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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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Indice Factors Primes Prime decomposition GCD and LCM Exercises 1 Factors 2 Prime and composite numbers 3 Prime decomposition 4 GCD and LCM 5 ExercisesAlberto Pardo Milan´s e Divisibility
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Indice Factors Primes Prime decomposition GCD and LCM Exercises FactorsAlberto Pardo Milan´s e Divisibility
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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
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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
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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
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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
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Indice Factors Primes Prime decomposition GCD and LCM Exercises Prime and composite numbersAlberto Pardo Milan´s e Divisibility
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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
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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
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Indice Factors Primes Prime decomposition GCD and LCM Exercises Prime decompositionAlberto Pardo Milan´s e Divisibility
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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
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Indice Factors Primes Prime decomposition GCD and LCM Exercises GCD and LCMAlberto Pardo Milan´s e Divisibility
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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
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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
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Indice Factors Primes Prime decomposition GCD and LCM Exercises ExercisesAlberto Pardo Milan´s e Divisibility
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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 193Alberto Pardo Milan´s e Divisibility
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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 XAlberto Pardo Milan´s e Divisibility
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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 XAlberto Pardo Milan´s e Divisibility
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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 XAlberto Pardo Milan´s e Divisibility
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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 XAlberto Pardo Milan´s e Divisibility
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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 XAlberto Pardo Milan´s e Divisibility
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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 XAlberto Pardo Milan´s e Divisibility
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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 XAlberto Pardo Milan´s e Divisibility
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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 XAlberto Pardo Milan´s e Divisibility
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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 XAlberto Pardo Milan´s e Divisibility
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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 XAlberto Pardo Milan´s e Divisibility
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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
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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) = 7Alberto Pardo Milan´s e Divisibility
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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) = 7Alberto Pardo Milan´s e Divisibility
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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) = 7Alberto Pardo Milan´s e Divisibility
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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) = 11Alberto Pardo Milan´s e Divisibility
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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) = 11Alberto Pardo Milan´s e Divisibility
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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) = 11Alberto Pardo Milan´s e Divisibility
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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
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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) = 24Alberto Pardo Milan´s e Divisibility
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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) = 24Alberto Pardo Milan´s e Divisibility
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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) = 24Alberto Pardo Milan´s e Divisibility
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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) = 13Alberto Pardo Milan´s e Divisibility
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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) = 13Alberto Pardo Milan´s e Divisibility
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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) = 13Alberto Pardo Milan´s e Divisibility
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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) = 13Alberto Pardo Milan´s e Divisibility
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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
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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
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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
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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
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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
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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
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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
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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
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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
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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
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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
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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
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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
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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
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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
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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
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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
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