Question

The number \((101)^{100}-1\) is divisible by
(A) \(10^{4}\)
(B) \(10^{6}\)
(C) \(10^{8}\)
(D) \(10^{12}\)

Answer 1

Ans: (A)
Hint : \((101)^{100}-1=(1+100)^{100}-1\)
$$
\begin{aligned}
&=\left[1+{ }^{100} \mathrm{C}_{1} 100+{ }^{100} \mathrm{C}_{2} 100^{2}+\ldots \ldots .+{ }^{100} \mathrm{C}_{100}(100)^{100}\right]-1=10^{4}\left(1+{ }^{100} \mathrm{C}_{2}+{ }^{100} \mathrm{C}_{3} 10^{2}+\ldots .+{ }^{100} \mathrm{C}_{100}(100)^{98}\right) \\
&=10^{4}(1+\text { an integer multiple of } 10)
\end{aligned}
$$

Answer 2

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Answer 3

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