Elementary Math Fractions, Percentages, Word Problems, Equations, Inequations, Factorization, Expansion

 November 24th, 2017, 12:27 PM #1 Newbie   Joined: Nov 2017 From: Romania Posts: 2 Thanks: 0 10 years old math problem - please help me solve it a, b, c are natural numbers and ab(a+b)+ac(a+c)+bc(b+c)=2013. To choose the correct value for product axbxc. A) 2013 B) 620 C) 324 D) 0 E) other number Thank you for your support.
 November 24th, 2017, 01:28 PM #2 Global Moderator   Joined: Dec 2006 Posts: 18,241 Thanks: 1438 Are you sure you typed the equation correctly? The expression ab(a+b)+ac(a+c)+bc(b+c) always has an even value, so it can't equal 2013. Thanks from mihaela
 November 24th, 2017, 01:48 PM #3 Newbie   Joined: Nov 2017 From: Romania Posts: 2 Thanks: 0 Yes, I'm sure that this is the request.
 November 24th, 2017, 03:17 PM #4 Senior Member   Joined: May 2016 From: USA Posts: 857 Thanks: 347 If a is even, then ab and ac are even and thus so are ab(a + b) and ac(a + c). So that sum is even. If either b or c is even, then bc is even as is bc(b + c). If neither b nor c is even, then b + c is even and so is bc(b + c). In short, if a is even, that sum is even no matter what b and c are. But we could do the same analysis for b and for c. So lets suppose a, b, and c is odd. Then (a + b), (a + c), and (b + c) are all even, so the sum is even. As skipjack said, there is no way that the sum can equal 2013. Thanks from mihaela

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