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M06-29

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Bunuel
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M06-29 [#permalink] New post   16 Sep 2014, 00:28
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If \(x\) and \(y\) are integers and \(x=\frac{y}{5}+2\), is \(xy\) an even number?


(1) \(5x - 10\) is an even number

(2) \(\frac{y}{x}\) is an even number
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Re M06-29 [#permalink] New post   16 Sep 2014, 00:28
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Official Solution:


If \(x\) and \(y\) are integers and \(x=\frac{y}{5}+2\), is \(xy\) an even number?

First, from the equation \(x=\frac{y}{5}+2\), we derive that \(y=5x-10\). Next, since \(x\) and \(y\) are integers, the product \(xy\) will be even if at least one of the factors is even. Therefore, the question essentially asks whether at least one of the unknowns, \(x\) or \(y\), is even.

(1) \(5x - 10\) is an even number.

Since \(y=5x-10\), then \(y=5x-10=\text{even}\). Sufficient.

(2) \(\frac{y}{x}\) is an even number.

Given: \(\frac{y}{x}=\text{even}\), so \(y=x*\text{even}=\text{even}\). Sufficient.


Answer: D
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Re: M06-29 [#permalink] New post   16 Dec 2016, 04:12
Bunuel wrote:
Official Solution:


Since \(x\) and \(y\) are integers then \(xy\) will be even if at least one of the multiple is even. From \(x=\frac{y}{5}+2\) we have that \(y=5x-10\)

(1) \(5x - 10\) is even. Since \(y=5x-10\) then \(y=5x-10=\text{even}\). Sufficient.

(2) \(\frac{y}{x}\) is even. Given: \(\frac{y}{x}=\text{even}\), so \(y=x*\text{even}=\text{even}\). Sufficient.


Answer: D


Hi Bunuel,

What happens if x is 0?
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Bunuel
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Re: M06-29 [#permalink] New post   16 Dec 2016, 04:20
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uvemdesalinas wrote:
Bunuel wrote:
Official Solution:


Since \(x\) and \(y\) are integers then \(xy\) will be even if at least one of the multiple is even. From \(x=\frac{y}{5}+2\) we have that \(y=5x-10\)

(1) \(5x - 10\) is even. Since \(y=5x-10\) then \(y=5x-10=\text{even}\). Sufficient.

(2) \(\frac{y}{x}\) is even. Given: \(\frac{y}{x}=\text{even}\), so \(y=x*\text{even}=\text{even}\). Sufficient.


Answer: D


Hi Bunuel,

What happens if x is 0?


For which statement? For the second statement x cannot be 0 because in this case y/x would be undefined not even.
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Re: M06-29 [#permalink] New post   16 Dec 2016, 06:51
Bunuel wrote:
uvemdesalinas wrote:
Bunuel wrote:
Official Solution:


Since \(x\) and \(y\) are integers then \(xy\) will be even if at least one of the multiple is even. From \(x=\frac{y}{5}+2\) we have that \(y=5x-10\)

(1) \(5x - 10\) is even. Since \(y=5x-10\) then \(y=5x-10=\text{even}\). Sufficient.

(2) \(\frac{y}{x}\) is even. Given: \(\frac{y}{x}=\text{even}\), so \(y=x*\text{even}=\text{even}\). Sufficient.


Answer: D


Hi Bunuel,

What happens if x is 0?


For which statement? For the second statement x cannot be 0 because in this case y/x would be undefined not even.


For the first one. Y would be even (Y=-10) but X*Y=0
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M06-29 [#permalink] New post   16 Dec 2016, 12:29
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uvemdesalinas wrote:

For the first one. Y would be even (Y=-10) but X*Y=0


ZERO:

1. Zero is an INTEGER.

2. Zero is an EVEN integer.

3. Zero is neither positive nor negative (the only one of this kind).

4. Zero is divisible by EVERY integer except 0 itself (\(\frac{x}{0} = 0\), so 0 is a divisible by every number, x).

5. Zero is a multiple of EVERY integer (\(x*0 = 0\), so 0 is a multiple of any number, x).

6. Zero is NOT a prime number (neither is 1 by the way; the smallest prime number is 2).

7. Division by zero is NOT allowed: anything/0 is undefined.

8. Any non-zero number to the power of 0 equals 1 (\(x^0 = 1\))

9. \(0^0\) case is NOT tested on the GMAT.

10. If the exponent n is positive (n > 0), \(0^n = 0\).

11. If the exponent n is negative (n < 0), \(0^n\) is undefined, because \(0^{negative}=0^n=\frac{1}{0^{(-n)}} = \frac{1}{0}\), which is undefined. You CANNOT take 0 to the negative power.

12. \(0! = 1! = 1\).
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Re: M06-29 [#permalink] New post   05 Apr 2021, 13:55
What if X was a negative
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Re: M06-29 [#permalink] New post   05 Apr 2021, 13:59
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shanellejp18 wrote:
What if X was a negative


It won't change the answer. For example, if x = -3, then for (2) it would mean that y/(-3) = even --> y = (-3)*even = even.
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Re: M06-29 [#permalink] New post   05 May 2023, 12:29
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I have edited the question and the solution by adding more details to enhance its clarity. I hope it is now easier to understand.
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Re: M06-29 [#permalink]
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