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

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09 Nov 2005, 19:42

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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

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.

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.

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.

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.

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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