Here are some RMO level 1 problems. One who makes his place in the leader board will be then promoted to level 2. The one who answers maximum of them will be the winner! Post full solutions! Questions!: Number Theory 1. Prove that 4 does not divide \( m^{2} + 2 \) for any integer m. 2.Find the last two digits of \(3^{1997}\).

This is not the end of the contest! Around 100's of questions are coming soon! First solve these. A piece of cake I guess! Points will be granted according to number of questions solved and clarity of solutions. Best of Luck!

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TopNewestSo, let m be any positive integer. Now 'm' can be written as \( m= 4b+r\) where \(( 0 \leq r < 4)\) Now this means all the positive integers in the world can be represented in this form.

Now the value of \( r = 0,1,2,3\)

\(1. If \) \(r=0\) then \( m= 4b+r\)

\( m= 4b\)

\( m^{2} = 4b^{2}\)

\( m^{2} = 16b^{2}\)

But it is given that whether 4 is divided by \( m^{2}+2\)

Now, \( m^{2}+2\) = \(16b^{2} +2\)= \( 4(4b^{2})+2\)

Now, \( 4(4b^{2})+2 \equiv 2 \pmod{4} \)

So, we are getting remainder \(2\). So for first case it is not divisible.

\(2. If \) \(r=1\) then \( m= 4b+r\)

\( m= 4b+1\)

\(m^{2}= (4b+1)^{2}\)

\(m^{2} = 16b^{2} + 8b +1\)

\(m^{2}= 4(4b^{2} + 2b) +1\)

But it is given that whether 4 is divided by \( m^{2}+2\)

Now, \( m^{2}+2\) = \(4(4b^{2} + 2b) +1+2\)

\( m^{2}+2\) = \(4(4b^{2} + 2b) + 3\)

Now, \( 4(4b^{2}+2b)+3 \equiv 3 \pmod{4} \)

So, we are getting remainder \(3\). So for second case it is not divisible.

\(3. If \) \(r=2\) then \( m= 4b+r\) \( m= 4b+2\)

\(m^{2}= (4b+2)^{2}\)

\(m^{2} = 16b^{2} + 16b +4\)

\(m^{2}= 4(4b^{2} + 4b+1)\)

But it is given that whether 4 is divided by \( m^{2}+2\)

Now, \( m^{2}+2\) = \(4(4b^{2} + 4b+1) +2\)

Now, \( 4(4b^{2}+4b+1)+2 \equiv 2 \pmod{4} \)

So, we are getting remainder \(2\). So for third case it is not divisible.

\(4. If\) \(r=3\) then \( m= 4b+r\)

\( m= 4b+3\)

\(m^{2}= (4b+3)^{2}\)

\(m^{2} = 16b^{2} + 24b +9\)

\(m^{2} = 16b^{2} + 24b +8+1\)

\(m^{2}= 4(4b^{2} + 6b+2)+1\)

But it is given that whether 4 is divided by \( m^{2}+2\)

Now, \( m^{2}+2\) = \(4(4b^{2} + 6b+2)+1+2\)

\(= 4(4b^{2} + 6b+2)+3\)

Now, \( 4(4b^{2}+6b+2)+3 \equiv 3 \pmod{4} \)

So, we are getting remainder \(3\). So for forth case it is not divisible.

As in all fours cases if \(m^{2}+2\) is divided by 4 we are getting some remainders . That means 4 does not divide \(m^{2}+2\)for any integer m – Abhisek Mohanty · 1 year, 6 months ago

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– Swapnil Das · 1 year, 6 months ago

U r of which class?Log in to reply

– Abhisek Mohanty · 1 year, 6 months ago

Just passed ninth and studying in 10th. And you !!!!!!!!!!!!!!Log in to reply

– Swapnil Das · 1 year, 6 months ago

Your age, 15?Log in to reply

– Swapnil Das · 1 year, 6 months ago

class 9, bhaiya!Log in to reply

– Abhisek Mohanty · 1 year, 6 months ago

Fine brother. But your age is same as of me , 14 yearsLog in to reply

– Swapnil Das · 1 year, 6 months ago

Ur birthdate?Log in to reply

– Abhisek Mohanty · 1 year, 6 months ago

18 mayLog in to reply

– Swapnil Das · 1 year, 6 months ago

Ur gonna be 15, so U are six months older than me, whoo! I thought I was studying late!Log in to reply

– Abhisek Mohanty · 1 year, 6 months ago

Fine. Are you in whatsappLog in to reply

– Swapnil Das · 1 year, 6 months ago

Nope, on google, but not frequent. I usually spend time in research and solving problems.Log in to reply

– Bryan Lee Shi Yang · 1 year, 6 months ago

You wanna ask my birthdate?Log in to reply

– Swapnil Das · 1 year, 6 months ago

Ya, if U would like to tell!Log in to reply

– Bryan Lee Shi Yang · 1 year, 6 months ago

DD/MM/YY=16/12/2003Log in to reply

For the 2nd question, my solution is as follows :

3^1997 = (3^6)^332 * 3^5 (mod 100)

Therefore, the last 2 digits of 3^1997 are 63 – Manish Dash · 1 year, 6 months ago

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\(3^{1}\)=3, \(3^{2}\)=9, \(3^{3}\)=27, \(3^{4}\)=81, \(3^{5}\)=43, \(3^{6}\)=29, \(3^{7}\)=87, \(3^{8}\)=61, \(3^{9}\)=83, \(3^{10}\)=49, \(3^{11}\)=47, \(3^{12}\)=41, \(3^{13}\)=23, \(3^{14}\)=69, \(3^{15}\)=07, \(3^{16}\)=21, \(3^{17}\)=63, \(3^{18}\)=89, \(3^{19}\)=67, \(3^{20}\)=01. And so on......

\(\frac{1997}{20}=99......17.\)

Refer to \(3^{17}\).

So, the last 2 digits of \(3^{1997}\) = 63. LoL

It's a VERY VERY old method...... XD – Bryan Lee Shi Yang · 1 year, 6 months ago

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– Bryan Lee Shi Yang · 1 year, 6 months ago

Um...... what reply do I get?......Log in to reply

– Swapnil Das · 1 year, 6 months ago

Upvoted!Log in to reply

– Bryan Lee Shi Yang · 1 year, 6 months ago

Oh........Log in to reply

– Bryan Lee Shi Yang · 1 year, 6 months ago

What does it mean? Correct, or how many points?Log in to reply

– Swapnil Das · 1 year, 6 months ago

2Log in to reply

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– Swapnil Das · 1 year, 6 months ago

2 points!Log in to reply

– Bryan Lee Shi Yang · 1 year, 6 months ago

TQVM!Log in to reply

\(The\) Last two digits of \(3^{1997}\) is 63. Just see the pattern. – Abhisek Mohanty · 1 year, 6 months ago

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– Swapnil Das · 1 year, 6 months ago

Corect answer ! 1 point!Log in to reply

– Abhisek Mohanty · 1 year, 6 months ago

But why only 1 point to me. And another thing do you go to FIITJEE bhubaneswar for coaching!!!!!!!!!!!!! If yes we can be friends because I also go there!!!!!!!!!Log in to reply

– Swapnil Das · 1 year, 6 months ago

Hi Abhisek, 1 point to U because U solved the easy problem. And no, I do not go to FIITJEE, but my first VIDYAMANDIR class is gonna start tommorrow. And no doubt we are good friends!Log in to reply

– Abhisek Mohanty · 1 year, 6 months ago

Ya thats for sure . And see my solution. I hope you would like it............................................Log in to reply

\(m^2 + 2 = 2 or 3 (mod4)\)

but \(4 = 0 (mod4)\)

A contradiction. – Dev Sharma · 1 year, 1 month ago

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Hey bros, I live very very far away from you. When would you like to come to my country? – Bryan Lee Shi Yang · 1 year, 6 months ago

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– Swapnil Das · 1 year, 6 months ago

if given a chance, sure!Log in to reply

– Bryan Lee Shi Yang · 1 year, 6 months ago

Um...... When?Log in to reply

– Swapnil Das · 1 year, 6 months ago

when given a chance or when u seriously need it!Log in to reply

– Bryan Lee Shi Yang · 1 year, 6 months ago

Haha......Log in to reply