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, $13^{99}=\left(1 + 12\right)^{99}=1+^{99}C_{1}\cdot 12+^{99}C_{2}\cdot 12^{2}+^{99}C_{3}\cdot 12^{3}+^{99}C_{4}\cdot 12^{4}+....$
$=1+99\times 12+\frac{99 \times 98}{2}\cdot 12^{2}+\frac{33 \cancel{99} \times 49 \cancel{98} \times 97}{\cancel{3} \times \cancel{2}}\times 12^{3}+81Ι_{1}$
$=1+99\times 12+99\times 49\cdot 12^{2}+3\times 11\times 49\times 97\times 12^{3}+81Ι_{1}$
$=1+99\times 12+81\left\{11 \times 49 \times 16 + 11 \times 49 \times 97 \times 64 + Ι_{1}\right\}$
$13^{99}=1+99\times 12+81Ι_{2}$
$19^{93}=\left(1 + 18\right)^{93}=1+^{93}C_{1}\cdot 18+^{93}C_{2}\cdot 18^{2}+....$
$=1+93\times 18+81Ι_{3}$
$\Rightarrow 13^{99}-19^{93}=99\times 12-93\times 18+81\left(Ι_{2} - Ι_{3}\right)$
$=27\left\{11 \times 4 - 31 \times 2\right\}+81Ι_{4}$
$=27\left(44 - 62\right)+81Ι_{4}=27\times \left(- 18\right)+81Ι_{4}$
$=81\left\{- 6 + Ι_{4}\right\}=81Ι_{5}$
Hence, it is also divisible by $81$ .
So, it is divisible by $162\Rightarrow $ Remainder $=0.$