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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]  16 Nov 2009, 07:01
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
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Re: Is OG 12th wrong? [#permalink]  26 May 2010, 04:20
michigancat wrote:
For statement 1, you could have m, p, and t be 3, 5, and 7 respectively, in addition to 2, 3, and 4. So mpt can be odd or even. Insufficient.

.
Ooops you are wrong a bit cause from (2) t – m = 16, - > m=t-16 , and not as you stated m=16-t

clear, I am wrong as for (1).

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Last edited by PTK on 26 May 2010, 04:25, edited 1 time in total.
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Re: Could a DS question from OG 12th be wrong? [#permalink]  26 May 2010, 04:25
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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, 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
[Reveal] Spoiler:
E

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 betweenp 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.
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Re: Could a DS question from OG 12th be wrong? [#permalink]  26 May 2010, 04:35
Bunuel wrote:
Pkit wrote:
DS , Q. 76 page 313

Hope it's clear.

Thank you Bunuel, now it is clear for me.
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Re: Could a DS question from OG 12th be wrong? [#permalink]  11 Sep 2011, 02:22
Really good question..
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Re: DS Q 76 OG12 [#permalink]  18 Jan 2012, 08:18
I thought the same thing, I'm glad this post is still around.
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Re: If m, p, and t are positive integers and m<p<t, is the [#permalink]  19 Jan 2012, 02:21
Expert's post
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.
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Save $100 on Veritas Prep GMAT Courses And Admissions Consulting Enroll now. Pay later. Take advantage of Veritas Prep's flexible payment plan options. Veritas Prep Reviews Manager Joined: 03 Jun 2010 Posts: 139 Location: Dubai, UAE Schools: IE Business School, Manchester Business School, HEC Paris, Rotterdam School of Management, Babson College Followers: 2 Kudos [?]: 4 [0], given: 4 Re: If m, p, and t are positive integers and m<p<t, is the [#permalink] 23 Jan 2012, 05:29 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. Posted from GMAT ToolKit Intern Joined: 14 Feb 2012 Posts: 40 Location: Germany Concentration: Technology, Strategy GMAT Date: 06-13-2012 Followers: 0 Kudos [?]: 8 [0], given: 13 Re: If m, p, and t are positive integers and m<p<t, is the [#permalink] 27 Mar 2012, 08:35 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?!?! Math Expert Joined: 02 Sep 2009 Posts: 16782 Followers: 2769 Kudos [?]: 17562 [1] , given: 2183 Re: If m, p, and t are positive integers and m<p<t, is the [#permalink] 27 Mar 2012, 08:57 1 This post received KUDOS Expert's post andih wrote: 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. _________________ Veritas Prep GMAT Instructor Joined: 16 Oct 2010 Posts: 4028 Location: Pune, India Followers: 857 Kudos [?]: 3610 [1] , given: 144 Re: If m, p, and t are positive integers and m<p<t, is the [#permalink] 27 Mar 2012, 09:46 1 This post received KUDOS Expert's post andih wrote: 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?!?! 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) _________________ Karishma Veritas Prep | GMAT Instructor My Blog Save$100 on Veritas Prep GMAT Courses And Admissions Consulting
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Re: Could a DS question from OG 12th be wrong? [#permalink]  03 Dec 2012, 06:27
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, 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
[Reveal] Spoiler:
E

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 betweenp 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, 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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Re: Could a DS question from OG 12th be wrong? [#permalink]  03 Dec 2012, 06:38
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Maxswe wrote:

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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DS: OG - 12TH EDITION - PAGE 313. Q.76 .Help! [#permalink]  07 Dec 2012, 23:48
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.

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Re: DS: OG - 12TH EDITION - PAGE 313. Q.76 .Help! [#permalink]  08 Dec 2012, 00:34
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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.

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

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Re: DS: OG - 12TH EDITION - PAGE 313. Q.76 .Help! [#permalink]  08 Dec 2012, 00:41
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.
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Re: If m, p, and t are positive integers and m<p<t, is the [#permalink]  17 Feb 2014, 01:00
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