Easy Proof Problems.

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

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

AND

Q2. Prove that 99n{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!

Note by Mehul Arora
4 years, 5 months ago

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

Q1:

99n=(1001)n100n(n1)100n1+(n2)100n21100199(mod100) 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:

99n=(1001)n100n+(n1)100n1+(n2)100n2++1100+11(mod100)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}

Arian Tashakkor - 4 years, 5 months ago

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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) 991=9999^{1}=99, so it is true for 11. Now, suppose 99(2n1)99^{(2n-1)} ends in 99.

  • Let 99(2n1)=100x+9999^{(2n-1)}=100x+99.

  • Then, 992n=9900x+9801=100(99x+98)+199^{2n}=9900x+9801=100(99x+98)+1,

  • So 992n99^{2n} ends in 0101.


  • 1) That implies that 99(2n+1)=9900(99x+98)+99=100(9801x+9702)+9999^{(2n+1)}=9900(99x+98)+99=100(9801x+9702)+99,

  • So 99(2n+1)99^{(2n+1)} ends in 9999. Therefore, by induction, it is true for all odd numbers.


Cheers! xD

Sravanth Chebrolu - 4 years, 5 months ago

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Cheers! @Sravanth Chebrolu

Mehul Arora - 4 years, 5 months ago

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

Sravanth Chebrolu - 4 years, 5 months ago

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

Arian Tashakkor - 4 years, 5 months ago

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@Arian Tashakkor Oh yes! It's absolutely right(sir).

Sravanth Chebrolu - 4 years, 5 months ago

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@Sravanth Chebrolu Thank you! Cheers! xD

Arian Tashakkor - 4 years, 5 months ago

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@Arian Tashakkor Welcome @arian tashakkor! cheers! xD

Sravanth Chebrolu - 4 years, 5 months ago

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

Akiva Weinberger - 4 years, 5 months ago

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well mehul here is a short and amazing proof of both questions 99=-1 mod 100 so99n{99}^{n} =1n{-1}^{n} =-1 =99 mod 100 if n is odd and 1 if n is even. so 99n99^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?

Kaustubh Miglani - 3 years, 9 months ago

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