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Level 1 Mathematics Olympiad contest! - By Swapnil

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!

Note by Swapnil Das
1 year, 9 months ago

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  1. Prove that 4 does not divide \( m^{2}+2\)
Sol- I think you must have read about Euclids Division Lemma.

So, 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, 9 months ago

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@Abhisek Mohanty U r of which class? Swapnil Das · 1 year, 9 months ago

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@Swapnil Das Just passed ninth and studying in 10th. And you !!!!!!!!!!!!!! Abhisek Mohanty · 1 year, 9 months ago

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@Abhisek Mohanty Your age, 15? Swapnil Das · 1 year, 9 months ago

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@Abhisek Mohanty class 9, bhaiya! Swapnil Das · 1 year, 9 months ago

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@Swapnil Das Fine brother. But your age is same as of me , 14 years Abhisek Mohanty · 1 year, 9 months ago

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@Abhisek Mohanty Ur birthdate? Swapnil Das · 1 year, 9 months ago

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@Swapnil Das 18 may Abhisek Mohanty · 1 year, 9 months ago

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@Abhisek Mohanty Ur gonna be 15, so U are six months older than me, whoo! I thought I was studying late! Swapnil Das · 1 year, 9 months ago

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@Swapnil Das Fine. Are you in whatsapp Abhisek Mohanty · 1 year, 9 months ago

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@Abhisek Mohanty Nope, on google, but not frequent. I usually spend time in research and solving problems. Swapnil Das · 1 year, 9 months ago

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@Swapnil Das You wanna ask my birthdate? Bryan Lee Shi Yang · 1 year, 9 months ago

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@Bryan Lee Shi Yang Ya, if U would like to tell! Swapnil Das · 1 year, 9 months ago

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@Swapnil Das DD/MM/YY=16/12/2003 Bryan Lee Shi Yang · 1 year, 9 months ago

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

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

             = (29)^332 * 43                 (mod 100)

             =   (29^2)^166 * 43          (mod 100)

             = (41)^166 * 43                  (mod 100)

             = (41^2)^83 * 43               (mod 100)

             = (81)^83 * 43                      (mod 100) 

             = (81^2)^41 * 81 * 43        (mod 100)

             = (61)^41 * 83                       (mod 100)

             = (61^2)^20 * 61 * 83         (mod 100)

             = (21)^20 * 63                        (mod 100)

             = (21^2)^10 * 63                   (mod 100)

             = ( 41)^10 *63                        (mod 100)

              = ( 41^2)^5 * 63                    (mod 100)

              = (81)^5 *63                           (mod 100)

              = (81^2)^2 * 81 *63             (mod 100)

              = (61)^2 *3                               (mod 100) 

              = 21 * 3                                       (mod 100)     

              = 63                                             (mod 100)

Therefore, the last 2 digits of 3^1997 are 63 Manish Dash · 1 year, 9 months ago

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  1. Find the last 2 digits of \(3^{1997}\)
A cycle of \(3^{n}\) shows like this for every \(3^{n}\):

\(3^{1}\)=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, 9 months ago

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@Bryan Lee Shi Yang Um...... what reply do I get?...... Bryan Lee Shi Yang · 1 year, 9 months ago

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@Bryan Lee Shi Yang Upvoted! Swapnil Das · 1 year, 9 months ago

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@Swapnil Das Oh........ Bryan Lee Shi Yang · 1 year, 9 months ago

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@Swapnil Das What does it mean? Correct, or how many points? Bryan Lee Shi Yang · 1 year, 9 months ago

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@Bryan Lee Shi Yang 2 Swapnil Das · 1 year, 9 months ago

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  1. For every integer \(m\), when it's square is divided by 4 , it will give out 1 or 0 respectively. So when these 2 numbers are each added by 2, it gives out 2 and 3 respectively. What's more, for every integer \(m\) which is an even number, its square has at least 2 2's, so for every integer \(m\) which is even, it has at least 2×2=\(\boxed{4}\). These 2 ways can prove that for every integer \(m\), \(m^{2}+2\) will never be divisible by 4.
Bryan Lee Shi Yang · 1 year, 9 months ago

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@Bryan Lee Shi Yang 2 points! Swapnil Das · 1 year, 9 months ago

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@Swapnil Das TQVM! Bryan Lee Shi Yang · 1 year, 9 months ago

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\(The\) Last two digits of \(3^{1997}\) is 63. Just see the pattern. Abhisek Mohanty · 1 year, 9 months ago

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@Abhisek Mohanty Corect answer ! 1 point! Swapnil Das · 1 year, 9 months ago

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@Swapnil Das 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!!!!!!!!! Abhisek Mohanty · 1 year, 9 months ago

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@Abhisek Mohanty 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! Swapnil Das · 1 year, 9 months ago

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@Swapnil Das Ya thats for sure . And see my solution. I hope you would like it............................................ Abhisek Mohanty · 1 year, 9 months ago

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

but \(4 = 0 (mod4)\)

A contradiction. Dev Sharma · 1 year, 4 months 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, 9 months ago

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@Bryan Lee Shi Yang if given a chance, sure! Swapnil Das · 1 year, 9 months ago

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@Swapnil Das Um...... When? Bryan Lee Shi Yang · 1 year, 9 months ago

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@Bryan Lee Shi Yang when given a chance or when u seriously need it! Swapnil Das · 1 year, 9 months ago

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@Swapnil Das Haha...... Bryan Lee Shi Yang · 1 year, 9 months ago

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