Practice Test - Number Theory(Part1) Answers: Name:
Class:
( Prime factorization, Least Common Multiple(LCM), Greatest Common Factor(GCF),
Divisibility, factors, prime, composite)
1) List the factors of 56 (1 mark)
1, 2, 4, 7, 8, 14, 28, 56
(1 x 56, 2 x 28, 4 x 14, 7 x 8 )
2) State whether the following numbers are divisible by 2, 3, 4, 5, 6, 9, 10, and if
so explain why in reference to the divisibility rules. (3 marks)
(i)
208 : 2 (the number is even), & 4 (the last two digits is divisible by 4)
Answer: 2 & 4
(ii)
13470 : 2(even), 3(sum of digits = 15, therefore it is divisible by 3),
therefore if divisible by 2, & 3 then it is also by 6. It is divisible by 5 &
10 as the last digit is 0.
Answer: 2, 3, 5, 6, 10
(iii)
2325616: The digits 2+3+2 + 5 +6+1+6 =25 not divisible by 3, the
number is even therefore divisible by 2, and the last two digits is 16
and is therefore divisible by 4.
Answer: 2 & 4
3) Use the divisibility rules to determine whether the following numbers are
composite, prime or neither. Explain.
(2 marks)
(i) 67 - prime(not divisible)
(ii) 441 - (divisible by 3 & 9 as sum of digits = 9 which is divisible by 3 &
9).Therefore number is composite.
4) How can you determine whether a number is divisible by 15? Explain.
(2 mark)
Factors of 15 are 3 & 5. If a number is divisible by both 3 & 5 then it should
also be divisible by 15 e.g 45 is divisible by 15, 3 and 5.
5) Write the prime factors of the following numbers: (2 marks)
(i) 28
(ii)
42
(iii)
2
14
2
2
7
21
3
5
7
65
13
So the answer is 2 x 2 x 7
So answer = 2 x 3 x 7
So answer = 5 x 13
6) Find the GCF of the following set of numbers: (2 marks)
(i)
42 , 90
(ii) 12, 18 , 32
3
42 90
2
14 30
7
2 l 12 18 30
3
6
9
15
2
3
5
15
So the GCF = 3 x 2 = 6
So the GCF = 2 x 3 = 6
7) Write 2 numbers that have a GCF of 30. Explain. (2 mark)
So 30 = 2 x 3 x 5. Therefore working backwards - if we place 2, 3, & 5 on
the outside of the ladder method & start with 1 & 2 on the inside we can then
find the 2 numbers with a GCF of 30.
Sample Answer:
2
1 2
(2 x 1=2 , 2 x 2 = 4)
3
2
4
(3 x 2 = 6, 3 x 4 =12)
5
6
12
(5 x 6 = 30 , 5 x 12 = 60 )
30 60
Sample Answer = 30 & 60
8) Mr. Olivas has 27 red pencils, 51 blue pencils and 57 yellow pencils to sell at
the SAS fair. He wants to put the pencils into boxes so that there is the same
number of colored pencils in each box without mixing the pencils. Show your
working.
(4 marks)
(i)
What is the greatest number of colored pencils he can place into each
box without mixing the colors? Answer 3 l 27r 51b 57y
9r 17b 19y
Answer: GCF = 3 pencils
(ii) How many boxes of each colored pencil can he sell without mixing them? .
Therefore he would have 9 boxes of red pencils, 17 Boxes blue pencils and 19 of
yellow pencils.
9) Find the LCM for each set of numbers: Show your working. (3 marks)
(i)
18, 24
(ii)
14, 18
2
18
24
2 14
18
3
9
12
7
9
3
4
So the LCM = 2 x 3 x 3 x 4 = 72
LCM = 2 x 7 x 9 = 126
10) The LCM of two numbers is 2 3 x 3 2. Find the pair of numbers that fit this
description. Working backwards - Using any combination of 2 x 2 x 2 x 3 x 3
= 2 3 x 3 2 I’ll use: (working backwards)
2
3
2
2
3
-> Using the ladder method 2 x 3 x 2 x 2 x 3 = LCM
Using the ladder method all the numbers on the outside are multiplied
together to get the LCM. So…
2
3
2
4
6
( 2 x 2 = 4, 3 x 2 = 6 )
2
3
2
3
12
18 (4 x 3 = 12, 6 x 3 = 18)
2
4
6
2
3
2
24
36
3
12
18
2
4
6
2
3
(12 x 2 =24, 18 x 2 =36 )
Sample answer = 24 & 36
Or Given 2 3 x 3 2 = 72. Find the Factors of 72: 1, 2, 3, 4, 6, 8, 9, 12, 24, 36, 72
Then we determine which 2 factors will give us the LCM of 72 given the
numbers 2 3 x 3 2
11)
One ribbon is 80cm long, and another ribbon is 32 cm long. The ribbons
are to be cut into pieces of equal length. What is the longest length possible
for the pieces, if there is to be no left over ribbon? Show your working. (3
marks).
We know that the string can’t be longer than the smaller of the two
measurements 32 cm. So we must be looking for the GCF.
2 I 32
80
2 l 16
40
2 l 8
20
2 l 4
10
2
5
Answer GCF = 2 x 2 x 2 x 2 = 16cm