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If \(x\) and \(y\) are integers and \(12x + 5y^2 = 0,\) which of the following must be true?

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If \(x\) and \(y\) are integers and \(12x + 5y^2 = 0,\) which of the following must be true?

by VJesus12 » Thu Jun 04, 2020 7:45 am

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If \(x\) and \(y\) are integers and \(12x + 5y^2 = 0,\) which of the following must be true?

(A) \(x\) is a multiple of 10

(B) \(x\) is a multiple of 15

(C) \(x\) is a multiple of 25

(D) \(x\) is not a multiple of 7

(E) \(x\) is negative

[spoiler]OA=B[/spoiler]

Source: Veritas Prep
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Re: If \(x\) and \(y\) are integers and \(12x + 5y^2 = 0,\) which of the following must be true?

by Jay@ManhattanReview » Thu Jun 04, 2020 8:29 pm
VJesus12 wrote: ↑
Thu Jun 04, 2020 7:45 am
 If \(x\) and \(y\) are integers and \(12x + 5y^2 = 0,\) which of the following must be true?

(A) \(x\) is a multiple of 10

(B) \(x\) is a multiple of 15

(C) \(x\) is a multiple of 25

(D) \(x\) is not a multiple of 7

(E) \(x\) is negative

[spoiler]OA=B[/spoiler]

Source: Veritas Prep
Given that \(12x + 5y^2 = 0,\) we have \(x = \dfrac{-5y^2}{12}\)

Since \(x\) is an integer and 5 and 12 are co-prime to each other, \(y^2\) must be a multiple of 12.

Say \(y^2 = 12p\), where \(p\) is an integer

Again, we have \(y^2 = 12p\), thus, \(y = 2√{3p}\). Since \(y\) is an integer, \(√{3p}\) must be an integer.

Thus, \(y^2=12*3p=36p\)

Again, from \(x = \dfrac{-5y^2}{12}=\dfrac{-5*36p}{12}= -15p\)

=> \(x\) must be a multiple of 15.

The correct answer: B

Hope this helps!

-Jay
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Re: If \(x\) and \(y\) are integers and \(12x + 5y^2 = 0,\) which of the following must be true?

by deloitte247 » Sat Jun 06, 2020 2:07 am
$$12x+5y^2=0$$
Express the equation in terms of x
Note: x and y are integers of whole numbers
$$\frac{12x}{12}=\frac{-5y^2}{12}$$
$$x=\frac{-5y^2}{12}$$
$$If\ \ y\ =\ 1$$
$$then\ x\ =\frac{-5\left(1\right)^2}{12}=\frac{-5}{12}=-0.42$$
$$x\ is\ not\ an\ integer\ so\ y\ne1$$
$$If\ \ y\ =\ 2$$
$$then\ x\ =\frac{-5\left(2\right)^2}{12}=\frac{-20}{12}=-1.67$$
$$x\ is\ not\ an\ integer\ so\ y\ne2$$
$$If\ \ y\ =\ 3$$
$$then\ x\ =\frac{-5\left(3\right)^2}{12}=\frac{-45}{12}=-3.75$$
$$x\ is\ not\ an\ integer\ so\ y\ne3$$
$$If\ \ y\ =\ 4$$
$$then\ x\ =\frac{-5\left(4\right)^2}{12}=\frac{-80}{12}=-6.62$$
$$x\ is\ not\ an\ integer\ so\ y\ne4$$
$$If\ \ y\ =\ 5$$
$$then\ x\ =\frac{-5\left(5\right)^2}{12}=\frac{-125}{12}=-10.42$$
$$x\ is\ not\ an\ integer\ so\ y\ne5$$
$$If\ \ y\ =\ 6$$
$$then\ x=\frac{-5\left(6\right)^2}{12}=\frac{-180}{12}=-15$$
$$x\ is\ an\ integer\ so\ y=6\ or\ multiple\ of\ 6$$
$$and\ x=-15\ or\ multiple\ of\ 15$$

Answer =B
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Re: If \(x\) and \(y\) are integers and \(12x + 5y^2 = 0,\) which of the following must be true?

by Scott@TargetTestPrep » Mon Jun 08, 2020 1:01 pm
VJesus12 wrote: ↑
Thu Jun 04, 2020 7:45 am
 If \(x\) and \(y\) are integers and \(12x + 5y^2 = 0,\) which of the following must be true?

(A) \(x\) is a multiple of 10

(B) \(x\) is a multiple of 15

(C) \(x\) is a multiple of 25

(D) \(x\) is not a multiple of 7

(E) \(x\) is negative

[spoiler]OA=B[/spoiler]

Solution:


Let’s write:

x = 5y^2/12

So, x is a multiple of 5. Further, notice that 12 has a factor of 3 and 5 is not divisible by 3; thus y^2 must be a multiple of 3. Finally, notice that every prime factor in the prime decomposition of y^2 appears an even number of times; thus y^2 has at least two factors of 3. After one of the factors of 3 cancels with the factor of 3 in the denominator, we still have at least one more factor of 3. Thus, y^2/12 has a factor of 3, which implies that x has a factor of 3.

Since x has both a factor of 5 and a factor of 3, x must be a multiple of 15.

Answer: B

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