132379

Mathematica

n=1;While[Union[First@IntegerDigits@#&/@{13^n,23^n,79^n}]!={9},n++];n

returns 5529

There is no easy way to solve this problem. The most efficient way is to write a computer program (it literally does what would would normally take you months to compute). Here is a Python 3.4 program that I used to solve this problem.

$a^n$ can be written as $10^{n \cdot lg(a)}$

$13^5 = 10^{5 \cdot lg(13)} = 10^{5.569}$

$10^{5.569} = 10^5 \cdot 10^{0.569}$

Notice the powers whole part only adds 0's to the number.

Because we are only interested in the first digit we only look at the fractional part of the power.

$pow = n \cdot lg(a) - floor(n \cdot lg(a))$

$pow$ is in range between 0 and 1, much easier to compute than doing large integer multiplication.

First digit of number $a^n$ is the same as the first digit of $10^{pow}$

$10^{0.569} = 3.71293 => 3$

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for(int n = 1; true; n++){
        double pow1 = n * log10(13);
        double pow2 = n * log10(23);
        double pow3 = n * log10(79);
        pow1 -= floor(pow1);
        pow2 -= floor(pow2);
        pow3 -= floor(pow3);

        int digit1 = pow(10, pow1);
        int digit2 = pow(10, pow2);
        int digit3 = pow(10, pow3);

        if(digit1 == 9 && digit2 == 9 && digit3 == 9){
            cout << n;
            break;
        }
    }