If $n>k$, $n \mid km$, and $m \nmid n$, does this imply that $k \mid n$?

Question

I think I am missing something very obvious here, but I've got the following question:

Be $k,m,n \in \mathbb{Z}$ with $n \ge m$, $n > k$, $n \mid k \cdot m$ and $m \nmid n$

Does this imply $k \mid n$?

If you take $k = 6$, $m = 10$ and $n = 15$, you have a counter-example. — Esperluet, Nov 28 '15 at 15:25

score 2 · Answer 1 · answered Oct 23 '17 at 22:09

2

No. For a counterexample, you can take $k = 6, m = 10$, and $n = 15$.

It is possible you are misremembering the following similar statement.

If $a \mid bc$ and $\gcd(a,b) = 1$, then $a \mid c$. This is stated and proved on MSE elsewhere.

answered Oct 23 '17 at 22:09

davidlowryduda

91,687

score 0 · Answer 2 · answered Oct 23 '17 at 22:22

Having a computer helps with this sort of problems.

The following code checks if the conjecture holds for $1\leq n,k,m\leq \max$

int main(){
        int MAX=10;
        for(int n=1;n<=MAX;n++){
                for(int k=1;k < n;k++){
                        for(int m=1;m<=n;m++){
                                if( (m*k)%n==0 && n%m != 0 && n%k!=0){
                                        printf("n = %d, k =  %d, m =  %d is a counterexample \n",n,k,m);
                                }
                        }
                }
        }
}

The output is:

$n = 9, k = 6, m = 6$ is a counterexample

score -1 · Answer 3 · answered Nov 28 '15 at 15:37

-1

the statement shows us that $k*m = r*n$ for some $r$. if $gcd(k,r) = d$, $d > 1$and $d$ doesn't divide $n$, then clearly $k$ doesn't divide $n$

answered Nov 28 '15 at 15:37

Robert S

1,144

If $n>k$, $n \mid km$, and $m \nmid n$, does this imply that $k \mid n$?

3 Answers3