If m,p, and t are positive integers and m<p<t, is the product mpt an even integer?
(I) t-p=p-m
(II) t-m=16
(I) t-p=p-m
(II) t-m=16
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adi_800 wrote:well..i am confused..
Consider statement 1..
t-p=p-m
=> t + m /2 = p
=> So, t + m has to be even so as to make the value of (t + m) /2 an even integer..
=> t and m are both odd or both even..
Case I:
t and m are both even ->product mpt even. Statement 1 sufficient.
Case II:
t and m are both odd -> value of p is even -> one even in product of three -> the product mpt is even..
Statement 1 is sufficient..
Statement II is insufficient..
So, I think answer is A..
where am i going wrong??
why are you assuming (t + m) /2 to be an even integer?=> So, t + m has to be even so as to make the value of (t + m) /2 an even integer..
if t and m are both odd, for e.g, 7 and 3, 7+3 = 10 is divisible by 2 and p will be 5 , i.e all ODDs.Case II:
t and m are both odd -> value of p is even -> one even in product of three -> the product mpt is even..
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