2
$\begingroup$

$a,b,c$ are cube roots of $p$ ,($p<0$) then for any permissible value of $x,y,z$ which is given by

$$\frac{|xa+yb+zc|}{|xb+yc+za|} + (a_1^2-2b_1^2)\omega + \omega^2([x]+[y]+[z]) = 0 $$

$\omega$ is cube root of unity $a_1,b_1$ are real positive numbers and $b_1$ is prime . We have to find the value of $[x+a_1] + [y+b_1] + [z]$ . where $[\cdot ]$ denotes the greatest integer function .

I don't have any idea , where to start !

$\endgroup$
  • $\begingroup$ You mean, $x, y, z$ is permissible is defined by saying "the relation holds"? $\endgroup$ – Igor Rivin Dec 3 '13 at 17:30
  • $\begingroup$ So, this result is independent of $p$? $\endgroup$ – Doc Dec 3 '13 at 17:40
  • $\begingroup$ I think the result is just a number . $\endgroup$ – abkds Dec 3 '13 at 17:53
  • $\begingroup$ Yes, the result is just a number. But note, that number is independent of $p$, $a$, $b$, and $c$. Okay, I guess it must depend on these values implicitly, so never mind. $\endgroup$ – Doc Dec 3 '13 at 18:23
  • $\begingroup$ Let $N$ be the number you want. I get $N=b_1(b_1+1)+\frac{a_1-1}{2}$. However, (1) I assumed that $a_1$ was an integer; (2) I didn't use the fact that $b_1$ is a prime; (3) I could well have messed up as the calculation was painful. Note also, that my "result" implies that $a_1$ is odd (which looks a bit strange, but may actually have to be the case, given the equality restriction). $\endgroup$ – Doc Dec 3 '13 at 19:13
2
$\begingroup$

We know that one of the following holds:

  • $b = a\omega$ and $c = b\omega$ and $a = c\omega$.
  • $b = c\omega$ and $a = b\omega$ and $c = a\omega$.

In each of these cases, the ratio

$$\frac{xa+yb+zc}{xb+yc+za}$$

has a definite value independent of $p$, namely $\omega^2$ or $\omega$ respectively. In both cases,

$$\frac{|xa+yb+zc|}{|xb+yc+za|} = \left|\frac{xa+yb+zc}{xb+yc+za}\right| = 1$$

Now we have an equation of the form $p + q\omega + r\omega^2 = 0$. From this we may deduce $p = q = r$. In the case in point:

$$a_1^2 - 2b_1^2 = 1$$ $$[x]+[y]+[z] = 1$$

At this point I can make no progress without assuming that $a_1$ is an integer. I will assume that that's what the question intended.

Then $a_1$ is odd, say $a_1 = 2k+1$, so

$$4k^2 + 4k + 1 - 2b_1^2 = 1$$ $$2k(k+1) = b_1^2$$

Therefore $b_1$ is even, and since it is prime, it must be $2$, and $a_1$ must therefore be $3$. Then

$$[x+a_1] + [y+b_1] + [z]$$ $$= [x] + 2 + [y] + 3 + [z]$$ $$= ([x] + [y] + [z]) + 5$$ $$= 6$$

$\endgroup$
  • $\begingroup$ Indeed that is the answer . Very well done :) . Thanks $\endgroup$ – abkds Dec 6 '13 at 5:00
  • $\begingroup$ Where did you find this question? It is clearly a deliberately obfuscated puzzle! $\endgroup$ – apt1002 Dec 6 '13 at 5:09
  • $\begingroup$ Its a question from a book ( for preparing for competitive exam ) . I was just solving those problems $\endgroup$ – abkds Dec 6 '13 at 5:12

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.