Find the remainder when $2^{561}$ is divided by $561$ using simple congruence properties. $2^{561}\equiv ? \pmod{561}$
Few observations :
$561 = 3\times 11\times 17$
So Fermat's little theorem is not useful here. Any hints ?
If possible, kindly avoid carmichael numbers/group theory/euler theorems as this is from elementary number theory and we have just started congruences chapter.

We have $2^2=4\equiv1\pmod3,$
$2^5=32\equiv-1\pmod{11}\implies2^{10}\equiv(-1)^2\equiv1\pmod{11}$
$2^4=16\equiv-1\pmod{17}\implies2^8\equiv{-1}^2\equiv1\pmod{17}$
$\displaystyle\implies2^{\text{lcm}(2,10,8)}\equiv1\pmod{3\cdot11\cdot17}$
i.e., $\displaystyle2^{40}\equiv1\pmod{561}$
$\displaystyle\implies2^{561}=2(2^{40})^{14}\equiv2(1)^{14}\pmod{561}$
Reference : Carmichael function
