Do you know your bases?

To get the answer we need to first convert the numbers into their algebraic forms. We only have to convert a few of them since we know that:

$169_{a + 1} = 144_{a + 2} = 121_{a + 3} + 100_{a + 4}$

So let's convert $100_{a + 4}$ since it's the easiest to convert.

$100_{a + 4} = (a + 4)^2 = a^2 + 8a + 16$

Now we need to convert $18G_{a}$ and $196_{a}$ .

$18G_{a} = a^2 + 8a + 16$

So we now know that $18G_{a} = 100_{a + 4}$ , so what about $196_{a}$ ?

$196_{a} = a^2 + 9a + 6$

$196_{a} \neq 100_{a + 4}$

So since $18G_{a} = 100_{a + 4}$ we can say that $a \geq 17$ since the highest value used is $G$ which is equal to $16$ . This means that the second equation holds true.

$18G_{a} = 169_{a + 1} = 144_{a + 2} = 121_{a + 3} + 100_{a + 4},\text{ } a \geq 17 \therefore \boxed{\text{Bella is right}}$

We wrote $169_{a+1}$ to base 10, which gives $9+6(a+1)+(a+1)^2 = 16+8a+a^2$ , then wrote $196_a$ t as $6+9a+a^2$ and $18G_a = 16+8a+a^2$

Note that $196_a-18G_a = a-10$

Here we can conclude that $196_a \geq 18G_a$ for $a \geq 10$ . As $G = 16$ , we can conclude that $196_a > 18G_a$ for $a \geq 17$ , and since $18G_a = 169_{a+1}$ , therefore $18G_a \neq 196_a$ . Bella is correct.