For positive integers $n$, let $f(n)$ return the smallest positive integer $k$ such that $\frac{1}{k}$ has exactly $n$ digits after the decimal point. how many positive integer divisors does $f(2010)$ have?

There are a total of 2011 integer divisors.

Step-by-step explanation:

The only primes p such that 1/p has finite spaces after the coma are 2 and 5. If we divide a number with last digit odd we will obtain 1 extra digit after the decimal point and if we divide a number by 5 we will obtain 1 more digit if that number has a last digit in the decimal which is not a multiple of 5.

If we take powers of those primes we will obtian one more digit each time. In order to obtain more digits it is convinient to divide by a power of 2 instead of a power of 5, because the resulting number will be smaller.

If we want 2010 digits after the decimal point, we need to divide 1 by 2 a total of 2010 times, hence f(2010) = 2²⁰¹⁰, which has as positive integer divisors every power of 2 between 0 and 2010, hence there are a total of 2011 integer divisors of f(2010).

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