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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?

Re: Is [m]x^2 y^4[/m] an integer divisible by 9 ? [#permalink]
13 Sep 2010, 13:32

4

This post received KUDOS

2

This post was BOOKMARKED

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... _________________

If you like my answers please +1 kudos!

Last edited by shaselai on 13 Sep 2010, 13:34, edited 1 time in total.

Re: Is [m]x^2 y^4[/m] an integer divisible by 9 ? [#permalink]
13 Sep 2010, 13:33

3

This post received KUDOS

Expert's post

1

This post was BOOKMARKED

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.

Re: Is [m]x^2 y^4[/m] an integer divisible by 9 ? [#permalink]
13 Sep 2010, 13:58

3

This post received KUDOS

Expert's post

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

One more thing about this question:

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".

Re: Is x^2*y^4 an integer divisible by 9 ? [#permalink]
10 Oct 2013, 00:28

1

This post received KUDOS

Expert's post

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. _________________

Re: Is [m]x^2 y^4[/m] an integer divisible by 9 ? [#permalink]
13 Sep 2010, 13:23

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.

Re: Is x^2 y^4 an integer divisible by 9 ? 1. x is an integer [#permalink]
17 Aug 2012, 11:19

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?

Re: Is x^2 y^4 an integer divisible by 9 ? 1. x is an integer [#permalink]
21 Aug 2012, 00:19

Expert's post

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).

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. _________________

Re: Is x^2*y^4 an integer divisible by 9 ? [#permalink]
09 Oct 2013, 21:23

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.

Is x^2*y^4 an integer divisible by 9 ? [#permalink]
28 Sep 2014, 10:43

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.

Answer: E.

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.

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 _________________

Click +1 Kudos if my post helped

gmatclubot

Re: Is x^2*y^4 an integer divisible by 9 ?
[#permalink]
30 Jan 2015, 06:15

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