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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] New post   Updated on: 05 Jul 2013, 15:19
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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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D

Originally posted by Bunuel on 19 Nov 2009, 13:45.
Last edited by Bunuel on 05 Jul 2013, 15:19, edited 1 time in total.
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Re: Three distinct positive integers [#permalink] New post   19 Nov 2009, 16:36
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IanStewart wrote:
Bunuel 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.


Bunuel, is this a problem you created, or is it from an existing source? It's a cleverly designed question. One can use trial and error to show that Statement 2 is also sufficient, but it's not required, as shown below:

Let our numbers be a, b, c, where 0 < a < b < c. Then if sqrt(abc) = c, it must be that ab = c, so to find ab (and thus answer the question) we only need to find the largest of our three numbers, and Statement 1 is sufficient.

For Statement 2, if the mean is 20/3, then the sum of our three numbers is 20, so we know: a + b + ab = 20, and a + b + ab is certainly even. If a and b were both odd, or if one were odd and the other even, then a + b + ab would be odd, which is not the case. So a and b must both be even. Thus a and b are two different positive even numbers which give a product less than 20. If a were greater than 2, then a would be at least 4 and b would be at least 6 (since b > a), which gives too large a product. So a must be 2, and using the equation above, substituting a=2, we have 2 + b + 2b = 20, or 3b = 18, and b = 6.

So Statement 2 not only lets us find the value of ab, it actually lets us find both a and b individually, and is sufficient, and the answer is D.


I took this question from some blog. And the OA given was A. But I disagreed. I think the answer should be D.

The way I solved it was slightly different from yours:

\(0<a<b<c\), \(\sqrt{abc}=c\) --> \(ab=c\)

(1) Clearly sufficient.

(2) \(a+b+c=20\) --> \(a+b+ab=20\) --> \((a+1)(b+1)=21\)

\(a\), \(b\), and \(c\) are integers, so:

\(a+1=1\) and \(b+1=21\), doesn't work, as \(a\) becomes 0 and we know that integers are more than 0.

OR
\(a+1=3\) and \(b+1=7\). \(a=2\) and \(b=6\) --> \(ab=12\)

Hence sufficient.

Answer: D.
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Re: Three distinct positive integers [#permalink] New post   19 Nov 2009, 16:20
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Bunuel 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.


Bunuel, is this a problem you created, or is it from an existing source? It's a cleverly designed question. One can use trial and error to show that Statement 2 is also sufficient, but it's not required, as shown below:

Let our numbers be a, b, c, where 0 < a < b < c. Then if sqrt(abc) = c, it must be that ab = c, so to find ab (and thus answer the question) we only need to find the largest of our three numbers, and Statement 1 is sufficient.

For Statement 2, if the mean is 20/3, then the sum of our three numbers is 20, so we know: a + b + ab = 20, and a + b + ab is certainly even. If a and b were both odd, or if one were odd and the other even, then a + b + ab would be odd, which is not the case. So a and b must both be even. Thus a and b are two different positive even numbers which give a product less than 20. If a were greater than 2, then a would be at least 4 and b would be at least 6 (since b > a), which gives too large a product. So a must be 2, and using the equation above, substituting a=2, we have 2 + b + 2b = 20, or 3b = 18, and b = 6.

So Statement 2 not only lets us find the value of ab, it actually lets us find both a and b individually, and is sufficient, and the answer is D.
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Re: Three distinct positive integers [#permalink] New post   19 Nov 2009, 16:16
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Bunuel 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.


0<x<y<z

sqrt (xyz) = z
xy = z

1: z = 12. Suff..

2: x + y + z = 20
x + y + xy = 20
The smallest integer cannot be 1.

After trail and error, x = 2, y = 6 and z =12. Suff..


D.
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Re: If the square root of the product of three distinct positive [#permalink] New post   29 Jun 2020, 02:34
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Lipun wrote:
Can you please confirm if the below approach is correct?

From ST1 we obtained ab = 12. If both statements are individually sufficient to answer a question, then the values obtained in both the statements should be same.
Thus, can we directly check for factors of ab=12 which satisfy the equation (a + b + ab) = 20?

Say, the factors of ab don't satisfy the above equation, then we can term ST2 as insufficient.


No, that approach would not be correct. It sounds like you're trying to prove Statement 2 is false, but Statement 2 is a fact, so it can't be false.

In any official DS question, if you find that Statement 1 gives only one solution, then because Statement 1 and Statement 2 must agree, that solution will always be a solution when you use only Statement 2. In this question, the answer is 12 if you use Statement 1 alone. That absolutely guarantees that 12 is one possible solution when you use Statement 2 alone. That observation is sometimes useful, but on its own it doesn't tell us anything about how to answer the question: what we need to know in DS is whether Statement 2 gives us only one solution (and is sufficient), or if it gives us more than one solution (and is not sufficient). So what you want to find out in this question, when looking at Statement 2 alone, is if 12 is the only solution for ab, or if there might be another solution for ab that is different from 12.

So what you wrote above (the part I highlighted in red) will never happen, except when you're solving badly-designed prep company DS questions (if it happened on an official question, it would mean you'd made a mistake somewhere).
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Re: If the square root of the product of three distinct positive [#permalink] New post   30 Oct 2012, 05:36
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sanjoo wrote:
statment 1: take largest num as Z

x>y>z..square root xyz=square root z

if z=12..

that means xyz must b 144?? xy must b is equal to 12..12*12=144..square root of 144 is equal to 12..

Statement 2..

Bunuel wrote:

The way I solved it was slightly different from yours:

\(0<a<b<c\), \(\sqrt{abc}=c\) --> \(ab=c\)

(1) Clearly sufficient.

(2) \(a+b+c=20\) --> \(a+b+ab=20\) --> \((a+1)(b+1)=21\)

\(a\), \(b\), and \(c\) are integers, so:

\(a+1=1\) and \(b+1=21\), doesn't work, as \(a\) becomes 0 and we know that integers are more than 0.

OR
\(a+1=3\) and \(b+1=7\). \(a=2\) and \(b=6\) --> \(ab=12\)

Hence sufficient.

Answer: D.


bunuel i didnt get statment 2. bold part..


You mean this part: \(a+b+c=20\) --> \(a+b+ab=20\) --> \((a+1)(b+1)=21\)?

We know that \(ab=c\). Now, substitute \(c\) to get: \(a+b+ab=20\). Add 1 to each part: \(a+b+ab+1=21\). Finally, \(a+b+ab+1\) can be written as \((a+1)(b+1)\), thus we have that \((a+1)(b+1)=21\).

Hope it's clear.
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Re: If the square root of the product of three distinct positive [#permalink] New post   30 Oct 2012, 05:42
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Yeah Bunuel this part..

was confused how 20 changed to 21..well its clear now..

Thank u :)
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Re: Three distinct positive integers [#permalink] New post   19 Nov 2009, 16:05
IMO D.

Statement 1) The largest number of the three distinct numbers is 12.
Consider three distinct positive numbers as x, y and z.
If z = 12, then xy = 12.

Statement 2) The average (arithmetic mean) of the three numbers is 20/3.
(x+y+z)/3 = 20/3. So, x+y+z = 20.
So z = xy and z = 20-(x+y)
xy = 20-(x+y)

Both the numbers are positive and thus value of xy must be less than 20.

Solving we can find the two numbers which satisfies the conditions:
xy = z and 20-(x+y) = z.
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Re: Three distinct positive integers [#permalink] New post   19 Nov 2009, 16:44
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Bunuel wrote:
IanStewart wrote:

I took this question from some blog. And the OA given was A. But I disagreed. I think the answer should be D.



Well, it's not a very clever question if the OA given is A :) Nice solution by the way.
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Re: Three distinct positive integers [#permalink] New post   19 Nov 2009, 19:18
IanStewart wrote:
Bunuel wrote:
IanStewart wrote:

I took this question from some blog. And the OA given was A. But I disagreed. I think the answer should be D.



Well, it's not a very clever question if the OA given is A :) Nice solution by the way.


I'm getting D so the answer must be right..just kidding..but yes, I think it's d
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Re: Three distinct positive integers [#permalink] New post   20 Nov 2009, 07:55
For Once I thought, For a change Bunuel has given a simple question for us.

Its definitely tricky and knowing the fact that its a question from Bunuel. I have been extra cautious in ruling the answer as A and found answer as D

Thanks Bunuel for sharpening our brains ......

Thanks all for different approaches.. WOW one fruit different knives.... and all cut it
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Re: If the square root of the product of three distinct positive [#permalink] New post   30 Oct 2012, 05:09
statment 1: take largest num as Z

x>y>z..square root xyz=square root z

if z=12..

that means xyz must b 144?? xy must b is equal to 12..12*12=144..square root of 144 is equal to 12..

Statement 2..

Bunuel wrote:

The way I solved it was slightly different from yours:

\(0<a<b<c\), \(\sqrt{abc}=c\) --> \(ab=c\)

(1) Clearly sufficient.

(2) \(a+b+c=20\) --> \(a+b+ab=20\) --> \((a+1)(b+1)=21\)

\(a\), \(b\), and \(c\) are integers, so:

\(a+1=1\) and \(b+1=21\), doesn't work, as \(a\) becomes 0 and we know that integers are more than 0.

OR
\(a+1=3\) and \(b+1=7\). \(a=2\) and \(b=6\) --> \(ab=12\)

Hence sufficient.

Answer: D.


bunuel i didnt get statment 2. bold part..
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Re: If the square root of the product of three distinct positive [#permalink] New post   31 Aug 2014, 02:43
(1) Clearly sufficient.

(2) a+b+c=20 --> a+b+ab=20 --> (a+1)(b+1)=21

a, b, and c are integers, so:

a+1=1 and b+1=21, doesn't work, as a becomes 0 and we know that integers are more than 0.

OR
a+1=3 and b+1=7. a=2 and b=6 --> ab=12

Hence sufficient.

Answer: D.
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Re: If the square root of the product of three distinct positive [#permalink] New post   21 Feb 2017, 03:54
Bunuel wrote:
IanStewart wrote:
Bunuel 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.


Bunuel, is this a problem you created, or is it from an existing source? It's a cleverly designed question. One can use trial and error to show that Statement 2 is also sufficient, but it's not required, as shown below:

Let our numbers be a, b, c, where 0 < a < b < c. Then if sqrt(abc) = c, it must be that ab = c, so to find ab (and thus answer the question) we only need to find the largest of our three numbers, and Statement 1 is sufficient.

For Statement 2, if the mean is 20/3, then the sum of our three numbers is 20, so we know: a + b + ab = 20, and a + b + ab is certainly even. If a and b were both odd, or if one were odd and the other even, then a + b + ab would be odd, which is not the case. So a and b must both be even. Thus a and b are two different positive even numbers which give a product less than 20. If a were greater than 2, then a would be at least 4 and b would be at least 6 (since b > a), which gives too large a product. So a must be 2, and using the equation above, substituting a=2, we have 2 + b + 2b = 20, or 3b = 18, and b = 6.

So Statement 2 not only lets us find the value of ab, it actually lets us find both a and b individually, and is sufficient, and the answer is D.


I took this question from some blog. And the OA given was A. But I disagreed. I think the answer should be D.

The way I solved it was slightly different from yours:

\(0<a<b<c\), \(\sqrt{abc}=c\) --> \(ab=c\)

(1) Clearly sufficient.

(2) \(a+b+c=20\) --> \(a+b+ab=20\) --> \((a+1)(b+1)=21\)

\(a\), \(b\), and \(c\) are integers, so:

\(a+1=1\) and \(b+1=21\), doesn't work, as \(a\) becomes 0 and we know that integers are more than 0.

OR
\(a+1=3\) and \(b+1=7\). \(a=2\) and \(b=6\) --> \(ab=12\)

Hence sufficient.

Answer: D.


Hi Bunuel,

Thank you for this solution, it can save really plenty of time in real conditions. I got D, but it took me > 4 min. to make 100% sure that 2-6-12 is the only combination that meets stem condition and (2).
Please advise if the way you rationalized 2nd condition can be used in inequalities?
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Re: If the square root of the product of three distinct positive [#permalink] New post   28 Jun 2020, 22:26
IanStewart wrote:

Bunuel, is this a problem you created, or is it from an existing source? It's a cleverly designed question. One can use trial and error to show that Statement 2 is also sufficient, but it's not required, as shown below:

Let our numbers be a, b, c, where 0 < a < b < c. Then if sqrt(abc) = c, it must be that ab = c, so to find ab (and thus answer the question) we only need to find the largest of our three numbers, and Statement 1 is sufficient.

For Statement 2, if the mean is 20/3, then the sum of our three numbers is 20, so we know: a + b + ab = 20, and a + b + ab is certainly even. If a and b were both odd, or if one were odd and the other even, then a + b + ab would be odd, which is not the case. So a and b must both be even. Thus a and b are two different positive even numbers which give a product less than 20. If a were greater than 2, then a would be at least 4 and b would be at least 6 (since b > a), which gives too large a product. So a must be 2, and using the equation above, substituting a=2, we have 2 + b + 2b = 20, or 3b = 18, and b = 6.

So Statement 2 not only lets us find the value of ab, it actually lets us find both a and b individually, and is sufficient, and the answer is D.


Hi IanStewart sir,

I acknowledge the conceptual approach you have taken for ST2 here. Can you please confirm if the below approach is correct?

From ST1 we obtained ab = 12. If both statements are individually sufficient to answer a question, then the values obtained in both the statements should be same.
Thus, can we directly check for factors of ab=12 which satisfy the equation (a + b + ab) = 20?

Say, the factors of ab don't satisfy the above equation, then we can term ST2 as insufficient.

Thanks in advance!

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Lipun
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