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If x and y are integers greater than 1, is x 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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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.
General Discussion
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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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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]
Bunuel
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]
russ9
Bunuel
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]
Bunuel
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.

Can you please explain statement 2 again with better clarity? Bunuel
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Re: If x and y are integers greater than 1, is x a multiple of y? [#permalink]
Krishchamp
Bunuel
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.

Can you please explain statement 2 again with better clarity? Bunuel

Not sure what to explain. From (2) x could be a multiple of y (for example, x = 2 and y = 2) but x could also not be a multiple of y (for example, x = 3 and y = 2).
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Re: If x and y are integers greater than 1, is x a multiple of y? [#permalink]
1:
y(3y+7)=x
y*K (integer) is x, so x is a multiple of y

2:
x(x-1)=y
so x*K(integer) is y, so y is a multiple of X, but not the opposite
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Re: If x and y are integers greater than 1, is x a multiple of y? [#permalink]
Attached is a visual that should help.

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