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x and y are positive integers such that x < y. If

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Lucky2783
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x and y are positive integers such that x < y. If [#permalink] New post   Updated on: 16 Feb 2015, 03:59
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x and y are positive integers such that x < y. If \(x\sqrt{y} = 6\sqrt{6}\), then xy could equal

A. 36
B. 48
C. 54
D. 96
E. 108
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E

Originally posted by Lucky2783 on 15 Feb 2015, 23:24.
Last edited by Bunuel on 16 Feb 2015, 03:59, edited 1 time in total.
Renamed the topic and edited the question.
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KarishmaB
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Re: x and y are positive integers such that x < y. If [#permalink] New post   15 Feb 2015, 23:42
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Lucky2783 wrote:
x and y are positive integers such that x < y. If x sqrt(y) = 6 sqrt(6) , then xy could equal

36
48
54
96
108


First instinct, when you see
\(x*\sqrt{y} = 6*\sqrt{6}\), you say that x = 6, y = 6 will satisfy this. But note that x < y.

So y should be 6*Perfect square so that when you square root it, you are left with \(\sqrt{6}\).

Try y = 6*4. Then x will be 3 so that you get 3*2 = 6 outside the square root.
x*y = 3*24 = 72 (not in the options)

Try y = 6*9. Then x will be 2 so that you get 2*3 = 6 outside the square root.
x*y = 2*54 = 108

Answer (E)
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Re: x and y are positive integers such that x < y. If [#permalink] New post   16 Feb 2015, 19:38
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HI Lucky2783,

This question is a mix of Exponent rules and Number Properties.

We're told that X and Y are POSITIVE INTEGERS and that X < Y. We're also told that X(sqrt Y) = 6(sqrt 6).

In most cases, we're asked to "simplify" a radical...

For example:
(sqrt 50) = 5(sqrt 2)

Here, we have to go "in reverse" and put the 6 "back in" to the radical...

6(sqrt 6) = (sqrt 216)

The question asks for a possible value of XY...

We have (X)(X)(Y) = 216 as a reference

Since the answers are ALL integers, we're looking for one, that when multiplied by another positive integer, gives us 216... Notice how 108 is exactly HALF of 216....?

IF...
(X)(Y) = 108 and X=2, then (X)(X)(Y) = 216.

Final Answer:
Show Spoiler
E


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Re: x and y are positive integers such that x < y. If [#permalink] New post   16 Feb 2015, 19:50
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X and Y are positive integer, which eliminates all negative situation.

x\sqrt{y} = 6\sqrt{6} ->
\sqrt{x*x*y} = \sqrt{6*6*6}
6*6*6 = 3*2*3*2*3*2
since x<y
-> x= 2, xy = 3*2*3*2*3 = 108
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Re: x and y are positive integers such that x < y. If [#permalink] New post   16 Feb 2015, 20:02
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Lucky2783 wrote:
x and y are positive integers such that x < y. If \(x\sqrt{y} = 6\sqrt{6}\), then xy could equal

A. 36
B. 48
C. 54
D. 96
E. 108


hi,
since it is given
\(x\sqrt{y} = 6\sqrt{6}\)...
It is clear that \(\sqrt{y} = t\sqrt{6}\)...
and t can be 2 or 3 or 6 as 6=2*3..
1)let t=3 so x will be 2.. \(\sqrt{y} = 3\sqrt{6}\)...
so y=\((3\sqrt{6})^2\)
y=54 and xy=108.. ans E..
although we already have our ans, the two other possible values can be..
2) the other possible value is t=2 and x=3
xy=3*\((2\sqrt{6})^2\)=3*24=72.. not a choice
3) the other possible value is t=6 and x=1
xy=1*\((6\sqrt{6})^2\)=1*216=216.. not a choice
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x and y are positive integers such that x < y. If [#permalink] New post   16 Feb 2015, 20:08
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EMPOWERgmatRichC wrote:
HI Lucky2783,

This question is a mix of Exponent rules and Number Properties.

We're told that X and Y are POSITIVE INTEGERS and that X < Y. We're also told that X(sqrt Y) = 6(sqrt 6).

In most cases, we're asked to "simplify" a radical...

For example:
(sqrt 50) = 5(sqrt 2)

Here, we have to go "in reverse" and put the 6 "back in" to the radical...

6(sqrt 6) = (sqrt 216)

The question asks for a possible value of XY...

We have (X)(X)(Y) = 216 as a reference

Since the answers are ALL integers, we're looking for one, that when multiplied by another positive integer, gives us 216... Notice how 108 is exactly HALF of 216....?

IF...
(X)(Y) = 108 and X=2, then (X)(X)(Y) = 216.

Final Answer:
Show Spoiler
E


GMAT assassins aren't born, they're made,
Rich


hi,
the ans is correct but there are other values in choices which when multiplied by integer would give you 216..
A. 36.... 36*6=216
C. 54....54*4=216
E. 108...108*2=216...
so it is important to see which of these satisfies the equation x<y to get the correct answer...
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Re: x and y are positive integers such that x < y. If [#permalink] New post   16 Feb 2015, 20:19
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Hi chetan2u,

As noted in my explanation, X and Y are both INTEGERS and X < Y; we need an answer that is in the format (X)(X)(Y) = 216

Of the 3 values that you listed, 2 of them do NOT fit that pattern.

36(6) = 216, but you would end up with (6)(6)(6) which is NOT a match (since X is NOT less than Y).

54(4) = 216, but you would end up with either (4)(4)(13.5) or (root54)(root54)(4), NEITHER of which is a match (since they both include NON-integer values).

108(2) = 216, which gives us (2)(2)(54), which IS the correct answer.

GMAT assassins aren't born, they're made,
Rich
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x and y are positive integers such that x < y. If [#permalink] New post   14 Dec 2015, 19:57
x is smaller than y

\(x*sqrt(y) = 6*sqrt(6)\\
sqrt[(x^2)*y] = sqrt(36*6)\\
sqrt[(x^2)*y] = sqrt(216)\)

ok, let's find prime factorization of 216
ok, 216 = 2 * 108
108 = 2 * 54
stop right here!
we have two 2's, and 54. since x^2 * y = 216, it might be the case that x=2 and y = 54. 2*54 = 108, and it is in the answer choices. since it is a "could be" question, there is no need to check further.
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Re: x and y are positive integers such that x < y. If [#permalink] New post   08 Aug 2022, 11:13
I can see why 36 is the choice for wrong answers mostly. So here's how I did it:

Remember x and y are positive integers. and x is less than or y is greater than x.

we know that x^2 * y = 216

If we prime factorize 216 we get 3^3 * 2^3

Basically you take one of the 2's and give it to y, so Y = 3^3 * 2 = 54
and x^2 remains with 2^2 and thus x = 2.

xy = 54 * 2 = 108

you could take one of the 3's and do it the other way but that won't give any of the options mentioned. (72)
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x and y are positive integers such that x < y. If [#permalink] New post   08 Aug 2022, 13:49
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Lucky2783 wrote:
x and y are positive integers such that x < y. If \(x\sqrt{y} = 6\sqrt{6}\), then xy could equal

A. 36
B. 48
C. 54
D. 96
E. 108



Square both sides:
\(x^2y = 216\)

Let's just try a few values.
x = 1, y = 216, xy = 216 ... not an answer choice.
x = 2, y = 216/4 = 54, xy = 108 ... Yay!
That is ALL we need to do. No reason to get all fancy.

Answer choice E.

.
.
.

Bonus:
x = 3, y = 216/9 = 24, xy = 72 ... not an answer choice.
x = 4, y = 216/16 = 13.something ... invalid since y is not an integer.
x = 5, y = 216/25 = 8.something ... invalid since y is not an integer.
x = 6, y = 216/36 = 6 ... invalid since x is not less than y.
If we keep going x will be greater than y. We didn't need to test all of these, but even if we'd had to, it wasn't that bad. Brute force and trial-and-error are frequently (usually) faster than figuring out formulas.
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Re: x and y are positive integers such that x < y. If [#permalink] New post   11 Oct 2023, 08:22
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