For positive integers f(n), let n return the smallest positive integer k such that 1/k has exactly n digits after the decimal point. How many positive integer divisors does f(2010) have?
Suppose B is a number with a finite number of digits after the decimal point. here's his decimal representation:
where r1,r2,r3.....rk are digits. This means that we can write B as a quotient of 2 positive integers:
this means that if we have a rational number C, he has a finite number of digits after the decimal point IF AND ONLY IF C=x/(10n) where n is some natural number (think about it). C has EXACTLY n digits after the decimal point if n is the minimal natural number that satisfies this equation (if C is a number between 0 and 1 this means that x is not divisible by 10).
if C's simplified form as a fraction is p1/q1 this means that the only divisors q1 can have are the divisors of 10: meaning we can write q1 as the product 2a*5b where a and b are natural numbers. if q1=2a*5b this means that the smallest k that will satisfy the equation x/10k=C is k=max(a,b).
a fraction of the form 1/k is already simplified. the fraction having a finite number of digits after the decimal point means that k=2a*5b meaning that he has exactly max(a,b) digits after the decimal point. we need to find the minimal k that satisfies k=2a*5b and max(a,b)=n, and that k is k=2n. why? because if a=n this means that the equation max(a,b)=n is already satisfied- we don't need k to be divisible by 5's, and the smallest number that is divisible by 2n is 2n itself.
if a a*5 n>=5 n>=2 n
if f(n)=2n then f(n) has exactly n positive integer divisors (the divisors are: 21,22,23,.....,2n)
so the answer to your question is:
f(2010) has 2010 positive divisors