I wanted to know integer solutions to the equation
(1/k1) + (10/k2) + (100/k3) + ..... + (10^18/k19) = 1 (where k1,k2,k3.... are integers)
which I believe is unique and exists. I tried for n=1 ... k1=1; I tried for n=2 ... k1=6,k2=12; k1=11 and k2=11 are also possible but I needed k1 not equal to k2. Plz correct me if anyone proves that it is not necessary for the solution to exist and be unique (including the condition k1!=k2!=k3!=k4......)
There are many solutions. Let $\pi$ be any permutation of $\{1,\ldots,n\}$ and let $$ k_i=10^{i-1}\cdot2^{\min(\pi(i),n-1)}. $$ The $k_i$ are distinct because they are divisible by distinct powers of $5$. We have $$\begin{eqnarray*} \sum_{i=1}^n\frac{10^{i-1}}{k_i} &=&\sum_{i=1}^n\frac1{2^{\min(\pi(i),n-1)}}\\ &=&\sum_{i=1}^n\frac1{2^{\min(i,n-1)}}\\ &=&\frac1{2^{n-1}}+\sum_{i=1}^{n-1}\frac1{2^i}\\ &=&\frac1{2^{n-1}}+\left(1-\frac1{2^{n-1}}\right)\\ &=&1. \end{eqnarray*}$$