LFT says that for any prime modulus [tex]p[/tex] and any integer [tex]n[/tex], we have
[tex]n^p\equiv n\pmod p[/tex]
From this we immediately know that
[tex]a^{561}\equiv a^{3\times11\times17}\equiv\begin{cases}(a^{11\times17})^3\pmod3\\(a^{3\times17})^{11}\pmod{11}\\(a^{3\times11})^{17}\pmod{17}\end{cases}\equiv\begin{cases}a^{11\times17}\pmod3\\a^{3\times17}\pmod{11}\\a^{3\times11}\pmod{17}\end{cases}[/tex]
Now we apply the Euclidean algorithm. Outlining one step at a time, we have in the first case [tex]11\times17=187=62\times3+1[/tex], so
[tex]a^{11\times17}\equiv a^{62\times3+1}\equiv (a^{62})^3\times a\stackrel{\mathrm{LFT}}\equiv a^{62}\times a\equiv a^{63}\pmod3[/tex]
Next, [tex]63=21\times3[/tex], so
[tex]a^{63}\equiv a^{21\times3}=(a^{21})^3\stackrel{\mathrm{LFT}}\equiv a^{21}\pmod3[/tex]
Next, [tex]21=7\times3[/tex], so
[tex]a^{21}\equiv a^{7\times3}\equiv(a^7)^3\stackrel{\mathrm{LFT}}\equiv a^7\pmod3[/tex]
Finally, [tex]7=2\times3+1[/tex], so
[tex]a^7\equiv a^{2\times3+1}\equiv (a^2)^3\times a\stackrel{\mathrm{LFT}}\equiv a^2\times a\equiv a^3\stackrel{\mathrm{LFT}}\equiv a\pmod3[/tex]
We do the same thing for the remaining two cases:
[tex]3\times17=51=4\times11+7\implies a^{51}\equiv a^{4+7}\equiv a\pmod{11}[/tex]
[tex]3\times11=33=1\times17+16\implies a^{33}\equiv a^{1+16}\equiv a\pmod{17}[/tex]
Now recall the Chinese remainder theorem, which says if [tex]x\equiv a\pmod n[/tex] and [tex]x\equiv b\pmod m[/tex], with [tex]m,n[/tex] relatively prime, then [tex]x\equiv b{m_n}^{-1}m+a{n_m}^{-1}n\pmod{mn}[/tex], where [tex]{m_n}^{-1}[/tex] denotes [tex]m^{-1}\pmod n[/tex].
For this problem, the CRT is saying that, since [tex]a^{561}\equiv a\pmod3[/tex] and [tex]a^{561}\equiv a\pmod{11}[/tex], it follows that
[tex]a^{561}\equiv a\times{11_3}^{-1}\times11+a\times{3_{11}}^{-1}\times3\pmod{3\times11}[/tex]
[tex]\implies a^{561}\equiv a\times2\times11+a\times4\times3\pmod{33}[/tex]
[tex]\implies a^{561}\equiv 34a\equiv a\pmod{33}[/tex]
And since [tex]a^{561}\equiv a\pmod{17}[/tex], we also have
[tex]a^{561}\equiv a\times{17_{33}}^{-1}\times17+a\times{33_{17}}^{-1}\times33\pmod{17\times33}[/tex]
[tex]\implies a^{561}\equiv a\times2\times17+a\times16\times33\pmod{561}[/tex]
[tex]\implies a^{561}\equiv562a\equiv a\pmod{561}[/tex]
