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Math Expert V
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If a, b, and c are distinct positive integers where a < b < c and  [#permalink]

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If a, b, and c are distinct positive integers where a < b < c and $$\sqrt{abc} = c$$, what is the value of a?

(1) c = 8
(2) The average of a, b, and c is 14/3 .

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Re: If a, b, and c are distinct positive integers where a < b < c and  [#permalink]

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Bunuel wrote:
If a, b, and c are distinct positive integers where a < b < c and $$\sqrt{abc} = c$$, what is the value of a?

(1) c = 8
(2) The average of a, b, and c is 14/3 .

$$\sqrt{abc} = c$$ a*b*c=$$c^2$$ a*b=c
(1) a*b=8 (a<b and a,b>0) --> Possible cases 1*8 , 2*4 but we know that b<c and c=8 so it there is only one case possible 2*4 -> a=2 SUfficient
(2) a+b+c=14 and ab=c, a+b=14-ab let's test some cases, first of all neither a,b can equal to 1 because a*b=c 1*5=5 and we have distinct digits
a + b = 14-ab
2 + 3 = 8 Not true
2 + 4 = 6 True
a=2

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Re: If a, b, and c are distinct positive integers where a < b < c and  [#permalink]

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Bunuel wrote:
If a, b, and c are distinct positive integers where a < b < c and $$\sqrt{abc} = c$$, what is the value of a?

(1) c = 8
(2) The average of a, b, and c is 14/3 .

Given : $$\sqrt{abc} = c$$
i.e. abc = c^2
i.e. ab = c

Question : a = ?

Statement 1: c = 8
i.e. ab = 8
@b=8, a=1 Not acceptable because b=c whereas b should be less than c
@b=4, a=2 only acceptable case
SUFFICIENT

Statement 2: The average of a, b, and c is 14/3 .
a+b+c = 14
a+b+ab = 14
@b=4, a=2 only acceptable case
SUFFICIENT

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Math Expert V
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Posts: 58445
Re: If a, b, and c are distinct positive integers where a < b < c and  [#permalink]

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1
Bunuel wrote:
If a, b, and c are distinct positive integers where a < b < c and $$\sqrt{abc} = c$$, what is the value of a?

(1) c = 8
(2) The average of a, b, and c is 14/3 .

VERITAS PREP OFFICIAL SOLUTION:

Question Type: What Is the Value? This question asks for the specific value of positive integer a.

Given information in the question stem or diagram: There is a lot of information to leverage from this question stem. a, b, and c are distinct positive integers; a < b < c; and the square root of abc = c. You should first manipulate the last one algebraically by squaring both sides to see that abc = c^2. Divide both sides by c (you can do this because you know that c cannot be 0 from the question stem) and the equation becomes ab = c. So before you even go to the statements you know that ab = c and all of the variables are different positive integers.

Statement 1: c = 8. Combined with what you learned from the question stem, this means that ab = 8. Since a and b are distinct positive integers and a < b, the only possibility is a = 2 and b= 4. You might consider a = 1 and b = 8 but since the integers must be distinct, you cannot have b = 8 since c = 8. This is sufficient but you will only see that if you properly leverage every piece of information given in the question stem. Remember: When you are given even a small piece of information in the question stem it is usually very important. The correct answer is A or D.

Statement 2: The average of a, b, and c is 14/3 . This means that the total of a + b + c = 14. This statement is even trickier than the last but requires a similar leveraging of all available information. It may seem at first glance that there are many possibilities for the values of a, b, and c. However, the only way that ab = c and a + b + c = 14 is for a = 2, b= 4, and c = 8. There is no other way to have three distinct numbers add up to 14 and have ab = c. This statement is also sufficient and the correct answer is D. This question provides an excellent example of a phenomenon you will see often in Data Sufficiency: When a lot of information is given in the question stem, statements are usually sufficient with much less information than you might first think.
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Re: If a, b, and c are distinct positive integers where a < b < c and  [#permalink]

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Bunuel wrote:
If a, b, and c are distinct positive integers where a < b < c and $$\sqrt{abc} = c$$, what is the value of a?

(1) c = 8
(2) The average of a, b, and c is 14/3 .

$$\sqrt{abc} = c$$ --> ab=c
(1) c = 8 --> $$ab = 8 = 1*8=2*4=4*2=8*1$$--> only possibility is a=2, b=4 --> Sufficient.
(2) $$\frac{a+b+c}{3}= \frac{14}{3}$$ --> a+b+c = 14 --> a+b+ab = 14 --> (a+1)(b+1) = 15 =$$1*15 = 3*5 = 5*3 = 15*1$$--> only possibility is a+1=3, b+1=5 --> a=2 --> Sufficient.

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