Two Scales

$\overline{abc}\ ^\circ \text C = \overline{cab}\ ^\circ \text F$ $\implies {\color{#3D99F6}(100a+10b+c)\times \dfrac 95} + 32 = 100c+10a+b$ . Since the RHS is an integer, the LHS must also be an integer. Then $\color{#3D99F6}(100a+10b+c)\times \dfrac 95$ must be an integer and $\color{#3D99F6}(100a+10b+c)$ is divisible by 5 hence $c=0$ or $c=5$ .

When $c=0$ ,

$\begin{aligned} (100a+10b)\times \frac 95 + 32 & = 10a+b \\ 180a + 18b + 32 & = 10a+b \\ 170a + 17b & = -32 \\ \implies \color{#D61F06} a, b & \color{#D61F06}< 0 & \small \color{#D61F06} \text{No solution.} \end{aligned}$

When $c=5$ ,

$\begin{aligned} (100a+10b + 5)\times \frac 95 + 32 & = 500 + 10a + b \\ 180a + 18b + 9 + 32 & = 500+10a+b \\ 170a + 17b & = 459 & \small \color{#3D99F6} \text{Divide both sides by }17 \\ 10a + b & = 27 \\ \implies a & = 2 \\ b & = 7 \end{aligned}$

Therefore, $a+b+c = 2+7+5 = \boxed{14}$ .

here are two ways to solve this using $Mathematica$

a) using conversion formula
Select[Range[100,999],Floor[1.8*#+32]==FromDigits[IntegerDigits[#][[{3,1,2}]]]&]

b) using built-ins
Select[Range[100,999],#&@@UnitConvert[Quantity[#,"DegreesCelsius"],"DegreesFahrenheit"]==FromDigits[IntegerDigits[#][[{3,1,2}]]]&]

both return $275$

Use the conversion :(100a+10b+c) . 9/5 + 32 = 100c+10a+b,then 170a+17b – 98.2c + 32 = 0, So c=5 10a + b = 27, b=27 – 10a, a=2,b=7, 275