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

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U r of which class?

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

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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

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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

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Um...... what reply do I get?......

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Upvoted!

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2 points!

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TQVM!

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\(The\) Last two digits of \(3^{1997}\) is 63. Just see the pattern.

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Corect answer ! 1 point!

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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!!!!!!!!!

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

but \(4 = 0 (mod4)\)

A contradiction.

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Hey bros, I live very very far away from you. When would you like to come to my country?

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if given a chance, sure!

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Um...... When?

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