Bunuel wrote:
If m, p, and t are positive integers and m < p < t, is the product mpt an even integer?
For \(mpt\) to be even at least one should be even (as m, p, and t are integers).
(1) \(t-p=p-m\) --> \(\frac{t+m}{2}=p\) --> this algebraic expression means that \(p\) is halfway between \(t\) and \(m\) on the number line: \(----m-------p-------t----\)
So m, p, and t are evenly spaced. Does this imply that any integer must be even? Not necessarily. If \(p\) is odd and \(m\) and \(t\) are some even constant below and above it, then all three will be odd. So we can have an YES as well as a NO answer. For example:
If \(m=1\), \(p=3\), \(t=5\) the answer is NO;
If \(m=2\), \(p=4\), \(t=6\) the answer is YES.
Not sufficient.
(2) \(t-m=16\). Clearly not sufficient. No info about \(p\).
(1)+(2) Second statement says that the distance between \(m\) and \(t\) is 16, so as from (1) \(m\), \(p\), and \(t\) are evenly spaced, then the distance between \(m\) and \(p\) and the distance between\(p\) and \(t\) must 8. But again we can have two different answers:
\(m=0\), \(p=8\), \(t=16\) --> \(mpt=even\);
\(m=1\), \(p=9\), \(t=17\) --> \(mpt=odd\).
Two different answers. Not sufficient.
Answer: E.
Hope it's clear.
Bunuel,
In the question prompt, \(m\) is the
positive number. So, can we use \(m=0\) here?
If we take \(m=0\), then we can also think of \(p=1\) (as \(p>0\)), but this will be seriously flaws of the condition (\(m<p<t\)). The reason is that p \(must\) be greater or equal to 2. If m=0, then someone can also take \(p=1\), which is not right at all.
Thanks__