# Number divides Least Common Multiple(LCM)

Given n and m find the smallest k such that:

n divides LCM(m,k) ; m divides LCM(n,k)

My Solution :

If:

   (m==n) then k=1


Else:

   k = LCM(m,n) / GCD(m,n)

If : k divides MIN(m,n) then k=MAX(m,n)


MIN(x,y) : Gives the minimum of x and y .

MAX(x,y) : Gives the maximum of x and y .

GCD(x,y) : Gives the Greatest Common Divisor of x and y .

LCM(x,y) : Gives the Least Common Multiple of x and y .

My solution gives wrong answer , Could anybody suggest me the correct answer ??


Suppose $n=p_1^{e_{n1}}p_2^{e_{n2}}p_3^{e_{n3}}\cdots$ and $m=p_1^{e_{m1}}p_2^{e_{m2}}p_3^{e_{m3}}\cdots$ then you need $k=p_1^{e_{k1}}p_2^{e_{k2}}p_3^{e_{k3}}\cdots$
where $e_{ki}=0$ when $e_{ni}=e_{mi}$ and $e_{ki}=\max(e_{ni},e_{mi})$ when $e_{ni}\not = e_{mi}$.
As an example, if $n=360$ and $m=270$, I think the smallest $k$ is $216$ while your suggestion seems to give $12$.