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Let $ a_{1},\ldots,a_{n}$ are pairwise coprime integers. Let $c$ be an integer that is divisible by each $a_{i}$.

Prove that $c$ is divisible by the product $$a_{1}\cdot\cdot\cdot a_{n}$$

I tried induction, saying that since $a_{1}|c$, $a_{2}|c$, and $gcd(a_{1},a_{2})=1$, that $lcm(a_{1},a_{2}) = a_{1}*a{2}$, so this is clearly divisible by the product. I then tried the induction step, but I don't think it holds that any subset product of integers is necessarily coprime to another subset product of the same integers. If not, then I'm not sure how to solve this.

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Don't you mean that only the $a_i$ are pairwise coprime, not the $a_i$ and also $c$? – Hagen von Eitzen Oct 24 '12 at 10:11
up vote 1 down vote accepted

The clearest if you use the Fundamental Theorem of Number theory, and write each $a_i$ as $a_i=\prod_j{p_{i,j}}^{\alpha_{i,j}} \ $ where $p_{i,j}$ is prime, $\alpha_{i,j}\in\Bbb N$ and $\alpha_{i,j}\ge 1$. Since these are pairwise coprime, for distinct $i\ne i'$ we have $p_{i,j}\ne p_{i',j'}$ for any $j,j'$.

From this, it easily follows (eithor with or without induction) that $lcm(a_1,a_2,..,a_n)=a_1a_2..a_n$.

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So I just want to say that each of these $a_{i}$ is a product of unique primes by the FTOArithmetic...And that from this product, I can say that the whole product of each $a_{i}$ is a unique representation of primes. Then $lcm(a_{1},a_{2},...,a_{n})$ is clearly the product, c. Thus c is divisible by all $a_{i}$. End? – Gbean Oct 24 '12 at 10:20

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