Unit2

Indice Factors Primes Prime decomposition GCD and LCM Exercises

Divisibility

Matematicas 1o E.S.O.Alberto Pardo Milanes

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1 Factors

2 Prime and composite numbers

3 Prime decomposition

4 GCD and LCM

5 Exercises

Alberto Pardo Milanes Divisibility


Factors



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 dmultiplied by another number.d divides n ⇔ d is a factor of n ⇔ n is divisible by d ⇔ n is amultiple of d.

Example: 14 : 2 = 7 so Two divides fourteen. Two is a factor offourteen. Fourteen is divisible by two.Fourteen is a multiple of two.



Factors








Factors








Factors








Prime and composite numbers




Prime and composite

A prime number is a number that has only two factors 1 and thenumber 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 and2.

A number with more than two factors it is called compositenumber.

Example: 6 is a composite number because has four factors:1, 2, 3and 6.




Prime and composite

A prime number is a number that has only two factors 1 and thenumber 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 and2.

A number with more than two factors it is called compositenumber.

Example: 6 is a composite number because has four factors:1, 2, 3and 6.



Prime decomposition



Prime decomposition

Prime decomposition

To factorize a number you have to express the number as aproduct of its prime factors.

Factorize a number by finding its prime decomposition. Primedecomposition is to find the set of prime factors of a number.

Example: 90 = 2 · 45 = 2 · 3 · 15 = 2 · 3 · 3 · 5We can simplify the product using powers properties 90 = 2 · 32 · 5.



GCD and LCM



GCD and LCM

Greatest Common Divisor

The Greatest Common Divisor (GCD) is the greatest number thatis 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.



GCD and LCM

Least Common Multiple

The Least Common Multiple (LCM) is the lowest number that is acommon 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 . . .



Exercises



Exercises

Exercise 1

Find every prime number and prime decomposition of eachcomposite number.

Prime Composite Neither Prime decomposition

12417259197121131193



Exercises

Exercise 1



1 X2 X4 X 4 = 22

17 X25 X 25 = 52

91 X 91 = 34

97 X121 X 121 = 112

131 X193 X



Exercises

Exercise 1



1 X2 X4 X 4 = 22

17 X25 X 25 = 52

91 X 91 = 34

97 X121 X 121 = 112

131 X193 X



Exercises

Exercise 1



1 X2 X4 X 4 = 22

17 X25 X 25 = 52

91 X 91 = 34

97 X121 X 121 = 112

131 X193 X



Exercises

Exercise 1



1 X2 X4 X 4 = 22

17 X25 X 25 = 52

91 X 91 = 34

97 X121 X 121 = 112

131 X193 X



Exercises

Exercise 1



1 X2 X4 X 4 = 22

17 X25 X 25 = 52

91 X 91 = 34

97 X121 X 121 = 112

131 X193 X



Exercises

Exercise 1



1 X2 X4 X 4 = 22

17 X25 X 25 = 52

91 X 91 = 34

97 X121 X 121 = 112

131 X193 X



Exercises

Exercise 1



1 X2 X4 X 4 = 22

17 X25 X 25 = 52

91 X 91 = 34

97 X121 X 121 = 112

131 X193 X



Exercises

Exercise 1



1 X2 X4 X 4 = 22

17 X25 X 25 = 52

91 X 91 = 34

97 X121 X 121 = 112

131 X193 X



Exercises

Exercise 1



1 X2 X4 X 4 = 22

17 X25 X 25 = 52

91 X 91 = 34

97 X121 X 121 = 112

131 X193 X



Exercises

Exercise 1



1 X2 X4 X 4 = 22

17 X25 X 25 = 52

91 X 91 = 34

97 X121 X 121 = 112

131 X193 X



Exercises

Exercise 2

Find the GCF of these numbers using prime decompositions (writethe prime factorization of each number and multiply only commonfactors)12 and 20:

22 and 45:

14, 21 and 70:

30 and 42:

28 and 98:

121 and 99:



Exercises

Exercise 2

Find the GCF of these numbers using prime decompositions (writethe prime factorization of each number and multiply only commonfactors) 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



Exercises

Exercise 2


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



Exercises

Exercise 2


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



Exercises

Exercise 2

Find the GCF of these numbers using prime decompositions (writethe prime factorization of each number and multiply only commonfactors) 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



Exercises

Exercise 2


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



Exercises

Exercise 2


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



Exercises

Exercise 3

Find the LCM of these numbers (write the prime factorization ofeach number, identify all common prime factors and multiplythem, find the product of the prime factors multiplying eachcommon prime factor only once and any remaining factors).2 and 15:

12 and 20:

8 and 12:

3, 6 and 15:

9 and 33:

13 and 39:



Exercises

Exercise 3

Find the LCM of these numbers (write the prime factorization ofeach number, identify all common prime factors and multiplythem, find the product of the prime factors multiplying eachcommon 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) = 3012 and 20:12 = 1© · 2© · 2© · 3 = 22 · 3,20 = 1© · 2© · 2© · 5 = 22 · 5,=⇒ LCM(12, 20) = 608 and 12:8 = 1© · 2© · 2© · 2 = 23,

12 = 1© · 2© · 2© · 3 = 22 · 3,=⇒ LCM(8, 12) = 24



Exercises

Exercise 3


15 = 1© · 3 · 5 ,=⇒ LCM(2, 15) = 3012 and 20:12 = 1© · 2© · 2© · 3 = 22 · 3,20 = 1© · 2© · 2© · 5 = 22 · 5,=⇒ LCM(12, 20) = 608 and 12:8 = 1© · 2© · 2© · 2 = 23,

12 = 1© · 2© · 2© · 3 = 22 · 3,=⇒ LCM(8, 12) = 24



Exercises

Exercise 3


15 = 1© · 3 · 5 ,=⇒ LCM(2, 15) = 3012 and 20:12 = 1© · 2© · 2© · 3 = 22 · 3,20 = 1© · 2© · 2© · 5 = 22 · 5,=⇒ LCM(12, 20) = 608 and 12:8 = 1© · 2© · 2© · 2 = 23,

12 = 1© · 2© · 2© · 3 = 22 · 3,=⇒ LCM(8, 12) = 24



Exercises

Exercise 3

Find the LCM of these numbers (write the prime factorization ofeach number, identify all common prime factors and multiplythem, find the product of the prime factors multiplying eachcommon 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) = 309 and 33:9 = 1© · 3© · 3 = 32,

33 = 1© · 3© · 11 = 3 · 11,=⇒ LCM(9, 33) = 9913 and 39:13 = 1© · 13© = 13,39 = 1© · 3 · 13© = 3 · 13,=⇒ LCM(13, 39) = 13



Exercises

Exercise 3


33 = 1© · 3© · 11 = 3 · 11,=⇒ LCM(9, 33) = 9913 and 39:13 = 1© · 13© = 13,39 = 1© · 3 · 13© = 3 · 13,=⇒ LCM(13, 39) = 13



Exercises

Exercise 3


33 = 1© · 3© · 11 = 3 · 11,=⇒ LCM(9, 33) = 9913 and 39:13 = 1© · 13© = 13,39 = 1© · 3 · 13© = 3 · 13,=⇒ LCM(13, 39) = 13



Exercises

Exercise 3


33 = 1© · 3© · 11 = 3 · 11,=⇒ LCM(9, 33) = 9913 and 39:13 = 1© · 13© = 13,39 = 1© · 3 · 13© = 3 · 13,=⇒ LCM(13, 39) = 13



Exercises

Exercise 4

Anna sells bags of different kinds of cookies. She earns e6 sellingbags of butter cookies, e12 selling chocolate cookies, and e15selling bags of honey cookies. Each bag of cookies costs the sameamount. What is the most that Anna could charge for each bag ofcookies?



Exercises

Exercise 4


Data:She earns:Butter c. e6.Chocolate c. e12.Honey c. e15.Each bag of co-okies costs the sa-me amount.

We needGCF (6, 12, 15) :6 = 1© · 2 · 3©,

12 = 1© · 2 · 2 · 3©15 = 1© · 3© · 5

=⇒GCF (6, 12, 15) = 3

Answer: Themost thatAnna couldcharge foreach bag ofcookies ise3.



Exercises

Exercise 4



We needGCF (6, 12, 15) :6 = 1© · 2 · 3©,

12 = 1© · 2 · 2 · 3©15 = 1© · 3© · 5

=⇒GCF (6, 12, 15) = 3




Exercises

Exercise 4



We needGCF (6, 12, 15) :6 = 1© · 2 · 3©,

12 = 1© · 2 · 2 · 3©15 = 1© · 3© · 5

=⇒GCF (6, 12, 15) = 3




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 ofpens and the same number of felt pens.What is the maximumnumber of packages you can make using all of them? How manypencils has each package?



Exercises

Exercise 5


Data: You have: 60 pencils, 90 pens, 120 felt pens. We needthe maximum number of packages.

We need GCF (60, 90, 120) :60 = 1© · 2© · 2 · 3© · 5©,90 = 1© · 2© · 3 · 3© · 5©,

120 = 1© · 2© · 2 · 2 · 3© · 5©=⇒ GCF (60, 90, 120) = 30

60 : 30 = 2

Answer: The ma-ximum number ofpackages we canmake is 30. Eachpackage has 2pencils.



Exercises

Exercise 5



We need GCF (60, 90, 120) :60 = 1© · 2© · 2 · 3© · 5©,90 = 1© · 2© · 3 · 3© · 5©,

120 = 1© · 2© · 2 · 2 · 3© · 5©=⇒ GCF (60, 90, 120) = 30

60 : 30 = 2




Exercises

Exercise 5



We need GCF (60, 90, 120) :60 = 1© · 2© · 2 · 3© · 5©,90 = 1© · 2© · 3 · 3© · 5©,

120 = 1© · 2© · 2 · 2 · 3© · 5©=⇒ GCF (60, 90, 120) = 30

60 : 30 = 2




Exercises

Exercise 6

Fred, Eva, and Teresa each have the same amount of money. Fredhas only 5 euro cents coins, Eva has only 20 euro cents coins, andTeresa has only 50 euro cents coins. What is the least amount ofmoney that each of them could have?



Exercises

Exercise 6


Data: They all have the same amount of money.Fred has only e.05 coins, Eva has only e.20 coins, Teresa hasonly e.50 coins. We need the least amount of money theycould have.

We need LCM(5, 20, 50) :5 = 1© · 5©,

20 = 1© · 2 · 2 · 5©,

50 = 1© · 2 · 5 · 5©,=⇒ LCM(5, 20, 50) = 100

Answer: The leastamount of moneythat each of themcould have is 100 eu-ro cents=e1.



Exercises

Exercise 6



We need LCM(5, 20, 50) :5 = 1© · 5©,

20 = 1© · 2 · 2 · 5©,

50 = 1© · 2 · 5 · 5©,=⇒ LCM(5, 20, 50) = 100




Exercises

Exercise 6



We need LCM(5, 20, 50) :5 = 1© · 5©,

20 = 1© · 2 · 2 · 5©,

50 = 1© · 2 · 5 · 5©,=⇒ LCM(5, 20, 50) = 100




Exercises

Exercise 7

Mandy, Lucy, and Danny each have bags of candy that have thesame total weight. Mandy’s bag has candy bars that each weigh 4ounces, Lucy’s bag has candy bars that each weigh 6 ounces, andDanny’s bag has candy bars that each weigh 9 ounces. What is theleast total weight that each of them could have?



Exercises

Exercise 7


Data: They all have the same total weight. Mandy has only4 ounces bars, Lucy has only 6 ounces bars, Danny has only 9ounces bars, We need the least total weight they could have.

We need LCM(4, 6, 9) :4 = 1© · 2 · 2 ,

6 = 1© · 2 · 3 ,

9 = 1© · 3 · 3 ,=⇒ LCM(4, 6, 9) = 36

Answer: The leasttotal weight thateach of them couldhave is 36 ounces.



Exercises

Exercise 7



We need LCM(4, 6, 9) :4 = 1© · 2 · 2 ,

6 = 1© · 2 · 3 ,

9 = 1© · 3 · 3 ,=⇒ LCM(4, 6, 9) = 36




Exercises

Exercise 7



We need LCM(4, 6, 9) :4 = 1© · 2 · 2 ,

6 = 1© · 2 · 3 ,

9 = 1© · 3 · 3 ,=⇒ LCM(4, 6, 9) = 36




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? Whatis the greatest divisor of a beat you need?



Exercises

Exercise 8


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

divisor of a beat you need LCM(2,4,8,16)

2 = 1© · 2©,4 = 1© · 2© · 2 ,

8 = 1© · 2© · 2 · 2 ,

16 = 1© · 2© · 2 · 2 · 2 ,=⇒ LCM(2, 4, 8, 16) = 16

Answer: Thegreatest divisor ofa beat you needis 16, the leastamount of timeyou can write isa beat divided by16.



Exercises

Exercise 8






2 = 1© · 2©,4 = 1© · 2© · 2 ,

8 = 1© · 2© · 2 · 2 ,

16 = 1© · 2© · 2 · 2 · 2 ,=⇒ LCM(2, 4, 8, 16) = 16




Exercises

Exercise 8






2 = 1© · 2©,4 = 1© · 2© · 2 ,

8 = 1© · 2© · 2 · 2 ,

16 = 1© · 2© · 2 · 2 · 2 ,=⇒ LCM(2, 4, 8, 16) = 16



Education

Unit2