Hi Everyone! This was my First experience Creating a proof problem. So this is what I came up with.

Q1. Prove that ${99}^{n}$ Ends with a "99" For N is an Odd natural Number

AND

Q2. Prove that ${99}^{n}$ Ends with a "01" For N is an even natural Number

Both These problems Are One Liners. They Can be Proved in one line. So, I would like You to Provide VERY Brief solutions i.e. One or Two lines. Try it!

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## Comments

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TopNewestI'm not exactly sure about my solution being actually "Single or double -lined" but here goes:(Both questions are solved using binomial theorem)

Q1:

$99^n=(100-1)^n \equiv 100^n - {n \choose 1} 100^{n-1} + {n \choose 2} 100^{n-2} - \dots - 1 \equiv 100 -1 \equiv 99 \pmod {100}$

Q2:Exactly goes the same:

$99^n=(100-1)^n \equiv 100^n + {n \choose 1}100^{n-1} + {n \choose 2}100^{n-2} + \dots + 1 \equiv 100 +1 \equiv 1 \pmod {100}$

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I wasn't able to figure it out in one line but here's what I think it should be;The answers are in reverse, i.e, 2 first and then 1.

2) $99^{1}=99$, so it is true for $1$. Now, suppose $99^{(2n-1)}$ ends in 99.

Let $99^{(2n-1)}=100x+99$.

Then, $99^{2n}=9900x+9801=100(99x+98)+1$,

So $99^{2n}$ ends in $01$.

1) That implies that $99^{(2n+1)}=9900(99x+98)+99=100(9801x+9702)+99$,

So $99^{(2n+1)}$ ends in $99$. Therefore, by induction, it is true for all odd numbers.

Cheers! xD

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Cheers! @Sravanth Chebrolu

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Cheers! what d'you think about my solution???

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I wonder if my solution is also correct.Do you mind.checking it out as well?

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@arian tashakkor! cheers! xD

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Mod 100, 99*99=01, and 01*99=99. You just end up in a loop from there.

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well mehul here is a short and amazing proof of both questions 99=-1 mod 100 so${99}^{n}$ =${-1}^{n}$ =-1 =99 mod 100 if n is odd and 1 if n is even. so $99^n$ Ends with a "01" For N is an even natural Number and ({99}^{n}) Ends with a "99" For N is an Odd natural Number as remainder by 100 is last 2 digits. well i hope u understand this . is this good?

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