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Re: Is x a multiple of y [#permalink]
29 Jan 2012, 15:31

16

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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: Is x multiple of y? If: 3y^2 + 7y = x? If: x^2-x is mult? [#permalink]
31 Dec 2012, 20:27

8

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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. _________________

Re: If x and y are integers greater than 1, is x a multiple of y [#permalink]
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: If x and y are integers greater than 1, is x a multiple of y [#permalink]
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 - xis a multiple ofy" must be rewritten as "x (x - 1)is a multiple ofy" 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]
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: Is x multiple of y? If: 3y^2 + 7y = x? If: x^2-x is mult? [#permalink]
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 _________________

“If you can't fly then run, if you can't run then walk, if you can't walk then crawl, but whatever you do you have to keep moving forward.”

Re: Is x multiple of y? If: 3y^2 + 7y = x? If: x^2-x is mult? [#permalink]
31 Dec 2012, 21:59

Expert's post

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. _________________

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

Re: If x and y are integers greater than 1, is x a multiple of y [#permalink]
19 Sep 2014, 01:31

Expert's post

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.

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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