$\large a = 1^{1x-1} \times 2^{2x-2} \times 2^{2x-2} \times 3^{3x-3} \times 3^{3x-3} \times 3^{3x-3} \times ... \times 10^{10x-10}$

Consider $a$ as defined above, assuming that $x$ is an integer greater than 1, $6^{n}k=a$ and $n=px-p$ . Which is the greatest integer $p$ , for which the equations above are true and for which $k$ is an integer?

The answer is 207.

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$6^{n}=2^{n} \times 3^{n}$

Let $f(q) = (q^{qx-q})^{q}$ .

You only need to consider values of q, which have a factor of 2 or 3.

$f(2), f(3), f(4), f(6), f(8), f(9), f(10)$

It is safe to say, that $2^{n}$ will be greater as $3^{n}$ .

Therefore you can ignore all factors of 2. You only need to consider $f(3)$ , $f(6)$ and $f(9)$ .

$f(3) = 3^{9x-9}$

$f(6) = 6^{36x-36} = 3^{36x-6} \times 2^{36x-6}$

Here you can ignore the factor of $2^{36x-6}$ , as it is a factor of 2 and not 3.

$f(9) = (3^{2})^{81x-81} = 3^{161x-162}$

$n = 3^{9x-9} \times 3^{36-36} \times 3^{162x-162} = 3^{(9+36+161)x-(9+36+161)} = 3^{207x - 207}$

Therefore p = 207