Arithmetic operations in a programming language

Operation	Symbol	Example	Result
Addition	+	7 + 3	10
Subtraction	−	7 − 3	4
Multiplication	*	7 * 3	21
Real division	/	7 / 2	3.5
Integer division	DIV	7 DIV 2	3
Remainder (modulus)	MOD	7 MOD 2	1

The first three need no explanation. Everything examinable in this subtopic lives in the difference between the three kinds of division.

The three divisions, side by side

Take 11 ÷ 2:

Operation	Question it answers	Result
11 / 2	"What is 11 divided by 2, exactly?"	5.5 (a real)
11 DIV 2	"How many whole 2s fit into 11?"	5 (an integer)
11 MOD 2	"What is left over after taking out those whole 2s?"	1 (an integer)

The relationship that ties them together:

$\text{DIV} \times \text{divisor} + \text{MOD} = \text{original number}$

Check it: $5 \times 2 + 1 = 11$. ✓

This identity is a superb self-check in the exam — if your DIV and MOD answers do not rebuild the original number, one of them is wrong.

AQA gives this exact example in the specification:

"the calculation 11/2 would generate the following values: Integer division: the integer quotient of 11 divided by 2 (11 DIV 2) = 5. Remainder: the remainder when 11 is divided by 2 (11 MOD 2) = 1."

AQA tip. The result of DIV and MOD is always an integer; the result of / may be a real. That type difference matters when the value is used as an array index — which is exactly why binary search uses (low + high) DIV 2 rather than / 2.

DIV always rounds DOWN — it never rounds to nearest

This is the single most common error in the subtopic.

Calculation	Real division	DIV	✗ Common wrong answer
7 DIV 2	3.5	3	4 ("3.5 rounds to 4")
9 DIV 2	4.5	4	5
19 DIV 4	4.75	4	5
3 DIV 5	0.6	0	1

DIV discards the fractional part entirely. It does not look at whether it is above or below .5. Think of it as "how many whole ones fit?" — 0.6 of a 5 does not fit even once, so 3 DIV 5 = 0.

MOD's answer is always less than the divisor

n MOD d gives the remainder, which must always be between 0 and d − 1.

• n MOD 2 can only ever be 0 or 1
• n MOD 3 can only ever be 0, 1 or 2
• n MOD 10 can only ever be 0 to 9

If your MOD answer is bigger than or equal to the divisor, it is wrong. That is a second reliable self-check.

The negative-number caution

At GCSE you will almost always meet DIV and MOD with positive numbers. Different languages handle negatives inconsistently, and AQA does not examine that case — so if a question uses negatives, work from the "how many whole ones fit, what is left over" model rather than assuming a rule.

AQA tip. In a trace table, write DIV and MOD results as whole numbers immediately. Writing 5.5 and then "correcting" it later is where most errors creep in — do the discard at the moment you calculate.

MOD looks like a curiosity until you see what it does. These four applications cover almost every exam use.

1. Testing whether a number is even or odd

IF number MOD 2 = 0 THEN
    OUTPUT 'Even'
ELSE
    OUTPUT 'Odd'
ENDIF

Why it works: every even number divides by 2 with no remainder, so MOD 2 gives 0. Every odd number leaves 1. This is the standard test and appears constantly in trace-table questions.

The same idea generalises: number MOD 3 = 0 tests whether a number is a multiple of 3; number MOD 5 = 0 tests for multiples of 5.

2. Cycling / wrapping around a range

# Move to the next player in a 4-player game, wrapping back to 0
currentPlayer ← (currentPlayer + 1) MOD 4

The result is always 0, 1, 2 or 3 — it can never escape the range. When currentPlayer is 3, (3 + 1) MOD 4 = 0 and it wraps neatly back to the start. This is how turn-taking, circular buffers and clock arithmetic work.

3. Extracting digits from a number

number ← 4729

lastDigit  ← number MOD 10       # 9  — the units digit
remaining  ← number DIV 10       # 472 — everything except the last digit

MOD 10 peels off the last digit; DIV 10 removes it. Repeat the pair in a loop and you can process every digit of a number in turn — the basis of digit-sum algorithms, check-digit validation and manual base conversion.

# Sum the digits of a number
total ← 0
WHILE number > 0
    total  ← total + (number MOD 10)
    number ← number DIV 10
ENDWHILE

Trace with 4729: adds 9, then 2, then 7, then 4 → 22.

4. Converting units

totalSeconds ← 275

minutes ← totalSeconds DIV 60    # 4  — whole minutes
seconds ← totalSeconds MOD 60    # 35 — seconds left over
OUTPUT minutes, ' min ', seconds, ' sec'    # 4 min 35 sec

Check: 4 × 60 + 35 = 275 ✓

The identical pattern converts pence to pounds-and-pence (DIV 100 / MOD 100), hours to days-and-hours (DIV 24 / MOD 24), and inches to feet-and-inches (DIV 12 / MOD 12).

AQA tip. The unit-conversion pattern is the most examined MOD application. Learn the shape: DIV gives the big unit, MOD gives the leftover small unit — and they always use the same divisor.

Programming languages follow the same precedence rules as mathematics:

Priority	Operations
1 (highest)	Brackets ( )
2	Multiplication, division — *, /, DIV, MOD
3 (lowest)	Addition, subtraction — +, −

Operations of equal priority are carried out left to right.

Worked examples

2 + 3 * 4          = 2 + 12   = 14    (not 20 — multiplication first)
(2 + 3) * 4        = 5 * 4    = 20    (brackets force the addition first)
10 - 4 - 3         = 6 - 3    = 3     (left to right, not 10 − 1)
20 / 4 * 2         = 5 * 2    = 10    (left to right, not 20 / 8)
7 + 10 MOD 3       = 7 + 1    = 8     (MOD binds tighter than +)
(7 + 10) MOD 3     = 17 MOD 3 = 2     (brackets change everything)

The MOD one catches people out. MOD has the same priority as multiplication, so it happens before any addition. 7 + 10 MOD 3 is 7 + (10 MOD 3) = 7 + 1 = 8, not 17 MOD 3 = 2.

The practical rule

When in doubt, use brackets. They cost nothing, they remove all ambiguity, and they make your intention clear to an examiner. Compare:

average ← total / count + 1         # ambiguous — probably a bug
average ← total / (count + 1)       # unmistakable

The first divides total by count and then adds 1 — almost certainly not what was intended.

The binary-search midpoint, revisited

mid ← (low + high) DIV 2

The brackets here are essential. Without them, low + high DIV 2 would compute high DIV 2 first and then add low — giving a completely wrong midpoint. With low = 0 and high = 8: the correct version gives (0+8) DIV 2 = 4; the unbracketed version gives 0 + (8 DIV 2) = 4 too, by coincidence. But with low = 4 and high = 8: correct gives (4+8) DIV 2 = 6; unbracketed gives 4 + 4 = 8. Wrong.

AQA tip. In a calculation question, write out the intermediate steps in precedence order. Examiners award method marks for correct working, and showing 7 + (10 MOD 3) = 7 + 1 demonstrates you understood the precedence even if the final arithmetic slips.

Exams test whether you can pick the right division for a scenario. The question to ask is: does a fractional answer make sense here?

Scenario	Correct choice	Why
Average of a set of test scores	Real division /	An average is meaningfully 74.5
How many full boxes of 12 eggs from 100 eggs	DIV	You cannot have 8.33 full boxes → 8
How many eggs are left over	MOD	100 MOD 12 = 4
The midpoint index of an array	DIV	An index must be a whole number
Price per person splitting a bill	Real division /	£10.75 each is perfectly sensible
How many complete minutes in 275 seconds	DIV	Whole minutes → 4
Testing whether a number is even	MOD	You need the remainder, not the quotient
Converting a percentage	Real division /	47/60 = 0.7833… must keep its decimals

The "full boxes" family of questions

This is the most common scenario shape in the whole subtopic:

"Eggs are packed into boxes of 12. A farm collects 100 eggs. How many full boxes can be packed, and how many eggs are left over?"

fullBoxes  ← 100 DIV 12     # 8
eggsLeft   ← 100 MOD 12     # 4

Check: 8 × 12 + 4 = 100 ✓

The word "full" or "complete" in a question is the signal for DIV. The words "left over", "remaining" or "remainder" signal MOD.

A caution about real division and data types

average ← total / count

The result of real division may have a decimal part, so average should be declared as a real, not an integer (3.2.1). Storing it in an integer would truncate the answer — 74.5 becomes 74 — silently losing precision.

AQA tip. Look for the signal words. "How many complete/full…" → DIV. "How many are left over / remaining" → MOD. "What is the average / exact value" → real division. Getting the operation right is usually worth more marks than the arithmetic itself.