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is b^a+1 – ba^b odd?

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is b^a+1 – ba^b odd?

by venmic » Fri Jul 29, 2011 6:13 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


Supposedly D

but in case of A a= 3 and b =3 so may not be possible

please explain
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by akhilsuhag » Fri Jul 29, 2011 7:38 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

Statement 1:

Can be rewritten as (5a - 8) is odd. Now this gives us that a is odd.

Looking at the prompt:

b^a +1 - b. a^b

Now a is odd, so a^b is odd. We have two sub cases(either b is even or b is odd)

Case1: B is even
Makes b^a +1 odd and b.a^b Even;

We know Odd - Even = Odd

Case2: B is Odd
Makes b^a +1 even and b.a^b (odd * odd) is odd.

now Even - Odd = Odd again.

So we get odd in both cases and thus STATEMENT 1 IS SUFFICIENT.

Statement2:
b^3 + 3b^2 + 5b + 7 is odd. The only way this is possible is if B is EVEN. You can plug in odd and even numbers to check.

Again look at the prompt.

b^a +1 will always be odd (even +1) and b. a^b will always be even ( Even* anything = Even)

So, Odd - Even = Odd.

STATEMENT IS SUFFICIENT.

Therefore both statements are SUFFICIENT.

I was hoping for feedback and if experts can tell me if the approach is correct.

Thanks.
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by Tani » Sat Jul 30, 2011 5:19 am
Picking numbers can help. You need four pair, both odd, both even, a odd and b even, b odd and a even.
Using 2 and 3, we see that the only way to get an even number is to have b be odd and a be even.

Checking our statements, the first tells us a is odd: sufficient.
The second tells us b must be even: sufficient
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by Warlock007 » Tue Aug 09, 2011 4:24 am
akhilsuhag 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

Statement 1:

Can be rewritten as (5a - 8) is odd. Now this gives us that a is odd.

Looking at the prompt:

b^a +1 - b. a^b

Now a is odd, so a^b is odd. We have two sub cases(either b is even or b is odd)

Case1: B is even
Makes b^a +1 odd and b.a^b Even;
We know Odd - Even = Odd
[spoiler]it means even^(even+1)- odd= what?

now my question is how can you say that Even^anything can be odd [/spoiler]

Case2: B is Odd
Makes b^a +1 even and b.a^b (odd * odd) is odd.

now Even - Odd = Odd again.

So we get odd in both cases and thus STATEMENT 1 IS SUFFICIENT.

Statement2:
b^3 + 3b^2 + 5b + 7 is odd. The only way this is possible is if B is EVEN. You can plug in odd and even numbers to check.

Again look at the prompt.

b^a +1 will always be odd (even +1) and b. a^b will always be even ( Even* anything = Even)

So, Odd - Even = Odd.

STATEMENT IS SUFFICIENT.

Therefore both statements are SUFFICIENT.

I was hoping for feedback and if experts can tell me if the approach is correct.

Thanks.
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by wasim4gmat » Sat Aug 13, 2011 3:43 am
I think the answer is A.Let me explain my logic

question can be rephrased as

b(b^a-a^b)

Now lets examine Statement1:
a + (a + 4) + (a - 8) + (a + 6) + (a - 10)= odd?
means 5a-8= odd
>if a is even...then result wont be odd
>so a must be odd
lets apply odd number(eg.7) to question
7(7^a-a^7) can be odd or even depending on value of a.

so statement 1 is not sufficient.
Similarly, examining statement 2)b^3 + 3b^2 + 5b + 7 is odd gives us value of b as even

So Statement 2 is sufficient.

Please clarify my logic is correct
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by Tani » Sat Aug 13, 2011 4:32 pm
You say:

"Now lets examine Statement1:
a + (a + 4) + (a - 8) + (a + 6) + (a - 10)= odd?
means 5a-8= odd
>if a is even...then result wont be odd
>so a must be odd
lets apply odd number(eg.7) to question
7(7^a-a^7) can be odd or even depending on value of a.

Incorrect: You have shown that a is odd, therefore the example above gives us odd(odd-odd) = odd*even = even.

so statement 1 is sufficient.
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by gaurav_chopra04 » Sun Aug 21, 2011 5:42 am
What if we take a= 3 and b = 3, the answer is 0. It doesnt say anywhere that a and b are two different integers.. now when we combine the two we get that a odd and b is even..
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by Tani » Sun Aug 21, 2011 1:54 pm
I think the problem is reading the question. My understanding is that the original equation is:
(b^a)+1 - b(a^b) and I think you are reading it as b^(a+1) - b(a^b). Typing algebra can be prolbematic.
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by gaurav_chopra04 » Mon Aug 22, 2011 9:55 am
Tani Wolff - Kaplan wrote:I think the problem is reading the question. My understanding is that the original equation is:
(b^a)+1 - b(a^b) and I think you are reading it as b^(a+1) - b(a^b). Typing algebra can be prolbematic.
The question asks for b^(a+1) - b(a^b)itself and that was the reason I was not sure why the answer is D and not C. I agree to the fact that had it been (b^a)+1 - b(a^b), the answer would have been D..

I think the person who asked this question should have clarified it a bit using ()..thanks anyways for your reply..:)
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by Tani » Mon Aug 22, 2011 11:33 am
I agree. It's a good idea when typing to be overly specific.
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