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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
2 years, 1 month ago

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1. Prove that 4 does not divide $$m^{2}+2$$

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 · 2 years, 1 month ago

U r of which class? · 2 years, 1 month ago

Just passed ninth and studying in 10th. And you !!!!!!!!!!!!!! · 2 years, 1 month ago

Your age, 15? · 2 years, 1 month ago

class 9, bhaiya! · 2 years, 1 month ago

Fine brother. But your age is same as of me , 14 years · 2 years, 1 month ago

Ur birthdate? · 2 years, 1 month ago

18 may · 2 years, 1 month ago

Ur gonna be 15, so U are six months older than me, whoo! I thought I was studying late! · 2 years, 1 month ago

Fine. Are you in whatsapp · 2 years, 1 month ago

Nope, on google, but not frequent. I usually spend time in research and solving problems. · 2 years, 1 month ago

You wanna ask my birthdate? · 2 years, 1 month ago

Ya, if U would like to tell! · 2 years, 1 month ago

DD/MM/YY=16/12/2003 · 2 years, 1 month ago

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 · 2 years, 1 month ago

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 · 2 years, 1 month ago

Um...... what reply do I get?...... · 2 years, 1 month ago

Upvoted! · 2 years, 1 month ago

Oh........ · 2 years, 1 month ago

What does it mean? Correct, or how many points? · 2 years, 1 month ago

2 · 2 years, 1 month ago

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.
· 2 years, 1 month ago

2 points! · 2 years, 1 month ago

TQVM! · 2 years, 1 month ago

$$The$$ Last two digits of $$3^{1997}$$ is 63. Just see the pattern. · 2 years, 1 month ago

Corect answer ! 1 point! · 2 years, 1 month 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!!!!!!!!! · 2 years, 1 month 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! · 2 years, 1 month ago

Ya thats for sure . And see my solution. I hope you would like it............................................ · 2 years, 1 month ago

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

but $$4 = 0 (mod4)$$

A contradiction. · 1 year, 9 months ago

Hey bros, I live very very far away from you. When would you like to come to my country? · 2 years, 1 month ago

if given a chance, sure! · 2 years, 1 month ago

Um...... When? · 2 years, 1 month ago

when given a chance or when u seriously need it! · 2 years, 1 month ago

Haha...... · 2 years, 1 month ago

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