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# There are exactly N distinct rational numbers k such that |k| < 200 an

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There are exactly N distinct rational numbers k such that |k| < 200 an  [#permalink]

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19 Mar 2019, 09:11
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Difficulty:

95% (hard)

Question Stats:

14% (03:31) correct 86% (03:10) wrong based on 14 sessions

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There are exactly N distinct rational numbers k such that $$|k| < 200$$ and $$5x^2 + kx + 12 = 0$$
has at least one integer solution for x. What is N?

(A) 6

(B) 12

(C) 24

(D) 48

(E) 78
Math Expert
Joined: 02 Aug 2009
Posts: 8320
Re: There are exactly N distinct rational numbers k such that |k| < 200 an  [#permalink]

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20 Mar 2019, 19:53
There are exactly N distinct rational numbers k such that $$|k| < 200$$ and $$5x^2 + kx + 12 = 0$$
has at least one integer solution for x. What is N?

(A) 6

(B) 12

(C) 24

(D) 48

(E) 78

Let us take k to one side => $$5x^2 + kx + 12 = 0........k=-5x-\frac{12}{x}............$$
Now |k|<200 or -200<k<200, that is $$-200<-5x-\frac{12}{x}<200$$
-5x will become equal to 200, the moment x is 40, so x has to be less than 40..
Similarly -5x will become greater than 200, the moment x is -40, so x has to be greater than -40..
Thus x are from -39 to 39, which equals 2*39+1=79, which includes k as 0.
However k as 0 is not possible as the equation is not defined at that point =>$$5x^2 + kx + 12 = 0..........5x^2+12=0.....5x^2=-12$$ NOT possible as $$5x^2\geq{0}$$
Thus 79-1=78

E
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Re: There are exactly N distinct rational numbers k such that |k| < 200 an  [#permalink]

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20 Mar 2019, 20:09
chetan2u wrote:
There are exactly N distinct rational numbers k such that $$|k| < 200$$ and $$5x^2 + kx + 12 = 0$$
has at least one integer solution for x. What is N?

(A) 6

(B) 12

(C) 24

(D) 48

(E) 78

Let us take k to one side => $$5x^2 + kx + 12 = 0........k=-5x-\frac{12}{x}............$$
Now |k|<200 or -200<k<200, that is $$-200<-5x-\frac{12}{x}<200$$
-5x will become equal to 200, the moment x is 40, so x has to be less than 40..
Similarly -5x will become greater than 200, the moment x is -40, so x has to be greater than -40..

Thus x are from -39 to 39, which equals 2*39+1=79, which includes k as 0.
However k as 0 is not possible as the equation is not defined at that point =>$$5x^2 + kx + 12 = 0..........5x^2+12=0.....5x^2=-12$$ NOT possible as $$5x^2\geq{0}$$
Thus 79-1=78

E

Chetan2u can you explain the highlighted part.
why does -5x becoming 400 is a problem. Since K will still b -5x-12/x which is less than 400
Similarly for next stmt
Math Expert
Joined: 02 Aug 2009
Posts: 8320
Re: There are exactly N distinct rational numbers k such that |k| < 200 an  [#permalink]

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20 Mar 2019, 20:43
globaldesi wrote:
chetan2u wrote:
There are exactly N distinct rational numbers k such that $$|k| < 200$$ and $$5x^2 + kx + 12 = 0$$
has at least one integer solution for x. What is N?

(A) 6

(B) 12

(C) 24

(D) 48

(E) 78

Let us take k to one side => $$5x^2 + kx + 12 = 0........k=-5x-\frac{12}{x}............$$
Now |k|<200 or -200<k<200, that is $$-200<-5x-\frac{12}{x}<200$$
-5x will become equal to 200, the moment x is 40, so x has to be less than 40..
Similarly -5x will become greater than 200, the moment x is -40, so x has to be greater than -40..

Thus x are from -39 to 39, which equals 2*39+1=79, which includes k as 0.
However k as 0 is not possible as the equation is not defined at that point =>$$5x^2 + kx + 12 = 0..........5x^2+12=0.....5x^2=-12$$ NOT possible as $$5x^2\geq{0}$$
Thus 79-1=78

E

Chetan2u can you explain the highlighted part.
why does -5x becoming 400 is a problem. Since K will still b -5x-12/x which is less than 400
Similarly for next stmt

-5x will become 200 when x is -40, but -12/x will also become positive (-12)/(-40)=12/40, so sum is 200+(12/40) >200
Similarly for the other case..
-12/x will have SAME sign as -5x, so will get added, either both as negative, then <-200 or both as positive, then >200
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Re: There are exactly N distinct rational numbers k such that |k| < 200 an   [#permalink] 20 Mar 2019, 20:43
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