COLUMN - I		COLUMN -  II
A)	$5^{97}$ is divided by 52, then the remainder is	p)	5
B)	${53}^{53}-{33}^3$ is divided by 10, then the remainder is	q)	6
C)	${27}^{10}+7^{51}$ is divided by 10, then the remainder is	r)	2
D)	${13}^{99}-{19}^{93}$ is divided by 162, then the remainder is	s)	0

 

 

 

 

 

 

$\begin{array}{c} For the Row A\\ 5^{97}=5·5^{96}\\ =5\cdot (25{)}^{48}\\ =5·{\left(26-1\right)}^{48}\\ =5\left[(26{)}^{48}-{}^{48}C_1(26{)}^{47}+...{-}^{48}{\mathrm{C}}_{47}\left(26\right)+1\right]\\ =5\left(52\mathrm{\lambda }\right)+5\end{array}$

Hence the remainder is 5.

B) ${53}^{53}-{33}^3=\left({53}^{53}-3^{53}\right)+3^{53}-\left[\left({33}^3-3^3\right)+3^3\right]$

$\Rightarrow {53}^{53}-3^{53}$ is divisible by 50

Then it is divisible by 10

$\Rightarrow {33}^3-3^3$ is divisible by 30 Hence by 10

$\begin{array}{l}\Rightarrow 3^{53}-3^3=3^3\left(3^{50}-1\right)\\ \Rightarrow =3^3\left[(10-1{)}^{25}-1\right]={23}^3[10\mathrm{K}-2]\Rightarrow =270\mathrm{k}-54\\ \Rightarrow =270\mathrm{k}-60+6\Rightarrow =10\mathrm{\lambda }+6\text{ reminder }6\end{array}$

C) ${27}^{10}+7^{51}=\left({27}^{10}-7^{10}\right)+7^{10}+7^{51}$ divisible by 20

Hence by ${107}^{10}+7^{51}={\left(7^2\right)}^5+(50-1{)}^{25}\cdot 7$

$\begin{array}{l}=(50-1{)}^5+(50-1{)}^{25}.7\\ \Rightarrow 10\mathrm{\lambda }-1+7(10\mathrm{k}-1)\Rightarrow =10\mathrm{\mu }-8\text{ remider }2\end{array}$

$\text{ D) }{13}^{99}-{19}^{33}=(1+12{)}^{99}-(1+18{)}^{93}$

$=99\times 12-93\times 18$is divisible by 81

also ${13}^{99}-{19}^{93}$ is divisible by 162