Tuesday, October 29, 2013

Factors, Multiples, GCF & LCM: Exercise problem and solution

Factors, Multiples, GCF & LCM: Exercise problem and solution

Problem: 1 The difference between the number factors of 32 & 64 is?
Option: (a) 3 (b) 2 (c) 1 (d) 0 (e) None

Explore: Let us write down all the factors of 32 and 64
32 : 1, 2, 4, 8, 16, 32
64 : 1, 2, 4, 8, 16, 32, 64
:. 64 has 1 more factor than 32.
Answer: (c)

Problem: 2 What is the greatest number that divides 84, 144 or 18 without any remainder?
Option: (a) 18 (b) 12 (c) 24 (d) 6 (e) None

Explore: The question asks us to find the greatest common divider of 84, 144 & 18. First we will break them up into prime factors.
84 = 2x2x3x7
144 = 2x2x2x2x3x3
18 = 2x3x3
:. GCD = 2x3 = 6
Andwer: (d)

Problem: 3 Find the smallest number of oranges that can be distributed completely & equally among 4, 6, 10 or 18 children.
Option: (a) 16 (b) 60 (c) 240 (d) 180 (e) None

Explore: To be able to divide the oranges completely & equally among 4, 6, 10 or 18 children, we need to find a number that is completely divisible by all of 4, 6, 10 or 18. i.e. we have to find the LCM of these numbers.
4 = 2x2
6 = 2x3
10 = 2x5
18 = 2x3x3
:. LCM = 2x2x3x3x5 = 4x9x5 = 180
Answer: (d)

Problem: 4 What is the smallest integer having only three different odd factors?
Option: (a) 75 (b) 90 (c) 105 (d) 120 (e) None

Explore: Let us list the factors of each number & mark the odd ones.
75 : 1, 3, 5, 15, 25, 75 = 6 odd factors.
90 : 1, 2, 3, 5, 6, 9, 10, 15, 18, 30, 45, 90 = 6 odd factors
105 : 1, 3, 5, 7, 15, 21, 35, 105 = 8 odd factors
120 : 1, 2, 3, 4, 5, 6, 8, 10, 12, 15, 20, 24....... more than 3 odd factors.
:. All of the options have more than 3 odd factors.
Answer: (e)

Problem: 5 Find the largest number of apples not exceeding 1000 which can be divided among 6, 15, 20 or24 boys?
Option: (a) 990 (b) 960 (c) 930 (d) 900 (e)None

Explore: LCM of 6, 15, 20 & 24 = ?
6 = 2x3
15 = 3x5
20 = 2x2x5
24 = 2x2x2x3
:. LCM = 2x2x2x3x5 = 120
960 is a multiple of 120. So, that is the required largest number.
Answer: (b)

Problem: 6 A man has 72 green marbles and 108 red marbles. He decides to pack them into packets of the same size, each containing either all red or all green marbles. What is the maximum number of marbles he can put in each packet?

Explore: Since each packet is of the size, we have to find a number that divides both 72 & 108, & since we have been asked to gibe the greatest one, we need the GCF.
GCF of 72 & 108
= 2x2x3x3
=  36
:. Maximum number of marbles that can be put in each packet is 36

Please also check second and third post:

Factors, Multiples, GCF & LCM: Part: 2 Exercise problem and solution

Factors, Multiples, GCF & LCM: Part: 3 Exercise problem and solution

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