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# m03 Q 22

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Intern
Joined: 12 Feb 2010
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m03 Q 22 [#permalink]

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16 Mar 2010, 13:52
Is the product of three integers P, Q, and R even?

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

a. Statement (1) ALONE is sufficient, but Statement (2) ALONE is not sufficient
b. Statement (2) ALONE is sufficient, but Statement (1) ALONE is not sufficient
c. BOTH statements TOGETHER are sufficient, but NEITHER statement ALONE is sufficient
d. EACH statement ALONE is sufficient
e. Statements (1) and (2) TOGETHER are NOT sufficient

[Reveal] Spoiler: OA
D

Let me know your reasoning too.

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Intern
Joined: 14 Mar 2010
Posts: 49

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Re: m03 Q 22 [#permalink]

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16 Mar 2010, 21:42
PQR is even if any of P, Q, or R is even and none of them are 0.

A: (p-1)*(r+1) = odd either if p and r are both even (since only even+-odd = odd and only odd*odd = odd) or if p and/or r is 0. NOT SUFFICIENT

B: (q-r)^2 = odd only if either q or r is even and the other is odd or if one of them is 0 and the other is odd. NOT SUFFICIENT

Together: r is restricted to 0 or even. Can't determine PQR.

Not sure why answer is C

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Intern
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Re: m03 Q 22 [#permalink]

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17 Mar 2010, 08:25
I chose e too with 0 value as deciding factor for my choice but it turns out the 0 is considered an even number. I just googled it and found it out. Hope the understanding is correct.

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Math Expert
Joined: 02 Sep 2009
Posts: 41666

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Re: m03 Q 22 [#permalink]

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17 Mar 2010, 09:18
Ironduke wrote:
Is the product of three integers P, Q, and R even?

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

[Reveal] Spoiler:
OA c

Let me know your reasoning too.

First of all zero is an even number.
Second answer can not be C, it should be D.

The product of three integers P, Q, and R is 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.

Answer: D.
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Intern
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Re: m03 Q 22 [#permalink]

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17 Mar 2010, 10:43
Excellent. Good reasoning.

Bunuel wrote:
Ironduke wrote:
Is the product of three integers P, Q, and R even?

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

[Reveal] Spoiler:
OA c

Let me know your reasoning too.

First of all zero is an even number.
Second answer can not be C, it should be D.

The product of three integers P, Q, and R is 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.

Answer: D.

_________________

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Intern
Joined: 14 Mar 2010
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Re: m03 Q 22 [#permalink]

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17 Mar 2010, 13:21
Ironduke wrote:
I chose e too with 0 value as deciding factor for my choice but it turns out the 0 is considered an even number. I just googled it and found it out. Hope the understanding is correct.

Then it's D. Dang math rules

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CIO
Joined: 02 Oct 2007
Posts: 1218

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Re: m03 Q 22 [#permalink]

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18 Mar 2010, 03:32
The OA is D, not C as was indicated in the first post. I've edited the first post to include a correct OA.
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Manager
Joined: 13 Dec 2009
Posts: 249

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Re: m03 Q 22 [#permalink]

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19 Mar 2010, 07:36
Ironduke wrote:
Is the product of three integers P, Q, and R even?

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

a. Statement (1) ALONE is sufficient, but Statement (2) ALONE is not sufficient
b. Statement (2) ALONE is sufficient, but Statement (1) ALONE is not sufficient
c. BOTH statements TOGETHER are sufficient, but NEITHER statement ALONE is sufficient
d. EACH statement ALONE is sufficient
e. Statements (1) and (2) TOGETHER are NOT sufficient

[Reveal] Spoiler: OA
D

Let me know your reasoning too.

product P Q R can be even only if one of them is even.
1. (p-1)(r+1)
both p-1 and r+1 are odd
means p and r are even => pqr is even suff.

2. (q-r)^2 is odd => q-r is odd.
there can be following cases for q-r to be odd
q is odd and r is even or q is even and r is odd
in any case one of either q or r are even so pqr will be even
hence d
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Manager
Joined: 02 May 2012
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Location: United Kingdom
WE: Account Management (Other)
Re: m03 Q 22 [#permalink]

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18 Sep 2012, 08:16
The key here is any multiplication by an even number results in an even.

So therefore we're on the hunt for even numbers:

1) (p-1)(r+1) is odd. Only odd x odd gives = odd, so both p-1 and r+1 are odd and hence both are even. An even number in qpr will given an even results

Sufficient

2)Similar to above, odd x odd = odd. So q-r must be odd, and therefore one of q or r must be even. An even number in qpr makes it eve

Sufficient
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Re: m03 Q 22   [#permalink] 18 Sep 2012, 08:16
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# m03 Q 22

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