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

Hey!

I need help with this one.

It's obvious that each statement alone is not sufficient, but I'm struggling with both statements together not sufficient.

From (1) we know that p=(t+m)/2
From (2) we know that t-m=16 -> following this info, you can say that either both t and m are even or both are odd.

Now putting it together, since t+m will be even and and saying that (t+m)/2 will be even, too, we can say that it will be even?!?!

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, great explanation but in my humble opinion I don't think you can take m=0 since it says that m p and t are positive integer

Hi,

Can someone help me find where I'm wrong in solving the following Q:

Q: If m,p and t are positive integers and m<p<t, is the product mpt an even integer?
1. t-p=p-m
2. t-m= 16

O.A (E)

My interpretation:

Q: For "mpt" to be an even integer atleast one of the three numbers should be even. to find if any one of m,p or t is even.

1. t-p=p-m

so, t+m=2p
t+m/2 = p

if t+m is divisible by 2 and results in an integer 'p', then 'p' has to be a multiple of '2' which is even. hence 'mpt' should be even. SUFFICIENT

we are down to Options A or D.

2. t-m=16.

this only says the diff of t and m is even. so t and m are either both 'odd' or both 'even'. INSUFFICIENT.

So the correct Answer ( in my opinion) is A.

kindly advise.

If m, p, and t are positive integers and m<p<t, is the product mpt an even integer?

(1) t - p = p - m
(2) t - m = 16

If m, p, and t are positive integers and m<p<t, is the product mpt an even integer?

(1) t - p = p - m
(2) t - m = 16

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.