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enigma123
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If x and y are integers greater than 1, is x a multiple of y? Updated on: 07 Jun 2019, 04:41
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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Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!

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66% (01:38) correct 34% (01:46) wrong based on 2630 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


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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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Bunuel
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If x and y are integers greater than 1, is x a multiple of y? 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.

Answer: A.

Hope it helps.
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If x and y are integers greater than 1, is x a multiple of y? 31 Dec 2012, 21:27
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Nadezda
x and y are integers > 1. Is x multiple of y?

(1): 3y^2 + 7y = x
(2): x^2 - x is multiple of y
Show more

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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If x and y are integers greater than 1, is x a multiple of y? 03 May 2014, 12:26
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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.
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If x and y are integers greater than 1, is x a multiple of y? 05 Sep 2012, 14:17
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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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If x and y are integers greater than 1, is x a multiple of y? 18 Sep 2014, 18:36
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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.

Answer: A.

Hope it helps.
Show more

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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If x and y are integers greater than 1, is x a multiple of y? 19 Sep 2014, 02:31
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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.

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?)
Show more

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? 23 Jun 2020, 23:33
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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.

Answer: A.

Hope it helps.
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Can you please explain statement 2 again with better clarity? Bunuel
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If x and y are integers greater than 1, is x a multiple of y? 23 Jun 2020, 23:46
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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.

Answer: A.

Hope it helps.

Can you please explain statement 2 again with better clarity? Bunuel
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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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If x and y are integers greater than 1, is x a multiple of y? 24 Jun 2020, 02:20
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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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If x and y are integers greater than 1, is x a multiple of y? 12 Sep 2023, 09:04
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Attached is a visual that should help.

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If x and y are integers greater than 1, is x a multiple of y? 29 Sep 2025, 06:46
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Hello from the GMAT Club BumpBot!

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