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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) \(x^2-x\) is a multiple of \(y\) --> \(x(x-1)\) is a multiple of \(y\) --> \(x\) can be multiple of \(y\) (\(x=2\) and \(y=2\)) OR \(x-1\) 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.

Re: If x and y are integers greater than 1, is x a multiple of y [#permalink]

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05 Sep 2012, 07:00

Hi Bunuel, I have not understood the answer to the (2) statement, infact we have that x(x-1) = 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;

Re: If x and y are integers greater than 1, is x a multiple of y [#permalink]

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05 Sep 2012, 13:17

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for(2)

x(x-1) is multiple of y. so either x is divisible by y or x-1 is divisible by y. if x-1 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.

Re: Is x multiple of y? If: 3y^2 + 7y = x? If: x^2-x is mult? [#permalink]

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31 Dec 2012, 20:27

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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 \(x(x-1)/y\) 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^2-x is mult? [#permalink]

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31 Dec 2012, 21:41

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^2-x is multiple of y ----> x(x-1)/y= Integer value Now we know x-1 and x are consecutive integers

X2-x is even because if x is odd then x2-x (odd-odd)=even and if x is even then x2-x 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^2-x is mult? [#permalink]

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31 Dec 2012, 21:59

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^2-x is multiple of y ----> x(x-1)/y= Integer value Now we know x-1 and x are consecutive integers

X2-x is even because if x is odd then x2-x (odd-odd)=even and if x is even then x2-x 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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01 Jan 2014, 23:27

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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, 11:26

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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 \(x-1\) is a multiple of \(y\)

the sub-statement \(x\) 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 sub-statement answers the question with the same answer as the first sub-statement.

the second sub-statement \(x-1\) is a multiple of \(y\) implies that

\(x-1 = q \cdot y\)

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 sub-statements 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.

Re: If x and y are integers greater than 1, is x a multiple of y [#permalink]

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18 Sep 2014, 17:36

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) \(x^2-x\) is a multiple of \(y\) --> \(x(x-1)\) is a multiple of \(y\) --> \(x\) can be multiple of \(y\) (\(x=2\) and \(y=2\)) OR \(x-1\) 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?)

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) \(x^2-x\) is a multiple of \(y\) --> \(x(x-1)\) is a multiple of \(y\) --> \(x\) can be multiple of \(y\) (\(x=2\) and \(y=2\)) OR \(x-1\) 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.
_________________

Re: If x and y are integers greater than 1, is x a multiple of y [#permalink]

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27 Sep 2015, 19:04

Hello from the GMAT Club BumpBot!

Thanks to another GMAT Club member, I have just discovered this valuable topic, yet it had no discussion for over a year. I am now bumping it up - doing my job. I think you may find it valuable (esp those replies with Kudos).

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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, 18:46

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^2-x is a multiple of y x(x-1) is a multiple of y i.e {x(x-1)}/y must be integer x/y say 4/2 =integer or 4-1/2=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, 00:59

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(x-1) is a multiple of y let x=7 x-1=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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