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Ok this is my take, take statement 1, t+m/2= p which means that t+m/2 equals an integer and the only way that is possible is when t+m are either both odd or both even since if they are odd and even that doesn't given an integer. Secondly, if t+m is either both odd or both even the result will be even hence p= even. But statement 1 insufficient because we dont know whether t+m are both odd or even.

There BCE
Now let's take statement 2

We have no information about p so insufficient.
Hence CE
Now both statements taken together
We still can't conclude that t and m are either odd or even since that will determine whether the product is even. So E.
Do let me know if my solution makes sense to people.

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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?!?!
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andih
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?!?!

Not sure understood your question correctly. Anyway, check the examples provided in this post: if-m-p-and-t-are-positive-integers-and-m-p-t-is-the-product-mpt-an-even-integer-126259.html#p729805 Hope they'll help to clear your doubts.
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Bunuel
Pkit
DS , Q. 76 page 313

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

My solution is:
(1) t+m=2p ->, \((t+m)/2=p\), p can not be odd since:
if t and m are even, then p is even , keep in mind that m < p < t, \(m<>t\)
if t and m are odd, p is even, keep in mind that m < p < t, \(m<>t\),[(5+11)/2=8, but (5+5)/2=5 odd,but m<>t]
if t is even and m is odd, then p can not be odd , since (even+odd)/2 must give us integer, so P could not be odd, thus it is even.

Then, if:
mt= even*even=even
mt= odd*odd=odd
mt=even*odd=even

Then we know that P must be even, so either of results m*t when multiplied by even number P give us Even product, so the product m*p*t= even
Sufficient.

(2) not sufficinet.

I choose A, however the OG12th's answer choice is Am I wrong?

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
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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

Correct, but it does not affect the answer. Consider:
\(m=2\), \(p=10\), \(t=18\) --> \(mpt=even\);
\(m=1\), \(p=9\), \(t=17\) --> \(mpt=odd\).
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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.
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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.

Hi. please post the OA inside the spoilers to give other a fair shot at the problem. Thanks.

you have said -> "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"

P can or cannot be a multiple of 2, t+m is a multiple of 2.
Take t=3 m =3 then t+m = 6, which is multiple of 2, and p can be 3 so that t+m = 2 (p) = 3+3 = 2 * 3
tmp = odd
if we take t = m = 6 then tmp = Even
t+m = 2p then only 2 things we can deduct
both t and m are both odd or Even, as E+O would result in Odd,which cannot be a multiple of 2

HTH

Cheers
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h2polo
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

1) t-m = 2p
or t-m = even
so, t,m = (even,even) or (odd,odd)
and p can be even or odd
so A alone is not sufficient.

2) t-m = 16
t,m = (17,1) or (18,2) ....
again the same condition and p is unknown.
so B alone is also insufficient.

(1)+(2)
t+m = 2p
t-m = 16
solving; t-p = 8
t,p = (100,92) or (101,93) ...
when (t,p) = (100,92) then m = even and m*t*p = even
when (t,p) = (101,93) then m = odd and m*t*p = odd
hence (1)+(2) also is insufficient.

Therefore, answer is E.

------------------
+1 if you liked my solution. Tx. :)
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h2polo
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

We are given that m, p, and t are positive integers with m < p < t, and we need to determine whether mpt is an even integer.

Statement One Alone:

t – p = p – m

Simplifying the equation in statement one, we have:

t – p = p – m

t + m = 2p

Since 2p must be even, we see that either t and m are both odd or t and m are both even. However, since we know nothing about p, p could be either odd or even. If t and m are both even, then regardless of whether p is odd or even, the product mpt will be even. However, if t and m are both odd and p is also odd, then mpt will be odd. Statement one alone is not sufficient to answer the question.

Statement Two Alone:

t – m = 16

Since t – m = 16, we see that either t and m are both odd or t and m are both even. However, since we know nothing about p, statement two alone is not sufficient to answer the question.

Statements One and Two Together:

Using statements one and two, we still know only that either t and m are both odd or t and m are both even, and we still have no information regarding p. Thus, the two statements together are not sufficient.

Answer: E
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E is the answer.

Let's see how...
Is the product mpt even?

To answer that, we need to know if atleast one of m, p or t is even because any integer (odd or even) multiplied with even is even.

1. t - p = p - m
= t + m = 2p
=> the sum of t and m is even. A sum or difference of two integers will be even if either both the integers are even or odd.
In this case, both t and m can be even or both t and m can be odd. Therefore, Insufficient.

2. t - m = 16
=> Both t and m can be even or odd as they were in Stat. 1.

We can easily solve the equations from both statements by adding them. That gives us 2t = 2p + 16 which means t = p + 8. As the obtained equation does not provide concluding results, the answer is E.
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Bunuel
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.


I have a question for option C:
It is given m, p, t are positive integers. In that case m cannot be 0 right? This will cancel the first option of 0, 8 and 16…?
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Given ­m<P<t
If m,p,t all are odd, product mpt is an odd integer, else product mpt an even integer
check both statements if it can conclude above statement.
(1) t - p = p - m
(2) t - m = 16

Option is E, because both cannot.
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­I am not a fan of assuming values, hence I tried solving this to the end.

S1) t-p=p-m
p = (t+m)/2
This gives us no info on the value of mpt. We dont know whether t,m or p are even/odd.

Reject S1) A,D are out.

S2) t-m = 16.
No info on p, or m or t.
Reject S2) B is out.

Combined:
We have 2 equations:
1) t+m=2p
2) t-m=16

Solving for t and m, we get:
t = p+8
m= p-8
so mpt= (p-8)(p)(p+8)
=p^3-64p.

This again gives no info. If p = odd then valus of the eqn is odd.
if p is even value is even.
 Hence E.
 
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