Consider the following results.

\(99^{1} = 99\)

\(99^{2} = 9801\)

\(99^{3} = 970299\)

\(99^{4} = 96059601\)

\(99^{5} = 9509900499\)

Prove that \(99^{n}\) ends in 99 for odd n.

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TopNewest\(99\equiv(-1)\pmod{100} \Rightarrow 99^{n}\equiv(-1)^{n}\pmod{100}.\)But,whenever \(n\) is an odd integer\(, (-1)^{n}=(-1) \Rightarrow 99^{n}\equiv(-1)\equiv(99)\pmod{100}\) for all odd \(n.\)

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Seriously. And Obviously -_- This was easy. Welll Done Adarsh Kumar

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For this , let us consider two cases when \(n\) is odd and when it is even ,

When \(n\) is even :

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This is similar to prove that

\((100-1)^n\equiv 99 \mod 100\)

For \(n\) is an odd number.

Noticed that every term of the expansion is divisible by \(100\) except for the last term, \(-1\). Hence we get

\((100-1)^n\equiv -1 \mod 100\)

\((100-1)^n\equiv 99 \mod 100\)

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While for \(n\) is an even number, a similar approach can prove that \(99^n\) ends in \(01\).

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