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# Is x^2*y^4 an integer divisible by 9 ?

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Is x^2*y^4 an integer divisible by 9 ? [#permalink]

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13 Sep 2010, 14:15
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Is x^2*y^4 an integer divisible by 9 ?

(1) x is an integer divisible by 3
(2) xy is an integer divisible by 9

[Reveal] Spoiler:
Statements (1) and (2) TOGETHER are NOT sufficient
Statements (1) and (2) combined are insufficient. If and then both S1 and S2 hold, but is not divisible by 9.

Why is s1 not sufficient. If x is divisible by 3 then the minimum value of x will be 3 and$$x^2 = 9$$ which is divisible by 9?
[Reveal] Spoiler: OA

Last edited by Bunuel on 18 Aug 2012, 08:37, edited 1 time in total.
Edited the question.
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Re: Is [m]x^2 y^4[/m] an integer divisible by 9 ? [#permalink]

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13 Sep 2010, 14:33
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seekmba wrote:
Is $$x^2 y^4$$ an integer divisible by 9 ?

1. x is an integer divisible by 3
2. xy is an integer divisible by 9

Statement (1) ALONE is sufficient, but Statement (2) ALONE is not sufficient
Statement (2) ALONE is sufficient, but Statement (1) ALONE is not sufficient
BOTH statements TOGETHER are sufficient, but NEITHER statement ALONE is sufficient
EACH statement ALONE is sufficient

Statements (1) and (2) TOGETHER are NOT sufficient
Statements (1) and (2) combined are insufficient. If and then both S1 and S2 hold, but is not divisible by 9.

Why is s1 not sufficient. If x is divisible by 3 then the minimum value of x will be 3 and$$x^2 = 9$$ which is divisible by 9?

Note that we are not told that $$x$$ and $$y$$ are integers.

(1) x is an integer divisible by 3 --> $$x=6$$ and $$y=any \ integer$$ then the answer would be YES but if $$x=6$$ and $$y=\frac{3}{2}$$ then the answer would be NO. Not sufficient.

(2) xy is an integer divisible by 9 --> the same example: $$x=6$$ and $$y=any \ integer$$ then the answer would be YES but if $$x=6$$ and $$y=\frac{3}{2}$$ then the answer would be NO. Not sufficient.

(1)+(2) We can use the above examples again:

If $$x=6$$ (divisible by 3) and $$y=any \ integer$$ ($$xy=multiple \ of \ 3$$) then the answer would be YES;
If $$x=6$$ (divisible by 3) and $$y=\frac{3}{2}$$ ($$xy=9$$, hence divisible by 9) then the answer would be NO --> $$x^2*y^4=(xy)^2*y^2=81*\frac{9}{4}$$ not an integer hence not divisible by 9.

Not sufficient.

Hope it's clear.
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Re: Is [m]x^2 y^4[/m] an integer divisible by 9 ? [#permalink]

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13 Sep 2010, 14:32
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seekmba wrote:
Is $$x^2 y^4$$ an integer divisible by 9 ?

1. x is an integer divisible by 3
2. xy is an integer divisible by 9

Statement (1) ALONE is sufficient, but Statement (2) ALONE is not sufficient
Statement (2) ALONE is sufficient, but Statement (1) ALONE is not sufficient
BOTH statements TOGETHER are sufficient, but NEITHER statement ALONE is sufficient
EACH statement ALONE is sufficient

Statements (1) and (2) TOGETHER are NOT sufficient
Statements (1) and (2) combined are insufficient. If and then both S1 and S2 hold, but is not divisible by 9.

Why is s1 not sufficient. If x is divisible by 3 then the minimum value of x will be 3 and$$x^2 = 9$$ which is divisible by 9?

the question ask if x^2y^4 is an INTEGER divisible by 9 but nowhere it said x or y has to be integer themselves!

1. Y could be non -integer so INSUFF
2. xy is divisible by 9. so it is reasonable to say x*x*y*y divisible by 9 right? so that we know x^2y^2 is divisible by 9. HOWEVER, we dont know what the extra y^2 is - it could be 9 or could be 16 what not so INSUFF

c. lets combine - we know x^2*y^2 div by 9 with y^2 unknown. also we know x is div by 3 BUT we can have some situations like:
3^2 * 1^2(the first part satisfies 2) *1^2 = 9 which is SUFF X=3, Y=1
OR
9^2 * (1/3)^2 (again satisfies 2) * (1/3)^2 = 1 which is INSUFF X=9 , Y = 1/3

E
hope it helps...
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Last edited by shaselai on 13 Sep 2010, 14:34, edited 1 time in total.
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Re: Is [m]x^2 y^4[/m] an integer divisible by 9 ? [#permalink]

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13 Sep 2010, 14:58
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seekmba wrote:
Is $$x^2 y^4$$ an integer divisible by 9 ?

1. x is an integer divisible by 3
2. xy is an integer divisible by 9

On GMAT when we are told that $$a$$ is divisible by $$b$$ (or which is the same: "$$a$$ is multiple of $$b$$", or "$$b$$ is a factor of $$a$$"), we can say that:
1. $$a$$ is an integer;
2. $$b$$ is an integer;
3. $$\frac{a}{b}=integer$$.

So the terms "divisible", "multiple", "factor" ("divisor") are used only about integers (at least on GMAT).

Which means that we could omit words in red in the question. For example "is $$x^2 y^4$$ an integer divisible by 9" is the same as "is $$x^2 y^4$$ divisible by 9" as well as "xy is an integer divisible by 9" is the same as "xy is divisible by 9".

Hope it helps.
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Re: Is [m]x^2 y^4[/m] an integer divisible by 9 ? [#permalink]

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13 Sep 2010, 20:16
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I took 81 and 1/9 and concluded my answer as E
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Re: Is x^2*y^4 an integer divisible by 9 ? [#permalink]

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10 Oct 2013, 01:28
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mahendru1992 wrote:
but if we take y as a fraction then $$x^2y^4$$ wouldn't be an integer. But it's stated in the question "is $$x^2y^4$$ an integer divisible by 9?" So how can we take y as a fraction when the whole expression$$x^2y^4$$ has to be an integer. I'm totally confused.

The question asks IS x^2*y^4 an integer divisible by 9. So, we are not told that x^2*y^4 is an integer. It has to be an integer to be divisible by 9, but it's not given.
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Re: Is [m]x^2 y^4[/m] an integer divisible by 9 ? [#permalink]

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13 Sep 2010, 14:23
In fact, even B is sufficient.

since xy is divisible by 9, 18, 27, 36 .
xy can have (1,9), (9,1), (3,3) , (2,9), (9,2), (6,3), (3,6) etc.
all these pairs when substituted for x^2 y^4, are divisible by 9.

I feel ANS should be D.
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Re: Is x^2 y^4 an integer divisible by 9 ? 1. x is an integer [#permalink]

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17 Aug 2012, 12:19
Bunuel, when you use y= any integer in your explanation, what if y was zero (in your example)? x^2=36 and y^2=0 (any integer) would not be divisible by 9, right?
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Re: Is x^2 y^4 an integer divisible by 9 ? 1. x is an integer [#permalink]

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21 Aug 2012, 01:19
jmuduke08 wrote:
Bunuel, when you use y= any integer in your explanation, what if y was zero (in your example)? x^2=36 and y^2=0 (any integer) would not be divisible by 9, right?

If y=0, then x^2*y^4=0={even integer}, because zero is an even integer. Zero is nether positive nor negative, but zero is definitely an even number.

An even number is an integer that is "evenly divisible" by 2, i.e., divisible by 2 without a remainder and as zero is evenly divisible by 2 then it must be even (in fact zero is divisible by every integer except zero itself).

For more check Number Theory chapter of Math Book: math-number-theory-88376.html

Hope it helps.
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Re: Is x^2*y^4 an integer divisible by 9 ? [#permalink]

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17 Jul 2013, 00:22
From 100 hardest questions
Bumping for review and further discussion.
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Re: Is x^2*y^4 an integer divisible by 9 ? [#permalink]

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08 Aug 2013, 09:16
Bunuel wrote:
From 100 hardest questions
Bumping for review and further discussion.

Do you have a set of these 100 hardest questions? Would love to get my hands on em'.

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Re: Is x^2*y^4 an integer divisible by 9 ? [#permalink]

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08 Aug 2013, 22:19
I feel the answer is D coz each statement can alone give the ans as yes.
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Re: Is x^2*y^4 an integer divisible by 9 ? [#permalink]

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08 Aug 2013, 23:09
I feel the answer is D coz each statement can alone give the ans as yes.

i think you are missing on the part that x and y can be fractions also.
is-x-2-y-4-an-integer-divisible-by-100947.html#p780992

hope this helps
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Re: Is x^2*y^4 an integer divisible by 9 ? [#permalink]

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09 Aug 2013, 14:37
seekmba wrote:
Is x^2*y^4 an integer divisible by 9 ?

(1) x is an integer divisible by 3
(2) xy is an integer divisible by 9

[Reveal] Spoiler:
Statements (1) and (2) TOGETHER are NOT sufficient
Statements (1) and (2) combined are insufficient. If and then both S1 and S2 hold, but is not divisible by 9.

Why is s1 not sufficient. If x is divisible by 3 then the minimum value of x will be 3 and$$x^2 = 9$$ which is divisible by 9?

.........
without any doubt the Answer is (E).........................
consider fractions, then you will find x^2*y^4 is divisible by 9 but still is a fraction due to y being a fraction.
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Re: Is x^2*y^4 an integer divisible by 9 ? [#permalink]

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09 Oct 2013, 22:23
but if we take y as a fraction then $$x^2y^4$$ wouldn't be an integer. But it's stated in the question "is $$x^2y^4$$ an integer divisible by 9?" So how can we take y as a fraction when the whole expression$$x^2y^4$$ has to be an integer. I'm totally confused.
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Re: Is x^2*y^4 an integer divisible by 9 ? [#permalink]

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26 Dec 2013, 17:05
vabhs192003 wrote:
Bunuel wrote:
From 100 hardest questions
Bumping for review and further discussion.

Do you have a set of these 100 hardest questions? Would love to get my hands on em'.

100-hardest-data-sufficiency-questions-162413.html
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Is x^2*y^4 an integer divisible by 9 ? [#permalink]

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28 Sep 2014, 11:43
Bunuel wrote:
seekmba wrote:
Is $$x^2 y^4$$ an integer divisible by 9 ?

1. x is an integer divisible by 3
2. xy is an integer divisible by 9

Statement (1) ALONE is sufficient, but Statement (2) ALONE is not sufficient
Statement (2) ALONE is sufficient, but Statement (1) ALONE is not sufficient
BOTH statements TOGETHER are sufficient, but NEITHER statement ALONE is sufficient
EACH statement ALONE is sufficient

Statements (1) and (2) TOGETHER are NOT sufficient
Statements (1) and (2) combined are insufficient. If and then both S1 and S2 hold, but is not divisible by 9.

Why is s1 not sufficient. If x is divisible by 3 then the minimum value of x will be 3 and$$x^2 = 9$$ which is divisible by 9?

Note that we are not told that $$x$$ and $$y$$ are integers.

(1) x is an integer divisible by 3 --> $$x=6$$ and $$y=any \ integer$$ then the answer would be YES but if $$x=6$$ and $$y=\frac{3}{2}$$ then the answer would be NO. Not sufficient.

(2) xy is an integer divisible by 9 --> the same example: $$x=6$$ and $$y=any \ integer$$ then the answer would be YES but if $$x=6$$ and $$y=\frac{3}{2}$$ then the answer would be NO. Not sufficient.

(1)+(2) We can use the above examples again:

If $$x=6$$ (divisible by 3) and $$y=any \ integer$$ ($$xy=multiple \ of \ 3$$) then the answer would be YES;
If $$x=6$$ (divisible by 3) and $$y=\frac{3}{2}$$ ($$xy=9$$, hence divisible by 9) then the answer would be NO --> $$x^2*y^4=(xy)^2*y^2=81*\frac{9}{4}$$ not an integer hence not divisible by 9.

Not sufficient.

Hope it's clear.

Hi Bunuel,

The explanation is crystal clear here. However, i have a doubt with regard to the concept (Exponentiation does not produce primes) -> I will be referring to it as concept E for convenience.

So as per Concept E if x is divisible by 9 then x^2 will also be divisible by 9.

In the above question: x is an integer divisible by 3, then x^2 y^4 (assuming y is an integer) is divisible by 9 is via some other practical logic but not CONCEPT E right?
If yes, then i am guessing it is done through basic plugging in of numbers.
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Re: Is x^2*y^4 an integer divisible by 9 ? [#permalink]

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30 Jan 2015, 07:15
seekmba wrote:
Is x^2*y^4 an integer divisible by 9 ?

(1) x is an integer divisible by 3
(2) xy is an integer divisible by 9

[Reveal] Spoiler:
Statements (1) and (2) TOGETHER are NOT sufficient
Statements (1) and (2) combined are insufficient. If and then both S1 and S2 hold, but is not divisible by 9.

Why is s1 not sufficient. If x is divisible by 3 then the minimum value of x will be 3 and$$x^2 = 9$$ which is divisible by 9?

Dear Bunuel, for statement 2 I chose x=1/2 and y = 18 and the result of x^2*y^4 was divisible by 9 how i can find

smart numbers as that you used under 2 minutes
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Re: Is x^2*y^4 an integer divisible by 9 ? [#permalink]

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24 Aug 2015, 06:21
Bunuel wrote:
seekmba wrote:
Is $$x^2 y^4$$ an integer divisible by 9 ?

1. x is an integer divisible by 3
2. xy is an integer divisible by 9

Statement (1) ALONE is sufficient, but Statement (2) ALONE is not sufficient
Statement (2) ALONE is sufficient, but Statement (1) ALONE is not sufficient
BOTH statements TOGETHER are sufficient, but NEITHER statement ALONE is sufficient
EACH statement ALONE is sufficient

Statements (1) and (2) TOGETHER are NOT sufficient
Statements (1) and (2) combined are insufficient. If and then both S1 and S2 hold, but is not divisible by 9.

Why is s1 not sufficient. If x is divisible by 3 then the minimum value of x will be 3 and$$x^2 = 9$$ which is divisible by 9?

Note that we are not told that $$x$$ and $$y$$ are integers.

(1) x is an integer divisible by 3 --> $$x=6$$ and $$y=any \ integer$$ then the answer would be YES but if $$x=6$$ and $$y=\frac{3}{2}$$ then the answer would be NO. Not sufficient.

(2) xy is an integer divisible by 9 --> the same example: $$x=6$$ and $$y=any \ integer$$ then the answer would be YES but if $$x=6$$ and $$y=\frac{3}{2}$$ then the answer would be NO. Not sufficient.

(1)+(2) We can use the above examples again:

If $$x=6$$ (divisible by 3) and $$y=any \ integer$$ ($$xy=multiple \ of \ 3$$) then the answer would be YES;
If $$x=6$$ (divisible by 3) and $$y=\frac{3}{2}$$ ($$xy=9$$, hence divisible by 9) then the answer would be NO --> $$x^2*y^4=(xy)^2*y^2=81*\frac{9}{4}$$ not an integer hence not divisible by 9.

Not sufficient.

Hope it's clear.

i know in GMAT when they say x is divisible by 4 they assume x is an integer but here how did you come up with x=6 and y =3/2 ?
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Re: Is x^2*y^4 an integer divisible by 9 ? [#permalink]

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24 Aug 2015, 08:20
jimwild wrote:
Bunuel wrote:
seekmba wrote:
Is $$x^2 y^4$$ an integer divisible by 9 ?

1. x is an integer divisible by 3
2. xy is an integer divisible by 9

Statement (1) ALONE is sufficient, but Statement (2) ALONE is not sufficient
Statement (2) ALONE is sufficient, but Statement (1) ALONE is not sufficient
BOTH statements TOGETHER are sufficient, but NEITHER statement ALONE is sufficient
EACH statement ALONE is sufficient

Statements (1) and (2) TOGETHER are NOT sufficient
Statements (1) and (2) combined are insufficient. If and then both S1 and S2 hold, but is not divisible by 9.

Why is s1 not sufficient. If x is divisible by 3 then the minimum value of x will be 3 and$$x^2 = 9$$ which is divisible by 9?

Note that we are not told that $$x$$ and $$y$$ are integers.

(1) x is an integer divisible by 3 --> $$x=6$$ and $$y=any \ integer$$ then the answer would be YES but if $$x=6$$ and $$y=\frac{3}{2}$$ then the answer would be NO. Not sufficient.

(2) xy is an integer divisible by 9 --> the same example: $$x=6$$ and $$y=any \ integer$$ then the answer would be YES but if $$x=6$$ and $$y=\frac{3}{2}$$ then the answer would be NO. Not sufficient.

(1)+(2) We can use the above examples again:

If $$x=6$$ (divisible by 3) and $$y=any \ integer$$ ($$xy=multiple \ of \ 3$$) then the answer would be YES;
If $$x=6$$ (divisible by 3) and $$y=\frac{3}{2}$$ ($$xy=9$$, hence divisible by 9) then the answer would be NO --> $$x^2*y^4=(xy)^2*y^2=81*\frac{9}{4}$$ not an integer hence not divisible by 9.

Not sufficient.

Hope it's clear.

i know in GMAT when they say x is divisible by 4 they assume x is an integer but here how did you come up with x=6 and y =3/2 ?

(2) says that xy is an integer divisible by 9. So, we are told that xy is an integer, not x and y separately. If $$x=6$$ and $$y=\frac{3}{2}$$, then $$xy=9=integer \ divisible \ by \ 9$$.
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Re: Is x^2*y^4 an integer divisible by 9 ?   [#permalink] 24 Aug 2015, 08:20

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