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the sum of all

Problem Solving — algebra and arithmetic (GMAT Focus Edition)
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by sanju09 » Sun Apr 29, 2012 11:54 pm
The product P of three positive integers is 6 times their sum, and one of these three positive integers is the sum of the other two, then what is the sum of all possible values of P?

[spoiler]Reproduced by Sanjeev K Saxena for www.avenuesabroad.com and posted to this sub forum in order to determine the remaining 4 incorrect answer choices[/spoiler]
The mind is everything. What you think you become. -Lord Buddha



Sanjeev K Saxena
Quantitative Instructor
The Princeton Review - Manya Abroad
Lucknow-226001

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by mathbyvemuri » Mon Apr 30, 2012 2:01 am
Let two of the numbers be x, y. As one of the numbers is sum of the other two, the third number will be (x+y).
Given that, Product of the numbers = 6 (Sum of the numbers)
=>x*y*(x+y) = 6 (x+y+x+y)
=> xy(x+y) = 12(x+y)
=> (x+y)(xy-12) = 0
As the given numbers are positive integers, x+y shall not be zero
=> xy-12 = 0 => xy = 12
The possible number sets are:
1,12,13 with a product of 1*12*13 = 156
(or)
2,6,8 with a product of 2*6*8 = 96
(or)
3,4,7 with a product of 3*4*7 = 84

The sum of all posible products = 156+96+84 = 336
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by sanju09 » Mon Apr 30, 2012 2:21 am
mathbyvemuri wrote:Let two of the numbers be x, y. As one of the numbers is sum of the other two, the third number will be (x+y).
Given that, Product of the numbers = 6 (Sum of the numbers)
=>x*y*(x+y) = 6 (x+y+x+y)
=> xy(x+y) = 12(x+y)
=> (x+y)(xy-12) = 0
As the given numbers are positive integers, x+y shall not be zero
=> xy-12 = 0 => xy = 12
The possible number sets are:
1,12,13 with a product of 1*12*13 = 156
(or)
2,6,8 with a product of 2*6*8 = 96
(or)
3,4,7 with a product of 3*4*7 = 84

The sum of all posible products = 156+96+84 = 336
Excellent! Are you the same vemuri who used to post here in the gone days, and now joined again with a different name? Just curious...
The mind is everything. What you think you become. -Lord Buddha



Sanjeev K Saxena
Quantitative Instructor
The Princeton Review - Manya Abroad
Lucknow-226001

www.manyagroup.com
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by mathbyvemuri » Mon Apr 30, 2012 6:12 am
Hi Sanjeev, I am new to this forum. I am a math enthusiast.
Thanks for the complements, Have a good day.
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