you shall not pass the test

$\color{#D61F06} \text{Note that the answer should be }\boxed{0} \text{. See the solution below.}$

Given that $a_0 = a_1 = 11$ and $a_n = a_{n-1} - a_{n-2}$ for $n\ge 2$ , therefore,

$\begin{aligned} a_2 & = a_1-a_0 = 1-1 = 0 \\ a_3 & = a_2-a_1 = 0 - 1 = - 1 \\ a_4 & = a_3-a_2 = -1 - 0 = - 1 \\ a_5 & = a_4-a_3 = -1 + 1 = 0 \\ a_6 & = a_5-a_4 = 0 + 1 = 1 \\ a_7 & = a_6-a_5 = 1 - 0 = 1 \end{aligned}$

We note that $a_6=a_7=a_0=a_1=1$ , the sequence repeats with a period of 6. That is $a_{n+6} = a_n$ . Therefore.

$a_n = \begin{cases} 1 & \text{if }n \bmod 6 = 0 \text{ or } 1 \\ 0 & \text{if }n \bmod 6 = 2 \text{ or } 5 \\ -1 & \text{if }n \bmod 6 = 3 \text{ or } 4 \end{cases}$

Since the first number is $a_0$ , the 1248th number is $a_{1247} = \boxed{0}$ , because $1247 \bmod 6 = 5$ .