Divisibility, divisor, multiple, GCF, LCM,

1. Indice Factors Primes Prime decomposition GCD and LCM Exercises
Divisibility
Matem´ticas 1o E.S.O.
a
Alberto Pardo Milan´s
e
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2. Indice Factors Primes Prime decomposition GCD and LCM Exercises
1 Factors
2 Prime and composite numbers
3 Prime decomposition
4 GCD and LCM
5 Exercises
Alberto Pardo Milan´s
e Divisibility

3. Indice Factors Primes Prime decomposition GCD and LCM Exercises
Factors
Alberto Pardo Milan´s
e Divisibility

4. Indice Factors Primes Prime decomposition GCD and LCM Exercises
Factors
A factor/divisor of a number n is a number d which divides n.
d divides n ⇔ d is a factor of n.
(read ⇔ if and only if)
Example: 14 : 2 = 7 so two divides fourteen.
A number n is divisible by a number d if d divides n.
d divides n ⇔ d is a factor of n ⇔ n is divisible by d.
A number n is a multiple of a number d if n is equal to d
multiplied by another number.
d divides n ⇔ d is a factor of n ⇔ n is divisible by d ⇔ n is a
multiple of d.
Example: 14 : 2 = 7 so Two divides fourteen. Two is a factor of
fourteen. Fourteen is divisible by two.
Fourteen is a multiple of two.
Alberto Pardo Milan´s
e Divisibility

5. Indice Factors Primes Prime decomposition GCD and LCM Exercises
Factors
A factor/divisor of a number n is a number d which divides n.
d divides n ⇔ d is a factor of n.
(read ⇔ if and only if)
Example: 14 : 2 = 7 so two divides fourteen.
A number n is divisible by a number d if d divides n.
d divides n ⇔ d is a factor of n ⇔ n is divisible by d.
A number n is a multiple of a number d if n is equal to d
multiplied by another number.
d divides n ⇔ d is a factor of n ⇔ n is divisible by d ⇔ n is a
multiple of d.
Example: 14 : 2 = 7 so Two divides fourteen. Two is a factor of
fourteen. Fourteen is divisible by two.
Fourteen is a multiple of two.
Alberto Pardo Milan´s
e Divisibility

6. Indice Factors Primes Prime decomposition GCD and LCM Exercises
Factors
A factor/divisor of a number n is a number d which divides n.
d divides n ⇔ d is a factor of n.
(read ⇔ if and only if)
Example: 14 : 2 = 7 so two divides fourteen.
A number n is divisible by a number d if d divides n.
d divides n ⇔ d is a factor of n ⇔ n is divisible by d.
A number n is a multiple of a number d if n is equal to d
multiplied by another number.
d divides n ⇔ d is a factor of n ⇔ n is divisible by d ⇔ n is a
multiple of d.
Example: 14 : 2 = 7 so Two divides fourteen. Two is a factor of
fourteen. Fourteen is divisible by two.
Fourteen is a multiple of two.
Alberto Pardo Milan´s
e Divisibility

7. Indice Factors Primes Prime decomposition GCD and LCM Exercises
Factors
A factor/divisor of a number n is a number d which divides n.
d divides n ⇔ d is a factor of n.
(read ⇔ if and only if)
Example: 14 : 2 = 7 so two divides fourteen.
A number n is divisible by a number d if d divides n.
d divides n ⇔ d is a factor of n ⇔ n is divisible by d.
A number n is a multiple of a number d if n is equal to d
multiplied by another number.
d divides n ⇔ d is a factor of n ⇔ n is divisible by d ⇔ n is a
multiple of d.
Example: 14 : 2 = 7 so Two divides fourteen. Two is a factor of
fourteen. Fourteen is divisible by two.
Fourteen is a multiple of two.
Alberto Pardo Milan´s
e Divisibility

8. Indice Factors Primes Prime decomposition GCD and LCM Exercises
Prime and composite numbers
Alberto Pardo Milan´s
e Divisibility

9. Indice Factors Primes Prime decomposition GCD and LCM Exercises
Prime and composite numbers
Prime and composite
A prime number is a number that has only two factors 1 and the
number itself.
1 is not considered a prime number as it only has one factor.
Example: 2 is a prime number because has only two factors: 1 and
2.
A number with more than two factors it is called composite
number.
Example: 6 is a composite number because has four factors:1, 2, 3
and 6.
Alberto Pardo Milan´s
e Divisibility

10. Indice Factors Primes Prime decomposition GCD and LCM Exercises
Prime and composite numbers
Prime and composite
A prime number is a number that has only two factors 1 and the
number itself.
1 is not considered a prime number as it only has one factor.
Example: 2 is a prime number because has only two factors: 1 and
2.
A number with more than two factors it is called composite
number.
Example: 6 is a composite number because has four factors:1, 2, 3
and 6.
Alberto Pardo Milan´s
e Divisibility

11. Indice Factors Primes Prime decomposition GCD and LCM Exercises
Prime decomposition
Alberto Pardo Milan´s
e Divisibility

12. Indice Factors Primes Prime decomposition GCD and LCM Exercises
Prime decomposition
Prime decomposition
To factorize a number you have to express the number as a
product of its prime factors.
Factorize a number by finding its prime decomposition. Prime
decomposition is to find the set of prime factors of a number.
Example: 90 = 2 · 45 = 2 · 3 · 15 = 2 · 3 · 3 · 5
We can simplify the product using powers properties 90 = 2 · 32 · 5.
Alberto Pardo Milan´s
e Divisibility

13. Indice Factors Primes Prime decomposition GCD and LCM Exercises
GCD and LCM
Alberto Pardo Milan´s
e Divisibility

14. Indice Factors Primes Prime decomposition GCD and LCM Exercises
GCD and LCM
Greatest Common Divisor
The Greatest Common Divisor (GCD) is the greatest number that
is a common factor of two or more numbers.
Example: GCD(4, 14) = 2, because
factors of 4 are 1, 2, 4, and
factors of 14 are 1, 2, 7, 14.
Alberto Pardo Milan´s
e Divisibility

15. Indice Factors Primes Prime decomposition GCD and LCM Exercises
GCD and LCM
Least Common Multiple
The Least Common Multiple (LCM) is the lowest number that is a
common multiple of two or more numbers.
Example: LCM (6, 9) = 18, because
multiples of 6 are 6, 12, 18, 24 . . . and
multiples of 9 are 9, 18, 27 . . .
Alberto Pardo Milan´s
e Divisibility

16. Indice Factors Primes Prime decomposition GCD and LCM Exercises
Exercises
Alberto Pardo Milan´s
e Divisibility

17. Indice Factors Primes Prime decomposition GCD and LCM Exercises
Exercises
Exercise 1
Find every prime number and prime decomposition of each
composite number.
Prime Composite Neither Prime decomposition
1
2
4
17
25
91
97
121
131
193
Alberto Pardo Milan´s
e Divisibility

18. Indice Factors Primes Prime decomposition GCD and LCM Exercises
Exercises
Exercise 1
Find every prime number and prime decomposition of each
composite number.
Prime Composite Neither Prime decomposition
1 X
2 X
4 X 4 = 22
17 X
25 X 25 = 52
91 X 91 = 34
97 X
121 X 121 = 112
131 X
193 X
Alberto Pardo Milan´s
e Divisibility

19. Indice Factors Primes Prime decomposition GCD and LCM Exercises
Exercises
Exercise 1
Find every prime number and prime decomposition of each
composite number.
Prime Composite Neither Prime decomposition
1 X
2 X
4 X 4 = 22
17 X
25 X 25 = 52
91 X 91 = 34
97 X
121 X 121 = 112
131 X
193 X
Alberto Pardo Milan´s
e Divisibility

20. Indice Factors Primes Prime decomposition GCD and LCM Exercises
Exercises
Exercise 1
Find every prime number and prime decomposition of each
composite number.
Prime Composite Neither Prime decomposition
1 X
2 X
4 X 4 = 22
17 X
25 X 25 = 52
91 X 91 = 34
97 X
121 X 121 = 112
131 X
193 X
Alberto Pardo Milan´s
e Divisibility

21. Indice Factors Primes Prime decomposition GCD and LCM Exercises
Exercises
Exercise 1
Find every prime number and prime decomposition of each
composite number.
Prime Composite Neither Prime decomposition
1 X
2 X
4 X 4 = 22
17 X
25 X 25 = 52
91 X 91 = 34
97 X
121 X 121 = 112
131 X
193 X
Alberto Pardo Milan´s
e Divisibility

22. Indice Factors Primes Prime decomposition GCD and LCM Exercises
Exercises
Exercise 1
Find every prime number and prime decomposition of each
composite number.
Prime Composite Neither Prime decomposition
1 X
2 X
4 X 4 = 22
17 X
25 X 25 = 52
91 X 91 = 34
97 X
121 X 121 = 112
131 X
193 X
Alberto Pardo Milan´s
e Divisibility

23. Indice Factors Primes Prime decomposition GCD and LCM Exercises
Exercises
Exercise 1
Find every prime number and prime decomposition of each
composite number.
Prime Composite Neither Prime decomposition
1 X
2 X
4 X 4 = 22
17 X
25 X 25 = 52
91 X 91 = 34
97 X
121 X 121 = 112
131 X
193 X
Alberto Pardo Milan´s
e Divisibility

24. Indice Factors Primes Prime decomposition GCD and LCM Exercises
Exercises
Exercise 1
Find every prime number and prime decomposition of each
composite number.
Prime Composite Neither Prime decomposition
1 X
2 X
4 X 4 = 22
17 X
25 X 25 = 52
91 X 91 = 34
97 X
121 X 121 = 112
131 X
193 X
Alberto Pardo Milan´s
e Divisibility

25. Indice Factors Primes Prime decomposition GCD and LCM Exercises
Exercises
Exercise 1
Find every prime number and prime decomposition of each
composite number.
Prime Composite Neither Prime decomposition
1 X
2 X
4 X 4 = 22
17 X
25 X 25 = 52
91 X 91 = 34
97 X
121 X 121 = 112
131 X
193 X
Alberto Pardo Milan´s
e Divisibility

26. Indice Factors Primes Prime decomposition GCD and LCM Exercises
Exercises
Exercise 1
Find every prime number and prime decomposition of each
composite number.
Prime Composite Neither Prime decomposition
1 X
2 X
4 X 4 = 22
17 X
25 X 25 = 52
91 X 91 = 34
97 X
121 X 121 = 112
131 X
193 X
Alberto Pardo Milan´s
e Divisibility

27. Indice Factors Primes Prime decomposition GCD and LCM Exercises
Exercises
Exercise 1
Find every prime number and prime decomposition of each
composite number.
Prime Composite Neither Prime decomposition
1 X
2 X
4 X 4 = 22
17 X
25 X 25 = 52
91 X 91 = 34
97 X
121 X 121 = 112
131 X
193 X
Alberto Pardo Milan´s
e Divisibility

28. Indice Factors Primes Prime decomposition GCD and LCM Exercises
Exercises
Exercise 2
Find the GCF of these numbers using prime decompositions (write
the prime factorization of each number and multiply only common
factors)
12 and 20: 30 and 42:
22 and 45: 28 and 98:
14, 21 and 70: 121 and 99:
Alberto Pardo Milan´s
e Divisibility

29. Indice Factors Primes Prime decomposition GCD and LCM Exercises
Exercises
Exercise 2
Find the GCF of these numbers using prime decompositions (write
the prime factorization of each number and multiply only common
factors) Common factors are circled:
12 and 20:
12 = 1 · 2 · 2 · 3 = 22 · 3,
20 = 1 · 2 · 2 · 5 = 22 · 5, =⇒ GCF (12, 20) = 4
22 and 45:
22 = 1 · 2 · 11 = 2 · 11,
45 = 1 · 32 · 5 = 32 · 5, =⇒ GCF (22, 45) = 1
14, 21 and 70:
14 = 1 · 2 · 7 = 2 · 7,
21 = 1 · 3 · 7 = 3 · 7,
70 = 1 · 2 · 5 · 7 = 2 · 5 · 7, =⇒ GCF (14, 21, 70) = 7
Alberto Pardo Milan´s
e Divisibility

30. Indice Factors Primes Prime decomposition GCD and LCM Exercises
Exercises
Exercise 2
Find the GCF of these numbers using prime decompositions (write
the prime factorization of each number and multiply only common
factors) Common factors are circled:
12 and 20:
12 = 1 · 2 · 2 · 3 = 22 · 3,
20 = 1 · 2 · 2 · 5 = 22 · 5, =⇒ GCF (12, 20) = 4
22 and 45:
22 = 1 · 2 · 11 = 2 · 11,
45 = 1 · 32 · 5 = 32 · 5, =⇒ GCF (22, 45) = 1
14, 21 and 70:
14 = 1 · 2 · 7 = 2 · 7,
21 = 1 · 3 · 7 = 3 · 7,
70 = 1 · 2 · 5 · 7 = 2 · 5 · 7, =⇒ GCF (14, 21, 70) = 7
Alberto Pardo Milan´s
e Divisibility

31. Indice Factors Primes Prime decomposition GCD and LCM Exercises
Exercises
Exercise 2
Find the GCF of these numbers using prime decompositions (write
the prime factorization of each number and multiply only common
factors) Common factors are circled:
12 and 20:
12 = 1 · 2 · 2 · 3 = 22 · 3,
20 = 1 · 2 · 2 · 5 = 22 · 5, =⇒ GCF (12, 20) = 4
22 and 45:
22 = 1 · 2 · 11 = 2 · 11,
45 = 1 · 32 · 5 = 32 · 5, =⇒ GCF (22, 45) = 1
14, 21 and 70:
14 = 1 · 2 · 7 = 2 · 7,
21 = 1 · 3 · 7 = 3 · 7,
70 = 1 · 2 · 5 · 7 = 2 · 5 · 7, =⇒ GCF (14, 21, 70) = 7
Alberto Pardo Milan´s
e Divisibility

32. Indice Factors Primes Prime decomposition GCD and LCM Exercises
Exercises
Exercise 2
Find the GCF of these numbers using prime decompositions (write
the prime factorization of each number and multiply only common
factors) Common factors are circled:
30 and 42:
30 = 1 · 2 · 5 · 3 = 2 · 5 · 3,
42 = 1 · 2 · 3 · 7 = 2 · 3 · 7, =⇒ GCF (30, 42) = 6
28 and 98:
28 = 1 · 2 · 2 · 7 = 22 · 7,
98 = 1 · 2 · 7 · 7 = 2 · 72 , =⇒ GCF (28, 98) = 14
121 and 99:
121 = 1 · 11 · 11 = 112 ,
99 = 1 · 3 · 3 · 11 = 32 · 11, =⇒ GCF (121, 99) = 11
Alberto Pardo Milan´s
e Divisibility

33. Indice Factors Primes Prime decomposition GCD and LCM Exercises
Exercises
Exercise 2
Find the GCF of these numbers using prime decompositions (write
the prime factorization of each number and multiply only common
factors) Common factors are circled:
30 and 42:
30 = 1 · 2 · 5 · 3 = 2 · 5 · 3,
42 = 1 · 2 · 3 · 7 = 2 · 3 · 7, =⇒ GCF (30, 42) = 6
28 and 98:
28 = 1 · 2 · 2 · 7 = 22 · 7,
98 = 1 · 2 · 7 · 7 = 2 · 72 , =⇒ GCF (28, 98) = 14
121 and 99:
121 = 1 · 11 · 11 = 112 ,
99 = 1 · 3 · 3 · 11 = 32 · 11, =⇒ GCF (121, 99) = 11
Alberto Pardo Milan´s
e Divisibility

34. Indice Factors Primes Prime decomposition GCD and LCM Exercises
Exercises
Exercise 2
Find the GCF of these numbers using prime decompositions (write
the prime factorization of each number and multiply only common
factors) Common factors are circled:
30 and 42:
30 = 1 · 2 · 5 · 3 = 2 · 5 · 3,
42 = 1 · 2 · 3 · 7 = 2 · 3 · 7, =⇒ GCF (30, 42) = 6
28 and 98:
28 = 1 · 2 · 2 · 7 = 22 · 7,
98 = 1 · 2 · 7 · 7 = 2 · 72 , =⇒ GCF (28, 98) = 14
121 and 99:
121 = 1 · 11 · 11 = 112 ,
99 = 1 · 3 · 3 · 11 = 32 · 11, =⇒ GCF (121, 99) = 11
Alberto Pardo Milan´s
e Divisibility

35. Indice Factors Primes Prime decomposition GCD and LCM Exercises
Exercises
Exercise 3
Find the LCM of these numbers (write the prime factorization of
each number, identify all common prime factors and multiply
them, find the product of the prime factors multiplying each
common prime factor only once and any remaining factors).
2 and 15: 3, 6 and 15:
12 and 20: 9 and 33:
8 and 12: 13 and 39:
Alberto Pardo Milan´s
e Divisibility

36. Indice Factors Primes Prime decomposition GCD and LCM Exercises
Exercises
Exercise 3
Find the LCM of these numbers (write the prime factorization of
each number, identify all common prime factors and multiply
them, find the product of the prime factors multiplying each
common prime factor only once and any remaining factors).
Common factors are circled, remaining factors are squared:
2 and 15:
2= 1 · 2,
15 = 1 · 3 · 5 , =⇒ LCM (2, 15) = 30
12 and 20:
12 = 1 · 2 · 2 · 3 = 22 · 3,
20 = 1 · 2 · 2 · 5 = 22 · 5, =⇒ LCM (12, 20) = 60
8 and 12:
8 = 1 · 2 · 2 · 2 = 23 ,
12 = 1 · 2 · 2 · 3 = 22 · 3, =⇒ LCM (8, 12) = 24
Alberto Pardo Milan´s
e Divisibility

37. Indice Factors Primes Prime decomposition GCD and LCM Exercises
Exercises
Exercise 3
Find the LCM of these numbers (write the prime factorization of
each number, identify all common prime factors and multiply
them, find the product of the prime factors multiplying each
common prime factor only once and any remaining factors).
Common factors are circled, remaining factors are squared:
2 and 15:
2= 1 · 2,
15 = 1 · 3 · 5 , =⇒ LCM (2, 15) = 30
12 and 20:
12 = 1 · 2 · 2 · 3 = 22 · 3,
20 = 1 · 2 · 2 · 5 = 22 · 5, =⇒ LCM (12, 20) = 60
8 and 12:
8 = 1 · 2 · 2 · 2 = 23 ,
12 = 1 · 2 · 2 · 3 = 22 · 3, =⇒ LCM (8, 12) = 24
Alberto Pardo Milan´s
e Divisibility

38. Indice Factors Primes Prime decomposition GCD and LCM Exercises
Exercises
Exercise 3
Find the LCM of these numbers (write the prime factorization of
each number, identify all common prime factors and multiply
them, find the product of the prime factors multiplying each
common prime factor only once and any remaining factors).
Common factors are circled, remaining factors are squared:
2 and 15:
2= 1 · 2,
15 = 1 · 3 · 5 , =⇒ LCM (2, 15) = 30
12 and 20:
12 = 1 · 2 · 2 · 3 = 22 · 3,
20 = 1 · 2 · 2 · 5 = 22 · 5, =⇒ LCM (12, 20) = 60
8 and 12:
8 = 1 · 2 · 2 · 2 = 23 ,
12 = 1 · 2 · 2 · 3 = 22 · 3, =⇒ LCM (8, 12) = 24
Alberto Pardo Milan´s
e Divisibility

39. Indice Factors Primes Prime decomposition GCD and LCM Exercises
Exercises
Exercise 3
Find the LCM of these numbers (write the prime factorization of
each number, identify all common prime factors and multiply
them, find the product of the prime factors multiplying each
common prime factor only once and any remaining factors).
Common factors are circled, remaining factors are squared:
3, 6 and 15:
3 = 1 · 3 = 3,
6 = 1 · 2 · 3 = 2 · 3,
15 = 1 · 3 · 5 = 3 · 5, =⇒ LCM (3, 6, 15) = 30
9 and 33:
9 = 1 · 3 · 3 = 32 ,
33 = 1 · 3 · 11 = 3 · 11, =⇒ LCM (9, 33) = 99
13 and 39:
13 = 1 · 13 = 13,
39 = 1 · 3 · 13 = 3 · 13, =⇒ LCM (13, 39) = 13
Alberto Pardo Milan´s
e Divisibility

40. Indice Factors Primes Prime decomposition GCD and LCM Exercises
Exercises
Exercise 3
Find the LCM of these numbers (write the prime factorization of
each number, identify all common prime factors and multiply
them, find the product of the prime factors multiplying each
common prime factor only once and any remaining factors).
Common factors are circled, remaining factors are squared:
3, 6 and 15:
3 = 1 · 3 = 3,
6 = 1 · 2 · 3 = 2 · 3,
15 = 1 · 3 · 5 = 3 · 5, =⇒ LCM (3, 6, 15) = 30
9 and 33:
9 = 1 · 3 · 3 = 32 ,
33 = 1 · 3 · 11 = 3 · 11, =⇒ LCM (9, 33) = 99
13 and 39:
13 = 1 · 13 = 13,
39 = 1 · 3 · 13 = 3 · 13, =⇒ LCM (13, 39) = 13
Alberto Pardo Milan´s
e Divisibility

41. Indice Factors Primes Prime decomposition GCD and LCM Exercises
Exercises
Exercise 3
Find the LCM of these numbers (write the prime factorization of
each number, identify all common prime factors and multiply
them, find the product of the prime factors multiplying each
common prime factor only once and any remaining factors).
Common factors are circled, remaining factors are squared:
3, 6 and 15:
3 = 1 · 3 = 3,
6 = 1 · 2 · 3 = 2 · 3,
15 = 1 · 3 · 5 = 3 · 5, =⇒ LCM (3, 6, 15) = 30
9 and 33:
9 = 1 · 3 · 3 = 32 ,
33 = 1 · 3 · 11 = 3 · 11, =⇒ LCM (9, 33) = 99
13 and 39:
13 = 1 · 13 = 13,
39 = 1 · 3 · 13 = 3 · 13, =⇒ LCM (13, 39) = 13
Alberto Pardo Milan´s
e Divisibility

42. Indice Factors Primes Prime decomposition GCD and LCM Exercises
Exercises
Exercise 3
Find the LCM of these numbers (write the prime factorization of
each number, identify all common prime factors and multiply
them, find the product of the prime factors multiplying each
common prime factor only once and any remaining factors).
Common factors are circled, remaining factors are squared:
3, 6 and 15:
3 = 1 · 3 = 3,
6 = 1 · 2 · 3 = 2 · 3,
15 = 1 · 3 · 5 = 3 · 5, =⇒ LCM (3, 6, 15) = 30
9 and 33:
9 = 1 · 3 · 3 = 32 ,
33 = 1 · 3 · 11 = 3 · 11, =⇒ LCM (9, 33) = 99
13 and 39:
13 = 1 · 13 = 13,
39 = 1 · 3 · 13 = 3 · 13, =⇒ LCM (13, 39) = 13
Alberto Pardo Milan´s
e Divisibility

43. Indice Factors Primes Prime decomposition GCD and LCM Exercises
Exercises
Exercise 4
Anna sells bags of different kinds of cookies. She earns e6 selling
bags of butter cookies, e12 selling chocolate cookies, and e15
selling bags of honey cookies. Each bag of cookies costs the same
amount. What is the most that Anna could charge for each bag of
cookies?
Alberto Pardo Milan´s
e Divisibility

44. Indice Factors Primes Prime decomposition GCD and LCM Exercises
Exercises
Exercise 4
Anna sells bags of different kinds of cookies. She earns e6 selling
bags of butter cookies, e12 selling chocolate cookies, and e15
selling bags of honey cookies. Each bag of cookies costs the same
amount. What is the most that Anna could charge for each bag of
cookies?
Data: We need
She earns: GCF (6, 12, 15) : Answer: The
Butter c. e6. 6= 1 ·2· 3, most that
Chocolate c. e12. 12 = 1 · 2 · 2 · 3 Anna could
Honey c. e15. 15 = 1 · 3 · 5 charge for
Each bag of co- =⇒ each bag of
okies costs the sa- GCF (6, 12, 15) = 3 cookies is
me amount. e3.
Alberto Pardo Milan´s
e Divisibility

45. Indice Factors Primes Prime decomposition GCD and LCM Exercises
Exercises
Exercise 4
Anna sells bags of different kinds of cookies. She earns e6 selling
bags of butter cookies, e12 selling chocolate cookies, and e15
selling bags of honey cookies. Each bag of cookies costs the same
amount. What is the most that Anna could charge for each bag of
cookies?
Data: We need
She earns: GCF (6, 12, 15) : Answer: The
Butter c. e6. 6= 1 ·2· 3, most that
Chocolate c. e12. 12 = 1 · 2 · 2 · 3 Anna could
Honey c. e15. 15 = 1 · 3 · 5 charge for
Each bag of co- =⇒ each bag of
okies costs the sa- GCF (6, 12, 15) = 3 cookies is
me amount. e3.
Alberto Pardo Milan´s
e Divisibility

46. Indice Factors Primes Prime decomposition GCD and LCM Exercises
Exercises
Exercise 4
Anna sells bags of different kinds of cookies. She earns e6 selling
bags of butter cookies, e12 selling chocolate cookies, and e15
selling bags of honey cookies. Each bag of cookies costs the same
amount. What is the most that Anna could charge for each bag of
cookies?
Data: We need
She earns: GCF (6, 12, 15) : Answer: The
Butter c. e6. 6= 1 ·2· 3, most that
Chocolate c. e12. 12 = 1 · 2 · 2 · 3 Anna could
Honey c. e15. 15 = 1 · 3 · 5 charge for
Each bag of co- =⇒ each bag of
okies costs the sa- GCF (6, 12, 15) = 3 cookies is
me amount. e3.
Alberto Pardo Milan´s
e Divisibility

47. Indice Factors Primes Prime decomposition GCD and LCM Exercises
Exercises
Exercise 5
You have 60 pencils, 90 pens and 120 felt pens to make packages.
Every pack has the same number of pencils, the same number of
pens and the same number of felt pens.What is the maximum
number of packages you can make using all of them? How many
pencils has each package?
Alberto Pardo Milan´s
e Divisibility

48. Indice Factors Primes Prime decomposition GCD and LCM Exercises
Exercises
Exercise 5
You have 60 pencils, 90 pens and 120 felt pens to make packages.
Every pack has the same number of pencils, the same number of
pens and the same number of felt pens.What is the maximum
number of packages you can make using all of them? How many
pencils has each package?
Data: You have: 60 pencils, 90 pens, 120 felt pens. We need
the maximum number of packages.
We need GCF (60, 90, 120) :
60 = 1 · 2 · 2 · 3 · 5 , Answer: The ma-
90 = 1 · 2 · 3 · 3 · 5 , ximum number of
120 = 1 · 2 · 2 · 2 · 3 · 5 packages we can
=⇒ GCF (60, 90, 120) = 30 make is 30. Each
60 : 30 = 2 package has 2
pencils.
Alberto Pardo Milan´s
e Divisibility

49. Indice Factors Primes Prime decomposition GCD and LCM Exercises
Exercises
Exercise 5
You have 60 pencils, 90 pens and 120 felt pens to make packages.
Every pack has the same number of pencils, the same number of
pens and the same number of felt pens.What is the maximum
number of packages you can make using all of them? How many
pencils has each package?
Data: You have: 60 pencils, 90 pens, 120 felt pens. We need
the maximum number of packages.
We need GCF (60, 90, 120) :
60 = 1 · 2 · 2 · 3 · 5 , Answer: The ma-
90 = 1 · 2 · 3 · 3 · 5 , ximum number of
120 = 1 · 2 · 2 · 2 · 3 · 5 packages we can
=⇒ GCF (60, 90, 120) = 30 make is 30. Each
60 : 30 = 2 package has 2
pencils.
Alberto Pardo Milan´s
e Divisibility

50. Indice Factors Primes Prime decomposition GCD and LCM Exercises
Exercises
Exercise 5
You have 60 pencils, 90 pens and 120 felt pens to make packages.
Every pack has the same number of pencils, the same number of
pens and the same number of felt pens.What is the maximum
number of packages you can make using all of them? How many
pencils has each package?
Data: You have: 60 pencils, 90 pens, 120 felt pens. We need
the maximum number of packages.
We need GCF (60, 90, 120) :
60 = 1 · 2 · 2 · 3 · 5 , Answer: The ma-
90 = 1 · 2 · 3 · 3 · 5 , ximum number of
120 = 1 · 2 · 2 · 2 · 3 · 5 packages we can
=⇒ GCF (60, 90, 120) = 30 make is 30. Each
60 : 30 = 2 package has 2
pencils.
Alberto Pardo Milan´s
e Divisibility

51. Indice Factors Primes Prime decomposition GCD and LCM Exercises
Exercises
Exercise 6
Fred, Eva, and Teresa each have the same amount of money. Fred
has only 5 euro cents coins, Eva has only 20 euro cents coins, and
Teresa has only 50 euro cents coins. What is the least amount of
money that each of them could have?
Alberto Pardo Milan´s
e Divisibility

52. Indice Factors Primes Prime decomposition GCD and LCM Exercises
Exercises
Exercise 6
Fred, Eva, and Teresa each have the same amount of money. Fred
has only 5 euro cents coins, Eva has only 20 euro cents coins, and
Teresa has only 50 euro cents coins. What is the least amount of
money that each of them could have?
Data: They all have the same amount of money.
Fred has only e.05 coins, Eva has only e.20 coins, Teresa has
only e.50 coins. We need the least amount of money they
could have.
We need LCM (5, 20, 50) :
5= 1 · 5, Answer: The least
20 = 1 · 2 · 2 · 5 , amount of money
50 = 1 · 2 · 5 · 5 , that each of them
=⇒ LCM (5, 20, 50) = 100 could have is 100 eu-
ro cents=e1.
Alberto Pardo Milan´s
e Divisibility

53. Indice Factors Primes Prime decomposition GCD and LCM Exercises
Exercises
Exercise 6
Fred, Eva, and Teresa each have the same amount of money. Fred
has only 5 euro cents coins, Eva has only 20 euro cents coins, and
Teresa has only 50 euro cents coins. What is the least amount of
money that each of them could have?
Data: They all have the same amount of money.
Fred has only e.05 coins, Eva has only e.20 coins, Teresa has
only e.50 coins. We need the least amount of money they
could have.
We need LCM (5, 20, 50) :
5= 1 · 5, Answer: The least
20 = 1 · 2 · 2 · 5 , amount of money
50 = 1 · 2 · 5 · 5 , that each of them
=⇒ LCM (5, 20, 50) = 100 could have is 100 eu-
ro cents=e1.
Alberto Pardo Milan´s
e Divisibility

54. Indice Factors Primes Prime decomposition GCD and LCM Exercises
Exercises
Exercise 6
Fred, Eva, and Teresa each have the same amount of money. Fred
has only 5 euro cents coins, Eva has only 20 euro cents coins, and
Teresa has only 50 euro cents coins. What is the least amount of
money that each of them could have?
Data: They all have the same amount of money.
Fred has only e.05 coins, Eva has only e.20 coins, Teresa has
only e.50 coins. We need the least amount of money they
could have.
We need LCM (5, 20, 50) :
5= 1 · 5, Answer: The least
20 = 1 · 2 · 2 · 5 , amount of money
50 = 1 · 2 · 5 · 5 , that each of them
=⇒ LCM (5, 20, 50) = 100 could have is 100 eu-
ro cents=e1.
Alberto Pardo Milan´s
e Divisibility

55. Indice Factors Primes Prime decomposition GCD and LCM Exercises
Exercises
Exercise 7
Mandy, Lucy, and Danny each have bags of candy that have the
same total weight. Mandy’s bag has candy bars that each weigh 4
ounces, Lucy’s bag has candy bars that each weigh 6 ounces, and
Danny’s bag has candy bars that each weigh 9 ounces. What is the
least total weight that each of them could have?
Alberto Pardo Milan´s
e Divisibility

56. Indice Factors Primes Prime decomposition GCD and LCM Exercises
Exercises
Exercise 7
Mandy, Lucy, and Danny each have bags of candy that have the
same total weight. Mandy’s bag has candy bars that each weigh 4
ounces, Lucy’s bag has candy bars that each weigh 6 ounces, and
Danny’s bag has candy bars that each weigh 9 ounces. What is the
least total weight that each of them could have?
Data: They all have the same total weight. Mandy has only
4 ounces bars, Lucy has only 6 ounces bars, Danny has only 9
ounces bars, We need the least total weight they could have.
We need LCM (4, 6, 9) :
4= 1 · 2 · 2,
6= 1 · 2 · 3, Answer: The least
9= 1 · 3 · 3, total weight that
=⇒ LCM (4, 6, 9) = 36 each of them could
have is 36 ounces.
Alberto Pardo Milan´s
e Divisibility

57. Indice Factors Primes Prime decomposition GCD and LCM Exercises
Exercises
Exercise 7
Mandy, Lucy, and Danny each have bags of candy that have the
same total weight. Mandy’s bag has candy bars that each weigh 4
ounces, Lucy’s bag has candy bars that each weigh 6 ounces, and
Danny’s bag has candy bars that each weigh 9 ounces. What is the
least total weight that each of them could have?
Data: They all have the same total weight. Mandy has only
4 ounces bars, Lucy has only 6 ounces bars, Danny has only 9
ounces bars, We need the least total weight they could have.
We need LCM (4, 6, 9) :
4= 1 · 2 · 2,
6= 1 · 2 · 3, Answer: The least
9= 1 · 3 · 3, total weight that
=⇒ LCM (4, 6, 9) = 36 each of them could
have is 36 ounces.
Alberto Pardo Milan´s
e Divisibility

58. Indice Factors Primes Prime decomposition GCD and LCM Exercises
Exercises
Exercise 7
Mandy, Lucy, and Danny each have bags of candy that have the
same total weight. Mandy’s bag has candy bars that each weigh 4
ounces, Lucy’s bag has candy bars that each weigh 6 ounces, and
Danny’s bag has candy bars that each weigh 9 ounces. What is the
least total weight that each of them could have?
Data: They all have the same total weight. Mandy has only
4 ounces bars, Lucy has only 6 ounces bars, Danny has only 9
ounces bars, We need the least total weight they could have.
We need LCM (4, 6, 9) :
4= 1 · 2 · 2,
6= 1 · 2 · 3, Answer: The least
9= 1 · 3 · 3, total weight that
=⇒ LCM (4, 6, 9) = 36 each of them could
have is 36 ounces.
Alberto Pardo Milan´s
e Divisibility

59. Indice Factors Primes Prime decomposition GCD and LCM Exercises
Exercises
Exercise 8
To write music in the stave you must divide beats by 2, 4, 8 or 16.
What is the least amount of time you can write on a stave? What
is the greatest divisor of a beat you need?
Alberto Pardo Milan´s
e Divisibility

60. Indice Factors Primes Prime decomposition GCD and LCM Exercises
Exercises
Exercise 8
To write music in the stave you must divide beats by 2, 4, 8 or 16.
What is the least amount of time you can write on a stave? What
is the greatest divisor of a beat you need?
Data: You divide beats by 2, 4, 8 or 16.
To find the least amount of time you can
write on a stave and to find the greatest Answer: The
divisor of a beat you need LCM(2,4,8,16) greatest divisor of
2= 1 · 2, a beat you need
4= 1 · 2 · 2, is 16, the least
8= 1 · 2 · 2 · 2, amount of time
16 = 1 · 2 · 2 · 2 · 2 , you can write is
=⇒ LCM (2, 4, 8, 16) = 16 a beat divided by
16.
Alberto Pardo Milan´s
e Divisibility

61. Indice Factors Primes Prime decomposition GCD and LCM Exercises
Exercises
Exercise 8
To write music in the stave you must divide beats by 2, 4, 8 or 16.
What is the least amount of time you can write on a stave? What
is the greatest divisor of a beat you need?
Data: You divide beats by 2, 4, 8 or 16.
To find the least amount of time you can
write on a stave and to find the greatest Answer: The
divisor of a beat you need LCM(2,4,8,16) greatest divisor of
2= 1 · 2, a beat you need
4= 1 · 2 · 2, is 16, the least
8= 1 · 2 · 2 · 2, amount of time
16 = 1 · 2 · 2 · 2 · 2 , you can write is
=⇒ LCM (2, 4, 8, 16) = 16 a beat divided by
16.
Alberto Pardo Milan´s
e Divisibility

62. Indice Factors Primes Prime decomposition GCD and LCM Exercises
Exercises
Exercise 8
To write music in the stave you must divide beats by 2, 4, 8 or 16.
What is the least amount of time you can write on a stave? What
is the greatest divisor of a beat you need?
Data: You divide beats by 2, 4, 8 or 16.
To find the least amount of time you can
write on a stave and to find the greatest Answer: The
divisor of a beat you need LCM(2,4,8,16) greatest divisor of
2= 1 · 2, a beat you need
4= 1 · 2 · 2, is 16, the least
8= 1 · 2 · 2 · 2, amount of time
16 = 1 · 2 · 2 · 2 · 2 , you can write is
=⇒ LCM (2, 4, 8, 16) = 16 a beat divided by
16.
Alberto Pardo Milan´s
e Divisibility