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 ${99}^{n}$ 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!

#Easy #Proof #99 #OneLiner

Easy Math Editor

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Math	Appears as
Remember to wrap math in `$` ... `$` or `\[` ... `\]` to ensure proper formatting.
`2 \times 3`	$2 \times 3$
`2^{34}`	$2^{34}$
`a_{i-1}`	$a_{i-1}$
`\frac{2}{3}`	$\frac{2}{3}$
`\sqrt{2}`	$\sqrt{2}$
`\sum_{i=1}^3`	$\sum_{i=1}^3$
`\sin \theta`	$\sin \theta$
`\boxed{123}`	$\boxed{123}$

Comments

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:

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

Q2:Exactly goes the same:

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

Arian Tashakkor - 6 years, 1 month ago

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

Sravanth C. - 6 years, 1 month ago

Cheers! @Sravanth Chebrolu

Mehul Arora - 6 years, 1 month ago

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

Sravanth C. - 6 years, 1 month ago

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

Arian Tashakkor - 6 years, 1 month ago

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

Sravanth C. - 6 years, 1 month ago

@Sravanth C. – Thank you! Cheers! xD

Arian Tashakkor - 6 years, 1 month ago

@Arian Tashakkor – Welcome @arian tashakkor! cheers! xD

Sravanth C. - 6 years, 1 month ago

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

Akiva Weinberger - 6 years, 1 month ago

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?

Kaustubh Miglani - 5 years, 5 months ago

Math	Appears as
Remember to wrap math in `\(` ... `\)` or `\[` ... `\]` to ensure proper formatting.
`2 \times 3`	$2 \times 3$
`2^{34}`	$2^{34}$
`a_{i-1}`	$a_{i-1}$
`\frac{2}{3}`	$\frac{2}{3}$
`\sqrt{2}`	$\sqrt{2}$
`\sum_{i=1}^3`	$\sum_{i=1}^3$
`\sin \theta`	$\sin \theta$
`\boxed{123}`	$\boxed{123}$