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If x and y are integers greater than 1, is x a multiple of y [#permalink]
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Updated on: 30 Oct 2012, 01:44
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Question Stats:
62% (00:40) correct 38% (00:43) wrong based on 1119 sessions
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If x and y are integers greater than 1, is x a multiple of y? (1) 3y^2+7y=x (2) x^2x is a multiple of y For me its D. Unless someone else thinks otherwise. This is how I solved this:
Statement 1
y is a multiple of 3 and 7
So when y = 2 then x will be 26 and YES x is a multiple of y. when y = 5 then x will be 110 and YES x is a multiple of y.
Therefore, statement 1 is sufficient to answer this questions.
Statement 2
x^2x is a multiple of y
x(x1) is a multiple of y. Sufficient to answer.
Therefore D for me i.e. both statements alone are sufficient to answer this question. Any thoughts guys?
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Originally posted by enigma123 on 29 Jan 2012, 16:21.
Last edited by Bunuel on 30 Oct 2012, 01:44, edited 2 times in total.
Added the OA



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Re: Is x a multiple of y [#permalink]
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29 Jan 2012, 16:31
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Re: If x and y are integers greater than 1, is x a multiple of y [#permalink]
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05 Sep 2012, 08:00
Hi Bunuel, I have not understood the answer to the (2) statement, infact we have that x(x1) = Y*F (because is a multiple), doesn't it depend on the number F if X is a multiple of y? For example: If x=2, then y could be either 2 or 1; so the answer would be yes; If x=3, then y could be either 2 or 3; so the answer would be no;



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Re: If x and y are integers greater than 1, is x a multiple of y [#permalink]
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05 Sep 2012, 14:17
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for(2)
x(x1) is multiple of y. so either x is divisible by y or x1 is divisible by y. if x1 is divisible by y, then x is not divisble by y.
plug in y=8, x=16>16*15 is divisible by 8 > yes x is a multiple of y. plug in y=8,x=17> 17*16 is still divisible by 8 > But x is not a multiple of y.
So insufficient.



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Re: Is x multiple of y? If: 3y^2 + 7y = x? If: x^2x is mult? [#permalink]
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31 Dec 2012, 21:27
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Nadezda wrote: x and y are integers > 1. Is x multiple of y?
(1): 3y^2 + 7y = x (2): x^2  x is multiple of y The question is asking whether \(x/y\) is an integer or not. Answer is A. Statement 1 can be written as \(y(3y +7)=x\) or \(3y+7=x/y\). Since \(y\) is an integer, therefore 3y+7 must also be an integer. Hence \(x/y\) will be an integer or x is a multiple of y. Sufficient. Statement 2 can be written as \(x(x1)/y\) is an integer. Notice that we can't deduce that x is a multiple of y because it is quite possible that the product is a multiple of y, but not the individual entities. ex. 2*3/6 is a mltiple of 6 BUT neither 2 nor 3 is a multiple of 6. Insufficient. +1A Please do add the OA while posting questions.
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Re: Is x multiple of y? If: 3y^2 + 7y = x? If: x^2x is mult? [#permalink]
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31 Dec 2012, 22:41
Nadezda wrote: x and y are integers > 1. Is x multiple of y?
(1): 3y^2 + 7y = x (2): x^2  x is multiple of y Hi Nadezda, The Questions asks whether x/y is an integer or not. from St 1 we have 3y^2 +7y=x > x/y= 3y+7 if y=2, x=13 and 13/2 is not an integer. If y=7, then x/y=4 which is an integer. So St1 alone not sufficient From St 2 we have x^2x is multiple of y > x(x1)/y= Integer value Now we know x1 and x are consecutive integers X2x is even because if x is odd then x2x (oddodd)=even and if x is even then x2x is even. Now x can be odd or even and hence alone not sufficient. Combining both statement,we get for x/y to be an integer y has to be 7 and x will be have to be multiple of 7. But we are not sure from above statement so ans should be E. What is OA?? Thanks Mridul
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Re: Is x multiple of y? If: 3y^2 + 7y = x? If: x^2x is mult? [#permalink]
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31 Dec 2012, 22:59
mridulparashar1 wrote: Nadezda wrote: x and y are integers > 1. Is x multiple of y?
(1): 3y^2 + 7y = x (2): x^2  x is multiple of y Hi Nadezda, The Questions asks whether x/y is an integer or not. from St 1 we have 3y^2 +7y=x > x/y= 3y+7 if y=2, x=13 and 13/2 is not an integer. If y=7, then x/y=4 which is an integer. So St1 alone not sufficient From St 2 we have x^2x is multiple of y > x(x1)/y= Integer value Now we know x1 and x are consecutive integers X2x is even because if x is odd then x2x (oddodd)=even and if x is even then x2x is even. Now x can be odd or even and hence alone not sufficient. Combining both statement,we get for x/y to be an integer y has to be 7 and x will be have to be multiple of 7. But we are not sure from above statement so ans should be E. What is OA?? Thanks Mridul if x/y=3y+7, then taking y as 2 will yield x/y as 13 and not 13/2.
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Re: If x and y are integers greater than 1, is x a multiple of y [#permalink]
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03 May 2014, 12:26
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FACTS ESTABLISHED BY STEMS  \(x\) is an integer
 \(y\) is an integer
 \(x > 1\)
 \(y > 1\)
QUESTION TRANSLATED Since both \(x\) and \(y\) are integers the question " is \(x\) a multiple of \(y\)?" really asks if \(x = q \cdot y + r\) where \(q\) is an integer (the quotient) and \(r\) is the remainder when you divide \(x\) by \(y\) so that \(r = 0\) which means that \(x = q \cdot y + 0\) which implies that \(x = q \cdot y\) So the question translates to: is \(x = q \cdot y\) when \(x\), \(q\) and \(y\) are all integers, and when \(x>1\) and \(y > 1\)?STATEMENT 1 \(3y^2 + 7y = x\) can be rewritten as \(y(3y + 7) = x\) however, from the question stems we know that \(y\) is an integer, which implies \(3y\) is an integer and that \(3y+7\) is also an integer: \(y \cdot integer = x\) which can be rewritten as \(x = y \cdot integer\) or better yet, as \(x = integer \cdot y\) which if you notice is an equation of the form \(x = q \cdot y\) since \(q\) is an integer which implies that \(x\) is indeed a multiple of \(y\). Therefore, Statement A is sufficient. STATEMENT 2 the statement "\(x^2  x\) is a multiple of \(y\)" must be rewritten as "\(x (x  1)\) is a multiple of \(y\)" which implies that either \(x\) is a multiple of \(y\) or \(x1\) is a multiple of \(y\) the substatement \(x\) is a multiple of \(y\) implies that \(x = q \cdot y\) which would answer the question, but we still need to figure out if the second substatement answers the question with the same answer as the first substatement. the second substatement \(x1\) is a multiple of \(y\) implies that \(x1 = q \cdot y\) which can be rewritten as \(x = q \cdot y + 1\) which is an equation of the form \(x = q \cdot y + r\) when \(r=1\) which implies that when \(x\) is divided by \(y\) we get a remainder of 1, meaning that \(x\) is NOT a multiple of \(y\). Since both substatements are contradictory, this means that when \(x^2  x\) is a multiple of \(y\), \(x\) is not always a multiple of \(y\). meaning that we cannot determine whether \(x\) is always a multiple of \(y\) according to Statement 2 alone. Therefore, Statement 2 is not sufficient. ANSWER (A) Statement (1) ALONE is sufficient, but statement (2) alone is not sufficient.



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Re: If x and y are integers greater than 1, is x a multiple of y [#permalink]
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18 Sep 2014, 18:36
Bunuel wrote: If x and y are integers great than 1, is x a multiple of y?
(1) \(3y^2+7y=x\) > \(y(3y+7)=x\) > as \(3y+7=integer\), then \(y*integer=x\) > \(x\) is a multiple of \(y\). Sufficient.
(2) \(x^2x\) is a multiple of \(y\) > \(x(x1)\) is a multiple of \(y\) > \(x\) can be multiple of \(y\) (\(x=2\) and \(y=2\)) OR \(x1\) can be multiple of \(y\) (\(x=3\) and \(y=2\)) or their product can be multiple of \(y\) (\(x=3\) and \(y=6\)). Not sufficient.
Answer: A.
Hope it helps. Hi Bunuel, Two questions: 1) If x = 2 and y = 2, does that mean that x is a multiple of y? I always thought that x needed to be 1 multiple greater, meaning, x=4 and y=2 etc? 2) In situations like these, I usually stumble with either starting off with factoring(like you did above) or some other method. What do you suggest? 3) Lastly, the way I solved statement 1 was  I saw it as (y)(multiple of y) = x, and the rule states that if i multiply two numbers that are multiples of y, then the result will be a multiple of y. Is that correct to assume?(I know that we can assume that with addition, right?)



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Re: If x and y are integers greater than 1, is x a multiple of y [#permalink]
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19 Sep 2014, 02:31
russ9 wrote: Bunuel wrote: If x and y are integers great than 1, is x a multiple of y?
(1) \(3y^2+7y=x\) > \(y(3y+7)=x\) > as \(3y+7=integer\), then \(y*integer=x\) > \(x\) is a multiple of \(y\). Sufficient.
(2) \(x^2x\) is a multiple of \(y\) > \(x(x1)\) is a multiple of \(y\) > \(x\) can be multiple of \(y\) (\(x=2\) and \(y=2\)) OR \(x1\) can be multiple of \(y\) (\(x=3\) and \(y=2\)) or their product can be multiple of \(y\) (\(x=3\) and \(y=6\)). Not sufficient.
Answer: A.
Hope it helps. Hi Bunuel, Two questions: 1) If x = 2 and y = 2, does that mean that x is a multiple of y? I always thought that x needed to be 1 multiple greater, meaning, x=4 and y=2 etc? 2) In situations like these, I usually stumble with either starting off with factoring(like you did above) or some other method. What do you suggest? 3) Lastly, the way I solved statement 1 was  I saw it as (y)(multiple of y) = x, and the rule states that if i multiply two numbers that are multiples of y, then the result will be a multiple of y. Is that correct to assume?(I know that we can assume that with addition, right?) 1. The least multiple of a positive integer is that integer itself. So, the least multiple of 2 is 2. 2. I suggest the way I used. 3. 3y+7 is not necessarily a multiple of y but it does not need to be. y*integer=x implies that x is a multiple of y.
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If x and y are integers greater than 1, is x a multiple of y [#permalink]
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20 Aug 2016, 19:46
x,y are integers>1, is x multiple of y? x multiple of y is same as x/y=integer 1. 3y^2+7y=x y(3y+7)=x dividing both sides by y, we get: (3y+7)=x/y x/y=int as question stem says y is integer statement 1. is sufficient 2. x^2x is a multiple of y x(x1) is a multiple of y i.e {x(x1)}/y must be integer x/y say 4/2 =integer or 41/2=3/2 not integer Thus, statement 2 is insufficient. Hence, A
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Re: If x and y are integers greater than 1, is x a multiple of y [#permalink]
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21 Aug 2016, 01:59
Awesome Question . Here we need to prove if x=y*I for some integer I Statement 1=> x=y[3y+y]=y*I => Sufficient Statement 2 => x(x1) is a multiple of y let x=7 x1=6 and y =6 here clearly the answer is NO now let y=7 => the answer will be YES hence Insufficient Smash that A
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Re: If x and y are integers greater than 1, is x a multiple of y [#permalink]
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26 Jan 2017, 19:34
Statement 1: x = y(3y + 7) = y*(Some integer) Therefore, x is a multiple of y Sufficient Statement 2: x(x  1) is a multiple of y. Therefore, either x or (x  1) or none of them is multiple of y. Not sufficient The correct answer is A.
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Re: If x and y are integers greater than 1, is x a multiple of y [#permalink]
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16 Oct 2017, 06:13
What if we change the question stem to factors... is B the answer?
If x and y are integers greater than 1, is x a factor of y?
(1) 3y^2+7y=x (2) x^2x is a multiple of y



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Re: If x and y are integers greater than 1, is x a multiple of y [#permalink]
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21 Nov 2017, 12:36
enigma123 wrote: If x and y are integers greater than 1, is x a multiple of y?
x,y > 1 (Integers only) is \(\frac{x}{y}\) = integer ? Quote: (1) 3y^2+7y=x (2) x^2x is a multiple of y
S1) 3\(y^2\) + 7y = x => y(3y+7) = x => \(\frac{x}{y}\) = 3y+7 => 3y + 7 will always be an integer value => x is a multiple of y Sufficient. S2) \(x^2\)  x = y*n => x(x1) = y*n x = 3, y = 2 => x is not a multiple of y x = 4, y = 2 => x is a multiple of y Inconsistent results. Insufficient. A is the answer.
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Re: If x and y are integers greater than 1, is x a multiple of y
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