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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]  09 Nov 2005, 18:42
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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 of the three distinct numbers is 12.
(2) The average (arithmetic mean) of the three numbers is 20/3
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Re: DS: exponents [#permalink]  09 Nov 2005, 18:57
marth750 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 of the three distinct numbers is 12.
(2) The average (arithmetic mean) of the three numbers is 20/3

a b c = c^2
a b = c

1. C = 12 a*b = 12. == suff.
2. a+ b + c = 20
a+b+ab=20 == suff by using values.
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sqrt(abc) = c
ab = c

1) abc = c^2 = 12^ = 144, ab = 144/12 = 12 => suff.

2) (a+b+c) = 20, a+b+ab = 20, ab = 20-a-b, => insuff?

i think A
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From the question stem we have

sqrt(xyz) = x, where x is the largest number. Solving we have

yz = x, where we are trying to solve for yz

(1) Sufficient. This implies yz = x =12

(2) Insufficient.

(x+y+z) = 20 and we can't do anything with this

I pick A.
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Re: DS: exponents [#permalink]  09 Nov 2005, 22:53
Let x,y,z be 3 disting numbers. Let z be the max.
From qn sqrt(xyz) = z

From stmt1. z = 12 -> sqrt(xyz) = 12
=> xyz=144 => xy = 12. So sufficient.

From stmt2 we get x+y+z = 20, and substituting the value of z from qn stem will also not help.

Hence A.
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can we have the OA for this one please?
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The OA is D.

Though, I don't see how this could be. I think the correct answer is A.

sorry for delay.
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Re: DS: exponents [#permalink]  15 Nov 2005, 05:07
marth750 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 of the three distinct numbers is 12.
(2) The average (arithmetic mean) of the three numbers is 20/3

uhm, let's consider (2)
a+b+c= 20/3*3 =20
we have sqrt(abc)= c ---> abc=c^2 ---> ab=c
--> a+b+ab= 20 ---> a(b+1)+ b+ 1= 20+1 --->
(a+1)(b+1)=21 . Since a and b are positive integer, we can obtain 4 cases:
Case 1: a+1=1 and b+1=21 ---> impossible coz a can be 0 since a is a positive integer
Case 2: a+1=21 and b+1=1 , similar to case 1
Case 3: a+1=3, b+1=7 ---> a=2 and b=6 --->ab=12
Case 4: a+1=7, b+1=3 ----> a=6, b=2 -----> ab=12
Thus from statement (2), we obtain ab=12 ----> suff
(1) is also suff as others proved.

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Re: DS: exponents [#permalink]  15 Nov 2005, 09:43
the reason why I thought that the B is not sufficient is
it gives you two equations

a+b+c=20
bc =a

and 3 Variables

I have read somewhere that you need same number of equations as the number of variables to get the values of all the variables...!!!

I guess that's not always true as you just proved it ....or am I missing something.

laxieqv wrote:
marth750 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 of the three distinct numbers is 12.
(2) The average (arithmetic mean) of the three numbers is 20/3

uhm, let's consider (2)
a+b+c= 20/3*3 =20
we have sqrt(abc)= c ---> abc=c^2 ---> ab=c
--> a+b+ab= 20 ---> a(b+1)+ b+ 1= 20+1 --->
(a+1)(b+1)=21 . Since a and b are positive integer, we can obtain 4 cases:
Case 1: a+1=1 and b+1=21 ---> impossible coz a can be 0 since a is a positive integer
Case 2: a+1=21 and b+1=1 , similar to case 1
Case 3: a+1=3, b+1=7 ---> a=2 and b=6 --->ab=12
Case 4: a+1=7, b+1=3 ----> a=6, b=2 -----> ab=12
Thus from statement (2), we obtain ab=12 ----> suff
(1) is also suff as others proved.

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Re: DS: exponents [#permalink]  15 Nov 2005, 14:49
cool_jonny009 wrote:
the reason why I thought that the B is not sufficient is
it gives you two equations

a+b+c=20
bc =a

and 3 Variables

I have read somewhere that you need same number of equations as the number of variables to get the values of all the variables...!!!

I guess that's not always true as you just proved it ....or am I missing something.

No, You are correct. It is not posible to find the individual values with absolute certanity. But the questions asks for the product of two variables. In such cases, the statement has to be analyzed very carefully. You can find the product but not the individual values.
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thanx amy_v for the clarification....i got it
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Re: DS: exponents [#permalink]  15 Nov 2005, 19:04
cool_jonny009 wrote:
I have read somewhere that you need same number of equations as the number of variables to get the values of all the variables...!!!

I guess that's not always true as you just proved it ....or am I missing something.

You are right if you want to solve for three variables you need three equations. In other words, two equations would not be sufficient to determine our answer, in this question. However, notice that there is one more restriction in the question stem, that both a and b are positive integers.

We know that ab=c
and a+b+c=20
we can derive a=(20-b)/(1+b)
Here clearly there are infinite pairs of a and b that would satisfy this equation. However we can only find two pairs of a and b that are positive integers (2,6) or (6,2) which give us a definite product of ab=12.
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Re: DS: exponents   [#permalink] 15 Nov 2005, 19:04
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