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Re: Could a DS question from OG 12th be wrong? [#permalink]
26 May 2010, 04:25

5

This post received KUDOS

Expert's post

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 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]
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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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.

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

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

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: 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.

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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