Why Couldn't I Count?

Since:

• the sum of the exterior angles is always 360° (independently from the number of vertices the polygon has),

• external angle = 180° - internal angle, and

• each exterior angle in a regular polygon with $n$ vertices can be calculated as:

$ext = \frac {360°}{n} \iff n = \frac {360°}{ext}$ ,

therefore, if the interior angle is:

$\text {• 120° , then } ext = 180° - 120° = 60° \text { and } n = \frac {360°}{60°} = 6$

We have a hexagon.

$\text {• 130° , then } ext = 180° - 130° = 50° \text { and } n = \frac {360°}{50°} = 7.2$

Since 7.2 is not an integer, we don't have a polygon here (as a polygon cannot have 7.2 vertices).

$\text {• 140° , then } ext = 180° - 140° = 40° \text { and } n = \frac {360°}{40°} = 9$

We have a nonagon.

$\text {• 150° , then } ext = 180° - 150° = 30° \text { and } n = \frac {360°}{30°} = 12$

We have a dodecagon.

Hence, our answer is:

$\boxed {130°}$

For any regular polygon of ' $n$ ' sides,

$\text{Sum of all interior angles} = (n-2)180^{\circ}$

So,

$\text{Magnitude of each interior angle} = \dfrac{(n-2)180^{\circ}}{n}$

Let the magnitude of each interior angle be $x$ .

$\begin{aligned} \therefore & ~~~~~~~~~~~ \dfrac{(n-2)180^{\circ}}{n} = x \\ \\ & \implies (180^{\circ}-x)n=360^{\circ} \\ \\ & \implies n=\dfrac{360^{\circ}}{180^{\circ}-x} \end{aligned}$

Since, $n$ represents number of sides of the polygon and must be an integer.

So, out of the given options, only $130^{\circ}$ does not give us an integer value for $n$ as $\dfrac{360^{\circ}}{180^{\circ}-130^{\circ}} = 7.2$ which is not an integer.

So, the required answer is $\boxed{130^{\circ}}$ .