Question Number 82991 by ahmadshahhimat775@gmail.com last updated on 26/Feb/20
Commented by mr W last updated on 26/Feb/20
$$\mathrm{11}^{\mathrm{800}} \:{mod}\:\mathrm{9} \\ $$$$=\mathrm{2}^{\mathrm{800}} \:{mod}\:\mathrm{9} \\ $$$$=\mathrm{7}^{\mathrm{200}} \:{mod}\:\mathrm{9} \\ $$$$=\mathrm{4}^{\mathrm{100}} \:{mod}\:\mathrm{9} \\ $$$$=\mathrm{7}^{\mathrm{50}} \:{mod}\:\mathrm{9} \\ $$$$=\mathrm{4}^{\mathrm{25}} \:{mod}\:\mathrm{9} \\ $$$$=\mathrm{4}×\mathrm{7}^{\mathrm{12}} \:{mod}\:\mathrm{9} \\ $$$$=\mathrm{4}×\mathrm{4}^{\mathrm{6}} \:{mod}\:\mathrm{9} \\ $$$$=\mathrm{4}×\mathrm{7}^{\mathrm{3}} \:{mod}\:\mathrm{9} \\ $$$$=\mathrm{4}×\mathrm{7}×\mathrm{4}\:{mod}\:\mathrm{9} \\ $$$$=\mathrm{4}×\mathrm{1}\:{mod}\:\mathrm{9} \\ $$$$=\mathrm{4} \\ $$
Commented by mr W last updated on 26/Feb/20
$${why}\:{don}'{t}\:{you}\:{type}\:{the}\:{question}\:{instead} \\ $$$${of}\:{posting}\:{an}\:{image}? \\ $$
Answered by TANMAY PANACEA last updated on 27/Feb/20
![(((9+2)^(800) )/9)=((9^(800) +800c_1 9^(799) 2+...+2^(800) )/9)=((f(9)+2^(800) )/9) ((f(9)+(2^8 )^(100) )/9)=((f(9)+(256)^(100) )/9)=((f(9)+(28×9+4)^(100) )/9) =((f(9)+g(9)+4^(100) )/9)=((f(9)+g(9)+(4^4 )^(25) )/9)=((f(9)+g(9)+(252+4)^(25) )/9) =((f(9)+g(9)+h(9)+4^(25) )/9) =((f(9)+g(9)+h(9)+4×(4^3 )^8 )/9) =((f(9)+g(9)+h(9)+4×(63+1)^8 )/9) =((f(9)+g(9)+h(9)+4[k(9)+1])/9) [note 63 is divisible by 9] and k(9)=(63+1)^8 =(63^8 )+8c_1 (63)^7 +...+1 so =((f(9)+g(9)+h(9)+4k(9)+4)/9) remainder=4](Q83027.png)
$$\frac{\left(\mathrm{9}+\mathrm{2}\right)^{\mathrm{800}} }{\mathrm{9}}=\frac{\mathrm{9}^{\mathrm{800}} +\mathrm{800}{c}_{\mathrm{1}} \mathrm{9}^{\mathrm{799}} \mathrm{2}+...+\mathrm{2}^{\mathrm{800}} }{\mathrm{9}}=\frac{\boldsymbol{{f}}\left(\mathrm{9}\right)+\mathrm{2}^{\mathrm{800}} }{\mathrm{9}} \\ $$$$\frac{\boldsymbol{{f}}\left(\mathrm{9}\right)+\left(\mathrm{2}^{\mathrm{8}} \right)^{\mathrm{100}} }{\mathrm{9}}=\frac{\boldsymbol{{f}}\left(\mathrm{9}\right)+\left(\mathrm{256}\right)^{\mathrm{100}} }{\mathrm{9}}=\frac{{f}\left(\mathrm{9}\right)+\left(\mathrm{28}×\mathrm{9}+\mathrm{4}\right)^{\mathrm{100}} }{\mathrm{9}} \\ $$$$=\frac{{f}\left(\mathrm{9}\right)+{g}\left(\mathrm{9}\right)+\mathrm{4}^{\mathrm{100}} }{\mathrm{9}}=\frac{{f}\left(\mathrm{9}\right)+{g}\left(\mathrm{9}\right)+\left(\mathrm{4}^{\mathrm{4}} \right)^{\mathrm{25}} }{\mathrm{9}}=\frac{{f}\left(\mathrm{9}\right)+{g}\left(\mathrm{9}\right)+\left(\mathrm{252}+\mathrm{4}\right)^{\mathrm{25}} }{\mathrm{9}} \\ $$$$=\frac{{f}\left(\mathrm{9}\right)+{g}\left(\mathrm{9}\right)+{h}\left(\mathrm{9}\right)+\mathrm{4}^{\mathrm{25}} }{\mathrm{9}} \\ $$$$=\frac{{f}\left(\mathrm{9}\right)+{g}\left(\mathrm{9}\right)+{h}\left(\mathrm{9}\right)+\mathrm{4}×\left(\mathrm{4}^{\mathrm{3}} \right)^{\mathrm{8}} }{\mathrm{9}} \\ $$$$=\frac{{f}\left(\mathrm{9}\right)+{g}\left(\mathrm{9}\right)+{h}\left(\mathrm{9}\right)+\mathrm{4}×\left(\mathrm{63}+\mathrm{1}\right)^{\mathrm{8}} }{\mathrm{9}} \\ $$$$=\frac{{f}\left(\mathrm{9}\right)+{g}\left(\mathrm{9}\right)+{h}\left(\mathrm{9}\right)+\mathrm{4}\left[{k}\left(\mathrm{9}\right)+\mathrm{1}\right]}{\mathrm{9}}\:\left[{note}\:\mathrm{63}\:{is}\:{divisible}\:{by}\:\mathrm{9}\right] \\ $$$${and}\:{k}\left(\mathrm{9}\right)=\left(\mathrm{63}+\mathrm{1}\right)^{\mathrm{8}} =\left(\mathrm{63}^{\mathrm{8}} \right)+\mathrm{8}{c}_{\mathrm{1}} \left(\mathrm{63}\right)^{\mathrm{7}} +...+\mathrm{1} \\ $$$$\boldsymbol{{so}} \\ $$$$=\frac{{f}\left(\mathrm{9}\right)+{g}\left(\mathrm{9}\right)+{h}\left(\mathrm{9}\right)+\mathrm{4}{k}\left(\mathrm{9}\right)+\mathrm{4}}{\mathrm{9}}\:\:\boldsymbol{{remainder}}=\mathrm{4} \\ $$