Thank you for using the timer - this advanced tool can estimate your performance and suggest more practice questions. We have subscribed you to Daily Prep Questions via email.

Customized for You

we will pick new questions that match your level based on your Timer History

Track Your Progress

every week, we’ll send you an estimated GMAT score based on your performance

Practice Pays

we will pick new questions that match your level based on your Timer History

Not interested in getting valuable practice questions and articles delivered to your email? No problem, unsubscribe here.

It appears that you are browsing the GMAT Club forum unregistered!

Signing up is free, quick, and confidential.
Join other 500,000 members and get the full benefits of GMAT Club

Registration gives you:

Tests

Take 11 tests and quizzes from GMAT Club and leading GMAT prep companies such as Manhattan GMAT,
Knewton, and others. All are free for GMAT Club members.

Applicant Stats

View detailed applicant stats such as GPA, GMAT score, work experience, location, application
status, and more

Books/Downloads

Download thousands of study notes,
question collections, GMAT Club’s
Grammar and Math books.
All are free!

Thank you for using the timer!
We noticed you are actually not timing your practice. Click the START button first next time you use the timer.
There are many benefits to timing your practice, including:

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 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.

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:

Yes, number properties theory is simple but its application can get really tricky. The worst thing is that you don't even realize that there was a trick in the question and that you have messed up! You might very confidently mark A here and move on! This is a perfect example of trickery of number properties on GMAT. _________________

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

Show Tags

23 Jan 2012, 06:29

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.

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?!?!

We need to find whether at least one of m, p and t is even.

S1: p = (t+m)/2 tells us that (t+m) is even since p has to be integer. So all we know is that t and m are both either odd or both even (since their sum is even). It doesn't say anything about p i.e. whether p is even or odd. p could be odd e.g. (4+2)/2 = 3 or (5+1)/2 = 3 or it could be even e.g. (8+4)/2 = 6 or (7+5)/2 = 6 etc.

S2: t - m = 16 tells us that t and m are either both odd or both even (because the difference between them is even).We again don't know whether they are even. e.g. 18 - 2 = 16 or 17 - 1 = 16. We also don't know whether p is even.

Using both statements, 2p = t+m, 16 = t - m. Add them to get p+8 = t. If t is odd, p is also odd. If t is even, p is also even. So basically, all 3 variables are either odd or all three are even. But we do not know whether they are even. Hence not sufficient. Answer (E) _________________

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

Show Tags

03 Dec 2012, 07:27

Bunuel wrote:

Pkit wrote:

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 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.

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:

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: \(m=2\), \(p=10\), \(t=18\) --> \(mpt=even\); \(m=1\), \(p=9\), \(t=17\) --> \(mpt=odd\). _________________

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

Show Tags

08 Dec 2012, 01:34

1

This post received KUDOS

imk wrote:

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

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

Show Tags

08 Dec 2012, 01:41

Thank you. How silly of me to miss that point. I was thinking of the odd numerator and totally wiped out the possibility that even/even can be an odd integer. btw point noted. this is my first post. Sorry that I gave the answer away.

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

Show Tags

17 Feb 2014, 02:00

Hello from the GMAT Club BumpBot!

Thanks to another GMAT Club member, I have just discovered this valuable topic, yet it had no discussion for over a year. I am now bumping it up - doing my job. I think you may find it valuable (esp those replies with Kudos).

Want to see all other topics I dig out? Follow me (click follow button on profile). You will receive a summary of all topics I bump in your profile area as well as via email. _________________

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

Show Tags

05 Jul 2014, 01:23

f 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 betweenp and t must 8. But again we can have two different answers:

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

Show Tags

06 Jul 2014, 06:14

h2polo wrote:

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. _________________

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

Show Tags

15 Aug 2014, 23:58

Bunuel wrote:

Pkit wrote:

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 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.

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:

This is the kickoff for my 2016-2017 application season. After a summer of introspect and debate I have decided to relaunch my b-school application journey. Why would anyone want...

Check out this awesome article about Anderson on Poets Quants, http://poetsandquants.com/2015/01/02/uclas-anderson-school-morphs-into-a-friendly-tech-hub/ . Anderson is a great place! Sorry for the lack of updates recently. I...

“Oh! Looks like your passport expires soon” – these were the first words at the airport in London I remember last Friday. Shocked that I might not be...