A question about relatively prime numbers


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1
15 months ago by

Let $a$a and $b$b be relatively prime integers with  $a>b>0$a>b>0  and  $\frac{a^3-b^3}{(a-b)^3}=\frac{73}{3}$a3b3(ab)3 =733  . What is  $a-b$ab?

(A) 1  (B) 2  (C) 3  (D) 4  (E) 5

Any ideas?

3 Answers


3
15 months ago by

  $\frac{a^3-b^3}{\left(a-b\right)^3}=\frac{\left(a-b\right)\times\left(a^2+ab+b^2\right)^{ }}{\left(a-b\right)\times\left(a-b\right)^2}=\frac{a^2+ab+b^2}{a^2-2ab+b^2}=\frac{73}{3}$a3b3(ab)3 =(ab)×(a2+ab+b2)(ab)×(ab)2 =a2+ab+b2a22ab+b2 =733   

Let  $a^2+b^2=x$a2+b2=x and  $ab=y$ab=y , we have

  $\frac{x+y}{x-2y}=\frac{73}{3}$x+yx2y =733  

and simplifying gives  $\frac{x}{y}=\frac{149}{70}$xy =14970 . Since 149 and 70 are relatively prime, we let x = 149 and y = 70

 $a^2+b^2=149$a2+b2=149  and  $ab=70$ab=70 . Since  $a>b>0$a>b>0 , we find the only solution is a = 10 and b = 7. 

So the answer is 3.

 

0
15 months ago by

 $\frac{73}{3}$733  cannot be furture simplified so  $\left(a-b\right)^3$(ab)3 has to be a multiple of 3. In the given answers, only C satisfies that. So I would say C is the answer.

0
15 months ago by

I think C is the right answer

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