prime factorization of a 4- digit number with at least three distinct factors Need two examples. SHOW ALL STEPS

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prime factorization of a 4- digit number with at least three distinct factors
Need two examples. SHOW ALL STEPS

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Hải Đăng 6 months 2021-08-08T23:51:53+00:00 1 Answers 19 views 0

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    2021-08-08T23:53:52+00:00

    Answer:

    We know that every number can be written as a product of prime numbers.

    The method to find the factorized form of a number depends on the number, we just try to find the different factors by dividing by them, for example for the number 1000 we have:

    1000 is an even number, then we can divide it by 2 (2 is a prime number)

    1000 = 2*500   (so we already found a prime factor)

    500 is also an even number, so we can divide it by 2

    1000 = 2*500 = 2*2*250  (we found another prime factor)

    dividing by 2 again we get:

    1000 = 2*2*250 = 2*2*2*125

    1000 = (2*2*2)*125

    now we just need to factorize 125

    we know that 125 is a multiple of 5, such that:

    125 = 5*25 = 5*5*5

    (5 is a prime number, so it is completely factorized).

    Then the factorization of 1000 is:

    1000 = (2*2*2)*(5*5*5) = 2^3*5^3

    Now with another example, 1422

    1422 is an even number, so we again start using the factor 2:

    1422 = 2 = 711

    then:

    1422 = 2*711

    we already found a factor.

    711 is a multiple of 3 (the sum of its digits is a multiple of 3), then:

    711/3 = 237

    We can write our number as:

    1422 = 2*3*237

    237 is also a multiple of 3

    237/3 = 79

    then:

    1422 = 2*3*3*79

    and 79 is a prime number, so we already have 1422 completely factorized.

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