I come into a problem where I need to define the least common Odd multiple.

Say I have two integer $a,b \in \mathbb{N}$, I want to define $c$ such that $a|c$ and $b|c$ in an oddly number fashion, meaning $\exists i \in \mathbb{Z}, c = (2i+1)a$, and $\exists j \in \mathbb{Z}, c=(2j+1)b$.

I want to define the smallest $c \in \mathbb{N}$ that satisfie the above, i.e., $\exists i,j \in \mathbb{Z}$ such those are true.

Can it be converted to a normal LCM? Or I have to define something like odd multiple?

  • 1
    $\begingroup$ You basically ignore the prime $2$. You can define the lcm using the prime factorization $$lcm (p_1^{a_1} \cdots p_k^{a_k}, p_1^{b_1}\cdots p_k^{b_k}) = p_1^{\max\{a_1,b_1\}}\cdots p_k^{\max\{a_k,b_k\}}$$ and yours do the same just without $p_1$ $\endgroup$
    – Yanko
    Commented Sep 20, 2018 at 22:00
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    $\begingroup$ @Yanko This is genius! $\endgroup$ Commented Sep 20, 2018 at 22:04
  • $\begingroup$ @RossMillikan Sorry. I have corrected them. $\endgroup$ Commented Sep 20, 2018 at 22:04
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    $\begingroup$ you can define anything you want any way you want. But this can't be done unless $a$ and $b$ both have the same power of $2$ as a divisor. You can find the lowest common odd mulitple of the odd components of $a$ and $b$ if you want. $\endgroup$
    – fleablood
    Commented Sep 20, 2018 at 22:15

1 Answer 1


If $a$ and $b$ have the same number of powers of $2$ in their factorizations the normal $\operatorname{LCM}$ will satisfy this. If they do not, there is no $c$ that will satisfy your desires because you must multiply the one with fewer powers of $2$ by an even number to bring the number of powers of $2$ up to the other one. For example, given $a=6,b=10$ that each have one power of $2$, the normal $\operatorname{LCM}$ is $30$, which is an odd number times each. Given $a=12,b=10$ the normal $\operatorname{LCM}$ is $60$, which multiplies $10$ by $6$ to get two powers of $2$ to match $12$.

  • $\begingroup$ Thank you for pointing this out. I wasn't realizing this fact. Indeed, this is becoming more interesting than I previously thought. $\endgroup$ Commented Sep 20, 2018 at 22:24
  • $\begingroup$ The comment by Yanko is the standard way to determine the LCM of two numbers. It is one way to explain what I said above $\endgroup$ Commented Sep 20, 2018 at 22:46

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