Question

The number $(101)^{100} - 1$ is divisible by

• A. $10^4$
• B. $10^6$
• C. $10^8$
• D. $10^{12}$

Answer 1

Answer: (A)

$(101)^{100}-1 =(1+100)^{100}-1 $
$= \left(1+{ }^{100} C_{1} \cdot 100+{ }^{100} C_{2} 100^{2}+\ldots \ldots\right)-1 $
$= { }^{100} C_{1} 100+{ }^{100} C_{2}(100)^{2}+$
${ }^{100} C_{3}(100)^{3}+\ldots \ldots .+{ }^{100} C_{100}(100)^{100}$
$= 10^{4}\left(1+{ }^{100} C_{2}+{ }^{100} C_{3} 10^{2}+\ldots\right.$
$+{ }^{100} C_{100}(100)^{98}$
$= 10^{4}(1+$ an integer multiple of $ 10) $