Two Versus Three

By rearranging the terms in LHS,

$\begin{aligned} && f\left( x \right) +\frac { 1 }{ 2 } +\frac { f\left( x \right) }{ 4 } +\frac { 1 }{ 8 } +\frac { f\left( x \right) }{ 16 } +\frac { 1 }{ 32 } \cdots \\ &=& f(x) \left(1 + \dfrac14 + \dfrac1{16} + \dfrac1{64} + \cdots \right) + \left( \dfrac12 + \dfrac18 + \dfrac1{32} + \cdots \right), \text{ infinite geometric progression sum} \\ &=&f(x) \left( \dfrac1{1 - 1/4} \right) + \dfrac{1/2}{1- 1/4} \\ & =& \dfrac43 f(x) + \dfrac23 \\ \end{aligned}$

Similarly, by rearranging the terms in RHS,

$\begin{aligned} && 1+\frac { x }{ 3 } +\frac { 1 }{ 9 } +\frac { x }{ 27 } +\frac { 1 }{ 81 } +\frac { x }{ 243 } \dots \\ &=& \left(1 + \dfrac19 + \dfrac1{81} + \cdots \right) + x \left( \dfrac13 + \dfrac1{27} + \dfrac1{243} + \cdots \right) \\ &=& \dfrac1{1- 1/9} + x \cdot \dfrac{1/3}{1- 1/9} \\ &=& \dfrac98 + \dfrac38 x \end{aligned}$

Equating these two expressions:

$\begin{aligned} \dfrac43 f(x) + \dfrac23 &=& \dfrac98 + \dfrac38 x \\ 8(4f(x) + 2) &=& 3(9 + 3x) \\ 32f(x) + 16 &=& 27 + 9x \\ 9x &=& -11 + 32f(x). \end{aligned}$

And we want to find the smallest value of $x$ such that $9x + 11$ is divisible by 32.

With modulo 32, we have $9z \equiv -11 \Rightarrow 63z \equiv -77 \Rightarrow -z \equiv -77 \Rightarrow z \equiv 77 \equiv 13 \pmod{32}$ .

And thus the answer is $\boxed{13}$ .

Look at Amit's solution. We can get $32 f(x) - 9 x = 11 \text{; [1] }$ . We are looking for integers solutions for [1] , so [1] is a diophantine equation which solutions are $(f(x) = 22 + 9 \cdot \lambda, x = 77 + 32 \cdot \lambda)$ with $\lambda \in \mathbb{Z}$ , therefore the minimum value for x to be a natural number is got for $\lambda = - 2 \Rightarrow x = 13$

How did this get changed to number theory from algebra??