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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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Updated on: 07 Jun 2019, 04:41
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Question Stats:

63% (01:37) correct 37% (01:46) wrong based on 747 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) $$x^2-x$$ 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^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?

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Originally posted by enigma123 on 29 Jan 2012, 16:21.
Last edited by Bunuel on 07 Jun 2019, 04:41, edited 4 times in total.
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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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29 Jan 2012, 16:31
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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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31 Dec 2012, 21:27
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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

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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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4
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: 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
13
8

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, 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) $$x^2-x$$ 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, 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) $$x^2-x$$ 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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Re: If x and y are integers greater than 1, is x a multiple of y?  [#permalink]

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07 Jun 2019, 04:44
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Re: If x and y are integers greater than 1, is x a multiple of y?   [#permalink] 07 Jun 2019, 04:44
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