Let $\{a_1,a_2,\ldots,a_k\}$ be a set of integers and let $m=\text{lcm}(a_1,\ldots,a_k)$ Prove that if $a_1|n, a_2|n,\ldots$ and $a_k|n$ then $m|n$ Let $\{a_1,a_2,\ldots,a_k\}$ be a set of integers and let $m=\text{lcm}(a_1,\ldots ,a_k)$
What I am trying to prove is that if $a_1|n, a_2|n,\ldots, a_k|n$ then $m|n$. 
I understand that $m$ is the lowest common multiple, and that $n$ dividies into all $a_1, a_2, \ldots a_k$. 
What I am trying to prove is that $m$ divides into $n$, or in other words, that the lowest common denominator divides into $n$.
However, I have no idea how to prove this algebraically. What steps should I do?
 A: Suppose not.  Then we use the division algorithm, with remainder, to write $n=Qm+R$ (where $Q$ is the quotient, and $R$ the remainder.) Of course we have $0<R<m$.  But it is easy to see that $a_i$ divides $R$ for each $i$, hence $R$ is a common multiple of the $a_i$, contradicting the assumption that $m$ was the least such.
A: The intersection of the ideal $(a_1),...,(a_k)$ is $(m)$, $a_i|n, i=1,...,k$ implies $n\in (a_i)$ henceforth $n\in (m)$.
A: That $\ a_1,\ldots,a_k\mid m\,\Rightarrow\,{\rm lcm}(a_1,\ldots,a_k)\mid m\ $ is a prototypical  Euclidean descent.
The set $M$ of all positive common multiples of all $\,a_i$ is closed under positive subtraction, i.e. $\,m> n\in M$ $\Rightarrow$ $\,a_i\mid m,n\,\Rightarrow\, a_i\mid m\!-\!n\,\Rightarrow\,m\!-\!n\in M.\,$ So $\,M\,$ is closed under mod, i.e. remainder, by mod = repeated subtraction: $\,m\bmod n\, =\, m\!-\!qn = ((m\!-\!n)\!-\!n)-\cdots-n.\,$ Thus the least $\,\ell\in M\,$ divides every  $\,m\in M,\,$ else $\ 0\ne m\ {\rm mod}\ \ell\ $ is in $\,M\,$ and smaller than $\,\ell,\,$ contra minimality of $\,\ell.$
Remark $ $ The innate structure exploited above, that sets of integers closed under subtraction are multiples of their least positive element plays a fundamental role in number theory and algebra. Such (principal ideal) structure is highlighted further further in other posts here.

