\hspace{-16}\bf{(1)\;\; \mathbb{F}}$ind $\mathbb{L}$ast $\bf{2}$ Digit of $\bf{7^{7^{7^{7}}}}$\\\\\\ $\bf{(2)\;\; \mathbb{F}}$ind $\mathbb{R}$emainder When $\bf{2222^{5555}+5555^{2222}}$ is Divided by $\bf{7}$.
1057The 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!!!
39read this completely and then try solving them yourself
http://www.targetiit.com/iit-jee-forum/posts/congruences-a-z-11709.html
1708Thanks Sir for Giving a Nice Link.
actually This is Not my Doubt.
I have just post here for practise.
Thanks
2305\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}