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Oksana_A [137]
3 years ago
11

Problema 6 Descompongan los siguientes números en factores primos. ¿Es posible encontrar, para cada número, más de una descompos

ición en factores primos? a. 42 b. 31 c. 36 d. 45
Mathematics
1 answer:
Bogdan [553]3 years ago
8 0

Answer:

a: 42 =  2*3*7

b: 31 = 31*1

c: 36 = 2*2*3*3

d: 45 = 3*3*5

Cada descomposición es unica.

Step-by-step explanation:

Sabemos que todo número entero puede ser reescrito como un producto de números primos.

Esta descomposición es única, dado que una vez tenemos un número escrito como producto de primos, esos números primos no pueden descomponerse en otra cosa, por lo que se concluye que una descomposición en primos es única.

a: 42

Para obtener la descomposición, podemos comenzar dividiendo por primos, comenzando por los más bajos.

En este caso, podemos comenzar por 2:

42/2 = 21

asi, podemos reescribir:

42 = 2*21

Ahora ya tenemos un factor que es primo, el 2, y un factor que no lo es, el 21.

Asi que debemos reescribir el 21 como producto de primos.

Y sabemos que 3*7 = 21, donde ambos 3 y 7 son primos, entonces:

42 = 2*21 = 2*3*7

42 = 2*3*7

así hemos reescrito 42 como un producto entre números primos.

b. 31

31 es un número primo, por lo que su descomposición es:

31 = 31*1

c. 36

Comenzamos dividiendo por 2.

36/2 = 18

36 = 2*18

Volvemos a dividir por 2 el 18.

18/2 = 9

18 = 2*9

reemplazando eso en nuestra descomposición obtenemos:

36 = 2*18 = 2*2*9

9 es impar así que no podemos dividir por 2, pasamos al próximo número primo, el 3.

9/3 = 3

9 = 3*3

Reemplazando eso, obtenemos:

36 = 2*2*3*3

d. 45

no podemos dividir por 2, puesto que es un número impar, así que pasamos al próximo primo, el 3.

45/3 = 15

45 = 3*15

Ahora descompongamos el 15.

15/3 = 5

15 = 3*5  (notar que 3 y 5 son primos)

Reemplazando eso, obtenemos:

45 = 3*15 = 3*3*5

45 = 3*3*5

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