New posts in least-common-multiple

How to show that $\displaystyle [a,b,c] = \frac{abc}{(ab,bc,ca)}$ without prime factorization?

Can the identity $ab=\gcd(a,b)\text{lcm}(a,b)$ be recovered from this category?

Proving that $a \mathbb{Z} \cap b\mathbb{Z} = \operatorname{lcm}(a,b)\mathbb Z$ [duplicate]

Finite abelian group contains an element with order equal to the lcm of the orders of its elements

why $ \rm{lcm}[1,2,3,\cdots,n]\in (2^n,4^n)$

The series of reciprocals of the least common multiples of integers $X_1<X_2<\dots$ converges

Reason for LCM of all numbers from 1 .. n equals roughly $e^n$

Is $\mbox{lcm}(a,b,c)=\mbox{lcm}(\mbox{lcm}(a,b),c)$?

Proving that any common multiple of two numbers is a multiple of their least common multiple

Prove the LCM (Least Common Multiple).

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.

If $m=\operatorname{lcm}(a,b)$ then $\gcd(\frac{m}{a},\frac{m}{b})=1$

Show that $\rm lcm(a,b)=ab \iff gcd(a,b)=1$

$\text{lcm}(1,2,3,\ldots,n)\geq 2^n$ for $n\geq 7$

Beginner: How to complete the induction case in a proof that all multiples are a product of the least common multiple

For integers $a$ and $b$, $ab=\text{lcm}(a,b)\cdot\text{hcf}(a,b)$

Concise proof that every common divisor divides GCD without Bezout's identity?

prove that $\operatorname{lcm}(n,m) = nm/\gcd(n,m)$

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