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Here is the solution:
y3=x3+8x2−6x+8<x3+9x2+27x+27=(x+3)3becausex3=x38x2<9x2−6x<27x8<27.Now,let us see when y3 is greater than or equal to (x+2)3⟹x3+6x2+12x+8≤x3+8x2−6x+8⟹9≤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!
Consider (x+3)3−y3=x2+33x+19. Discriminant of this quadratic Δ<0 and since the coefficient of x2 is positive, it implies that (x+3)3−y3 is always positive. So, y3<(x+3)3. So, y<x+2.
Easy Math Editor
This discussion board is a place to discuss our Daily Challenges and the math and science related to those challenges. Explanations are more than just a solution — they should explain the steps and thinking strategies that you used to obtain the solution. Comments should further the discussion of math and science.
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Solution for 2nd question :
2000∣aabb⇒2∣a or b and 5∣a or b.Therefore, in every case 10∣ab.Thus, the minimum value of ab is obviously 10, possible cases 101011 and 111010.
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Done very well!!!
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Can you post solution of the first one ?
1)x=0,y=2
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x=9,y=11
Can you provide its solution ?
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Here is the solution: y3=x3+8x2−6x+8<x3+9x2+27x+27=(x+3)3becausex3=x38x2<9x2−6x<27x8<27.Now,let us see when y3 is greater than or equal to (x+2)3 ⟹x3+6x2+12x+8≤x3+8x2−6x+8⟹9≤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!
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y<(x+3).
The proof whyConsider (x+3)3−y3=x2+33x+19. Discriminant of this quadratic Δ<0 and since the coefficient of x2 is positive, it implies that (x+3)3−y3 is always positive. So, y3<(x+3)3. So, y<x+2.
x≤8,there are no solutions.
Actually I have left an integral part,I haven't proved that forSorry ,0 is not an positive integer!!