6 p x p 6^{px-p} as greatest factor of ( q q x q ) q (q^{qx-q})^{q}

Algebra Level 3

a = 1 1 x 1 × 2 2 x 2 × 2 2 x 2 × 3 3 x 3 × 3 3 x 3 × 3 3 x 3 × . . . × 1 0 10 x 10 \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 a as defined above, assuming that x x is an integer greater than 1, 6 n k = a 6^{n}k=a and n = p x p n=px-p . Which is the greatest integer p p , for which the equations above are true and for which k k is an integer?


The answer is 207.

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1 solution

Ron Lauterbach
Sep 17, 2017

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

Let f ( q ) = ( q q x q ) q 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 ) f(2), f(3), f(4), f(6), f(8), f(9), f(10)

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

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

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

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

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

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

n = 3 9 x 9 × 3 36 36 × 3 162 x 162 = 3 ( 9 + 36 + 161 ) x ( 9 + 36 + 161 ) = 3 207 x 207 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

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