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RMO adventure

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

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

Note by Dev Sharma
2 years, 2 months ago

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Solution for \(2^{nd}\) question :

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

Akshat Sharda - 2 years, 2 months ago

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Done very well!!!

Dev Sharma - 2 years, 2 months ago

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Can you post solution of the first one ?

Akshat Sharda - 2 years, 1 month ago

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1)x=0,y=2

Naitik Sanghavi - 2 years, 1 month ago

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x=9,y=11

Can you provide its solution ?

Akshat Sharda - 2 years, 1 month ago

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

Adarsh Kumar - 2 years, 1 month ago

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

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

Surya Prakash - 2 years, 1 month ago

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

Adarsh Kumar - 2 years, 1 month ago

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Sorry ,0 is not an positive integer!!

Naitik Sanghavi - 2 years, 1 month ago

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