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Re M1603
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15 Sep 2014, 23:58



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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.
Answer: E Hi BunuelThis 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, 02:44



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Re: M1603
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04 Mar 2015, 23:17
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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Sorry, I have some misreading.



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Re M1603
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26 Aug 2015, 04:51
I think this is a highquality question and I agree with explanation. A great trap for someone who is answering rapidly... Well done !



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Re: M1603
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20 Nov 2015, 08:46
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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Re: M1603
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20 Aug 2016, 22:03
Hi Bunuel, I'm wondering whether there is another way to solve this problem, instead of plugin method?



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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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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.
Answer: E 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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Re: M1603
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26 May 2017, 06:53
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, 07:35



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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, 03:49
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. Answer: E. Hope it helps.
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Re: M1603
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06 Feb 2018, 05:25
Hi Bunuel,
I didn't use the plugmethod 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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Re: M1603
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05 Jun 2018, 11:18
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 Can you please guide ? 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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Re: M1603
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05 Jun 2018, 19:30
Hi
I am not able to understand the solution, can anyone please explain the solution again.










