Alphabet's Problem

Note that $104976 = 2^4\times 3^8$ . For $\sqrt[C]{104976}$ to be an integer, $C$ can only be $2$ or $4$ . If $C = 2$ , then $\overline{B2}$ is not a power of any integer except when $B=3$ , then $\overline{B2} = 32 = 2^5$ , but $\sqrt[3]{32}$ is not an integer.

If $C=4$ , then

$\begin{aligned} \sqrt[A]{\overline{AB}} + \color{#3D99F6} \sqrt[B]{\overline{B4}} & = \sqrt[4]{104976} & \small \color{#3D99F6} \text{Possible power for }\overline{B4} = 64 = 2^6 \\ {\color{#D61F06}\sqrt[A]{\overline{A6}}} + \color{#3D99F6} \sqrt[6]{64} & = 18 & \small \color{#D61F06} \text{Possible powers for }\overline{A6} = 16, 36 \text{, but only 16 acceptable.} \\ {\color{#D61F06}\sqrt[1]{\overline{16}}} + 2 & = 18 \\ 16 + 2 & = 18 \end{aligned}$

Therefore, $A+B+C = 1+6+4 = \boxed {11}$ .

<?php
for($a=1;$a<=9;$a++){
for($b=1;$b<=9;$b++){
for($c=1;$c<=9;$c++){
if( pow((10 * $a)+$b,(1/$a)) + pow((10 * $b)+$c,(1/$b)) == pow(104976,(1/$c)) ){
echo $a+$b+$c; exit; // prints 11
}
}
}
}
?>

$104976 = 324^2 = 18^4$ Since, $\overline{AB}$ and $\overline{BC}$ are two digit numbers.

Therefore $\sqrt[C]{104976} = 18$

$\Rightarrow \boxed{C= 4}$

Now, $\sqrt[A]{AB} + \sqrt[B]{B4} = 18$

According to the question, $\overline{B4}$ is a two digit number which is also the perfect power of $B$

And the only 2-digit number ending with $4$ , which is also a perfect power is $64$ .

$\sqrt[2]{64} = 8$

$\sqrt[3]{64} = 4$

$\sqrt[6]{64} = 2$

We see that $\sqrt[6]{64} = 2$ satisfies the requirement.

$\Rightarrow \boxed{B=6}$

Now, $\sqrt[A]{A6} + \sqrt[6]{64} = \sqrt[4]{104976}$ $\Rightarrow \sqrt[A]{A6} + 2 = 18$ $\Rightarrow \sqrt[A]{A6} = 16$ Since $\overline{A6}$ is a 2-digit number.

Therefore $\boxed{ A=1}$

Now, Let $X= A + B + C$ $\Rightarrow X= 1 + 6 + 4$ $\Rightarrow \boxed{ X= 11}$