ziyuenlau wrote:
If a,b, and c are integers, what is value of \(a+b+c\)?
(1) \(a^2+b^2+c^2=27\)
(2) \(ab+ac+bc=11\)
Nice question.
Target question: What is the value of a + b + c? Given: a,b, and c are integers Statement 1: a² + b² + c² = 27 Since we're told that a,b, and c are INTEGERS, our options for a, b, and c are quite limited.
That said, there is more than one possible solution to the equation in statement 1. Here are two:
Case a: a = 1, b = 1 and c = 5 (notice that a² + b² + c² = 1² + 1² + 5² = 27). In this case,
a + b + c = 1 + 1 + 5 = 7Case b: a = -1, b = -1 and c = -5 (notice that a² + b² + c² = (-1)² + (-1)² + (-5)² = 27). In this case,
a + b + c = (-1) + (-1) + (-5)= -7Since we cannot answer the
target question with certainty, statement 1 is NOT SUFFICIENT
Statement 2: ab + ac + bc = 11 NOTE: Rather than try to find solutions to this equation, I'll start by checking whether the two solutions I found for statement 1 also satisfy statement 2.
Yes, it turns out they DO!
That is, the following solutions satisfy the equation in statement 2:
Case a: a = 1, b = 1 and c = 5
(notice that ab + ac + bc = (1)(1) + (1)(5) + (1)(5) = 11). In this case,
a + b + c = 1 + 1 + 5 = 7Case b: a = -1, b = -1 and c = -5
(notice that ab + ac + bc = (-1)(-1) + (-1)(-5) + (-1)(-5) = 11). In this case,
a + b + c = (-1) + (-1) + (-5)= -7Since we cannot answer the
target question with certainty, statement 2 is NOT SUFFICIENT
Statements 1 and 2 combined Since we were able to find the same counter-examples in both statement 1 AND statement 2, the same counter-examples will work when we COMBINE both statements.
That is, the following solutions satisfy BOTH statements:
Case a: a = 1, b = 1 and c = 5. In this case,
a + b + c = 1 + 1 + 5 = 7Case b: a = -1, b = -1 and c = -5. In this case,
a + b + c = (-1) + (-1) + (-5)= -7Since we cannot answer the
target question with certainty, the combined statements are NOT SUFFICIENT
Answer: E
Cheers,
Brent
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