A "Brilliant" Question

Let $\textcolor{#D61F06}{BRILLIANT}$ $= x$

Therefore, $\textcolor{#D61F06}{LIME}$ $+$ $\textcolor{#D61F06}{BRILLIANT}$ $= x + 21$

Or rearranging, $\textcolor{#D61F06}{(LIMEBRANT)}$ $+$ $\textcolor{#D61F06}{ILLI}$ $= x + 21$

Since, the sum of integers from 1 to 9 is equal to 45, and $\textcolor{#D61F06}{LIMEBRANT}$ includes 1 of each letter,

$45$ $+$ $\textcolor{#D61F06}{ILLI}$ $= x + 21$

$24$ $+$ $\textcolor{#D61F06}{ILLI}$ $= x$

As we have let $x$ be the value of $\textcolor{#D61F06}{BRILLIANT}$ , we want to maximise $x$ to answer the question. This in turn means maximising the values of $I$ and $L$ . The two largest (distinct) digits that can be used are 8 and 9. This makes $x = \boxed{58}$

Now that we have found the maximum case, all we need to do is show that it is possible and it can be done using any one of the combinations below:

Therefore the maximum possible value of $\textcolor{#D61F06}{BRILLIANT}$ $= \boxed{58}$