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# M16-03

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Math Expert
Joined: 02 Sep 2009
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16 Sep 2014, 00:58
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Difficulty:

95% (hard)

Question Stats:

26% (01:01) correct 74% (00:55) wrong based on 145 sessions

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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

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Joined: 02 Sep 2009
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16 Sep 2014, 00:58
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Official Solution:

Statements (1) and (2) combined are insufficient. If $$x = 81$$ and $$y = \frac{1}{9}$$ then both S1 and S2 hold, but $$x^2y^4 = 1$$ is not divisible by 9.

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22 Oct 2014, 03:19
Bunuel wrote:
Official Solution:

Statements (1) and (2) combined are insufficient. If $$x = 81$$ and $$y = \frac{1}{9}$$ then both S1 and S2 hold, but $$x^2y^4 = 1$$ is not divisible by 9.

Hi Bunuel

This is in relation to the concept: Exponentiation does not produce primes.

So had the Question been: Is $$x^2*y^2$$ an integer divisible by 9?

And the given statement (2) been: xy is an integer divisible by 9.

Then it would have been safe to say that the above statement is sufficient ?
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22 Oct 2014, 03:44
1
earnit wrote:
Bunuel wrote:
Official Solution:

Statements (1) and (2) combined are insufficient. If $$x = 81$$ and $$y = \frac{1}{9}$$ then both S1 and S2 hold, but $$x^2y^4 = 1$$ is not divisible by 9.

Hi Bunuel

This is in relation to the concept: Exponentiation does not produce primes.

So had the Question been: Is $$x^2*y^2$$ an integer divisible by 9?

And the given statement (2) been: xy is an integer divisible by 9.

Then it would have been safe to say that the above statement is sufficient ?

Yes, because if xy is divisible by 9, then so must be (xy)^2.
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05 Mar 2015, 00:17
Bunuel 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

This question exploits the fact that since xy is an integer, does not mean that x and y are integers as well.$$x = 81$$and $$y = 1/9$$

How does our response change if it is given that x,y are integers?
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01 Jul 2015, 22:55
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26 Aug 2015, 05:51
I think this is a high-quality question and I agree with explanation. A great trap for someone who is answering rapidly...
Well done !
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20 Nov 2015, 09:46
basically the question asks whether 2 conditions are met: integer? divisible by 9?

stems combined do not give a clue whether y^4 is an integer
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20 Aug 2016, 23:03
Hi Bunuel, I'm wondering whether there is another way to solve this problem, instead of plug-in method?
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02 Sep 2016, 08:22
hope this one will help everyone, very tricky question

choice 1 , x = 3, y = 1/2 so x^2y^4= 9*1/16 not an integer

choice 2- xy integer divisible by 9, let say x = 9/(1/7)to power 8 and y (1/7)to power 8 so xy = 9

but x^2y^4 will not be an integer now= it will be 81 * (1/7) to power 16, which is obviously not an integer

combining both 1 and 2 also doesn't solve the problem as in choice 2 x is divisible by 3

we can take those easy cases when it comes integer too and left that exercise for you
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09 Sep 2016, 19:05
Bunuel wrote:
Official Solution:

Statements (1) and (2) combined are insufficient. If $$x = 81$$ and $$y = \frac{1}{9}$$ then both S1 and S2 hold, but $$x^2y^4 = 1$$ is not divisible by 9.

Bunuel Can you explain why combining both condition is not sufficient . My take is that combining condition is sufficient. I will try to explain

condition 1 - x is divisible by 3
condition 2 - xy is divisible by 9

if we combine these two condition for condition 2 to be true either x = 3n and y =3m or x= 9n and y can be any integer not fraction . in both the cases x^2y^4 is divisible by 9 hence sufficient .

Sorry if I missed something but will appreciate if you can check my logic
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26 May 2017, 07:53
Mr. Brunel,

Is my assumption correct?

x^2*y^4 an integer.

So could be 0 and and any +ve integers since even exponents.

S1: x is an integer divisible by 3.

X could be zero or multiples of 3 (NS)

S2: XY is an integer divisible by 9.

X and Y could be zero or X=0 and y=1 or x or y could by multiples of 3. (NS)

S1 and S2: could be zero or multiples of 3. Hence (E).
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26 May 2017, 08:35
Manoraaju wrote:
Mr. Brunel,

Is my assumption correct?

x^2*y^4 an integer.

So could be 0 and and any +ve integers since even exponents.

S1: x is an integer divisible by 3.

X could be zero or multiples of 3 (NS)

S2: XY is an integer divisible by 9.

X and Y could be zero or X=0 and y=1 or x or y could by multiples of 3. (NS)

S1 and S2: could be zero or multiples of 3. Hence (E).

No.

0 is divisible by every integer except 0 itself.
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08 Jul 2017, 02:15
hi
what if ,we take xy from statement 2?
from question x2y4 is an integer
since xy an integer ,divisible by 9 as per stmt2.
x2y4(int)=xy(int).xy3(int)
isnt statement 2 alone sufficient
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08 Jul 2017, 04:49
digvijay99 wrote:
hi
what if ,we take xy from statement 2?
from question x2y4 is an integer
since xy an integer ,divisible by 9 as per stmt2.
x2y4(int)=xy(int).xy3(int)
isnt statement 2 alone sufficient

The highlighted part is not necessarily true. xy^2 = (xy)*y^2 may not be an integer because y^2 may not be an integer.

Check alternative solution below:

Is $$x^2 y^4$$ an integer 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 helps.
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06 Feb 2018, 06:25
Hi Bunuel,

I didn't use the plug-method as I found it too confusing...just wondered if the following rationale was correct?

x^2*y^4 = 3^2*int?

Simplified: (x*y^2)^2 = 3*int?

(1) x= 3*int
translates to 9*y^4=3*int?
Insufficient as we don't know whether y is an integer.

(2) x*y= 3*int
translates to (3*int*y)^2
Simplified further to: 9*int*y^2=3*int?
Again, Insufficient as we don't know whether y is an integer.

(1) AND (2) Together = still insufficient as we still don't know whether y is an integer...Hence E.
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05 Jun 2018, 12:18
Bunuel 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

Hi Bunuel

Question states , X^2Y^4 is an integer divisible by 9 , Y/N

Doesn't this mean that X and Y both are integers ?
I have done like this , X^2 * Y^4 , is this wrong ?

Statement B says , XY integer can be divided by 9 , so doesn't this already means that X & Y both are integers ? In this case by divisibility rules , Statement B is sufficient.
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05 Jun 2018, 20:30
Hi

I am not able to understand the solution, can anyone please explain the solution again.
Re: M16-03 &nbs [#permalink] 05 Jun 2018, 20:30
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# M16-03

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