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If the square root of the product of three distinct positive [#permalink]
 19 Nov 2009, 13:45
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Re: Three distinct positive integers [#permalink]
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]
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)=21a, 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=12Hence sufficient. Answer: D.
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Re: Three distinct positive integers [#permalink]
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: Three distinct positive integers [#permalink]
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]
19 Nov 2009, 16:44
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]
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]
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]
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]
30 Oct 2012, 05:36
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]
30 Oct 2012, 05:42
Yeah Bunuel this part.. was confused how 20 changed to 21..well its clear now.. Thank u
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Re: If the square root of the product of three distinct positive
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30 Oct 2012, 05:42
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