Showthat 7 power9 +9 power7 is divisible by 64

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$$\begin{gathered} \mathrm{We} \mathrm{have} \\ {7}^9+9^7 \\ =9^7+7^9 \\ ={\left(1+8\right)}^7+{\left(8-1\right)}^9 \\ ={\left(1+8\right)}^7-{\left(1-8\right)}^9 \\ \mathrm{Expand} \mathrm{using} \mathrm{binomial} \mathrm{theorem} \\ \Rightarrow {\left(1+8\right)}^7-{\left(1-8\right)}^9 \\ =1{+}^7{\mathrm{C}}_1\left(8\right){+}^7{\mathrm{C}}_2{\left(8\right)}^2{+}^7{\mathrm{C}}_3{\left(8\right)}^3+....{+}^7{\mathrm{C}}_7{\left(8\right)}^7-\left(1{-}^9{\mathrm{C}}_1\left(8\right){+}^9{\mathrm{C}}_2{\left(8\right)}^2{-}^9{\mathrm{C}}_3{\left(8\right)}^3+....{-}^9{\mathrm{C}}_9{\left(8\right)}^9\right) \\ =1+56{+}^7{\mathrm{C}}_2{\left(8\right)}^2{+}^7{\mathrm{C}}_3{\left(8\right)}^3+....{+}^7{\mathrm{C}}_7{\left(8\right)}^7-\left(1-72{+}^9{\mathrm{C}}_2{\left(8\right)}^2{-}^9{\mathrm{C}}_3{\left(8\right)}^3+....{-}^9{\mathrm{C}}_9{\left(8\right)}^9\right) \\ =1+56{+}^7{\mathrm{C}}_2{\left(8\right)}^2{+}^7{\mathrm{C}}_3{\left(8\right)}^3+....{+}^7{\mathrm{C}}_7{\left(8\right)}^7-1+72{+}^9{\mathrm{C}}_2{\left(8\right)}^2{+}^9{\mathrm{C}}_3{\left(8\right)}^3-....{+}^9{\mathrm{C}}_9{\left(8\right)}^9 \\ =128{+}^7{\mathrm{C}}_2{\left(8\right)}^2{+}^7{\mathrm{C}}_3{\left(8\right)}^3+....{+}^7{\mathrm{C}}_7{\left(8\right)}^7{+}^9{\mathrm{C}}_2{\left(8\right)}^2{+}^9{\mathrm{C}}_3{\left(8\right)}^3-....{+}^9{\mathrm{C}}_9{\left(8\right)}^9 \\ =2·8^2{+}^7{\mathrm{C}}_2{\left(8\right)}^2{+}^7{\mathrm{C}}_3{\left(8\right)}^3+....{+}^7{\mathrm{C}}_7{\left(8\right)}^7{+}^9{\mathrm{C}}_2{\left(8\right)}^2{+}^9{\mathrm{C}}_3{\left(8\right)}^3-....{+}^9{\mathrm{C}}_9{\left(8\right)}^9 \\ =8^2\left[2{+}^7{\mathrm{C}}_2{+}^7{\mathrm{C}}_3{\left(8\right)}^1+....{+}^7{\mathrm{C}}_7{\left(8\right)}^5{+}^9{\mathrm{C}}_2{+}^9{\mathrm{C}}_3{\left(8\right)}^1-....{+}^9{\mathrm{C}}_9{\left(8\right)}^7\right] \\ =64\left[2{+}^7{\mathrm{C}}_2{+}^7{\mathrm{C}}_3{\left(8\right)}^1+....{+}^7{\mathrm{C}}_7{\left(8\right)}^5{+}^9{\mathrm{C}}_2{+}^9{\mathrm{C}}_3{\left(8\right)}^1-....{+}^9{\mathrm{C}}_9{\left(8\right)}^7\right] \\ \mathrm{As} 65 \mathrm{is} \mathrm{one} \mathrm{of} \mathrm{the} \mathrm{factors}, \mathrm{hence} 7^9+9^7 \mathrm{is} \mathrm{divisible} \mathrm{by} 64. \end{gathered}$$

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