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?

Creeperhissboom
Mar 27, 2018

#1**-1 **

f(n)=2^{n}.

Proof-

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

0.r_{1}r_{2}r_{3}.....r_{k}0000000000......

where r_{1},r_{2},r_{3}.....r_{k} are digits. This means that we can write B as a quotient of 2 positive integers:

p=r_{1}*10^{k-1}+r_{2}*10^{k-2}+......r_{k-1}*10^{1}+r_{k}*10^{0}, 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/(10^{n}) 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 p_{1}/q_{1} this means that the only divisors q_{1} can have are the divisors of 10: meaning we can write q_{1 }as the product 2^{a}*5^{b} where a and b are natural numbers. if q_{1}=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=2^{n}. 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 2^{n} is 2^{n} itself.

if a a*5 ^{n}>=5 ^{n}>=2 ^{n}

if f(n)=2^{n} then f(n) has exactly n positive integer divisors (the divisors are: 2^{1},2^{2},2^{3},.....,2^{n})

so the answer to your question is:

f(2010) has 2010 positive divisors

Guest Mar 29, 2018

edited by
Guest
Mar 29, 2018