Consider the fraction:
$\frac{A}{41}$ where A is a positive integer.
$\frac{18}{41}=0.4390\overline{24390}$
$\frac{69660}{41}=1699.0\overline{24390}$
$\frac{92020}{41}=2244.390\overline{24390}$
What are the values of A for which $\frac{A}{41}$ ends with the period $\overline{24390}$?
$24390=29^3+1$
Moreover $\frac{24390}{41*271*9}=0.\overline{24390}$
On
Bear with me.
$41*2439 = 99,999$ and so $10^5 \equiv 1 \pmod {41}$.
$1,10,10^2, 10^3, 10^4$ are equivalent $\mod 41$ to $1,10, 18,16,37$.
Claim: If $x = a*41 + r$ where $r = 1,10,18,16,$ or $37$ then $\frac x{41}$ will have a repeating term $02439$.
If $x = a*41 + 1$ then $\frac x{41} = a + \frac 1{41}$ and will have the same repeating terms as $\frac 1{41}$.
Likewise with $x = a*41 + 10$ then $\frac x{41} = a + \frac {10}{41}$ will have the same repeating terms as $\frac {10}{41}$.
For $x = a*41 + 18 = (a-2)*41 + 100$ then $\frac x{41} = a-2 + \frac {100}{41}$ will have the same repeating terms as $\frac {100}{41}$
And by similar argument if $x = a*41 + 16 = b*41 + 1000$ and $x = a*41 + 37 = b*41 + 10^4$ we get the same repeating ters as $\frac {10^3}{41}$ and $\frac {10^4}{41}$
And as they are powers of ten they will all have the same repeating terms just begining at different points.
And $\frac 1{41} = 0.\overline {02439}$
$\frac {1}{41} = 0.\overline {02439}\\ \frac {10}{41} = 0.\overline {24390}\\ \frac {100}{41} = 2.\overline {43902}\\ \frac {100}{41} - 2 = \frac {18}{41} = 0.\overline {43902}$
The numerators are equivlatent to $10^n \pmod {41}$