Question

If $13^{99}-19^{93}$ is divided by $162,$ then the remainder is

• A. $3$
• B. $6$
• C. $5$
• D. $0$

Answer 1

Answer: (D)

$13^{99}-19^{93}=$ odd number $-$ odd number $=$ even number
$\Rightarrow$ It is divisible by $2.$
Also, $13^{99}={\left(1+12\right)}^{99}=1{+}^{99}{C}_1\cdot 12{+}^{99}{C}_2\cdot 12^2{+}^{99}{C}_3\cdot 12^3{+}^{99}{C}_4\cdot 12^4+....$
$=1+99\times 12+\frac{99\times 98}{2}\cdot 12^2+\frac{33\cancel{99}\times 49\cancel{98}\times 97}{\cancel{3}\times \cancel{2}}\times 12^3+81I_1$
$=1+99\times 12+99\times 49\cdot 12^2+3\times 11\times 49\times 97\times 12^3+81I_1$
$=1+99\times 12+81\left\{11\times 49\times 16+11\times 49\times 97\times 64+I_1\right\}$
$13^{99}=1+99\times 12+81I_2$
$19^{93}={\left(1+18\right)}^{93}=1{+}^{93}{C}_1\cdot 18{+}^{93}{C}_2\cdot 18^2+....$
$=1+93\times 18+81I_3$
$\Rightarrow 13^{99}-19^{93}=99\times 12-93\times 18+81\left(I_2-I_3\right)$
$=27\left\{11\times 4-31\times 2\right\}+81I_4$
$=27\left(44-62\right)+81I_4=27\times \left(-18\right)+81I_4$
$=81\left\{-6+I_4\right\}=81I_5$
Hence, it is also divisible by $81$ .
So, it is divisible by $162\Rightarrow$ Remainder $=0.$