GMAT Question of the Day - Daily to your Mailbox; hard ones only

 It is currently 20 Feb 2019, 16:44

### GMAT Club Daily Prep

#### 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

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

## Events & Promotions

###### Events & Promotions in February
PrevNext
SuMoTuWeThFrSa
272829303112
3456789
10111213141516
17181920212223
242526272812
Open Detailed Calendar
• ### Free GMAT Prep Hour

February 20, 2019

February 20, 2019

08:00 PM EST

09:00 PM EST

Strategies and techniques for approaching featured GMAT topics. Wednesday, February 20th at 8 PM EST

February 21, 2019

February 21, 2019

10:00 PM PST

11:00 PM PST

Kick off your 2019 GMAT prep with a free 7-day boot camp that includes free online lessons, webinars, and a full GMAT course access. Limited for the first 99 registrants! Feb. 21st until the 27th.

# If m, n, p are positive integers, is (m2 + n)(2m + p) odd

Author Message
TAGS:

### Hide Tags

Senior Manager
Joined: 21 Oct 2013
Posts: 418
If m, n, p are positive integers, is (m2 + n)(2m + p) odd  [#permalink]

### Show Tags

13 Jan 2014, 04:38
2
3
00:00

Difficulty:

25% (medium)

Question Stats:

77% (01:32) correct 23% (01:44) wrong based on 125 sessions

### HideShow timer Statistics

If m, n, and p are positive integers, is (m^2 + n)(2m + p) an odd integer?

(1) n is an odd integer.
(2) p is an even integer.

OE
(1) For (m^2 + n). Since don't know if m is odd or even, don't know if m^2 is odd or even, which means adding odd, n could make whole statement odd or even. (odd + odd) = even, but (even + odd) = odd. So don't know if (m^2 + n) is odd or even.
For (2m + p). Since m is integer, 2m must even. However, don't know if p is odd or even.
So don't know if (2m + p) is odd or even.
Since don't know if (m^2 + n) is odd or even and don't know if (2m + p) is odd or even, don't know if (m^2 + n)(2m + p) is odd or even.
Insufficient
(2) For (m^2 + n). Don't know if m is odd or even, so don't know if m^2 is odd or even. Don't know if n is odd or even. So don't know if (m^2 + n) is odd or even.
For (2m + p). Since m is integer, 2m is even integer. Since p is even, (2m + p) = (even integer + even integer) = even. (2m + p) is even.
Since (m^2 + n) is integer, (m^2 + n)(2m + p) = (integer x even integer) = even integer. Always No.
Sufficient
Math Expert
Joined: 02 Sep 2009
Posts: 53020
Re: If m, n, p are positive integers, is (m2 + n)(2m + p) odd  [#permalink]

### Show Tags

13 Jan 2014, 05:37
1
goodyear2013 wrote:
If m, n, and p are positive integers, is (m^2 + n)(2m + p) an odd integer?

(1) n is an odd integer.
(2) p is an even integer.

OE
(1) For (m^2 + n). Since don't know if m is odd or even, don't know if m^2 is odd or even, which means adding odd, n could make whole statement odd or even. (odd + odd) = even, but (even + odd) = odd. So don't know if (m^2 + n) is odd or even.
For (2m + p). Since m is integer, 2m must even. However, don't know if p is odd or even.
So don't know if (2m + p) is odd or even.
Since don't know if (m^2 + n) is odd or even and don't know if (2m + p) is odd or even, don't know if (m^2 + n)(2m + p) is odd or even.
Insufficient
(2) For (m^2 + n). Don't know if m is odd or even, so don't know if m^2 is odd or even. Don't know if n is odd or even. So don't know if (m^2 + n) is odd or even.
For (2m + p). Since m is integer, 2m is even integer. Since p is even, (2m + p) = (even integer + even integer) = even. (2m + p) is even.
Since (m^2 + n) is integer, (m^2 + n)(2m + p) = (integer x even integer) = even integer. Always No.
Sufficient

If m, n, and p are positive integers, is (m^2 + n)(2m + p) an odd integer?

(1) n is an odd integer --> (m^2 + n)(2m + p) = (m^2 + odd)(even + p) --> If m=odd, then m^2 + odd = odd + odd = even and the whole product is even but if m=even and p=odd, then both (m^2 + odd) and (even + p) are odd and the product is odd. Not sufficient.

(2) p is an even integer --> (m^2 + n)(2m + p) = (m^2 + n)(even + even) = (m^2 + n)*even = even. So, the product is NOT odd. Sufficient.

_________________
Director
Joined: 08 Jun 2013
Posts: 555
Location: France
GMAT 1: 200 Q1 V1
GPA: 3.82
WE: Consulting (Other)
Re: If m, n, p are positive integers, is (m2 + n)(2m + p) odd  [#permalink]

### Show Tags

12 Oct 2018, 20:57
We want to determine whether or not there is sufficiency to say that $$(m^2 + n)(2m + p)$$ even or that $$(m^2 + n)(2m + p)$$ is not even.

Statement (1) says that n is an odd integer. Let's consider $$m^2 + n$$. Since we don't know if m is odd or even, we don't know if $$m^2$$ is odd or even. We don't know if n is odd or even. So we don't know if $$m^2 + n$$ is odd or even. Let's consider $$2m + p$$. We see that since m is an integer, $$2m$$ must be even. However, we don't know if p is odd or even. So we don't know if $$2m + p$$ is odd or even. Since we don't know if $$m^2 + n$$ is odd or even and we don't know if $$2m + p$$ is odd or even, we don't know if $$(m^2 + n)(2m + p)$$ is odd or even. Statement (1) is insufficient. We can eliminate choices (A) and (D).

Statement (2) says that p is an even integer. Let's look at $$m^2 + n$$. We don't know if m is odd or even, so we don't know if $$m^2$$ is odd or even. We don't know if n is odd or even. So we don't know if $$m^2 + n$$ is odd or even. Now let's look at $$2m + p$$. Since m is an integer, $$2m$$ is an even integer. Since p is even, $$2m + p$$ is the sum of an even integer and an even integer. An even plus an even is even, so $$2m + p$$ is even. Since $$(m^2 + n)$$ is an integer, $$(m^2 + n)(2m + p)$$ is an integer times an even integer, which must be an even integer. The answer to the question is "no." Statement (2) is sufficient. Choice (B) is correct.
_________________

Everything will fall into place…

There is perfect timing for
everything and everyone.
Never doubt, But Work on
improving yourself,
Keep the faith and
It will all make sense.

Re: If m, n, p are positive integers, is (m2 + n)(2m + p) odd   [#permalink] 12 Oct 2018, 20:57
Display posts from previous: Sort by