Question

$\frac{6^{83}+8^{83}}{49}$, find remainder.

Answer 1

$6^{83}=(7-1{)}^{83}=-(1-7{)}^{83}=-({}^{83}{C}_o{-}^{83}{C}_1{7}^1{+}^{83}{C}_2{7}^2-...)$

$8^{83}=(7+1{)}^{83}=(1+7{)}^{83}=({}^{83}{C}_o{+}^{83}{C}_1{7}^1{+}^{83}{C}_2{7}^2+...)$

$6^{83}+8^{83}=2({}^{83}{C}_1{7}^1{+}^{83}{C}_3{7}^3+...)$

$⟹6^{83}+8^{83}=2\times 83\times 7+2({}^{83}{C}_3{7}^3{+}^{83}{C}_5{7}^5+...)$

$⟹6^{83}+8^{83}=35+49\times 23+2\times 49({}^{83}{C}_3{7}^1{+}^{83}{C}_5{7}^3+...)$

$⟹6^{83}+8^{83}=35+49(23+{2.}^{83}{C}_3{7}^1+{2.}^{83}{C}_5{7}^3+...)$

Hence on dividing $6^{83}+8^{83}$ by $49$ remainder comes $35$.