

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer

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.

share|cite|improve this question
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 $lcm(a_1,a_2,..,a_n)=a_1a_2..a_n$.

share|cite|improve this answer
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

Your Answer


