAEC tests this directly.