Let juans number be j, with k digits. Then $10j+1+10^{k+1}-j=14789$. $9j+10^{k+1}=14788$. Clearly, j must be somewhere around 3 digits, thus $k=3.$ This gives $j=532$. Plugging $j=532$ back into the original equation satisfies the conditions, thus it is our answer.