The abundancy ratio of a number is the sum of the numbers divisors divided by the number itself. Thus `a12=(1+2+3+4+6+12)/12=28/12=2.bar(3)`

It is clear that `an>1` for all `n in NN, n>1` since every number has at least 2 divisors -- the number itself and 1. thus you have `an>= (1+n)/n`

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The abundancy ratio of a number is the sum of the numbers divisors divided by the number itself. Thus `a12=(1+2+3+4+6+12)/12=28/12=2.bar(3)`

It is clear that `an>1` for all `n in NN, n>1` since every number has at least 2 divisors -- the number itself and 1. thus you have `an>= (1+n)/n`

Thus we have the following:

`an<1.001 ==> (1+n)/n<1.001` Solving for n we get n>1000.

Not every number greater than 1000 works, as all numbers **except primes** have more than 2 divisors. Searching a list of primes we find the smallest prime greater than 1000 to be 1009.

Thus `a1009=(1+1009)/1009~~1.00099108028<1.001`

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**So one possible answer is 1009.** Any prime bigger than this will work.

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** A good question is "Is this the smallest such number?"