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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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29 Sep 2018, 10:06
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41% (01:12) correct 59% (01:29) wrong based on 64 sessions
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Re: If [m] y = x + x[/m], where [m]y[/m] is an integer. Is[m] y = 0[/m]?
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Updated on: 29 Sep 2018, 11:06
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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Re: If [m] y = x + x[/m], where [m]y[/m] is an integer. Is[m] y = 0[/m]?
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29 Sep 2018, 22:23
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 SUFFICIENTHence, 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]?
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29 Sep 2018, 10:12
Please find the solution inside the image. IMO option B.
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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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Updated on: 30 Sep 2018, 07:57
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, 10:13.
Last edited by Manvi_nagpal on 30 Sep 2018, 07:57, 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]?
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29 Sep 2018, 10:14
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]?
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29 Sep 2018, 10:15
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]?
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29 Sep 2018, 10:34
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]?
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29 Sep 2018, 11:47
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]?
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29 Sep 2018, 14:51
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]?
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29 Sep 2018, 19:46
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 Sent from my iPhone using GMAT Club Forum mobile app




Re: If [m] y = x + x[/m], where [m]y[/m] is an integer. Is[m] y = 0[/m]? &nbs
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29 Sep 2018, 19:46






