CareerCup

Initially we can check whether n is a factor of k or not
if(k%n==0)
then only we need to check further, otherwise return false directly.

and if it returns true then we can check whether n is even or odd in that way we need to loop only for logarithmic time that is greater improvement .

if n is even then proceed in multiples of n
eg:
if(n%2==0){
int flag=n;
for(i=0;i<n;i++)
{
k=k/flag;
if(k==1) return true;
flag=flag*flag;
}
return false;
}
similar could be handled for n being odd only 1 more iteration required in that case.

so in this case we didn't need to check again and again for same value of n and value of n increases in each iteration.. that is n, n^2, n^4, n^16..... so the algorithm runs in logarithmic time, that is greater improvement than naive approach.

we can compute eccentricities of the nodes in the graph:
Let G be a graph and v be a vertex of G. The eccentricity of the vertex v is the maximum distance from v to any vertex. That is, e(v)=max{d(v,w):w in V(G)}.
then by finding minimum eccentricity ie radius of the graph will do our task:
The radius of G is the minimum eccentricity among the vertices of G. Therefore, radius(G)=min{e(v):v in V(G)}.
in that way we can find common meeting place.

let me know if it is fine or not?
I am not sure whether infinite number of nodes will affect it or not. but we can construct a graph for any number of nodes and once graph is available then we can proceed further to identify radius of the graph.