There are two ways to compute the abundancy number for an integer n; it is the sum of the divisors of n divided by n, or the sum of the reciprocals of the divisors.

For example: `A_(12)=(1+2+3+4+6+12)/12=2.bar(3)` or `A_(12)=1/1+1/2+1/3+1/4+1/6+1/(12)=2.bar(3)` .

(1)All integers greater than 1 have an abundancy number greater...

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There are two ways to compute the abundancy number for an integer n; it is the sum of the divisors of n divided by n, or the sum of the reciprocals of the divisors.

For example: `A_(12)=(1+2+3+4+6+12)/12=2.bar(3)` or `A_(12)=1/1+1/2+1/3+1/4+1/6+1/(12)=2.bar(3)` .

(1)All integers greater than 1 have an abundancy number greater than 1 since every number greater than 1 has at least two factors. It seems clear that to minimize the abundancy factor we should try to have the fewest factors possible -- primes have only two factors.

Thus to find an integer `n` such that `A_n<1.001` we need an n such that `(1+n)/n<1.001=>n>1000` . Thus we need a prime number greater than 1000; consulting a list of primes we find 1009 to be the smallest prime greater than 1000.

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`A_(1009)=(1+1009)/1009~~1.00099108<1.001` as required.

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Note that any larger prime will work. An interesting question is to find a non-prime number with this property.

(2) We need an integer n such that `1.9<A_n<2` . We know that all powers of 2 have as factors all of the smaller powers of 2; e.g. 16 has factors 1,2,4,8,16 or `2^0,2^1,2^2,2^3,2^4` .

Using the second formulation for `A_n` , and the fact that the factors of a power of 2 form a geometric progression we get :

`A_8=1+1/2+1/4+1/8=1.875`

`A_16=1+1/2+1/4+1/8+1/16=1.9375`

` `

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Thus `1.9<A_(16)=1.9375<2` as required.

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You can show that any higher power of 2 will work.