Problem 7 · AMC 8 Stretch
Stretch
Algebra & Patterns
Number Theory
use-a-key-identityguess-and-check-with-divisors
Three numbers add up to \(13\), multiply to \(-165\), and the sum of their squares is \(155\). What are the three numbers?
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Answer: The numbers are -3, 5, 11
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Hint 1 of 4
The product is negative (\(-165\)) and not too big, so at least one number is negative and they are probably small whole numbers. Factor \(165=3\times5\times11\) to get candidate sizes.
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Hint 2 of 4
Useful identity: \((a+b+c)^2=a^2+b^2+c^2+2(ab+ac+bc)\). You know the sum (13) and the sum of squares (155), so you can find the 'sum of products' \(ab+ac+bc\).
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Hint 3 of 4
Plug in: \(13^2=155+2(ab+ac+bc)\), so \(169=155+2(\ldots)\), giving \(ab+ac+bc=7\).
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Approach: Squaring identity to get the pairwise-product sum, then factor-based guess-and-check
- Because the product \(-165=3\times5\times11\) is small and negative, the three numbers are probably small whole numbers with one negative.
- Use the identity \((a+b+c)^2=a^2+b^2+c^2+2(ab+ac+bc)\). With \(a+b+c=13\) and \(a^2+b^2+c^2=155\): \(169=155+2(ab+ac+bc)\Rightarrow ab+ac+bc=7\).
- We need three numbers with sum \(13\), pairwise-product sum \(7\), and product \(-165\). The factors of \(165\) are \(3,5,11\). Try \(-3,\,5,\,11\).
- Check: sum \(-3+5+11=13\); product \((-3)(5)(11)=-165\); sum of squares \(9+25+121=155\). All three conditions match, so the numbers are \(-3,\ 5,\ 11\).
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