problem on remainder....

The basic idea is to recast the expression into a form that would result in xⁿ-yⁿ or x²ⁿ⁺¹+y²ⁿ⁺¹

2222⁵⁵⁵⁵ + 5555²²²²=(2222⁵⁵⁵⁵+4⁵⁵⁵⁵)+(5555²²²²-4²²²²)-(4⁵⁵⁵⁵-4²²²²)

2222⁵⁵⁵⁵ + 4⁵⁵⁵⁵ is divisible by 2222+4=2226=318 x 7
5555²²²² - 4²²²² is divisible by 5555-4=5551=793 x 7

and 4⁵⁵⁵⁵-4²²²²=4²²²²(4³³³³-1)=4²²²²(64¹¹¹¹-1) which is divisible by 63 and thus also by 7...

so the expression you gave is divisible by 7....
done so much for an answer as simple as 0..huh!!!

Thanks Sir for Giving a Nice Link.

actually This is Not my Doubt.

I have just post here for practise.

Thanks

\dpi{200} \fn_phv 7^{2}\equiv 1(mod 4)

\dpi{200} \fn_cs 7^{7}\equiv (7^{2})^{3}7\equiv 7\equiv 3(mod 4)

\dpi{200} \fn_cs \textit{Let }7^{7}= 4k+3

\dpi{200} \fn_cs \textit{Now }7^{4}\equiv 1(mod 100)

\dpi{200} \fn_cs \textit{Therefore }7^{7^{7}}=7^{4k+3}=(7^{4})^{k}\cdot 7^{3}\equiv 1\cdot 343\equiv 43(mod 100)

\dpi{200} \fn_cs \textit{Hence }7^{7^{7^{7}}}\equiv 7^{43}\equiv (7^{4})^{10}\cdot 7^{3}\equiv 343\equiv 43(mod 100)

\dpi{200} \fn_cs \textit{Last 2 digits - 43}

\dpi{200} \fn_cs \textit{Similarly last 2 digits of }7^{7^{7^{7^{7^{7...\textit{and so on}}}}}}\textit{is 43}