Find some natural number which the sum of their inverse is equal to 0.601
I find that for such a problems I should look for factor of 1000 which is the set {1, 2, 4, 5, 8, 10, 20, 25, 40, 50, 100, 125, 200, 250, 500, 1000} so I should look for factors that their sum became 601 which is for this problem the solutions is {1+100+500=601} and {250+200+125+25+1=601} now if I divide 1000 on these sets, those natural number bacame 1000/500,1000/100, 1000/1 so the natural number for this set is 2, 10, 1000 and for the other set is 1000/250, 1000/200, 1000/125, 1000/25, 1000/1 therefor the natural number for this set is 4, 5, 8, 40 & 1000.
Now if the number 0.601 change or suppose generalize like 0.abc...z (it has n decimals) then I should find 10^n factors and look for the sum of factors which is equal to abc.....z now this number become so big and so its factors now how could I count the solutions.