# Maths RMO Problems(1)........

Hey friends.

I mean Brilliantians I am back with some amazing problems which are generally asked in the RMO-INMO level examination .I am sharing the image of the paper containing the questions.

Please try and if possible send the solutions

.Also it would be great if you all participate in sharing the questions from your own. I would also be sharing problems based on NSEP level. Thanks Dont forget to share and like this............. Note by Abhisek Mohanty
3 years, 3 months ago

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Question 2

$(xy-7)^2=x^2+y^2\Rightarrow x^2y^2-14xy+49=x^2+y^2 \\ x^2y^2-12xy+36+13=x^2+y^2+2xy \\ (x+y+xy-6)(x+y-xy+6)=13=13×1=1×13=-13×-1=-1×-13 \\ (x,y)=(3,4),(4,3),(0,7),(7,0)$

- 3 years, 3 months ago

Nice solution................upvoted....

- 3 years, 3 months ago

Question 1 \begin{aligned} x &\equiv 0,1,2,3,4,5,6 \pmod{7}\\ x^3 &\equiv 0,1,-1\pmod{7} \end{aligned}

Assume that either of $a,b,c$ is a multiple of 7.

Then obviously $abc(a^3-b^3)(b^3-c^3)(c^3-a^3)$ is a multiple of 7.

So now WLOG assume that $a,b,c\not\equiv 0\pmod{7}$.Then $a^3,b^3,c^3\equiv 1,-1\pmod{7}$.There are $2\times 2\times 2=8$ different possible cases corresponding to the different values of $a^3,b^3$ and $c^3$ modulo 7,which are: $\begin{array}{c|c|c|c} \text{Values modulo 7} & a^3 & b^3 & c^3 \\ \hline \text{Case 1} & 1 & 1 & 1 \\ \hline \text{Case 2} & -1 & -1 & -1 \\ \hline \text{Case 3} & 1 & -1 & -1 \\ \hline \text{Case 4} & -1 & 1 & -1 \\ \hline \text{Case 5} & -1 & -1 & 1 \\ \hline \text{Case 6} & -1 & 1 & 1 \\ \hline \text{Case 7} & 1 & -1 & 1 \\ \hline \text{Case 8} & 1 & 1 & -1 \end{array}$

Observe that,because of the symmetry of the expression,Cases 3,4,5 and Cases 6,7,8 are equivalent.Therefore,we only need to check Case 1,2,3 and 6.Simply evaluate the cases to get that $abc(a^3-b^3)(b^3-c^3)(c^3-a^3)\equiv 0\pmod{7}\;\forall \;a,b,c\in \mathbb{Z}$

- 3 years, 3 months ago

Nice solution bro..........upvoted

- 3 years, 3 months ago

vmc questions

- 3 years, 3 months ago

Question 8(i)

Let the roots be $a,b,c,d$.(Note that the roots are positive).Then: \begin{aligned} p&=-(a+b+c+d)\\ q&=ab+ac+ad+bc+bd+cd\\ r&=-(abc+abd+acd+bcd)\\ s&=abcd \end{aligned} $pr-16s\geq 0\implies (a+b+c+d)(abc+abd+acd+bcd)\geq 16abcd$ which follows by applying AM-GM on each term.

I couldn't understand;what does the variable "a" denote in Q 8(ii)?

- 3 years, 3 months ago

Question 9 We shall analyze 2 cases. Case 1 Either of $a,b$ is even.

WLOG let $a$ be even.Then $ab(a-b)=45045$ is even.But 45045 is odd,contradiction. Hence no solutions exist in this case.

Case 2 Both $a,b$ are odd.

Then $a-b$ is even.Therefore $ab(a-b)=45045$ is even,contradiction.

Hence no solutions exist.

- 3 years, 3 months ago

- 3 years, 3 months ago

question 11 given 34x=43y =>34x+43x=43(y+x) =>77x=43(x+y) now 43 does not divide 77 hence x+y contains 77 i.e-11*7 hence x+y is not prime.

- 3 years, 3 months ago

That does not exclude the possibility of x+y being odd but not prime

- 3 years, 3 months ago

Can you please explain me what u are trying to say?

- 3 years, 3 months ago

Can anyone recommend me some good books for INMO and and other maths olympiad??????????

- 3 years, 3 months ago

Q13 The sum of the digits of any number formed using the given conditions is 1 + 4 + 9+ . . . . + 81 = 285 = 3(95) which implies the number is divisible by 3 but not by 3 squared which is 9. Therefore any number formed using the given conditions is not a perfect square

- 3 years, 1 month ago