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imqwerty
 one year ago
fun question :D
imqwerty
 one year ago
fun question :D

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imqwerty
 one year ago
Best ResponseYou've already chosen the best response.1How many integers 'x' are there such that \[x^2  3x 19 \] is divisible by 289? :)

imqwerty
 one year ago
Best ResponseYou've already chosen the best response.1nd hw did u get that :)

imqwerty
 one year ago
Best ResponseYou've already chosen the best response.1no thats not correct :)

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0so do u knoe correct answer huh ? ok give me options if u can

imqwerty
 one year ago
Best ResponseYou've already chosen the best response.1yes i knw the answer :D nd this is a subjective question

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0can u give me options ?

imqwerty
 one year ago
Best ResponseYou've already chosen the best response.1ok your options are  a)5 b)9 c)289 d)none of these

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0u are asking about numbers that are divisible by 289 or only a certain number for x?

imqwerty
 one year ago
Best ResponseYou've already chosen the best response.1m asking that for how many values of x the expression \[x^2 3x 19 \] is divisible by 289

imqwerty
 one year ago
Best ResponseYou've already chosen the best response.1:) can u give any reason to support your answer?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0yes because only 289 is divisible by 289 not any other number ...u can't get 289 by putting any value of x

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0ganes is typing a reply ...:)

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.3Suppose \( x^23x19\equiv 0\pmod{289}~~\color{red}{\star}\) Since \(289=17^2\), it must necessarily be the case that \(x^23x19 \equiv 0\pmod{17}\) multiply \(4\) through out, complete the square and simplifying gives : \[(2x3)^2\equiv 0\pmod{17}\] so the solution is \(x\equiv 10\pmod{17}\)

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.3plugging that solution in \(\color{red}{\star}\) : \(x^23x19\equiv (10+17k)^23(10+17k)19 \\ \equiv 100+340k+289k^23051k19\\ \equiv 51+289k+289k^2\\ \equiv 51\\ \not\equiv 0\pmod{289} \) contradiction \(\blacksquare\)

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.3im 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!

imqwerty
 one year ago
Best ResponseYou've already chosen the best response.1yes :) there is another simple method to solve it.

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.3il let others try :)

ParthKohli
 one year ago
Best ResponseYou've already chosen the best response.3\[x^2  3x  19 = 289t\]\[\Rightarrow x^2  3x  19  289t = 0\]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?

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.3That's pretty clever!

ParthKohli
 one year ago
Best ResponseYou've already chosen the best response.3In 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 oddcase.

ParthKohli
 one year ago
Best ResponseYou've already chosen the best response.3@Astrophysics I love discriminants.

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.3I 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...

ParthKohli
 one year ago
Best ResponseYou've already chosen the best response.3No, it actually means that the roots are not only rational, but integral. :P

imqwerty
 one year ago
Best ResponseYou've already chosen the best response.1clever and neat @ParthKohli

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.3how do you know the top is always even ?

ParthKohli
 one year ago
Best ResponseYou've already chosen the best response.3I 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.

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.3Ahh okay, the only rational roots of \(x^2+bx+c\) are integers by rational root theorem

ParthKohli
 one year ago
Best ResponseYou've already chosen the best response.3Oh, that's a good insight. Thanks. Why in the world didn't I think about that... >_<

imqwerty
 one year ago
Best ResponseYou've already chosen the best response.1my method  \[x^2 3x 19\]\[(x10)(x+7)+51\] 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 (x10)(x+7). Since 17 is prime it must divide (x10) or (x+7). nd we see that (x+7(x10) = 17 so when 17 divides one of (x10) or (x+7) it must also divide the other :) so this means 17^2=289 can divide (x10)(x+7) it follows that 289 can divide 51 :D which is impossible . So there is no integer x for which 289 divides \[x^2  3x 19\]

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.3Actually i was considering the case \(ax^2+bx+c\) when \(a\ne 1\) somehow and thought ur earlier statement must be wrong

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.3Wow that is a cool one too! now we have 3 different methods

ParthKohli
 one year ago
Best ResponseYou've already chosen the best response.3Hmm, it seems like you guys got the same contradiction in the end. Cool answers.

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.3yes first and third methods look more or less similar, but the 3rd solution looks better because it avoids all that congruence mess

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.32nd solution is my fav so far though

ParthKohli
 one year ago
Best ResponseYou've already chosen the best response.3Thanks, @geniusie8.
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