<s>[INST] Finding remainders of large number This question was asked in a competitive exam.

Find the remainder of dividing $6^{83}+ 8^{83}$ by $49$

Are there any theorems/rules to compute the answer ?
 [/INST] Using Euler's Totient theorem, 
As $(6,49)=1$  and $\phi(49)=42$
$6^{42}\equiv 1\pmod {49}\implies 6^{42.k}\equiv1\pmod {49}$ where is any integer.
$\implies 6^{84}\equiv 1\pmod{49}\implies 6^{83}\equiv 6^{-1}\pmod{49}$
By observation, $6\times 8=48\equiv -1\pmod{49}\implies 6\times (49-8)\equiv 1\pmod{49}\implies 41\equiv 6^{-1}\pmod{49}$
Similarly, $8^{84}\equiv 1\pmod{49}\implies 8^{83}\equiv 8^{-1}\pmod{49}$ as $(8,49)=1$
As, $6\times 8=48\equiv -1\pmod{49}$,  so $(49-6)\times 8\equiv 1\pmod {49}\implies 43\equiv 8^{-1}\pmod{49}$
So, the remainder be  $43+41=84\equiv 35\pmod{49}$.
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