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I know that this is true for two numbers, but does this also hold for more than two? I.e. if $m$ is a common multiple of several numbers $n_1, \ldots , n_k$, does it hold that lcm$(n_1,\ldots,n_k)$ divides $m$? If so, how would one go about proving that? Maybe considering pairs of numbers of the set?

Silver Pine
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  • What is the definition of lcm$(n_1,\dots,n_k)$? – fwd Apr 20 '21 at 18:13
  • Hint : Divide $m$ by the lcm of the numbers with remainder and use that the remainder divides as well all the numbers. This leads to an easy proof by contradiction. – Peter Apr 20 '21 at 18:16
  • Yes, of course. fwd asked "What is the definition of $lcm(n_1, ..., n_k)$?". If you know that, and you should if you ask about "lcm", that should be obvious. – user247327 Apr 20 '21 at 18:17
  • @user247327 That lcm is the least common multiplier does not immediately show that every common multiple must be a multiple of the lcm, just that it must be larger , if it is not the lcm. – Peter Apr 20 '21 at 18:19
  • @Peter Okay, so if $l$ is the lcm of the numbers, then by euclidean division we have $m=lk+r$ for some $k \in \mathbb{Z}$ and a remainder $r$. Assuming that $l$ does not divide $m$ yields $r>0$. But then $r$ is a common multiple of the numbers, and smaller than $l$, which is the contradiction? Would this be the proof? – Silver Pine Apr 20 '21 at 18:30
  • Exactly! You got it ! Congratulations ! – Peter Apr 20 '21 at 18:30
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    This really makes sense, thank you so much! – Silver Pine Apr 20 '21 at 18:31
  • Most proofs for two integers also work for any finite number of integers, e.g. see the linked dupes. Alternatively use associativity of LCM to reduce to (compositions) of LCM of two arguments – Bill Dubuque Apr 20 '21 at 18:38

2 Answers2

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Hint Consider the prime factor decomposition and note that the exponent $p(x)$ of a prime $p$ dividing the lcm $x$ of numbers $x_1,...,x_k$ is given by $\max \{p(x_1), ... , p(x_k)\}$. What can you say about $p(y)$ for a common multiple $y$ of $x_1,...,x_k$?

Jonas Linssen
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By definition, $\DeclareMathOperator{\lcm}{lcm}\lcm(n_1,\dots,n_k)$ is the positive generator of the (principal) ideal of common multiples of $n_1,\dots,n_k$: $$n_1\mathbf Z\cap\dots\cap n_k\mathbf Z=\lcm(n_1,\dots,n_k)\mathbf Z,$$ so the answer is obvious.

Bernard
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