Grade 8 - Number Theory #3

It is easy to see that $0.\overline{\dot{a}bc\dot{d}}+0.\overline{\dot{c}da\dot{b}}=1$ . Multiplying both sides with $10,000$ :

$\begin{aligned} 10,000 & =\overline{abcd.\dot{a}bc\dot{d}}+\overline{cdab.\dot{c}da\dot{b}} \\ & =\overline{abcd} + \overline{cdab} + 0.\overline{\dot{a}bc\dot{d}} + 0.\overline{\dot{c}da\dot{b}} \\ & =\overline{abcd}+\overline{cdab}+1\end{aligned}$

$\begin{aligned}\Rightarrow & \overline{abcd}+\overline{cdab}=9999 \\ & a+c=9; b+d=9; c+a=9; d+b=9 \\ & 2(a+b+c+d)=9\times 4 \\ & \boxed{a+b+c+d=18}\end{aligned}$

We will prove that $0.\overline{abcd}$ can always be expressed as $\displaystyle \frac{abcd}{9999}$ for distinct, nonzero, positive integers $a, b, c, d$ .

We can write $0.\overline{abcd}$ as $0.abcdabcdabcdabcd...$ . Now let $x = 0.\overline{abcd}$ . Then,

$\displaystyle \begin{aligned} x & = 0.abcd + 0.0000abcd + 0.00000000abcd + \cdots \\ & = \frac{abcd}{10000} + \frac{abcd}{100000000} + \frac{abcd}{1000000000000} + \cdots \\ & = abcd \times \left( \frac{1}{10^4} + \frac{1}{10^8} + \frac{1}{10^{12}} + \cdots \right) & \small \color{#3D99F6} S_{\infty} = \frac{a}{1 - r} \\ & = abcd \times \frac{\frac{1}{10^4}}{1 - \frac{1}{10^4}} \\ & = \frac{abcd}{9999} & \small \text{QED} \end{aligned}$

Observe that the maximum of $\displaystyle abcd + cdba$ occurs when $\{a, b, c, d\} = \{9, 7, 8, 6\}$ . We have

$\displaystyle \begin{aligned} \frac{abcd}{9999} + \frac{cdba}{9999} & = \frac{9786}{9999} + \frac{8697}{9999} \\ & = \frac{18483}{9999} < 2 & \small \color{#3D99F6} \frac{abcd}{9999} + \frac{cdab}{9999} \ \text{is a natural number} \\ \implies \frac{abcd}{9999} + \frac{cdab}{9999} & = 1 \\ \implies abcd + cdab & = 9999 \end{aligned}$

$a + c = 9$ and $b + d = 9$ , so $a + b + c + d = \boxed{18}$ .