Abundancy ratio, An, of a positive integer n is the sum of all its positive divisors divided by n.
E.G. A10 = 1+2+5+10/10 = 1.8
a) Find an integer n greater than 1 which An < 1.001
b) Find an integer n for which 1.9 < An < 2
c) Show that An is greater than or equal to 2, if n is a multiple of 6.
Just need some help explaining the working out process for getting the answer.
Thanks for the help :)
E.G. A10 = 1+2+5+10/10 = 1.8
a) Find an integer n greater than 1 which An < 1.001
b) Find an integer n for which 1.9 < An < 2
c) Show that An is greater than or equal to 2, if n is a multiple of 6.
Just need some help explaining the working out process for getting the answer.
Thanks for the help :)
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a) An becomes small if the number has few divisors.
So to minimize An, choose a very large prime number for n, say 1013, which will give An = 1.00098...
b) Choose a power of 2, where An will always be (2n-1)/n = 2-1/n, which is very close to 2 as desired.
c) For multiples of 6, which can be expressed as 6n, for n = any integer, 6n, 3n, 2n, n, 6, 3, 2, and 1 are sure factors of 6n.
Thus:
A(6n) = (6n + 3n + 2n + n + 6 + 3 + 2 + 1 + ....) / 6n
= (12n + 12 + ...) / 6n
= 12n / 6n + 12/6n + (...)/6n
= 2 + 2/n + (...)/6n <--- always greater than or equal to 2.
So to minimize An, choose a very large prime number for n, say 1013, which will give An = 1.00098...
b) Choose a power of 2, where An will always be (2n-1)/n = 2-1/n, which is very close to 2 as desired.
c) For multiples of 6, which can be expressed as 6n, for n = any integer, 6n, 3n, 2n, n, 6, 3, 2, and 1 are sure factors of 6n.
Thus:
A(6n) = (6n + 3n + 2n + n + 6 + 3 + 2 + 1 + ....) / 6n
= (12n + 12 + ...) / 6n
= 12n / 6n + 12/6n + (...)/6n
= 2 + 2/n + (...)/6n <--- always greater than or equal to 2.
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b