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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 · 1 year, 9 months ago

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U r of which class? · 1 year, 9 months ago

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Just passed ninth and studying in 10th. And you !!!!!!!!!!!!!! · 1 year, 9 months ago

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Your age, 15? · 1 year, 9 months ago

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class 9, bhaiya! · 1 year, 9 months ago

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

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Ur birthdate? · 1 year, 9 months ago

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18 may · 1 year, 9 months ago

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

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Fine. Are you in whatsapp · 1 year, 9 months ago

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

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You wanna ask my birthdate? · 1 year, 9 months ago

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Ya, if U would like to tell! · 1 year, 9 months ago

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DD/MM/YY=16/12/2003 · 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 · 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 · 1 year, 9 months ago

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Um...... what reply do I get?...... · 1 year, 9 months ago

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Upvoted! · 1 year, 9 months ago

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Oh........ · 1 year, 9 months ago

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What does it mean? Correct, or how many points? · 1 year, 9 months ago

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2 · 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.
· 1 year, 9 months ago

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2 points! · 1 year, 9 months ago

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TQVM! · 1 year, 9 months ago

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

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Corect answer ! 1 point! · 1 year, 9 months ago

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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!!!!!!!!! · 1 year, 9 months ago

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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! · 1 year, 9 months ago

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

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

but $$4 = 0 (mod4)$$

A contradiction. · 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? · 1 year, 9 months ago

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if given a chance, sure! · 1 year, 9 months ago

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Um...... When? · 1 year, 9 months ago

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when given a chance or when u seriously need it! · 1 year, 9 months ago

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Haha...... · 1 year, 9 months ago

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