Base 2014 Termination

Lemma: If $\dfrac{1}{n}$ is a terminating decimal in base $2014=2\cdot 19\cdot 53$ , then $n=2^a\cdot 19^b\cdot 53^c$ for non-negative integers $a,b,c$ .

Proof: If we can prove that after moving the decimal point of the base $2014$ representation enough places to the right, the fraction becomes an integer, then we would have proved that it is a terminating decimal. Note that multiplying a number in base $2014$ by $2014$ moves the decimal point one place to the right, just like in base $10$ , where when we multiply a number by $10$ it moves the decimal point one place to the right. Let $x=\text{max}(a,b,c)$ . We multiply our fraction $\dfrac{1}{2^a\cdot 19^b\cdot 53^c}$ by $2014^x$ (moving the decimal $x$ places to the right) to obtain the fraction $\dfrac{2014^x}{2^a\cdot 19^b\cdot 53^c}=\dfrac{2^x\cdot 19^x\cdot 53^x}{2^a\cdot 19^b\cdot 53^c}$

Clearly $2^a\mid 2^x$ , $19^b\mid 19^x$ , and $53^c\mid 53^x$ ; therefore $\dfrac{2^x\cdot 19^x\cdot 53^x}{2^a\cdot 19^b\cdot 53^c}$ is an integer and we are done.

Lemma: If $n$ contains a factor that is not $2$ , $19$ , or $53$ , then $\dfrac{1}{n}$ is not a terminating decimal in base $2014$ .

Proof: Let $n=(2^a\cdot 19^b\cdot 53^c)\cdot (3^{e_1}\cdot 5^{e_2}\cdot 7^{e_3}\cdots )$ , where $e_1,e_2,e_3,\ldots \in \mathbb{Z}^+$ (this is just the prime factorization form of the number $n$ ). We multiply $\dfrac{1}{n}$ by $2014^x$ for some positive integer $x$ to move the decimal point $x$ places to the right. However, notice that for all $x$ , $\dfrac{2014^x}{(2^a\cdot 19^b\cdot 53^c)\cdot (3^{e_1}\cdot 5^{e_2}\cdots)}$ will only be an integer if $e_1=e_2=e_3=\cdots = 0$ , because $3,5,7,\cdots \not\mid 2014$ .

Conclusion: Therefore, the only way for $\dfrac{1}{n}$ to be a terminating decimal is when $n=2^a\cdot 19^b\cdot 53^c$ .

Finishing the problem: We want to find the sum of all values of $\dfrac{1}{2^a\cdot 19^b\cdot 53^c}$ as $a,b,c$ ranges through all non-negative integers. You may calculate this the classic way, but here is a neat trick:

We see that $\left(\dfrac{1}{2^0}+\dfrac{1}{2^1}+\dfrac{1}{2^2}+\cdots\right)\left(\dfrac{1}{19^0}+\dfrac{1}{19^1}+\dfrac{1}{19^2}+\cdots\right)\left(\dfrac{1}{53^0}+\dfrac{1}{53^1}+\dfrac{1}{53^2}+\cdots\right)$ covers every single ordered pair $(a,b,c)$ for all $a,b,c\ge 0$ . (Convince yourself this is true with the distributive property.) We use the sum of geometric sequence formula for each of the geometric sums, to get the sum of all $\dfrac{1}{2^a\cdot 19^b\cdot 53^c}$ as $a,b,c$ ranges through all non-negative integers as: $\left(\dfrac{2}{1}\right)\left(\dfrac{19}{18}\right)\left(\dfrac{53}{52}\right)=\dfrac{1007}{468}$ . However, remember that $r<1$ , but we included the case where $r=\dfrac{1}{2^0\cdot 19^0\cdot 53^0}=1$ . Therefore, we subtract $1$ from $\dfrac{1007}{468}$ to get $\dfrac{539}{468}$ , and our answer is therefore $\boxed{539}$ . Phew! $\Box$