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, $13^{99}=(1+12)^{99}=1+{ }^{99} C_{1} \cdot 12+{ }^{99} C_{2} \cdot 12^{2}+{ }^{99} C_{3} \cdot 12^{3}+{ }^{99} C_{4}$
$\cdot 12^{4}+\ldots$
$=1+99 \times 12+\frac{99 \times 98}{2} \cdot 12^{2}+\frac{3399 \times 4998 \times 97}{3 \times 2} \times 12^{3}+81 I _{1}$
$=1+99 \times 12+99 \times 49 \cdot 12^{2}+3 \times 11 \times 49 \times 97 \times 12^{3}+81 I _{1}$
$=1+99 \times 12+81\left\{11 \times 49 \times 16+11 \times 49 \times 97 \times 64+ I _{1}\right\}$
$13^{99}=1+99 \times 12+81 I _{2}$
$19^{93}=(1+18)^{93}=1+{ }^{93} C_{1} \cdot 18+{ }^{93} C_{2} \cdot 18^{2}+\ldots$
$=1+93 \times 18+81 I _{3}$
$\Rightarrow 13^{99}-19^{93}=99 \times 12-93 \times 18+81\left( I _{2}- I _{3}\right)$
$=27\{11 \times 4-31 \times 2\}+81 I _{4}$
$=27(44-62)+81 I _{4}=27 \times(-18)+81 I _{4}$
$=81\left\{-6+ I _{4}\right\}=81 I _{5}$
Hence, it is also divisible by $81 .$ So, it is divisible by $162 \Rightarrow $ Remainder $=0$.