RMO adventure

1) Solve the equation y3=x3+8x26x+8y^3 = x^3 + 8x^2 - 6x + 8 for positive integer x and y.

2) Consider two positive integer a and b which are such that aabba^{a}b^{b} is divisible by 2000. What is minimum value of ab?

Note by Dev Sharma
4 years ago

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Solution for 2nd2^{nd} question :

2000aabb2a or b and 5a or b.Therefore, in every case 10ab.Thus, the minimum value of ab is obviously 10, possible cases 101011 and 111010.2000|a^{a}b^{b}\Rightarrow 2|a\text{ or }b \text{ and } 5|a \text{ or } b. \\ \text{Therefore, in every case } 10|ab. \\ \text{Thus, the minimum value of }ab \text{ is obviously }\boxed{10}\text{, possible cases }10^{10}1^{1} \text{ and }1^{1}10^{10}.

Akshat Sharda - 4 years ago

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Done very well!!!

Dev Sharma - 4 years ago

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Can you post solution of the first one ?

Akshat Sharda - 3 years, 11 months ago

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1)x=0,y=2

naitik sanghavi - 3 years, 11 months ago

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x=9,y=11

Can you provide its solution ?

Akshat Sharda - 3 years, 11 months ago

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Here is the solution: y3=x3+8x26x+8<x3+9x2+27x+27=(x+3)3becausex3=x38x2<9x26x<27x8<27y^3=x^3+8x^2-6x+8<x^3+9x^2+27x+27=(x+3)^3 \text{because}\\ x^3=x^3\\ 8x^2<9x^2\\ -6x<27x\\ 8<27.Now,let us see when y3y^3 is greater than or equal to (x+2)3(x+2)^3 x3+6x2+12x+8x3+8x26x+89x\Longrightarrow x^3+6x^2+12x+8 \leq x^3+8x^2-6x+8\\ \Longrightarrow 9 \leq x.Now,we have that for only one value of xx can the given expression be a cube and that happens when,x=9,y=(x+2)=11x=9,y=(x+2)=11.And done!

Adarsh Kumar - 3 years, 11 months ago

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@Adarsh Kumar The proof why y<(x+3)y < (x+3).

Consider (x+3)3y3=x2+33x+19(x+3)^3- y^3 = x^2 +33x +19 . Discriminant of this quadratic Δ<0\Delta < 0 and since the coefficient of x2x^2 is positive, it implies that (x+3)3y3(x+3)^3 -y^3 is always positive. So, y3<(x+3)3y^3 < (x+3)^3. So, y<x+2y < x+2.

Surya Prakash - 3 years, 11 months ago

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@Adarsh Kumar Actually I have left an integral part,I haven't proved that for x8x \leq 8,there are no solutions.

Adarsh Kumar - 3 years, 11 months ago

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Sorry ,0 is not an positive integer!!

naitik sanghavi - 3 years, 11 months ago

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