Summing Powers

What the equation $\overline{a_1a_2a_3\ldots a_n}=( a_1+ a_2+ a_3+\ldots+ a_n)^n$ actually says is that an $n$ -digit number equals to the digit sum of the $n$ -digit number to the power of $n$ .

Firstly, since we know that $\overline{a_1a_2a_3\ldots a_n}$ is an $n$ -digit number, hence $\overline{a_1a_2a_3\ldots a_n}=(a_1+a_2+a_3+\ldots+a_n)^n<10^n$ $a_1+a_2+a_3+\ldots+a_n<10$ So the digit sum of the $n$ -digit number could be any number from 1 to 9.

For convenience, let's just let the digit sum of the $n$ -digit number be $p$ .

Since $n>1$ , thus $\overline{a_1a_2a_3\ldots a_n}\geqslant10$ .

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Consider $p=1$ , then $\overline {a_1a_2a_3\ldots a_n}=1<10$ contradicts $\overline{a_1a_2a_3 \ldots a_n}\geqslant10$ , so $p$ cannot be 1.

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Consider $p=2$ , then we have to find a number which its digit sum is 2 and can be expressed as $2^n$ . But for any integer $n>1$ , the digit sum of $2^n$ must be greater than 2 (this can be easily proved), so $p$ cannot be 2.

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Consider $p=3$ , then we have to find a number which its digit sum is 3 and can be expressed as $3^n$ . But for any integer $n>1$ , the digit sum of $3^n$ must be greater than 3 (since $3^n$ is a multiple of 9 when $n>1$ , thus the sum of digits must be a multiple of 9), so $p$ also cannot be 3.

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$p$ also can't be 4 as the digit sum of $4^n$ is always greater than 4 when $n> 1$ .

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The same situation applies to when $p=5$ and $p=6$ .

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Now consider $p=7$ , plugging in $n=1,2,3,4$ and we find out when $n=4$ , $7^4=2401$ satisfies the equation, let's see the case when $n> 4$ , by some experiments we see that $7^5$ and $7^6$ doesn't satisfy the equation, when $n$ reaches 7, $7^7=823543$ is not a 7-digit number, so for $n>7$ , $7^n$ must not be an $n$ -digit number, therefore, there's no such number $7^n$ that satisfies the equation when $n> 4$ .

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Now consider $p=8$ , doing some experiments we find out when $n=3$ , $8^3=512$ satisfies the equation, when $3<n<11$ , $8^n$ doesn't satisfy the equation, when $n\geqslant11$ , $8^n$ is not an $n$ -digit number.

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As for $p=9$ , we can easily get that $9^2=81$ satisfies the equation, If $p>2$ , then the digit sum of $9^n$ would be greater than 9.

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Combining all the situations above, we find that the greatest value of $n$ is 4, so the greatest value of $k$ is 4.

java code:

public class brilliant201409092308{
    public static void main(String[] args){
        for(int n=2;;n++){
            boolean found = false;
            for(int i=2;i<10&&i<9*n;i++){
                long pow = 1;
                for(int j=0;j<n;j++) pow *= i;
                int sum = 0;
                long copy = pow;
                while(copy>9){
                    sum += ((int)(copy%10));
                    copy /= 10;
                }
                sum += copy;
                if(sum==i){
                    found = true;
                    System.out.println(sum+":"+pow);
                    break;
                }
            }
            if(!found){
                System.out.println(n-1);
                break;
            }
        }
    }
}

output:

9:81
8:512
7:2401
4
Press any key to continue . . .