Questions tagged [least-common-multiple]

For questions about the least common multiple of a collection of numbers (or more generally, elements of a commutative ring).

If $a, b \in \mathbb{N}$, write $a \mid b$ if $a$ divides $b$, i.e. there is $k \in \mathbb{N}$ such that $b = ka$.

The least (or lowest) common multiple of $a_1, \dots, a_k \in \mathbb{N}$ is the smallest positive integer $N$ such that $a_i \mid N$ for $i = 1, \dots, k$. We usually denote $N$ by $\operatorname{lcm}(a_1, \dots, a_k)$. Note that $\operatorname{lcm}(a_1, \dots, a_k)$ can be defined recursively from a binary definition. That is,

$$\operatorname{lcm}(a_1, \dots, a_k) = \operatorname{lcm}(\operatorname{lcm}(\dots\operatorname{lcm}(\operatorname{lcm}(a_1, a_2), a_3), \dots, a_{k-1}), a_k).$$

If $a, b \in \mathbb{N}$ and $a = p_1^{r_1}\dots p_m^{r_m}$, $b = p_1^{s_1}\dots p_m^{s_n}$ are their prime decompositions (where some of the $r_i$ and $s_j$ can be zero), we have

$$\operatorname{lcm}(a, b) = p_1^{\max(r_1, s_1)}\dots\ p_m^{\max(r_m, s_m)}.$$

Note that $\operatorname{lcm}(a, b)\operatorname{gcd}(a, b) = ab$ where $\operatorname{gcd}(a, b)$ is the greatest common divisor of $a$ and $b$.

All of these notions can be generalised to any commutative ring; the above is just the particular case of (positive elements of) the ring $\mathbb{Z}$.

342 questions
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If $a \mid m$ and $(a + 1) \mid m$, prove $a(a + 1) | m$.

Can anyone help me out here? Can't seem to find the right rules of divisibility to show this: If $a \mid m$ and $(a + 1) \mid m$, then $a(a + 1) \mid m$.
KaliKelly
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Concise proof that every common divisor divides GCD without Bezout's identity?

In the integers, it follows almost immediately from the division theorem and the fact that $a | x,y \implies a | ux + vy$ for any $u, v \in \mathbb{Z}$ that the least common multiple of $a$ and $b$ divides any other common multiple. In contrast,…
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$\text{lcm}(1,2,3,\ldots,n)\geq 2^n$ for $n\geq 7$

I can prove that $\text{lcm}(1,2,3,\ldots,n)\geq 2^{n-1}$. Newly, i read in a paper that for $n\geq 7$ we have: $$\text{lcm}(1,2,3,\ldots,n)\geq 2^n$$ Can you prove it? (This inequality is an interesting inequality. For example with this inequality…
user120269
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Reason for LCM of all numbers from 1 .. n equals roughly $e^n$

I computed the LCM for all natural numbers from 1 up to a limit $n$ and plotted the result over $n$. Due to the fast-raising numbers, I plotted the logarithm of the result and was surprised to find a (more or less) identity curve ($x=y$). In other…
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why $ \rm{lcm}[1,2,3,\cdots,n]\in (2^n,4^n)$

Let $n\ge 7$ be positive integers,show that $$f(n)=\rm{lcm}[1,2,3,\cdots,n]\in (2^n,4^n)$$ Anyone know this problem background?or maybe have best proof or best result?
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Can the identity $ab=\gcd(a,b)\text{lcm}(a,b)$ be recovered from this category?

Define the category $\mathcal{C}$ as follows. The objects are defined as $\text{Obj}(\mathcal{C})=\mathbb{Z}^+$, and a lone morphism $a\to b$ exists if and only if $a\mid b$. Otherwise $\text{Hom}(a,b)=\varnothing$. It can be shown that $\gcd(a,b)$…
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Prove $4^m < \operatorname{lcm} (m+1, m+2, \dots, 2m+1)$

I was looking at review sheet for my midterm for primality testing and factorization and saw this on it $$4^m < \operatorname{lcm} (m+1, m+2, \dots, 2m+1)$$ How can I prove this?
user7416
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The lowest number that is a multiple of both 60 and the integer n is 180. Find the smallest possible value of n.

I have one solution but I think it's just a wild guessed one. Tell me if I am correct and also if not, then how should it be done? What I have done is divided 180 by 60 to get 3. Then take lcm of 60 and multiples of 3 one by one until I get 180 as…
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If a divisor of $pq-1$ divides the LCM of $p-1$ and $q-1$, then it also divides the GCD of these two numbers

Suppose that $p,q$ are distinct odd primes. Suppose an integer $k$ divides $pq-1$ and also $k|\operatorname{lcm}(p-1,q-1)$. Show that $k|\operatorname{gcd}(p-1,q-1)$. I've spent ages looking at this problem and very little to show. Surely I would…
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Numbers on the blackboard

There's a list of $N$ natural numbers on the blackboard. At one step you can pick any two number, which do not divide each other. Instead of this pairs you write its gcd and lcm. Problem asks to prove that final result doesn't depend on order on…
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Finite abelian group contains an element with order equal to the lcm of the orders of its elements

I will quote a question from my textbook, to prevent misinterpretation: Let $G$ be a finite abelian group and let $m$ be the least common multiple of the orders of its elements. Prove that $G$ contains an element of order $m$. I figured that, if…
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Function that turns GCD and LCM into intersections and unions?: $f(a)\cap f(b)=f(\gcd(a,b))$, $f(a)\cup f(b)=f(\operatorname{lcm}(a,b))$

Is there a function $f:\Bbb N_+\to\cal P(\Bbb N_+)$ such that: $f(a)\cap f(b)=f(\gcd(a,b))$, $f(a)\cup f(b)=f(\operatorname{lcm}(a,b))$, $a\in f(a)$, and $f$ is injective? Without the third condition, the function that maps a number to the set of…
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LCM of Fibonacci numbers

$\newcommand{\lcm}{\operatorname{lcm}}$There is a nice property of Fibonacci numbers which says that: $$\gcd(F_{a_1}, \ldots, F_{a_n}) = F_{\gcd(a_1, \ldots, a_n)}$$ I am curious is there anything similar with respect to $\lcm$? Clearly…
Salvador Dali
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Find the least number which when divided by 2, 3, 4, 5, 6 leaves a remainder of 1 but it is divided by 7 completely.

I came across a question which is as follows: Find out the smallest number which leaves remainder of 1 when divided by 2, 3, 4, 5, 6 but divided by 7 completely. What I did is given below step wise. Step 1- Find out the LCM of 2, 3, 4, 5, 6 which is…
Deepak Uniyal
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