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kevincan
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GRE 1: Q170 V170
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GRE 1: Q170 V170
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Add all of them and you have two of each of ab, bc and ab added, that is 2(ab+bc+ac)=152+162+170=484
So ab+bc+ac=242.
Choose any of the given expression and subtract from above..
C(a+b)=ca+bc=170..
So ab =242-170=72
Thus ab*c will be a multiple of 72..
Hit 720..


If there were other multiples of 72 in options..
Subtract the other two expressions in similar manner to get ac and bc as 80 and 90..
Therefore ab*bc*ac =72*80*90
(abc)^2=72*8*10*9*10=72*72*10*10
So abc = 720


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kevincan
If a, b, and c are positive real numbers such that a(b + c) = 152, b(c + a) = 162, and c(a + b) = 170, what is the value of abc?

(A) 704
(B) 720
(C) 736
(D) 760
(E) 784
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please provide simplified and quicker solution for this question
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kevincan
If a, b, and c are positive real numbers such that a(b + c) = 152, b(c + a) = 162, and c(a + b) = 170, what is the value of abc?

(A) 704
(B) 720
(C) 736
(D) 760
(E) 784
Since all products are integers, I am thinking that likely a, b and c are integers. Also the products are similar, so it seems that a, b and c have values close to each other. All are distinct because the result in all 3 cases is different.

c(a+b) = 170 = 10*17
So a and b could be 8 and 9 to give a+b=17

b(c + a) = 162 = 9*18
a(b + c) = 152 = 8*19

So a = 8, b = 9, c = 10 works.
abc = 8*9*10 =720

This method works when you have a strong sense of numbers.
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