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# If x and y are integers greater than 1, is x 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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29 Jan 2012, 15:21
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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^2-x is a multiple of y

[Reveal] Spoiler:
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^2-x is a multiple of y

x(x-1) 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?
[Reveal] Spoiler: OA

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Last edited by Bunuel on 30 Oct 2012, 00:44, edited 2 times in total.

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Re: Is x a multiple of y [#permalink]

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29 Jan 2012, 15:31
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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.

Hope it helps.
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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, 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;

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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, 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.

So insufficient.

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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, 20:27
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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.

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
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 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:
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 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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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.

(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, 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.

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, 01:31
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) $$x^2-x$$ 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.

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

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26 Jan 2017, 18:34
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, 05:13
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^2-x 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, 11:36
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^2-x 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(x-1) = 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   [#permalink] 21 Nov 2017, 11:36
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