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Is x^2*y^4 an integer divisible by 9 ? [#permalink]
 13 Sep 2010, 14:15
Question Stats:
16% (02:36) correct
83% (00:27) wrong based on 1 sessions
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 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 andx^2 = 9 which is divisible by 9?
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]
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 [m]x^2 y^4[/m] an integer divisible by 9 ? [#permalink]
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 andx^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]
13 Sep 2010, 14:33
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 andx^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.
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Re: Is [m]x^2 y^4[/m] an integer divisible by 9 ? [#permalink]
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
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". Hope it helps.
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Re: Is [m]x^2 y^4[/m] an integer divisible by 9 ? [#permalink]
13 Sep 2010, 20:16
I took 81 and 1/9 and concluded my answer as E
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Re: Is [m]x^2 y^4[/m] an integer divisible by 9 ? [#permalink]
17 Sep 2010, 09:42
shaselai 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 andx^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... Wow! Great explanation. +1 man.
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Re: Is x^2 y^4 an integer divisible by 9 ? 1. x is an integer [#permalink]
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]
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.htmlHope it helps.
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Re: Is x^2 y^4 an integer divisible by 9 ? 1. x is an integer
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21 Aug 2012, 01:19
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