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If [m] y = |x| + x[/m], where [m]y[/m] is an integer. Is[m] y = 0[/m]?

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If [m] y = |x| + x[/m], where [m]y[/m] is an integer. Is[m] y = 0[/m]?  [#permalink]

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New post 29 Sep 2018, 09:06
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Weekly Quant Quiz Question -2



If \(y = |x| + x\), where \(y\) is an integer. Is \(y = 0\)?
a) \(x! + 7\) is prime
b) \(y < 2\)

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Re: If [m] y = |x| + x[/m], where [m]y[/m] is an integer. Is[m] y = 0[/m]?  [#permalink]

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New post Updated on: 29 Sep 2018, 10:06
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Statement I:

X = 3, x! + 7 is Prime. Hence, Suffiecient.
Minimum value of X => 3. Hence, Y will never be 0.

Statement II:

Y = 1, X =1/2
Y = 0, X = 0

Hence, A.
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Originally posted by rahul16singh28 on 29 Sep 2018, 09:27.
Last edited by rahul16singh28 on 29 Sep 2018, 10:06, edited 1 time in total.
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Re: If [m] y = |x| + x[/m], where [m]y[/m] is an integer. Is[m] y = 0[/m]?  [#permalink]

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New post 29 Sep 2018, 21:23

Official Solution:



Statement 1:
x! is defined only for non negative integers, but for x = 1, 2 : x!+7 is not prime
X = 3, x! + 7 is Prime. Hence, Sufficient.
Minimum value of X = 3 or 4 .... Hence, Y is definitely not 0.
SUFFICIENT

Statement 2:
Y = 1, X = 0.5
Y = 0, X = 0
NOT SUFFICIENT

Hence, Answer = A.
gmatbusters wrote:

Weekly Quant Quiz Question -2



If \(y = |x| + x\), where \(y\) is an integer. Is \(y = 0\)?
a) \(x! + 7\) is prime
b) \(y < 2\)

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Re: If [m] y = |x| + x[/m], where [m]y[/m] is an integer. Is[m] y = 0[/m]?  [#permalink]

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New post 29 Sep 2018, 09:12
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Please find the solution inside the image. IMO option B.
Attachments

File comment: Q2
Q2_sep29.jpeg
Q2_sep29.jpeg [ 117.39 KiB | Viewed 979 times ]


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If [m] y = |x| + x[/m], where [m]y[/m] is an integer. Is[m] y = 0[/m]?  [#permalink]

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New post Updated on: 30 Sep 2018, 06:57
since y is an integer 2nd statement says y is less then 2 therefore y can only be 0
if x is positive then y=2x which is not possible in this case
if x is -ve then y has to be 0
therefore 2nd statement is sufficient
B is correct

Originally posted by Manvi_nagpal on 29 Sep 2018, 09:13.
Last edited by Manvi_nagpal on 30 Sep 2018, 06:57, edited 1 time in total.
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Re: If [m] y = |x| + x[/m], where [m]y[/m] is an integer. Is[m] y = 0[/m]?  [#permalink]

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New post 29 Sep 2018, 09:14
y=|x|+x => y=0 when x<0 * y=2x when x>0
1) . Suff as x! is not defined when , x<0
2) Insuff ... as y is any integer lower than 2. Thus A.
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Re: If [m] y = |x| + x[/m], where [m]y[/m] is an integer. Is[m] y = 0[/m]?  [#permalink]

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New post 29 Sep 2018, 09:15
Consider 'A', x>0. This is the only possible case for x! + 7 to be prime. x = 3,4,5,..... satisfies the condition and y is clearly not equal to zero. So A is sufficient.
Consider B, y <2, x can be 0.5 -> y = 1, x can be -0.5 -> y= 0. So B is not sufficient.

Answer A.
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Re: If [m] y = |x| + x[/m], where [m]y[/m] is an integer. Is[m] y = 0[/m]?  [#permalink]

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New post 29 Sep 2018, 09:34
Question asks to prove whether x is 0 or negative then only y will be 0.

a) X!+ 7 is prime
If x=0, therefore 0!+7 is not prime out.
X=1, not prime,out,
X=2, not prime out,
X=3, 3!+7=13 prime,
Sufficient,
X is not negative therefore y is not zero.
b) Y<2
Means y=-1,0,1
Insufficient.
ANSWER=A
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Re: If [m] y = |x| + x[/m], where [m]y[/m] is an integer. Is[m] y = 0[/m]?  [#permalink]

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New post 29 Sep 2018, 10:47
1) x will be a integer >=0
If x = 0 then, x! + 7 = 7 which is a prime number
Therefore y = 0

If x = 3 then x! = 6, x! +7 = 13 which is a prime number. Thus, y = 6

Hence A is not sufficient

2) y < 2

If x <=0 then y = 0

If x> 0 then y = 2x <2
Thus, x <1, thus y is not equal to 0

Thus B is not sufficient

Both,
From A, x is >=0 and is an integer. From B x < 1. Therefore, x =0. Thus, y = 0. Hence, C is sufficient.

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Re: If [m] y = |x| + x[/m], where [m]y[/m] is an integer. Is[m] y = 0[/m]?  [#permalink]

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New post 29 Sep 2018, 13:51
stem is asking if x= 0 or x<0. If x<0. As y=0 only in these two cases.

From 1: x!+7 is prime, implies that x is not equal 0. Also x is not negative. so this statement is sufficient
from 2: y<2, if x=0, then y=0, if x= 0.5, then y=1, an integer. 2 different answers, Not sufficient

Answer is A
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Re: If [m] y = |x| + x[/m], where [m]y[/m] is an integer. Is[m] y = 0[/m]?  [#permalink]

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New post 29 Sep 2018, 18:46
In the solution shown above, I have a doubt.
What if in the second statement y=1 , x=1/2
It’s not mentioned that x has to be an int . Then there will be two cases
Y=1 when x=1/2
Y=0 when X<0


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Re: If [m] y = |x| + x[/m], where [m]y[/m] is an integer. Is[m] y = 0[/m]? &nbs [#permalink] 29 Sep 2018, 18:46
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