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Number Systems -Odds & Evens

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Number Systems -Odds & Evens

by sukhman » Wed Oct 16, 2013 9:41 am
If a and b are both positive integers, is b^(a+1) - ba^b odd?
(1) a + (a + 4) + (a - 8) + (a + 6) + (a - 10) is odd
(2) b^3 + 3b^2 + 5b + 7 is odd
Last edited by sukhman on Wed Oct 16, 2013 10:59 pm, edited 1 time in total.
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by Brent@GMATPrepNow » Wed Oct 16, 2013 11:36 am
sukhman wrote:If a and b are both positive integers, is ba+1 - bab odd?
(1) a + (a + 4) + (a - 8) + (a + 6) + (a - 10) is odd
(2) b3 + 3b2 + 5b + 7 is odd
Are you sure you transcribed this question correctly?
For example, it seems peculiar to write bab
Is there an exponent hiding here?

I ask because you seem reluctant to check your posts to ensure that any exponents are clearly evident (e.g., b3 + 3b2)

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by mainbhidhruv » Tue Oct 29, 2013 6:55 am
If a and b are both positive integers, is b^(a+1) - ba^b odd?
(1) a + (a + 4) + (a - 8) + (a + 6) + (a - 10) is odd
(2) b^3 + 3b^2 + 5b + 7 is odd

b^(a+1)-ba^b = b(b^a-a^b)

I. 5a-4 no reference of b - not sufficient
II. b^3 + 3b^2 + 5b + 7 is odd for all the even values of b and will make b(b^a-a^b) even
Sufficient.

OA B

Brent, Please correct me if I am wrong here !
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by Brent@GMATPrepNow » Tue Oct 29, 2013 7:18 am
mainbhidhruv wrote:If a and b are both positive integers, is b^(a+1) - ba^b odd?
(1) a + (a + 4) + (a - 8) + (a + 6) + (a - 10) is odd
(2) b^3 + 3b^2 + 5b + 7 is odd

b^(a+1)-ba^b = b(b^a-a^b)

I. 5a-4 no reference of b - not sufficient
II. b^3 + 3b^2 + 5b + 7 is odd for all the even values of b and will make b(b^a-a^b) even
Sufficient.

OA B

Brent, Please correct me if I am wrong here !
EDIT: I originally agreed with you, but mevicks' awesome solution showed me otherwise.
The correct answer is D

Great work, mevicks!

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Last edited by Brent@GMATPrepNow on Tue Oct 29, 2013 7:47 am, edited 1 time in total.
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by mevicks » Tue Oct 29, 2013 7:42 am
sukhman wrote:If a and b are both positive integers, is b^(a+1) - ba^b odd?
(1) a + (a + 4) + (a - 8) + (a + 6) + (a - 10) is odd
(2) b^3 + 3b^2 + 5b + 7 is odd
Umm, I think the answer should be D
Here is my reasoning:

Note:
Odd ± Even = Odd
Odd * Odd = Odd


Q: b^(a+1) - ba^b = odd?

St1:
a + (a + 4) + (a - 8) + (a + 6) + (a - 10) = odd
5a - 8 = odd
a = odd

If b = odd ---> b^(a+1) - ba^b = odd^even - odd*odd^odd = odd - odd = even
If b = even --> b^(a+1) - ba^b = even^even - even*odd^even = even
Thus b^(a+1) - ba^b is not odd.
SUFFICIENT

St2:
b^3 + 3b^2 + 5b + 7 = odd
(b^3 + 3b^2 + 5b) + 7 = odd
(b^3 + 3b^2 + 5b) = even

If b = odd ---> (b^3 + 3b^2 + 5b) = odd + odd + odd = odd ---> INVALID from the above deduction
If b = even --> Is the only VALID option

Now, original equation : b^(a+1) - ba^b with b = even becomes
Even - even = Even
Thus b^(a+1) - ba^b is not odd
SUFFICIENT

Answer D
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by theCodeToGMAT » Tue Oct 29, 2013 7:51 am
Yes, Answer must be [spoiler]{D} [/spoiler]

To find: (b^a)(b) - (b)(a^b) --> odd

b[ b^a - a^b ]

Statement 1:
a + (a + 4) + (a - 8) + (a + 6) + (a - 10) --> ODD
5a -8 = ODD
5A = ODD
A = ODD

If b= ODD
ODD [ ODD^ODD - ODD^ODD] => ODD [EVEN] = EVEN

If b= EVEn
EVEn {..} --> Always EVEN
SUFFICIENT
R A H U L
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by Brent@GMATPrepNow » Tue Oct 29, 2013 9:02 am
I thought I'd mention that, for these kinds of questions, it's useful to create a table that allows you to test various cases (e.g., a is even and b is even, or a is odd and b is even, etc)

If anyone is interested, we have a free video on this topic: https://www.gmatprepnow.com/module/gmat- ... ies?id=839

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by Matt@VeritasPrep » Wed Oct 30, 2013 11:04 pm
Nice question! Just want to sum up the above (and make it a little more legible).

OG question: If a and b are positive integers, is b * bᵃ - b * aᵇ odd?

Start by factoring out b and rewriting the question as "Is b * (bᵃ - aᵇ) odd?"

Now let's thinking about what we know. b is clearly an integer, and (bᵃ - aᵇ) is an integer minus an integer, so it's an integer too. If the product of two integers is odd, then BOTH integers being multiplied must be odd themselves, so our question is really

"Are both b and (bᵃ - aᵇ) odd?"

Now the statements should be a lot easier to work with.

S1:: 5a - 8 is odd, so a is odd.

If a is odd, then aᵇ is also odd, as it's just a bunch of odd numbers multiplied together.

If b is odd, then bᵃ is also odd, so (bᵃ - aᵇ) = (odd - odd) = even, and the answer to our question is NO, it is NOT the case that both b and (bᵃ - aᵇ) are odd.

If b is even, then clearly the answer again is NO, it is NOT the case that both b and (bᵃ - aᵇ) are odd.

So in either case, SUFFICIENT - one of the numbers is even!

S2::

b³ + 3b² + 5b + 7 is odd. If b is odd, we have (Odd + Odd) + (Odd + Odd) = (Even + Even) = Even ... but our sum is supposed to be odd, per the prompt, so b is even.

If b is even, then it is NOT the case that both b and (bᵃ - aᵇ) are odd. SUFFICIENT!

Good q, very 2013 GMAT.
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