Question 708707
Of all the pairs (x,n) whose product is the same {{{xn=N}}} number, the greatest {{{n}}} will be paired with the least positive {{{x}}}.
Since {{{N=1*N}}}, the greatest integer that could be {{{n}}} is {{{N}}}
So we have to find {{{N=(25)*(4^4)*(8^(8/3))*(16^(3/4))^or}}} {{{N}}} =(25)*(4^4)*(8^(8/3))*(16^(3/4))
Either way, that expression is hard to type.
Luckily, it can be simplified, because
{{{16=2^4}}} --> 16^(3/4)={{{(root(4,16))^3=(root(4,2^4))^3=2^3}}}
{{{8=2^3}}} --> 8^(8/3)={{{(root(3,8))^8=(root(3,2^3))^8=2^8}}}
and, while we are at it,
{{{4=2^2}}} --> {{{4^4=(2^2)^4=2^(2*4)=2^8}}}
So substituting all that, we get a large number that can be calculated with a calculator,
{{{N=(25)*(2^8)*(2^8)*(2^3)=highlight(25*2^19)}}} or with pencil and paper:
{{{N=(25)*(2^8)*(2^8)*(2^3)=25*2^19=25*2^2+2^17=25*4*2^17=100*2^17=100*2^10*2^7=100*1024*128=highlight(13107200)}}}
 
NOTE 1:
If it was specified that x, and n could not be 1,
then the least possible positive x would be 2,
and that would make {{{25^2^18}}} the greatest possible n.
 
NOTE 2:
Since {{{xn=25*2^19=2^19*5^2}}} the positive factors are all of the form {{{2^a*5^b}}}
with {{{a}}} being any of the 20 integers from 0 to 19 including 0 and 19,
and {{{b}}} being 0, 1, or 2 (just 3 choices).
That makes the total number of positive factors {{{3*20=60}}}.
That makes 60 (x,n) ordered pairs, or 30 {x,n} sets of factors (if order does not matter).
Considering the negative factors doubles the possibilities.