MOD: remainder (8.1c)
Name:
Date:
Class:
A
Quick Questions
Shade in the A or B circle for each question.
1. What does MOD(10, 3) return?
2. Which statement describes MOD(a, b)?
3. What does MOD(12, 4) return?
4. A student wants the leftover minutes after converting 130 minutes into hours. Which expression is correct?
5. If MOD(a, b) gives 0, what do we know?
B
Match the Pairs
Draw a line connecting each item on the left to its match on the right.
| Left column | Right column | |
|---|---|---|
| 1MOD(10, 3) | AGives 0, because 4 divides 12 exactly with nothing left over | |
| 2MOD(12, 4) | BGives 1, because 3 goes into 10 three times (9) and 1 is left over | |
| 3Find the leftover minutes after taking 130 minutes and removing whole hours | CIF MOD(Number, 2) = 0 THEN | |
| 4Check whether a whole number is even by testing if the remainder after dividing by 2 is 0 | DMOD needs two arguments, a and b. Only one was given, so it cannot work out a remainder. It should be MOD(10, 3). | |
| 5OUTPUT MOD(10) | Eleftover ← MOD(130, 60) | |
| 6OUTPUT MOD(3, 10) | FThe 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. |
C
Definitions
MOD(a, b)
D
Fill in the Blanks
Complete each line of code by filling in the missing parts.
1.
OUTPUT ___(10, 3)
2.
Minutes ← MOD(130, __)
3.
LeftoverMinutes ← MOD(130, 60)
OUTPUT _______________
4.
IF MOD(Number, 2) = _ THEN
OUTPUT "Even"
ENDIF
5.
Hours ← 130 DIV 60
Minutes ← ___(130, 60)
OUTPUT Hours, Minutes
E
Spot the Error
Shade in the letter of the line that contains an error.
Question 1
A
DECLARE TotalMins : INTEGER
B
TotalMins ← 130
C
DECLARE Left : INTEGER
D
Left ← MOD(60, TotalMins)
E
OUTPUT Left
Question 2
A
Num ← 10
B
Rem ← MOD(Num 3)
C
OUTPUT Rem
Question 3
A
DECLARE Number : INTEGER
B
Number ← 12
C
IF MOD(Number, 2) = 1 THEN
D
OUTPUT "Even"
E
ENDIF
Question 4
A
TotalMins ← 130
B
Mins ← TotalMins DIV 60
C
OUTPUT Mins
Question 5
A
A ← 17
B
B ← 5
C
R = MOD(A, B)
D
OUTPUT R
Answer Key
MOD: remainder (8.1c)
Section A: Quick Questions
Q1
A — 1
Q2
B — It returns the remainder left when a is divided by b
Q3
B — 0
Q4
A — MOD(130, 60)
Q5
A — b divides a exactly
Section B: Match the Pairs
Q1
MOD(10, 3) → Gives 1, because 3 goes into 10 three times (9) and 1 is left over
Q2
MOD(12, 4) → Gives 0, because 4 divides 12 exactly with nothing left over
Q3
Find the leftover minutes after taking 130 minutes and removing whole hours → leftover ← MOD(130, 60)
Q4
Check whether a whole number is even by testing if the remainder after dividing by 2 is 0 → IF MOD(Number, 2) = 0 THEN
Q5
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).
Q6
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.
Section C: Definitions
Q1
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.Section D: Fill in the Blanks
Q1
OUTPUT MOD(10, 3)Q2
Minutes ← MOD(130, 60)Q3
LeftoverMinutes ← MOD(130, 60)
OUTPUT LeftoverMinutesQ4
IF MOD(Number, 2) = 0 THEN
OUTPUT "Even"
ENDIFQ5
Hours ← 130 DIV 60
Minutes ← MOD(130, 60)
OUTPUT Hours, MinutesSection E: Spot the Error
Q1
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.
Q2
B — The two arguments of MOD must be separated by a comma. It should be MOD(Num, 3), which gives 1.
Q3
C — An even number divides exactly by 2, so the remainder is 0, not 1. The condition should be MOD(Number, 2) = 0.
Q4
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.
Q5
C — Assignment in IGCSE pseudocode uses the ← arrow, not =. It should be R ← MOD(A, B), which stores 2.