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?

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?

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TopNewestSolution 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 · 1 year ago

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– Dev Sharma · 1 year ago

Done very well!!!Log in to reply

– Akshat Sharda · 11 months, 3 weeks ago

Can you post solution of the first one ?Log in to reply

1)x=0,y=2 – Naitik Sanghavi · 11 months, 3 weeks ago

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Can you provide its solution ? – Akshat Sharda · 11 months, 3 weeks ago

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– Adarsh Kumar · 11 months, 3 weeks ago

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!Log in to reply

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 · 11 months, 3 weeks ago

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– Adarsh Kumar · 11 months, 3 weeks ago

Actually I have left an integral part,I haven't proved that for \(x \leq 8\),there are no solutions.Log in to reply

– Naitik Sanghavi · 11 months, 3 weeks ago

Sorry ,0 is not an positive integer!!Log in to reply