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I am reviewing some GRE/GMAT math and I don't have the books on me... but I'm trying to remember how prime factors related to multiples of a number.

Say we are trying to figure out if a number is a multiple of 10.

The prime factorization of 10 is $5^1 2^1$. So any multiple of 10 has to have at least a single 2 and a single 5 in its prime factorization right?

Let's take the number 20. That has a prime factorization of $2^2 5^1$ so it is a multiple of 10.

Let's take the number 38. That has a prime factorization of $19^1 2^1$. It is not a multiple of 10 because it does not share all the prime factors of 10 right? It is missing a single 5 right?

So if we then just add the 5 in its prime factorization $19^1 5^1 2^1$ which is the number 190, that is now a multiple of 10.

Is this how it works? What's the reason?

Jwan622
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1 Answers1

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Yes. Alternatively, a number $x$ is a multiple of $10$ if we may write $x=10k$ for an integer $k$.

TomGrubb
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    So to finish this thought (for OP) $x=2^1\cdot 5^1\cdot k$ where $k$ is any other integer (or product of primes). Thus the prime factorization of any multiple of $10$ will always contain at least a $2^1\cdot 5^1$. –  Nov 13 '15 at 18:00
  • Thanks @Bye_World. Mind taking a look at this? http://math.stackexchange.com/questions/1527488/gre-problem-involving-lcd-prime-factorization-and-sets/1527553#1527553 – Jwan622 Nov 13 '15 at 21:08