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# If m, p, and t are positive integers and m<p<t, is the

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If m, p, and t are positive integers and m<p<t, is the  [#permalink]

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07 Apr 2020, 14:48
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:

$$m=0$$, $$p=8$$, $$t=16$$ --> $$mpt=even$$;
$$m=1$$, $$p=9$$, $$t=17$$ --> $$mpt=odd$$.

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__
If m, p, and t are positive integers and m<p<t, is the   [#permalink] 07 Apr 2020, 14:48

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