← IGCSE Computer Science Paper 2

MOD: remainder (8.1c)

Returns the remainder left when a is divided by b. MOD(10, 3) is 1. The result is 0 when b divides a exactly. Example: MOD(130, 60) gives the 10 leftover minutes.

Part of: IGCSE Computer Science 0478 — 8.1c Procedures and Functions — Module 36

Understanding MOD: remainder (8.1c) in IGCSE Computer Science Paper 2

What it does: 130 minutes is 2 hours (120 minutes) with 10 minutes left, and MOD(130, 60) gives that remainder 10. An even number divides by 2 exactly, so MOD(Number, 2) is 0. Odd numbers leave a remainder of 1. 10 divided by 3 is 3 remainder 1. MOD gives the remainder, so the answer is 1. MOD gives the remainder. The whole-number quotient is given by DIV. MOD(130, 60) is 10, the leftover minutes after 2 whole hours. The first argument is the number being divided. MOD needs two arguments, a and b. MOD(17, 5) returns 2. 7 divided by 2 is 3 remainder 1. MOD returns the remainder 1, not the quotient 3.

Common mistakes: The arguments are the wrong way round. MOD(a, b) is the remainder when a is divided by b, so the total minutes must come first: MOD(TotalMins, 60) gives the 10 leftover minutes. The two arguments of MOD must be separated by a comma. It should be MOD(Num, 3), which gives 1.

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Learning

Duo

What does MOD(10, 3) return?

Matches

OUTPUT MOD(10)

↓

MOD needs two arguments, a and b. Only one was given, so it cannot work out a remainder. It should be MOD(10, 3).

Applying

Meanings

Term

MOD(a, b)

Meaning

Returns the remainder left when a is divided by b. MOD(10, 3) is 1. The result is 0 when b divides a exactly. Example: MOD(130, 60) gives the 10 leftover minutes.

Practice

TotalMinutes ← 130

Testing

Gaps

OUTPUT (10, 3)

Check Answer

Spots

3DECLARE Left : INTEGER

4Left ← MOD(60, TotalMins)

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Quick Questions

1. What does MOD(10, 3) return?

A. 1 ✓

B. 3

10 divided by 3 is 3 remainder 1. MOD gives the remainder, so the answer is 1.

2. Which statement describes MOD(a, b)?

A. It returns the whole-number part of a divided by b

B. It returns the remainder left when a is divided by b ✓

MOD gives the remainder. The whole-number quotient is given by DIV.

3. What does MOD(12, 4) return?

A. 3

B. 0 ✓

4 divides 12 exactly (3 times) with nothing left over, so the remainder is 0. The 3 would be the result of DIV(12, 4).

4. A student wants the leftover minutes after converting 130 minutes into hours. Which expression is correct?

A. MOD(130, 60) ✓

B. MOD(60, 130)

MOD(130, 60) is 10, the leftover minutes after 2 whole hours. The first argument is the number being divided.

5. If MOD(a, b) gives 0, what do we know?

A. b divides a exactly ✓

B. a is smaller than b

A remainder of 0 means b goes into a an exact number of times, with nothing left over.

Match the Pairs

MOD(10, 3) → Gives 1, because 3 goes into 10 three times (9) and 1 is left over

MOD(12, 4) → Gives 0, because 4 divides 12 exactly with nothing left over

Find the leftover minutes after taking 130 minutes and removing whole hours → leftover ← MOD(130, 60)

Check whether a whole number is even by testing if the remainder after dividing by 2 is 0 → IF MOD(Number, 2) = 0 THEN

OUTPUT MOD(10) → MOD needs two arguments, a and b. Only one was given, so it cannot work out a remainder. It should be MOD(10, 3).

OUTPUT MOD(3, 10) → The arguments are in the wrong order. To get the remainder of 10 divided by 3 it should be MOD(10, 3), which gives 1. MOD(3, 10) gives 3.

Key Term

MOD(a, b)

Returns the remainder left when a is divided by b. MOD(10, 3) is 1. The result is 0 when b divides a exactly. Example: MOD(130, 60) gives the 10 leftover minutes.

Fill in the Blanks

1. OUTPUT MOD(10, 3)

Hint: Type the library routine that returns the remainder after division.

2. Minutes ← MOD(130, 60)

Hint: Type the divisor: the number of minutes in one hour.

3. LeftoverMinutes ← MOD(130, 60) OUTPUT LeftoverMinutes

Hint: Type the variable that stores the remainder so it can be output.

4. IF MOD(Number, 2) = 0 THEN OUTPUT "Even" ENDIF

Hint: Type the value MOD returns when 2 divides the number exactly.

5. Hours ← 130 DIV 60 Minutes ← MOD(130, 60) OUTPUT Hours, Minutes

Hint: Type the routine that gives the leftover minutes after the whole hours.

Spot the Error

Question 1

A DECLARE TotalMins : INTEGER

B TotalMins ← 130

C DECLARE Left : INTEGER

D Left ← MOD(60, TotalMins)

E OUTPUT Left

Line D — The arguments are the wrong way round. MOD(a, b) is the remainder when a is divided by b, so the total minutes must come first: MOD(TotalMins, 60) gives the 10 leftover minutes.

Question 2

A Num ← 10

B Rem ← MOD(Num 3)

C OUTPUT Rem

Line B — The two arguments of MOD must be separated by a comma. It should be MOD(Num, 3), which gives 1.

Question 3

A DECLARE Number : INTEGER

B Number ← 12

C IF MOD(Number, 2) = 1 THEN

D OUTPUT "Even"

E ENDIF

Line C — An even number divides exactly by 2, so the remainder is 0, not 1. The condition should be MOD(Number, 2) = 0.

Question 4

A TotalMins ← 130

B Mins ← TotalMins DIV 60

C OUTPUT Mins

Line B — DIV gives the whole number of hours (2), not the leftover minutes. The leftover minutes need the remainder, so use MOD(TotalMins, 60), which gives 10.

Question 5

A A ← 17

B B ← 5

C R = MOD(A, B)

D OUTPUT R

Line C — Assignment in IGCSE pseudocode uses the ← arrow, not =. It should be R ← MOD(A, B), which stores 2.

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