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# m03#22

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m03#22 [#permalink]  01 Jan 2009, 09:44
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Is the product of three integers $$p$$, $$q$$, and $$r$$ even?

1. $$(p - 1)(r + 1)$$ is odd
2. $$(q - r)^2$$ is odd

[Reveal] Spoiler: OA
D

Source: GMAT Club Tests - hardest GMAT questions

explanation says D.
I think it should be E because p,q or r could be 0.
Am I being dense?
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Re: m03#22 [#permalink]  19 Jul 2012, 05:43
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Expert's post
suryans wrote:
Is the product of three integers $$p$$, $$q$$, and $$r$$ even?

1. $$(p - 1)(r + 1)$$ is odd
2. $$(q - r)^2$$ is odd

[Reveal] Spoiler: OA
D

Source: GMAT Club Tests - hardest GMAT questions

explanation says D.
I think it should be E because p,q or r could be 0.
Am I being dense?

First of all zero is an even integer. Zero is nether positive nor negative, but zero is definitely an even number.

An even number is an integer that is "evenly divisible" by 2, i.e., divisible by 2 without a remainder and as zero is evenly divisible by 2 then it must be even (in fact zero is divisible by every integer except zero itself).

Or in another way: an even number is an integer of the form $$n=2k$$, where $$k$$ is an integer. So for $$k=0$$ --> $$n=2*0=0$$.

BACK TO THE ORIGINAL QUESTION"

Is the product of three integers $$p$$, $$q$$, and $$r$$ even?

The product of three integers P, Q, and R will be even if either of them is even.

(1) $$(p-1)(r+1)=odd$$. Product of two integers is odd only if both are odd. Hence $$p-1=odd$$ and $$r+1=odd$$, which means $$p$$ and $$r$$ are even. Sufficient.

(2) $$(q-r)^2=odd$$ --> $$q-r=odd$$. Difference of two integers is odd only if one of them is odd and another is even. Hence $$p$$ or $$r$$ is even. Sufficient.

Hope it's clear.
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Re: m03#22 [#permalink]  01 Jan 2009, 10:21
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suryans wrote:
There are three integers: p , q and r . Is their product pqr even?

1.(p-1)(r+1) is odd
2.(q-r)^2 is odd

explanation says D.
I think it should be E because p,q or r could be 0.
Am I being dense?

D is correct since 0 is always even.
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Re: m03#22 [#permalink]  13 Jul 2010, 05:40
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I did with p=6,q=12,r=9.
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Re: m03#22 [#permalink]  19 Jul 2012, 20:51
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you're asked if q,r, or p is even.

You just need the foundations of evens and odds:
e+e=e exe=e
o+o=e oxo=o
e+o=o exo=e
* same rules apply for subtraction

A) you plug in these values and you find that both have to be even.
B) boils down to: (odd)x(odd) or q-r=odd. so you know that either q or r have to be even

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Re: m03#22 [#permalink]  19 Jun 2010, 06:44
My understanding was that a product of three integers will be even if one of the numbers is even.

1. p-1 * r+1 = odd

odd * odd => p is even, r is even hence pqr regardless of r is even
odd* even => p is even, r is odd again since one of the numbers is even product pqr is even
even *odd => r is even, p is odd .....ditto as above

Hence A is sufficient.

2. (q-r)^2 is odd

implies q -r is odd

odd - even (5 -2) (7-4) etc hence r is even. One of the numbers is even hence pqr is even.

Hence B is sufficient.

is my understanding correct?
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Re: m03#22 [#permalink]  13 Jul 2010, 13:41
for product of any numbers to be odd each one of them has to be odd , If one of them is also even then the resultant is even irrespective of the other numbers .
now as per stat 1 , since the product is odd either of them is even hence 1 suff
stat 2 , since the square of diff is odd it means that one is odd and one is even (the diff of both odd and both even can never be odd ), if one is even then the product of the three can never be even therefore stat 2 is suff

Hence D
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Re: m03#22 [#permalink]  13 Jul 2010, 15:29
GMAT TIGER wrote:
suryans wrote:
...

D is correct since 0 is always even.

It seems that the official guide is ambiguous regarding zero.

According to Official Guide 11: 4.1 Arithmetic: 1. Properties of Integers
Quote:
Any integer that is divisible by 2 is an even integer; the set of even integers is ...0...

Then further down the page under Properties of the Integer 0
Quote:
The integer 0 is neither positive nor negative.

Does anyone have Official Guide 12 to see if this has been resolved?

Does anyone have an acronym or shortcut for remembering the odd+odd=even, odd+even=odd odd*odd=even rules???

I can write them out now, but every time I take a sample test I don't trust myself so I do a couple examples 2*3 etc....
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Re: m03#22 [#permalink]  13 Jul 2010, 19:29
manjot77 wrote:
for product of any numbers to be odd each one of them has to be odd , If one of them is also even then the resultant is even irrespective of the other numbers .
now as per stat 1 , since the product is odd either of them is even hence 1 suff
stat 2 , since the square of diff is odd it means that one is odd and one is even (the diff of both odd and both even can never be odd ), if one is even then the product of the three can never be even therefore stat 2 is suff

Hence D

I agree with the explanation above. But a small correction. According to stat 1, since the product is odd, BOTH (p-1) & (r+1) are odd. Therefore, BOTH p & r are even.

Cheers,
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Re: m03#22 [#permalink]  13 Jul 2010, 19:36
TallJTinChina wrote:
GMAT TIGER wrote:
suryans wrote:
...

D is correct since 0 is always even.

It seems that the official guide is ambiguous regarding zero.

According to Official Guide 11: 4.1 Arithmetic: 1. Properties of Integers
Quote:
Any integer that is divisible by 2 is an even integer; the set of even integers is ...0...

Then further down the page under Properties of the Integer 0
Quote:
The integer 0 is neither positive nor negative.

Does anyone have Official Guide 12 to see if this has been resolved?

Does anyone have an acronym or shortcut for remembering the odd+odd=even, odd+even=odd odd*odd=even rules???

I can write them out now, but every time I take a sample test I don't trust myself so I do a couple examples 2*3 etc....

Hi
I think you are confused between even/odd & positive/negative.
Zero is NEITHER positive NOR negative. It lies in the center of the number line.
Zero is an even number. [Proof: i. Zero is surrounded on both sides by odd numbers (+1 & -1). ii. Zero is a sum of an integer with itself (0+0=0) iii. Zero is a multiple of 2 iv. even+odd=odd -> 0+7=7; even+even=even -> 0+10=10]

Cheers,
Venky
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Re: m03#22 [#permalink]  14 Jul 2010, 01:39
Wow！That is what happens when you do question of the day at 7AM without any coffee!
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Re: m03#22 [#permalink]  20 Jul 2012, 02:20
I narrowly missed it.. time was nearing to 3 min so hurried up and Did not evaluate option B well

managing time is a challenge.
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m03#22 [#permalink]  04 Jul 2014, 05:53

I forgot that product of even and odd is always even.

Each statement alone is sufficient.
m03#22   [#permalink] 04 Jul 2014, 05:53
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# m03#22

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