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Prove .....

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.

Note by A K
1 year, 11 months ago

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\(99\equiv(-1)\pmod{100} \Rightarrow 99^{n}\equiv(-1)^{n}\pmod{100}.\)But,whenever \(n\) is an odd integer\(, (-1)^{n}=(-1) \Rightarrow 99^{n}\equiv(-1)\equiv(99)\pmod{100}\) for all odd \(n.\) Adarsh Kumar · 1 year, 11 months ago

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@Adarsh Kumar Seriously. And Obviously -_- This was easy. Welll Done Adarsh Kumar Mehul Arora · 1 year, 6 months ago

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

When \(n\) is even :

\(99\) could be expressed as \(100-1\). So, we could write it as \((100-1)^{n}\). As in this case we have assumed \(n\) to be even us let us write \(n=2k\) . So we have \((100-1)^{2k}\) . Now , little bit expanding it , we have , \(10000+1-200)^{k}\) . Now taking \(p=10000+1\) and \(q=-200\) , we have \((p+q)^{n}\) . Now let us prove one more result , for any \((100+1)^{x}\) , the last two digits would be \(01\) . By expanding by binomial theory , \[(100+1)^{x} = 100^{x}+a_{1}100^{x-1} \times 1 + . . . . a_{y}100 \times 1^{x-1} + 1^{x}\] So, expanding it will have \(x+1\) terms. It is clear that the first \(x\) terms would have \(00\) in their last as they are multiples of some powers of \(100\). But the last term will be 1. So , returning to our expansion we, would have , \[p^{n} +a_{1}p^{n- 1} q^ {1} + . . . . . a_{y} p q^{n-1} + q ^{n}\] Now in this we have except \(p^{n}\) all other terms would be a multiple of \(100\) as \(q\) is a motile of it . As we have earlier proove that the last term of \(p^{n}\) would be 1 . And , as all other terms end in \(00\) , therefore 3rd two digits of the whole computation would be \(01\) . So we have thus profed that any power that is even would yield a result of 01 at the end when applied to \(100-1\). Case 2 could be proved in a similar way and after proving it we would have proved the result so instead doing that long proof for case 2 also we could say that any odd number can be expressed \(n=2k+1\) so just multiplying one more \(99\) to any even power we would obtain the odd power. And multiplying the end two digits which are \(01\) for any even power of 99 as we have proved , by 99 would give us 99 at the end. Hence , proved.

QED Utkarsh Dwivedi · 1 year, 11 months ago

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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

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

\((100-1)^n\equiv 99 \mod 100\) Christopher Boo · 1 year, 11 months ago

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@Christopher Boo While for \(n\) is an even number, a similar approach can prove that \(99^n\) ends in \(01\). Christopher Boo · 1 year, 11 months ago

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