Somewhat tricky algrebra question, help please? Can I use trial and error for question a below? I have a rough idea on how to do it, but I just need some help with the working out and evidence.   The abundancy ratio 'An' of a positive integer n is the sum of all its positive divisors divided by n. a) Find an integer n greater than 1 for which an <1.001 b) Find an integer n for which 1.9 < An < 2 c) Show that An is more than or equal to 2, if n is a multiple of 6 Just a query: Can I use trial and error for question a? and I have a rough idea on how to do it, but I just need some help with the working out and evidence. Thank you for all the help.

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The abundancy ratio of a number is defined as the sum of the number's divisors divided by the number itself.

An alternative definition is the sum of the reciprocals of the divisors.

(1) Using the first definition, we want a number with a small abundancy ratio. We recognize that the...

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The abundancy ratio of a number is defined as the sum of the number's divisors divided by the number itself.

An alternative definition is the sum of the reciprocals of the divisors.

(1) Using the first definition, we want a number with a small abundancy ratio. We recognize that the ratio must be greater than 1 as all integers greater than 1 have at least two divisors; 1 and the number.

It seems likely that primes have small abundancy ratios. The abundancy ratio for a prime is `(1+n)/n` .

Then `(1+n)/n<1.001==>n>1000` . So we search for the smallest prime greater than 1000 and find 1009.

A(1009)=`1010/1009~~1.00099108<1.001`

So 1009 works, as do all primes greater than 1009.

(2) Any power of 2 greater than 8 works. See link below

e.g. `A(16)=31/16=1.9375`

(3) We use the alternative definition. If n is a multiple of 6 then it has 1,2,3,6 as factors, at least. Then the sum of the reciprocals of the factors is `1/2+1/2+1/3+1/6=2` at least. (If the number is greater than 6, then there are other factors and the sum of the reciprocals of the divisors is greater.

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