Chapter 3 Numeric And Algebraic Expressions
Section 3.2: Find Greatest Common Factor And Least Common Multiple
Page 123 Exercise 1 Answer
Since Mark sets the table every two days and dries the dishes every three days, he will do both chores on the days which are multiples of both 2 and 3. The least common multiple of 2 and 3 is 6.
All multiples of 6 are also multiples of 2 and 3, so Mark will do both chores on days that are multiples of 6.
Result
He will do both chores on days which are multiples of both 2 and 3 so Mark will do both chores on days that are multiples of 6.
Page 123 Exercise 1 Answer
We know that Mark sets the dinner table ever 2 days and he sets the table of Day 2. He will then also set the table on Day 4, Day 6, Day 8, Day 10, ……
We know that Mark dries the dishes every 3 days and that he dries the dishes on Day 3. He also then dries on Day 6, Day 9, Day 12, ….
The table can then be completed as:
From the table, he will perform both chores on Day 6.
Read And Learn More: enVisionmath 2.0 Grade 6 Volume 1 Solutions
Result
Day 6
If every two days Marks sets the table and every 3 days he dries the dishes, he would do both chores on the day which is the least common multiple of 2 and 3.
Since both 2 and 3 are prime numbers, than their least common mulitple is their product.
2 × 3 = 6
Result
Mark would first perform both chores on the same day on Day 6.
Page 124 Exercise 1 Answer
To find the prime factorization of 56 divide the number by primes, start with two. When the remainder is no longer divisible by two, try with the next prime.
56 = 2 × 28 = 2 × 2 × 14 = 2 × 2 × 2 × 7
or
56 = 23 × 7
A composite number is by definition a number with more than two factors. Since the number in question is greater than 2 and it has 2 as a factor, it must have more than 2 factors since it at least has 1, 2, and the number itself as factors. It must then be a composite number.
Result
The prime factorization of 56 is 2 × 2 × 2 × 7 or 23 × 7.
Page 125 Exercise 2 Answer
Keesha can add an equal number of beads to each bag since there are 6 bags, 24 beads, and 24 is divisible by 6.
24 = 4 × 6
She adds 4 beads to each bag.
Result
Keesha can add the beads and have no supplies left over.
Page 125 Exercise 3 Answer
First, we find the GCF of 12 and 36.
12 = 2 × 6 = 2 × 2 × 3
36 = 2 × 18 = 2 × 2 × 9 = 2 × 2 × 3 × 3
The common factors are 2, 2, and 3, so the GCF is
2 × 2 × 3 = 12.
Next we write each number as a product using the GCF as a factor.
12 + 36 = 12 + 12 × 3 = 12(1 + 3) = 12(4) = 48
Result
The sum of 12 and 36 is 48.
Page 126 Exercise 4 Answer
To answer the question we need to find the least common multiple of 12 and 10.
12 = 2 × 6 = 2 × 2 × 3
10 = 2 × 5
The LCM of 12 and 10 is the product 2 × 2 × 3 × 5 or 60.
Since bottled water is sold in cases of 12 bottles and juice in boxes of 10, write 60 as a product of two number so that one of them is 12, and write 60 as s product of two number so that one of them is 10.
60 = 12 × 5
60 = 10 × 6
Result
Grant should buy 5 cases of bottled water and 6 boxes of juice.
Page 127 Exercise 1 Answer
Whole numbers greater than 1 are either prime or composite numbers. A composite number can be written as a product of its prime factors, called its prime factorization.
To find the prime factorization of a number, write the number as products of its factors. We start with the least prime factor and continue using prime factors until all factors are prime numbers.
For example, to find the prime factorization of 630 we first divide the number by 2 beacuse it is the least prime number which divides 630.
630 = 2 ⋅ 315
Than, we are further looking for the prime factorization of 315.
2 doesn’t divide 315 so we are looking at the next least prime number which is 3.
3 divides 315 so we use 3.
315 = 3 ⋅ 105
3 divides 105 so we use 3 again.
105 = 3 ⋅ 35
3 doesn’t divide 35 so we look at the next least prime number which is 5. 5 divides 35 so we use 5.
35 = 5 ⋅ 7
Since both factors in 5⋅7 are prime numbers we stop here.
630 = 2 ⋅ 315 = 2 ⋅ 3 ⋅ 105 = 2 ⋅ 3 ⋅ 3 ⋅ 35 = 2 ⋅ 3 ⋅ 3 ⋅ 5 ⋅ 7
The prime factorization of 630 is 2 ⋅ 3 ⋅ 3 ⋅ 5 ⋅ 7.
To find the greatest common factor and the least common multiple of two numbers first we need to find their prime factorizations.
For example, let’s find the greatest common factor and least common multiple of 42 and 60.
First, we find their prime factorizations.
42 = 2 ⋅ 21 = 2 ⋅ 3 ⋅ 7
60 = 2 ⋅ 30 = 2 ⋅ 2 ⋅ 15 = 2 ⋅ 2 ⋅ 3 ⋅ 5
Their greatest common factor is found by multiplying all their common prime factors which are, in this case, 2 and 3. The greatest common factor of 42 and 60 is 6. Their least common multiple is found by listing the greatest number of times each factor appeared in either prime factorization. We then multiply these factors to find the least common multiple.
22 ⋅ 3 ⋅ 5 ⋅ 7
The least common multiple of 42 and 60 is 420.
Result
To find the prime factorization of a number, write the number as products of its factors until all factors are prime numbers. To find the GCF and the LCM of two numbers, first we need to find their prime factorizations. Their GCF is found by multiplying all their common prime factors. Their LCM is found by listing the greatest number of times each factor appeared in either prime factorization.
Page 127 Exercise 2 Answer
In Example 1, we saw that there were two ways to find the prime factorization of a number.
The first way is to write its factors as a product by starting with the least prime factor and continuing until all the factors are prime. The second way is to make a factor tree by writing the number as the product of two factors and then writing each composite factor as a product of two factors until all of the factors are prime factors.
Result
The first way is to write its factors as a product by starting with the least prime factor and continuing until all the factors are prime. The second way is to make a factor tree by writing the number as the product of two factors and then writing each composite factor as a product of two factors until all of the factors are prime factors.
Page 127 Exercise 3 Answer
The greatest common factor of two numbers is the greatest positive number that is a factor of both numbers. Prime numbers, by definition, have only two factors, one and themselves. When looking for the GCF of two prime numbers we are looking for the greatest number which is a factor of both of them. The only factor they have in common is 1 so the GCF must be 1.
Result
The only common factor of two prime number is 1 so the GCF must be 1.
Page 127 Exercise 4 Answer
In Example 4, we had to find the LCM of 6 and 8 since the juice bottles came 6 per pack and the applesauce came 8 per pack. We found that the LCM was 24.
If the juice bottles come in packages of 3 instead of 6, then we need to find the LCM of 3 and 8. The prime factorization of the two numbers is:
3 = 3
8 = 2 × 2 × 2
The factor of 3 appears 1 time in the factorizations and the factor of 2 appears 3 times. The LCM is then:
3 × 2 × 2 × 2 = 24
This is the same LCM we found in Example 4 so the LCM did not change.
Result
The LCM did not change.
Page 127 Exercise 5 Answer
You can’t find the LCM of any two numbers by simply multiplying them together. This works if and only if those two numbers don’t have any common factors in their prime factorization.
For example, we can find the LCM of 5 and 23 by simply mulitplying them since they are both prime, however we can’t do the same with any number, for example we can’t find the LCM of 12 and 48 by multiplying them because they have common prime factors.
12 = 2 × 6 = 2 × 2 × 3,
48 = 2 × 24 = 2 × 2 × 12 = 2 × 2 × 2 × 6 = 2 × 2 × 2 × 2 × 3.
Result
You can’t find the LCM of any two numbers by simply multiplying them together. This works if and only if those two numbers don’t have any common factors in their prime factorization. For example, we can find the LCM of 5 and 23 by simply multiplying them since they are prime, however, we can’t do the same when finding the LCM of 12 and 48.
Page 127 Exercise 6 Answer
To find the prime factorization of 33 we write 33 as a product of prime numbers. We start with the least prime number 2, but since 2 doesn’t divide 33 we can’t use it in the product. We try with the next least prime number which is 3. 3 divides 33 so we can use it.
33 = 3 ⋅ 11
Since 11 is also a prime number we are done. The solution is 3 ⋅ 11.
Result
3 ⋅ 11
Page 127 Exercise 7 Answer
To find the prime factorization of 32 we write 32 as a product of prime numbers. We start with the least prime number 2, since it divides 32 we can use it in the product.
32 = 2 ⋅ 16 = 2 ⋅ 2 ⋅ 8 = 2 ⋅ 2 ⋅ 2 ⋅ 4 = 2 ⋅ 2 ⋅ 2 ⋅ 2 ⋅ 2
Prime factorization of 32 is 2 ⋅ 2 ⋅ 2 ⋅ 2 ⋅ 2.
Result
2 ⋅ 2 ⋅ 2 ⋅ 2 ⋅ 2
Page 127 Exercise 8 Answer
19 is a prime number.
Result
Prime.
Page 127 Exercise 9 Answer
To find the greatest common factor we first find prime factorizations of both numbers.
18 = 2 ⋅ 9 = 2 ⋅ 3 ⋅ 3 = 2 ⋅ 3 ⋅ 3
36 = 2 ⋅ 18 = 2 ⋅ 2 ⋅ 9 = 2 ⋅ 2 ⋅ 3 ⋅ 3 = 2 ⋅ 2 ⋅ 3 ⋅ 3
The GCF is the product of prime numbers which appear in both prime factorizations.
2 ⋅ 3 ⋅ 3 = 18
Result
18
Page 127 Exercise 10 Answer
To find the greatest common factor we first find prime factorizations of both numbers.
22 = 2 ⋅ 11 = 2 ⋅ 11
55 = 5 ⋅ 11 = 5 ⋅ 11
The GCF is the product of prime numbers which appear in both prime factorizations, but in this case it is only one number 11, which is then the GCF of 22 and 55.
Result
11
Page 127 Exercise 11 Answer
To find the greatest common factor we first find prime factorizations of both numbers.
100 = 2 ⋅ 50 = 2 ⋅ 2 ⋅ 25 = 2 ⋅ 2 ⋅ 5 ⋅ 5 = 2 ⋅ 2 ⋅ 5 ⋅ 5
48 = 2 ⋅ 24 = 2 ⋅ 2 ⋅ 12 = 2 ⋅ 2 ⋅ 2 ⋅ 6 = 2 ⋅ 2 ⋅ 2 ⋅ 2 ⋅ 3 = 2 ⋅ 2 ⋅ 2 ⋅ 2 ⋅ 3
The GCF is the product of prime numbers which appear in both prime factorizations.
2 ⋅ 2 = 4
Result
4
Page 127 Exercise 12 Answer
Both numbers, 2 and 5 are prime, so their least common multiple is simply their product.
2 ⋅ 5 = 10
Result
10
Page 127 Exercise 13 Answer
To find the least common multiple of 8 and 12 we first need to find their prime factorizations.
8 = 2 ⋅ 4 = 2 ⋅ 2 ⋅ 2
12 = 2 ⋅ 6 = 2 ⋅ 2 ⋅ 3
We then list the greatest number of times each factor appeared in either prime factorization.
2,2,2,3
Now, we multiply these factors as the last step.
2 ⋅ 2 ⋅ 2 ⋅ 3 = 24
Result
24
Page 127 Exercise 14 Answer
To find the least common multiple of 8 and 12 we first need to find their prime factorizations.
8 = 2 ⋅ 4 = 2 ⋅ 2 ⋅ 2
10 = 2 ⋅ 5
We then list the greatest number of times each factor appeared in either prime factorization.
2,2,2,5
Now, we multiply these factors as the last step.
2 ⋅ 2 ⋅ 2 ⋅ 5 = 40
Result
40
Page 128 Exercise 15 Answer
To find the prime factorization of 27 start dividing by prime numbers, start with 3.
27 = 3 × 9 = 3 × 3 × 3
Result
The prime factorization of 27 is 3 × 3 × 3.
Page 128 Exercise 16 Answer
To find the prime factorization of 30 start dividing by primes, start with 2.
30 = 2 × 15 = 2 × 3 × 5
Result
The prime factorization of 30 is 2 × 3 × 5.
Page 128 Exercise 17 Answer
Prime factorization of 26 is found by writing 26 as a product of prime numbers. When looking for prime factors, we start with the least prime number, which is 2. Since 2 divides 26, we can write 26 = 2 ⋅ 13 . 13 is a prime number so we stop here, because we have found the prime factors whose product give 26.
Result
Prime factorization of 26 is 2 ⋅ 13.
Page 128 Exercise 18 Answer
Number 47 is prime.
Result
Prime.
Page 128 Exercise 19 Answer
To find the greatest common factor of 21 and 49, we must first find prime factorizations of 21 and 49.
21 = 3 ⋅ 7
49 = 7 ⋅ 7
The greatest common factor is the product of prime numbers which are common to both 21 and 49. In this case, the GCF is 7.
Result
The GCF is 7.
Page 128 Exercise 20 Answer
To find the greatest common factor of 8 and 52, we must first find prime factorizations of 8 and 52.
8 = 2 ⋅ 2 ⋅ 2
52 = 2 ⋅ 26 = 2 ⋅ 2 ⋅ 13
The greatest common factor is the product of prime numbers which are common to both 8 and 52. In this case, the GCF is
2 ⋅ 2 = 4.
Result
The GCF is 4.
Page 128 Exercise 21 Answer
To find the greatest common factor of 32 and 81, we must first find prime factorizations of 32 and 81.
32 = 2 ⋅ 16 = 2 ⋅ 2 ⋅ 8 = 2 ⋅ 2 ⋅ 2 ⋅ 4 = 2 ⋅ 2 ⋅ 2 ⋅ 2 ⋅ 2
81 = 3 ⋅ 27 = 3 ⋅ 3 ⋅ 9 = 3 ⋅ 3 ⋅ 3 ⋅ 3
The greatest common factor is the product of prime numbers which are common to both 32 and 81, however, in the case of 32 and 81, they have no common prime factors. The only number which divides both 32 and 81 is 1 so the GCF is 1.
Result
The GCF is 1.
Page 128 Exercise 22 Answer
Using the GCF and the Distributive Property we need to find the sum 30+66.
First, we find prime factorizations of 30 and 66.
30 = 2 ⋅ 15 = 2 ⋅ 3 ⋅ 5
66 = 2 ⋅ 33 = 2 ⋅ 3 ⋅ 11
Then, we write the sum 30+60 as 2 ⋅ 3 ⋅ 5 + 2 ⋅ 3 ⋅ 11. The GCF is 2 ⋅ 3 = 6, the number which divides both 30 and 66. We can extract it and write a product of the GCF and the sum to simplify calculations.
30 + 60 = 2 ⋅ 3 ⋅ 5 + 2 ⋅ 3 ⋅ 11 = 2 ⋅ 3 ⋅ (5 + 11) = 6 ⋅ 16 = 96
Result
The sum is 96.
Page 128 Exercise 23 Answer
Using the GCF and the Distributive Property we need to find the sum 34 + 51.
First, we find prime factorizations of 34 and 51.
34 = 2 ⋅ 17
51 = 3 ⋅ 17
Then, we write the sum 34 + 51 as 2 ⋅ 17 + 3 ⋅ 17. The GCF is 17, the number which divides both 34 and 51. We can extract it and write a product of the GCF and the sum to simplify calculations.
34 + 51 = 2 ⋅ 17 + 3 ⋅ 17 = 17 ⋅ (2 + 3) = 17 ⋅ 5 = 85
Result
The sum is 85.
Page 128 Exercise 24 Answer
Using the GCF and the Distributive Property we need to find the sum 15 + 36.
First, we find prime factorizations of 15 and 36.
15 = 3 ⋅ 5
36 = 2 ⋅ 18 = 2 ⋅ 2 ⋅ 9 = 2 ⋅ 2 ⋅ 3 ⋅ 3
Then, we write the sum 15 + 36 as 3 ⋅ 5 + 2 ⋅ 2 ⋅ 3 ⋅ 3. The GCF is 3, the number which divides both 15 and 36. We can extract it and write a product of the GCF and the sum to simplify calculations.
15 + 36 = 3 ⋅ 5 + 2 ⋅ 2 ⋅ 3 ⋅ 3 = 3 ⋅ (5 + 2 ⋅ 2 ⋅ 3) = 3 ⋅ (5 + 12) = 3 ⋅ 17 = 51
Result
The sum is 51.
Page 128 Exercise 25 Answer
To find the least common multiple we must first find prime factorizations of 12 and 11.
12 = 2 ⋅ 6 = 2 ⋅ 2 ⋅ 3
11 is a prime number.
They don’t have any common factors, so to find the LCM we simply multiply 12 and 11.
12 ⋅ 11 = 132
Result
The LCM of 12 and 11 is 132.
Page 128 Exercise 26 Answer
To find the least common multiple of 4 and 12 we must first find prime factorizations of 4 and 11.
4 = 2 ⋅ 2
12 = 2 ⋅ 6 = 2 ⋅ 2 ⋅ 3
We can see that 12 is a multiple of 4. Since, every number is its least multiple, the least multiple of 12 is 12.
The LCM of 4 and 12 is 12.
Result
The LCM of 4 and 12 is 12.
Page 128 Exercise 27 Answer
To find the least common multiple of 5 and 8 we must first find prime factorizations of 5 and 8.
5 is a prime number.
8 = 2 ⋅ 4 = 2 ⋅ 2 ⋅ 2
Since 5 is a prime number and doesn’t divide 8, then to find the LCM we simply multiply 5 and 8.
5 ⋅ 8 = 40
Result
The LCM of 5 and 8 is 40.
Page 128 Exercise 28 Answer
Both Gabrielle and John got the same answer of 2 × 2 × 2 × 2 × 2 × 2. Neither student made a mistake in their work. We can verify that their answer is correct by multiplying it out and showing it equals 64:
2 × 2 × 2 × 2 × 2 × 2 = 4 × 2 × 2 × 2 × 2
= 8 × 2 × 2 × 2
= 16 × 2 × 2
= 32 × 2
= 64
Result
Neither student made a mistake in their work.
Page 128 Exercise 29 Answer
To celebrate its grand opening, a store is giving customers gift certificates. Every 8th customer gets a $50 gift certificate, and every 6th customer gets a $10 gift certificate. A $50 gift certificate will get 8th, 16th, 24th, … customer. We notice that these are multiples of 8. The same for a $10 gift certificate which will get 6th, 12th, 18th, … customer, and these are the multiples of 6. When looking for the first person who will get two gift certificates we are actually looking for the least common multiple of 8 and 6.
First, we find prime factorizations of 8 and 6.
8 = 2 ⋅ 4 = 2 ⋅ 2 ⋅ 2 = 23
6 = 2 ⋅ 3
The least common multiple is the product of multiplying the highest power of each prime number when looking at prime factorizations together. The LCM of 8 and 6 is 23 ⋅ 3 = 8 ⋅ 3 = 24.
The first customer to get two gift certificates is the 24th customer who comes to the store.
Result
The 24th customer is the first customer that will recieve two gift certificates.
Page 129 Exercise 3a Answer
The Venn diagram shows the factors of 24 and 40. The left circle contains factors of 24, which are 1,2,3,4,6,8,12 and 24. The right circle contains factors of 40, which are 1,2,4,5,8,10,20 and 40. Factors in the region which is common to both circles, their intersection, contains common factors – the numbers which divide both 24 and 40. They are 1,2,4 and 8.
Result
The left circle contains the factors of 24, the right circle contains the factors of 40, and the intersection of the circles contains the factors common to 24 and 40.
Page 129 Exercise 30b Answer
The Venn diagram shows the factors of 24 and 40 where the intersection of circles are the common factors. The intersection contains the factors: 1,2,4,8. Since 8 is the greatest number, it is the greatest common factor of 24 and 40.
Result
8 since the greatest number in the intersection is 8.
Page 129 Exercise 31 Answer
If we have 50 blueberry scones and 75 cranberry scones, how many identical bags can we make? Each bag should have an equal number of blueberry scones and an equal number of cranberry scones. What is the greatest number of bags we can fill?
First, we find the prime factorization of 50 and 75.
50 = 2 ⋅ 25 = 2 ⋅ 5 ⋅ 5
75 = 3 ⋅ 25 = 3 ⋅ 5 ⋅ 5
If we put 2 blueberry scones and 3 cranberry scones in each bag, we can make 25 identical bags. However, if we put 1 blueberry scones and 1 cranberry scones in each bag, we can fill 50 bags.
Result
50 bags if we put 1 blueberry scone and 1 cranberry scone in each bag.
Page 129 Exercise 32 Answer
The prime factorization of A and B are shown. Find the value n that needs to be listed as a prime factor of B so that the greatest common factor of A and B is 9.
Prime factorization of A: 3 ⋅ 3 ⋅ 3.
Prime factorization of B: 2 ⋅ 2 ⋅ 3 ⋅ n.
The greatest common factor of A and B is 9, which means that 9 is the greatest number which divides both A and B. 9 already divides A, but it doesn’t divide B. If we put n = 3, then 9 will divide B.
Also, when looking for the greatest common factor, we usually first find the prime factorization, which is here given, and than we take numbers which are common to both numbers. In this case what is common to A and B is 3 and possibly n, it depends on what we decide n is.
Prime factorization of A: 3 ⋅ 3 ⋅ 3.
Prime factorization of B: 2 ⋅ 2 ⋅ 3 ⋅ n.
Since the prime factorization of A contains only number 3, if we want n to be common to both, we must put n = 3.
Result
n = 3
Page 129 Exercise 33 Answer
Gena has 28 trading cards, Sam has 91 trading cards, and Tiffany has 49 trading cards. Using the GCF and the Distributive Property we have to find the total number of trading cards they all have together.
First, we find prime factorizations of 28,91 and 49.
28 = 2 ⋅ 14 = 2 ⋅ 2 ⋅ 7
91 = 7 ⋅ 13
49 = 7 ⋅ 7
28 + 91 + 49 = 2 ⋅ 2 ⋅ 7 + 7 ⋅ 13 + 7 ⋅ 7 = 7 ⋅ (2 ⋅ 2 + 13 + 7) = 7 ⋅ 24 = 168
Gena, Sam and Tiffany together have 168 trading cards.
Result
Gena, Sam and Tiffany together have 168 trading cards.
Page 129 Exercise 34 Answer
One of the species emerges every 13 years, and the other every 17 years.
Both 13 and 17 are prime numbers and the GCF of two prime numbers is always 1, so the GCF of the years is 1.
Result
The GCF of the years is 1.
Page 129 Exercise 35 Answer
People are waiting in line for a theater premiere. Every 5th person in line will receive a free theater ticket. Every 6th person will receive a gift card for $40.
The first person to receive both a free theater ticket and a gift card is the one whose number in the line is the least common multiple of 5 and 6.
The prime factorization of 5 is 5 and the prime factorization of 6 is 2 × 3. The LCM of 5 and 6 is then 5 × 2 × 3 = 30.
The 30th person will receive both prizes.
Result
The 30th person will recieve both prizes.
Page 129 Exercise 36 Answer
Group A plants trees in clusters of 3, Group B plants the trees in clusters of 10, and both groups plant the same number of trees. We need to find the least number of clusters that Group B plants.
Since both groups planted the same number of trees, the number of planted trees must be the lowest common multiple of 3 and 10. The prime factorization of 3 is 3 and the prime factorization of 10 is 10 = 2 × 5. The LCM of 3 and 10 is then 3 × 2 × 5 = 30.
Since Group B plants trees in clusters of 10 they must plant at 3 clusters to plant 30 trees. So, Group B had to plant at least 3 clusters.
Result
The least number of clusters that Group B plants is 3.
Page 130 Exercise 37 Answer
Since all three numbers are prime the LCM of each two is their product.
2 × 3 = 6
3 × 7 = 21
2 × 7 = 14
6 = F
21 = U
14 = N
Result
The word is FUN.
Page 130 Exercise 38 Answer
Rami has swimming lessons every 3 days, so the days on which he has swimming lessons are multiples of 3. He has guitar lessons every 8 days, so the days in which he has guitar lessons are multiples of 8. The days on which he has both lessons are multiples of both 3 and 8. Since 3 is a prime number and the prime factorization of 8 is 2 ⋅ 2 ⋅ 2, the LCM of 3 and 8 is their product 3 ⋅ 8 = 24.
Rami has both lessons every 24 days. So if he had both lessons on the first day of the month, he will have both lessons on the same day again in 24 days.
Result
Rami will have both lessons on the same day again in 24 days.
Page 130 Exercise 39 Answer
A number is between 58 and 68. It has prime factors of 2, 3 and 5. Let’s call that number n.
58 < n < 68
n is one of the following numbers 59,60,61,62,63,64,65,66,67. Since we now its prime factors are
5 we know it is divisible by 2,3 and 5, and is not a prime number.
59,61 and 67 are prime numbers, which leaves us 60,61,62,63,64,65,66.
However, most of these numbers do not have 2 as a factor since they are odd and only even number have 2 as a factor. That leaves us 60,62,64 and 66.
Furthermore, n must be divisible by 5, and the only number which is still on the list of potential candidates which is divisible by 5 is 60. Now, we must check if 60 is divisible by 3. Since, 60 = 3 ⋅ 20, 3 divides 60 so n = 60.
Result
The number is 60.
Page 130 Exercise 40 Answer
The shuttle departs from Dickson Hall every 12 minutes and from Lot B every 10 minutes. They first both depart from their location at 9:10 A.M, to find the next time when they both depart for campus at the same time we need to find the LCM of 10 and 12 and add that to 9:10.
12 = 2 × 6 = 2 × 2 × 3
10 = 2 × 5
The LCM of 12 and 10 is the product
2 × 2 × 3 × 5 = 60.
60 minutes is equal to one hour, so both shuttles will depart for campus at the same time every hour. Since the first time they departed at the same time is 9:10 A.M., the next is 10:10 A.M.
Result
The next time both shuttle will depart for campus at the same time is 10:10 A.M.
Page 130 Exercise 41 Answer
Mr. Jenkins bought the same number of blue pencils as green pencils. Blue pencils come in packs of 5 so the number of blue pencils he bought is a multiple of 5. Green pencils come in packs of 4 so the number of green pencils he bought is a multiple of 4. He bought the same number of blue and green pencils, which means that the number of pencils he bought for each color is the least common multiple of 4 and 5.
5 is a prime number.
4 = 2 ⋅ 2
So the least common multiple of 5 and 4 is 5 ⋅ 2 ⋅ 2 = 20.
Result
The least number of blue or green pencils Mr. Jenkins could have bought is 20.
Page 130 Exercise 42 Answer
Mr. Jenkins has 48 brushes and 60 tubes of paint to put into cups. The greatest common factor of the number of paintbrushes and the number of tubes of paint is equal to the number of cups Mr. Jenkins has.
To find the greatest common factor of 48 and 60 we first need to find their prime factorizations.
48 = 2 ⋅ 24 = 2 ⋅ 2 ⋅ 12 = 2 ⋅ 2 ⋅ 2 ⋅ 6 = 2 ⋅ 2 ⋅ 2 ⋅ 2 ⋅ 3
60 = 2 ⋅ 30 = 2 ⋅ 2 ⋅ 15 = 2 ⋅ 2 ⋅ 3 ⋅ 5
The common part of prime factorization of 48 and 60 is 2 ⋅ 2 ⋅ 3, thus the greatest common factor is 12.
Result
Mr. Jenkins has 12 cups.