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# If the square root of the product of three distinct positive

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If the square root of the product of three distinct positive [#permalink]

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02 Sep 2006, 10:17
This topic is locked. If you want to discuss this question please re-post it in the respective forum.

If the square root of the product of three distinct positive
integers is equal to the largest of the three numbers, what is the
product of the two smaller numbers?

(1) The largest number is 12.
(2) The average (arithmetic mean) of the three numbers is 20/3

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02 Sep 2006, 10:22
(D). Rephrasing the question:

(xyz)^0.5 = z (assume z is the largest)
xyz=z^2
z(z-xy)=0 (Grouping terms on one side)
Since z is the largest, it CANNOT BE ZERO SINCE THE NUMBERS ARE POSITIVE AND DISTINCT.
So, z=xy

(1) gives us the value of z, SUFFICIENT
(2) x+y+z=20; xy+x+y=20 , If x and y are 6 and 2, this statement is satisfied, SUFFICIENT. (Does any other combination sum to 20 ? )

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03 Sep 2006, 01:43
I also look at this problem and think...statement one is too easy, so there must be a trick. They want me to chose A...so it is probably D

No math here.

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04 Sep 2006, 11:54
ANYONE ELSE .............

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04 Sep 2006, 23:23
Square root of the product of three numbers ...

sqrt(xyz) = z

then xyz=z^2
or xy=z = 12 (Option 1)

Option 2 InSUFF

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04 Sep 2006, 23:32
seeing Mahidhar's expalanation - it's D indeed

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Retired Moderator
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05 Sep 2006, 02:21
If the square root of the product of three distinct positive
integers is equal to the largest of the three numbers, what is the
product of the two smaller numbers?

(1) The largest number is 12.
(2) The average (arithmetic mean) of the three numbers is 20/3

sqrt xyz = z therfore xyz = z^2 from st one if z^2 = 144 therfore xy = 12

st two

sum of three numbers = x+y+z = 20 but from question stem z = xy

therfore x+y+xy = 20 ie x(1+y)+y = 20 therfore x = 20-y/1+y

WE KNOW THAT ODD/EVEN CAN NEVER GIVE AN INTIGER THUS Y CAN NEVER BE ODD ie : if y is odd 20-y is odd and Y+1 IS EVEN AND THUS

x = 20-y/1+y WOULD BE A FRACTION ( BUT X IS A POSITIVE INTIGER)

THUS Y COULD BE {2,4,8,} ANY VALUE FOR Y > 8 WILL YIELD 20-Y/1+Y<1 ,THUS BLUGGING IN VALUES FOR Y WE SEE THAT ONLY 2 CAN YIELD A POSITIVE INTIGER X = 6

AND THUS Y = 2 AND Z = 12 .... SUFF

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05 Sep 2006, 07:12
defenestrate wrote:
I also look at this problem and think...statement one is too easy, so there must be a trick. They want me to chose A...so it is probably D

No math here.

Sometimes you just gotta "use the force Luke." Intuition can help to make educated guesses.

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05 Sep 2006, 08:30
No it is not reasonable at all ..come on

where did you get this from

Option 2

(X x Y x Z)/3 = 20/3 => X x Y x Z = 20
=> X x Y = 20 / Z
=> Z = 20/Z
=> Z ^ 2 = 20
=> Z = sqrt 20 (only positive value) (B is true)

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Re: SQUARE ROOT OF A PRODUCT [#permalink]

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05 Sep 2006, 08:49
yezz wrote:
If the square root of the product of three distinct positive
integers is equal to the largest of the three numbers, what is the
product of the two smaller numbers?

(1) The largest number is 12.
(2) The average (arithmetic mean) of the three numbers is 20/3

**EDIT***

(1) #'s are 3,4,12 Then product of least 2#'s=12
#'s are 2,6,12 Then product of least 2#'s=12

BD

(2) #'s are 2,6,12 Then product of least 2#'s=12

Hence D

Heman

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Re: SQUARE ROOT OF A PRODUCT   [#permalink] 05 Sep 2006, 08:49
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# If the square root of the product of three distinct positive

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