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imqwerty

  • one year ago

fun question :D

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  1. imqwerty
    • one year ago
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    How many integers 'x' are there such that \[x^2 - 3x -19 \] is divisible by 289? :)

  2. anonymous
    • one year ago
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    28

  3. imqwerty
    • one year ago
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    nd hw did u get that :)

  4. imqwerty
    • one year ago
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    thats wrng btw

  5. anonymous
    • one year ago
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    ok

  6. imqwerty
    • one year ago
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    no thats not correct :)

  7. imqwerty
    • one year ago
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    don't guess :)

  8. sohailiftikhar
    • one year ago
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    so do u knoe correct answer huh ? ok give me options if u can

  9. imqwerty
    • one year ago
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    yes i knw the answer :D nd this is a subjective question

  10. sohailiftikhar
    • one year ago
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    can u give me options ?

  11. imqwerty
    • one year ago
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    ok your options are - a)5 b)9 c)289 d)none of these

  12. sohailiftikhar
    • one year ago
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    u are asking about numbers that are divisible by 289 or only a certain number for x?

  13. imqwerty
    • one year ago
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    m asking that for how many values of x the expression \[x^2 -3x -19 \] is divisible by 289

  14. sohailiftikhar
    • one year ago
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    no one

  15. imqwerty
    • one year ago
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    :) can u give any reason to support your answer?

  16. sohailiftikhar
    • one year ago
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    yes because only 289 is divisible by 289 not any other number ...u can't get 289 by putting any value of x

  17. sohailiftikhar
    • one year ago
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    ganes is typing a reply ...:)

  18. ganeshie8
    • one year ago
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    Suppose \( x^2-3x-19\equiv 0\pmod{289}~~\color{red}{\star}\) Since \(289=17^2\), it must necessarily be the case that \(x^2-3x-19 \equiv 0\pmod{17}\) multiply \(4\) through out, complete the square and simplifying gives : \[(2x-3)^2\equiv 0\pmod{17}\] so the solution is \(x\equiv 10\pmod{17}\)

  19. ganeshie8
    • one year ago
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    plugging that solution in \(\color{red}{\star}\) : \(x^2-3x-19\equiv (10+17k)^2-3(10+17k)-19 \\ \equiv 100+340k+289k^2-30-51k-19\\ \equiv 51+289k+289k^2\\ \equiv 51\\ \not\equiv 0\pmod{289} \) contradiction \(\blacksquare\)

  20. imqwerty
    • one year ago
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    yes :D

  21. ganeshie8
    • one year ago
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    im pretty sure there are other simpler ways to solve this but what i showed you uses the standard method to solve any quadratic congruence... no tricks!

  22. imqwerty
    • one year ago
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    yes :) there is another simple method to solve it.

  23. ganeshie8
    • one year ago
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    il let others try :)

  24. imqwerty
    • one year ago
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    ok :)

  25. ParthKohli
    • one year ago
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    \[x^2 - 3x - 19 = 289t\]\[\Rightarrow x^2 - 3x - 19 - 289t = 0\]This equation has integral roots. So for that to happen, \(D\) is a perfect square.\[\Rightarrow 9 + 4(19 - 289t) = 85 - 4\cdot 289 t = 17 (5 - 68t)\]is a perfect square. The second factor must be divisible by 17 to make it a perfect square. (And that is only one condition.) But that's not possible, right?

  26. ganeshie8
    • one year ago
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    That's pretty clever!

  27. ParthKohli
    • one year ago
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    In general, for integral roots of the equation \(x^2 + bx + c = 0\), the discriminant must be a perfect square. Proof:\[ x= \frac{-b\pm \sqrt{b^2 - 4c}}{2}\]Now if \(b\) is even then \(D\) is also even, hence its root which is an integer is also even. So we've shown that the roots will be even. Similarly, we can think about the odd-case.

  28. ParthKohli
    • one year ago
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    @Astrophysics I love discriminants.

  29. ganeshie8
    • one year ago
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    I like your method more! the discriminant must be a perfect square for the roots to be rational, otherwise we get a nonsimplifying radical in the quadratic formula...

  30. ParthKohli
    • one year ago
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    No, it actually means that the roots are not only rational, but integral. :P

  31. imqwerty
    • one year ago
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    clever and neat @ParthKohli

  32. ganeshie8
    • one year ago
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    how do you know the top is always even ?

  33. ParthKohli
    • one year ago
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    I wrote the proof above. Take it casewise. First, take b to be even, then odd. If we know that D is a perfect square, then the top will be even. Surprising.

  34. ganeshie8
    • one year ago
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    Ahh okay, the only rational roots of \(x^2+bx+c\) are integers by rational root theorem

  35. ParthKohli
    • one year ago
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    Oh, that's a good insight. Thanks. Why in the world didn't I think about that... >_<

  36. imqwerty
    • one year ago
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    my method - \[x^2 -3x -19\]\[(x-10)(x+7)+51\] suppose 289 divides \[x^2 - 3x -19\] so for some integer x. then 17 divides it ...we knw that 17 divides 51 nd also 17 must divide (x-10)(x+7). Since 17 is prime it must divide (x-10) or (x+7). nd we see that (x+7-(x-10) = 17 so when 17 divides one of (x-10) or (x+7) it must also divide the other :) so this means 17^2=289 can divide (x-10)(x+7) it follows that 289 can divide 51 :D which is impossible . So there is no integer x for which 289 divides \[x^2 - 3x -19\]

  37. ganeshie8
    • one year ago
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    Actually i was considering the case \(ax^2+bx+c\) when \(a\ne 1\) somehow and thought ur earlier statement must be wrong

  38. ganeshie8
    • one year ago
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    Wow that is a cool one too! now we have 3 different methods

  39. imqwerty
    • one year ago
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    :)

  40. ParthKohli
    • one year ago
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    Hmm, it seems like you guys got the same contradiction in the end. Cool answers.

  41. ganeshie8
    • one year ago
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    yes first and third methods look more or less similar, but the 3rd solution looks better because it avoids all that congruence mess

  42. ganeshie8
    • one year ago
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    2nd solution is my fav so far though

  43. ParthKohli
    • one year ago
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    Thanks, @geniusie8.

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