Mathematical Expressions: Study Guide (Computer Science Principles 3.3)

Programs compute. A thermostat works out how far the temperature is from the target, a game adds points, a timer converts seconds into minutes and seconds. All of that uses mathematical expressions. This topic covers the arithmetic operators in the exam notation, the order in which an expression is evaluated, and the operator that surprises most students, MOD, which gives the remainder after division. MOD turns out to be one of the most useful ideas in the whole course: it detects even and odd numbers, tests divisibility, extracts digits and makes values “wrap around” like a clock.

Notation. The exam uses its own pseudocode, not a real programming language. The code in this pack is Marlbridge’s rendering of that reference notation, not a copy of the official sheet: ← is assignment, = tests equality, lists start at index 1, DISPLAY shows a value, and braces mark blocks. Use the Exam Reference Sheet from your teacher for the official wording.

Expressions and how they are evaluated

An expression can be a single value, a variable, an operator applied to values, or a procedure call that returns a value. Every expression is evaluated to produce a single value. For example 7 + 2 * 3 evaluates to 13, and count + 1 evaluates to whatever count holds, plus one.

The arithmetic operators available in the exam notation are:

Operator	Meaning	Example	Value
a + b	addition	9 + 4	13
a - b	subtraction	9 - 4	5
a * b	multiplication	9 * 4	36
a / b	division	17 / 5	3.4
a MOD b	remainder when a is divided by b	17 MOD 5	2

Notice that / is ordinary division, so 17 / 5 is 3.4 and not 3. For a MOD b the course assumes a is an integer greater than or equal to 0 and b is an integer greater than 0.

Order of operations. The usual rules from mathematics apply: brackets first, then *, / and MOD, then + and -. The important detail is that MOD has the same precedence as * and /. When operators of the same precedence meet, evaluate them from left to right.

MOD from first principles

Divide 17 by 5. Five goes into 17 three whole times (15), leaving 2 over. That leftover is the remainder, and 17 MOD 5 is 2. In general, a MOD b is the amount left when you take as many whole copies of b out of a as you can.

Some useful facts follow directly:

• If a is smaller than b, nothing whole fits, so a MOD b = a. For example 3 MOD 7 = 3.
• If b divides a exactly, the remainder is 0. For example 12 MOD 4 = 0.
• a MOD b is always between 0 and b − 1, so it can never reach b. That is the “wrap around” property.

Task	Expression	Why it works
Is n even?	n MOD 2 = 0	Even numbers leave remainder 0 when divided by 2
Is n odd?	n MOD 2 = 1	Odd numbers leave remainder 1
Is n divisible by d?	n MOD d = 0	No remainder means d divides n exactly
Last digit of n	n MOD 10	The remainder on division by 10 is the units digit
Hour after adding k hours (24-hour clock)	(h + k) MOD 24	The result wraps back to 0 after 23

Worked example 1: evaluating an expression with MOD

Find the value of 2 + 3 * 4 MOD 5 - 1.

result ← 2 + 3 * 4 MOD 5 - 1
DISPLAY(result)

Output: 3

Step 1. Find the highest-precedence operators. * and MOD have equal precedence and sit next to each other, so work left to right.

Step 2. 3 * 4 = 12. The expression is now 2 + 12 MOD 5 - 1.

Step 3. 12 MOD 5: five goes into 12 twice (10), leaving 2. The expression is now 2 + 2 - 1.

Step 4. + and - also share precedence: 2 + 2 = 4, then 4 - 1 = 3.

Answer: 3.

Check. A common wrong order is to do 4 MOD 5 first (giving 4), then 3 * 4 = 12, so 2 + 12 - 1 = 13. The result changes completely, which shows how much the order matters. Use brackets in your own code when you want to be sure: 2 + ((3 * 4) MOD 5) - 1.

A second expression, mixing division: 20 - 12 / 4 * 2 + 7 MOD 4. Here 12 / 4 = 3, then 3 * 2 = 6, 7 MOD 4 = 3, and finally 20 - 6 + 3 = 17.

DISPLAY(20 - 12 / 4 * 2 + 7 MOD 4)

Output: 17

Worked example 2: splitting minutes and wrapping the clock

A fitness app records a run of 135 minutes. It must display the time as hours and minutes. A second feature takes a start hour on a 24-hour clock and an elapsed number of hours, and displays the finishing hour.

Part A: hours and minutes. The exam notation has no whole-number division, so we use MOD to find the minutes left over and then divide the rest.

total ← 135
minutes ← total MOD 60
hours ← (total - minutes) / 60
DISPLAY(hours)
DISPLAY(minutes)

Output: 2 15

Line	Variable	Calculation	Value
2	minutes	135 MOD 60: 60 fits twice (120), remainder 15	15
3	hours	(135 − 15) / 60 = 120 / 60	2

So 135 minutes is 2 hours 15 minutes. Check: 2 × 60 + 15 = 135 ✓.

Why subtract the remainder before dividing? Because 135 / 60 is 2.25, not 2. Removing the leftover minutes first leaves an exact multiple of 60.

Part B: the clock wrap. A walk starts at 21:00 and lasts 8 hours.

start ← 21
elapsed ← 8
finish ← (start + elapsed) MOD 24
DISPLAY(finish)

Output: 5

21 + 8 = 29. Twenty-four fits once (24), leaving 5, so the walk finishes at 05:00 the next day. Without MOD the program would claim the time was “29:00”. MOD keeps the result between 0 and 23.

Part C: a digit. The last digit of 4,826 is 4826 MOD 10, which is 6. To drop the last digit, subtract it and divide by 10: (4826 - 6) / 10 = 482.

n ← 4826
last ← n MOD 10
rest ← (n - last) / 10
DISPLAY(last)
DISPLAY(rest)

Output: 6 482

Using variables and procedure calls in expressions

Anywhere a value can appear, a variable or procedure call can appear instead. If price holds 12 and qty holds 3, then price * qty + 4 evaluates to 40. When a procedure that returns a value is called inside an expression (Topic 3.12), the call is replaced by its returned value before the rest of the expression is evaluated.

Common misconceptions

• “MOD means divide.” a MOD b is the remainder, not the quotient. 17 MOD 5 is 2, not 3.
• “MOD is evaluated after +.” It shares precedence with * and /, so 10 + 7 MOD 4 is 10 + 3 = 13, not 17 MOD 4 = 1.
• “/ rounds down.” In the exam notation 17 / 5 is 3.4. Use the MOD pattern when you need the whole part.
• “Equal precedence means any order.” Same-precedence operators go left to right, which matters for division: 12 / 4 * 3 is 9, not 1.
• “a MOD b can equal b.” It cannot. The largest possible value is b − 1.
• “n MOD 2 = 1 is true for negative odd numbers too.” The course only defines MOD for a ≥ 0 and b > 0; do not use it outside that range.

Where this leads

MOD patterns appear again in conditionals (Topic 3.6, even or odd branches), in the “divisible by” algorithms of Topic 3.9 and in loops that process every digit or every n-th item (Topic 3.8).

Next steps: try the practice questions, skim the revision notes the night before, and tick off the topic checklist. The next topic is Topic 3.4, Strings.