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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
1 year ago

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Solution for $$2^{nd}$$ question :

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

Done very well!!! · 1 year ago

Can you post solution of the first one ? · 11 months, 3 weeks ago

1)x=0,y=2 · 11 months, 3 weeks ago

x=9,y=11

Can you provide its solution ? · 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! · 11 months, 3 weeks ago

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, $$y < x+2$$. · 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. · 11 months, 3 weeks ago