In the question how they got $\lambda = 2^{1008}$?
$$2\sum_{r=0}^n\binom{2n+1}r=(1+1)^{2n+1}$$
Now as $2^5\equiv-1\pmod{33},2^{10}\equiv1$
$1008\equiv8\pmod{10}\implies2^{1008}\equiv2^8\pmod{33}\equiv25$
$$2\sum_{r=0}^n\binom{2n+1}r=(1+1)^{2n+1}$$
Now as $2^5\equiv-1\pmod{33},2^{10}\equiv1$
$1008\equiv8\pmod{10}\implies2^{1008}\equiv2^8\pmod{33}\equiv25$
– lab bhattacharjee Mar 19 '17 at 09:46