Euler would love programming

Java:

public class brilliant{
    public static void main(String args[]){
        int min=-1;
        for(int i=100;i<111;i++){
            int phi = 0;
            for(int j=1;j<i;j++){
                if(gcd(i,j)==1) phi++;
            }
            if(min==-1||phi<min) min=phi;
        }
        System.out.println(min);
    }
    private static int gcd(int i,int j){
        if(i==0) return j;
        else return gcd(Math.abs(j-i),i);
    }
}

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import fractions
def phi(n):
    amount = 0
    for k in range(1, n + 1):
        if fractions.gcd(n, k) == 1:
            amount += 1
    return amount
array=[]
for i in range(100,111):
    array.append(phi(i))
print min(array)

Modified from source .

\( #include <iostream>

using namespace std;

int gcd(int a, int b){ int c; while(a!=0){ c=a; a=b%a; b=c; } return b;

} int phi(int d){ int p=0; for (int i=1; i<=d; i++){ if(gcd(d,i)==1){p=p+1;} } return p; }

int main() { int min=phi(100); for(int l=100; l<=110; l++) { //cout<<phi(l)<<endl; if(phi(l)<=min){min=phi(l);} } cout<<"minimum value is: "<<min; return 0; } \)

using $Mathematica$

Min@Table[EulerPhi@n,{n,100,110}]

returns $32$

Let p 1, p 2, p 3,......, p n be the distinct prime factors o n Then phi(n) = n [1 - 1/(p_1)] [1 - 1/(p 2)]*...........*[1 - 1/(p n)]. Ed Gray