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The number $(101)^{100}-1$ is divisible by
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Verified Answer
The correct answer is:
$10^{4}$
$(101)^{100}-1=(1+100)^{100}-1$
$=\left(1+{ }^{100} C_{1} \cdot 100+{ }^{100} C_{2} 100^{2}+\ldots \ldots\right)-1$
$={ }^{100} C_{1} 100+{ }^{100} C_{2}(100)^{2}+$
$\quad{ }^{100} C_{3}(100)^{3}+\ldots \ldots+{ }^{100} C_{100}(100)^{100}$
$=10^{4}\left(1+{ }^{100} C_{2}+{ }^{100} C_{3} 10^{2}+\ldots .\right.$
$\quad+{ }^{100} C_{100}(100)^{98}$
$=10^{4}(1+$ an integer multiple of 10$)$
$=\left(1+{ }^{100} C_{1} \cdot 100+{ }^{100} C_{2} 100^{2}+\ldots \ldots\right)-1$
$={ }^{100} C_{1} 100+{ }^{100} C_{2}(100)^{2}+$
$\quad{ }^{100} C_{3}(100)^{3}+\ldots \ldots+{ }^{100} C_{100}(100)^{100}$
$=10^{4}\left(1+{ }^{100} C_{2}+{ }^{100} C_{3} 10^{2}+\ldots .\right.$
$\quad+{ }^{100} C_{100}(100)^{98}$
$=10^{4}(1+$ an integer multiple of 10$)$
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