Double Exponential Function

Computer Science Level 4

For positive integer $x$ less than $99$ , let $\large f(x) = \left ( x^{x^x} \right ) \bmod {100}$ . What value of $x$ yields the largest $f(x)$ ?

The answer is 77.

3 solutions

Thaddeus Abiy
May 7, 2014

Modular Exponentiation .

Two lines of code will do the trick here.

1 2	`Dict={pow(x,pow(x,x,100),100):x for x in range(99)} print Dict[max(Dict)]`

Kenny Lau
Aug 29, 2014

Java:

public class brilliant{
    public static void main(String args[]){
        int max=-1,res=0;
        for(int i=0;i<99;i++){
            int exp = pow(i,pow(i,i,40),100);
            if(exp>max){
                max = exp;
                res = i;
            }
        }
        System.out.println(res);
    }
    private static int pow(int a,int b,int mod){
        int res = 1;
        for(int i=0;i<b;i++){
            res *= a;
            res %= mod;
        }
        return res;
    }
}

Aditya Sriram
Apr 15, 2014

I wrote this Java program for this:

public class Double_Exponential {

public static int f(int a, int b)
{
    int pro=a;
    for(int i=0;i<=b;i++)
    {
        pro = pro*b;
        pro = pro%100;
    }

    return f2(pro,b);
}
public static int f2(int pro, int b)
{
    for(int i=0;i<=b;i++)
    {
        pro = pro*b;
        pro = pro%100;
    }
    return pro;
}

public static void main(String []rgs)
{
    int l=1;
    for(int i = 1;i<99;i++)
    {
        if(f(i,i)>=f(l,l))
            l=i;
    }

    System.out.println("Answer is "+l);
}

}

And the output was "Answer is 77"

Basically, I figured that since only the last 2 digits matter, only the last 2 digits are required for each multiplication. Some of the stuff (like taking pro = a instead of using a) might be unnecessary but I tried a different method earlier for which that was required and was too lazy to make those changes :P

Aditya Sriram - 7 years, 1 month ago

Double Exponential Function

The answer is 77.

3 solutions

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