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# 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!

Note by Mehul Arora
1 year, 10 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:

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

Q2:Exactly goes the same:

$$99^n=(100-1)^n \equiv 100^n + {n \choose 1}100^{n-1} + {n \choose 2}100^{n-2} + \dots + 1 \equiv 100 +1 \equiv 1 \pmod {100}$$ · 1 year, 10 months 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? · 1 year, 2 months ago

Mod 100, 99*99=01, and 01*99=99. You just end up in a loop from there. · 1 year, 10 months 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 · 1 year, 10 months ago

Cheers! @Sravanth Chebrolu · 1 year, 10 months ago

I wonder if my solution is also correct.Do you mind.checking it out as well? · 1 year, 10 months ago

Oh yes! It's absolutely right(sir). · 1 year, 10 months ago

Thank you! Cheers! xD · 1 year, 10 months ago

Welcome @arian tashakkor! cheers! xD · 1 year, 10 months ago