Is x+y an even number?
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16 Jan 2022, 04:15
Is \(x + y\) an even number?
First of all notice that we are not given that \(x\) and \(y\) are integers
(1) \(5x + 4\) is an odd number
\(5x + 4 = odd\);
\(5x = odd - 4 = odd - even = odd\);
\(x = \frac{odd}{5}\). So, \(x\) is either a fraction of the form \(\frac{odd}{5}\) (..., -3/5, -1/5, 1/5, 3/5, ...) or an odd integer (..., -3, -1, 1, 3, ...).
If \(x = 1\) and \(y = 2\), then \(x + y = 3\), which is NOT even.
If \(x = 1\) and \(y = 1\), then \(x + y = 2\), which IS even.
Not sufficient.
(2) \(x - y\) is an odd number.
If \(x = 1.1\) and \(y = 0.1\), then \(x + y = 1.2\), which is NOT even.
If \(x = 1.5\) and \(y = 0.5\), then \(x + y = 2\), which IS even.
Not sufficient.
(1)+(2) From (1) we have that \(x = \frac{odd}{5}\), so from (2):
\(x - y = odd\);
\(\frac{odd}{5} - y = odd\);
\(y=\frac{odd}{5}-odd=\frac{odd-5*odd}{5}=\frac{even}{5}\). So, \(y\) is either a fraction of the form \(\frac{even}{5}\) (..., -4/5, -2/5, 2/5, 4/5, ...) or an even integer (..., -4, -2, -0, 2, 4, ...).
We got that \(x = \frac{odd}{5}\) and \(y=\frac{even}{5}\), thus \(x+y=\frac{odd}{5}+\frac{even}{5}=\frac{odd+even}{5}=\frac{odd}{5}\). An odd number divvied by 5 is either an odd number or not an integer, so \(x + y\) cannot be even. So, we have a definite NO answer to the question. Sufficient.
Answer: C.