By 11 and 13.

Let Mathias' number = $11x$ , where $x$ is an integer. Similarly, let Mathilde's number = $13y$ , $y$ being an integer. $x$ and $y$ must have the same sign. If $x$ and $y$ are both negative, then the two numbers must be negative, and thus the sum must also be negative, a contradiction. Thus, $x$ and $y$ must be both positive. Thus

Mathia's number + Mathilde's number = $11x + 13y = 316$

Simplifying, we have $x = \frac{316 - 13y}{11}$

Since $x$ is to be an integer, we must have

$13y \equiv 2y \equiv 316 \equiv 8 \text{ (mod 11) }$ .

The lowest value of $y$ that satisfies this condition will give the largest possible value of Mathias' number. In this case, $y = 4$ , so that $2(4)=8 \equiv 8 \text{ (mod 11)}$ , which checks.

Therefore, Mathias' number = 316 - Mathilde's number = $316 - 13y = 316 - 13(4) = \boxed{264}$ .

i try different numbers that are divisible by 11 and 13 who's sum is 316. i try different numbers until i tried to subtract 52 from 316 who's answer is 264. then i tried to divide it 11 if it is divisible by this number and luckily i got it :)