#include <stdio.h> //已知30=2*5*3
int min(int a, int b, int c){
if(a<b&&a<c){
return a;
}
else if(b<a&&b<c){
return b;
}
else{
return c;
}
}
int main() {
int n;
scanf("%d",&n);
int A[n];
int n_5=0,n_3=0,n_2=0;
for(int i=0;i<n;i++){
scanf("%d",&A[i]);
while(A[i]%2==0||A[i]%3==0||A[i]%5==0){ //判断这个数的最简因子有多少个2,3,5
if(A[i]%2==0){
n_2++;
A[i]/=2;
}
else if(A[i]%3==0){
n_3++;
A[i]/=3;
}
else if(A[i]%5==0){
n_5++;
A[i]/=5;
}
}
}
printf("%d",min(n_2,n_3,n_5));//最小值才能出现足够的2*5*3=30
return 0;
}