<s>[INST] Remainder when $2^{55}+1$ is divided by $33$ There's this problem I encountered in a math Olympiad for my country. 
Find the remainder when $2^{55}+1$ is divided by $33$. 
My approach was to consider $2^{55}$ as the sum of numbers in the 56th row of Pascal's triangle. Then I showed that apart from 1 and 55, all other numbers have at least one factor of 11 and 3 to spare, so one would consider only 1 and 55 for the answer. Doing this leads to $113 \mod 33=14$. But the choices for that question were between 0 and 5.
Is my reasoning wrong? Or am I missing something?
 [/INST] Sticking to Pascal's triangle, you can just start calculating $2^k \pmod{33}$ row by row until you see a 'pattern' and can then get lazy.
$\quad 2^0 \equiv \;\,1 \pmod{33}$
$\quad 2^1 \equiv \;\,2 \pmod{33}$
$\quad 2^2 \equiv \;\,4 \pmod{33}$
$\quad 2^3 \equiv \;\,8 \pmod{33}$
$\quad 2^4 \equiv 16 \pmod{33}$
$\quad 2^5 \equiv 32 \pmod{33}$
$\quad 2^6 \equiv 31 \pmod{33}$
$\quad 2^7 \equiv 29 \pmod{33}$
$\quad 2^8 \equiv 25 \pmod{33}$
$\quad 2^9 \equiv 17 \pmod{33}$
and then repeats since
$\quad 2^{10} \equiv \;\,1 \pmod{33}$
So the pattern repeats after a cycle of length $10$.
We want to get to the $56^{th}$ row ($0 \le k \le 55$), and since $56 = 5 \times 10 + 6$, we use the $6^{th}$ entry of the cycle, $2^5 \equiv 32 \pmod{33}$, so that also
$\tag 1 2^{55} \equiv 32 \pmod{33}$
and adding $1$ we can write the answer,
$\tag {ANS} 2^{55}+1 \equiv 0 \pmod{33}$
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