number theory

f(n)=2ⁿ.

Proof-

Suppose B is a number with a finite number of digits after the decimal point. here's his decimal representation:

0.r₁r₂r₃.....r_k0000000000......

 where r₁,r₂,r₃.....r_k are digits. This means that we can write B as a quotient of 2 positive integers:

p=r₁*10^k-1+r₂*10^k-2+......r_k-1*10¹+r_k*10⁰, q=10^k

B=p/q

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 2^a*5^b where a and b are natural numbers. if q₁=2^a*5^b this means that the smallest k that will satisfy the equation x/10^k=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=2^a*5^b meaning that he has exactly max(a,b) digits after the decimal point. we need to find the minimal k that satisfies k=2^a*5^b 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