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# If a and b are both positive integers, is b^(a + 1)– b*(a^b)

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If a and b are both positive integers, is b^(a + 1)– b*(a^b) [#permalink]

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10 Jul 2011, 06:42
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If a and b are both positive integers, is $$b^{a + 1}– b*(a^b)$$ odd?

(1) $$a + (a + 4) + (a – 8 ) + (a + 6) + (a – 10)$$ is odd
(2) $$b^3 + 3b^2 + 5b + 7$$ is odd
[Reveal] Spoiler: OA

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Last edited by chetan2u on 14 Aug 2017, 23:23, edited 2 times in total.
Edited the question.

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Re: If a and b are both positive integers, is b^(a + 1)– b*(a^b) [#permalink]

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10 Jul 2011, 07:33
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AnkitK wrote:
If a and b are both positive integers, is b^(a+1)– b*(a^b)odd?
(1) a + (a + 4) + (a – 8 ) + (a + 6) + (a – 10) is odd
(2) b^3+ 3b^2+ 5b + 7 is odd

1.
a + (a + 4) + (a – 8 ) + (a + 6) + (a – 10) is odd
5a-8 odd
5*Integer-even=odd
odd*Integer=odd
Integer=odd
a=odd

b^(a+1)– b*(a^b)
Integer^(odd+1)-Integer*(Odd^Integer)
Integer^(even)-Integer*(Odd)

If Integer=Even
Even^Even-Even*(Odd)=even-even=even

If Integer=Odd
Odd^Even-Odd*(Odd)=odd-odd=even

Sufficient.

(2) b^3+ 3b^2+ 5b + 7 is odd
Integer^Odd+Odd*Integer^Even+Odd*Integer+Odd is odd

If Integer=Even
Even^Odd+Odd*Even^Even+Odd*Even+Odd
Even+Even+Even+odd=odd
b could be EVEN.

If Integer=Odd
Odd^Odd+Odd*Odd^Even+Odd*Odd+Odd
Odd+Odd+Odd+odd=EVEN
Thus, b cannot be ODD.
b=Even

b^(a+1)– b*(a^b)
Even^(Integer+Odd)-Even*Integer^Even
Even-even=even
Sufficient.

Ans: "D"
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Re: If a and b are both positive integers, is b^(a + 1)– b*(a^b) [#permalink]

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10 Jul 2011, 12:55
AnkitK wrote:
If a and b are both positive integers, is b^(a+1)– b*(a^b)odd?
(1) a + (a + 4) + (a – 8 ) + (a + 6) + (a – 10) is odd
(2) b^3+ 3b^2+ 5b + 7 is odd

1. 5a-8 is odd
5a must be odd. a must be odd.
let a=1

Question: b^(a+1) - b(a^b)= b^a.b^1 - b. a^b = b(b^a-a^b) is odd?
let b=even=2
Expression: b(b^a-a^b)
2(2^1-1^2)=2=even
let b=odd=3
Expression : b(b^a-a^b)
3(3^1-1^3)=6=even.
Hence for b=even and b= odd, expression in question is even.
b(b^a-a^b) is odd? NO. Sufficient.

2. b(b^2+3b+5) + 7 is odd
even + odd = odd
hence expression b(b^2+3b+5) must be even. b must be even. let b=2
let a=even=2
Expression : b(b^a-a^b)
2(2^2-2^2)=0 = even
let a=odd=1
Expression: b(b^a-a^b)
2(2^1-1^2)=2=even
Hence for a=even and a= odd, expression in question is even.
b(b^a-a^b) is odd? NO. Sufficient.

OA D
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Re: If a and b are both positive integers, is b^(a + 1)– b*(a^b) [#permalink]

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01 Jul 2014, 21:23
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Re: If a and b are both positive integers, is b^(a + 1)– b*(a^b) [#permalink]

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30 Jan 2017, 18:47
Cool one thank you guys!!!

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If a and b are both positive integers, is b^(a + 1)– b*(a^b) [#permalink]

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14 Aug 2017, 23:02
Hello Moderators,

This is a good question and worth Practicing. Do you think this question has to be made Math friendly? It took be sometime to understand what exactly are the Base and exponents here

AnkitK wrote:
If a and b are both positive integers, is b^(a + 1)– b*(a^b) odd?

(1) a + (a + 4) + (a – 8 ) + (a + 6) + (a – 10) is odd
(2) b^3 + 3b^2 + 5b + 7 is odd

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Re: If a and b are both positive integers, is b^(a + 1)– b*(a^b) [#permalink]

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14 Aug 2017, 23:34
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Expert's post
susheelh wrote:
Hello Moderators,

This is a good question and wort Practicing. Do you think this question has to be made Math friendly? It took be sometime to understand what exactly are the Base and exponents here

AnkitK wrote:
If a and b are both positive integers, is b^(a + 1)– b*(a^b) odd?

(1) a + (a + 4) + (a – 8 ) + (a + 6) + (a – 10) is odd
(2) b^3 + 3b^2 + 5b + 7 is odd

Hi..
Edited...
$$b^{a+1}-b*a^b.......b(b^a-a^b)$$..
So cases..
I) if b is even, ans is NO as equation will be EVEN.
2) if b is odd, and can be yes when a is even AND no when a is odd.
3) if a is even, can be yes or no
4) if a is odd, always NO

Let's see the statements...

(1) a + (a + 4) + (a – 8 ) + (a + 6) + (a – 10) is odd
Or 5a-8 is odd...
Possible only when 5a is odd, thus a is odd.
Case 4 above.
$$b(b^a-a^b)$$
If b is odd..Odd(odd-odd)=odd*even=even
If b is even... Even (even-odd)=even*odf=even
Sufficient

(2) $$b^3 + 3b^2 + 5b + 7$$ is odd
So B is even..
Case I..
Sufficient

D
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Re: If a and b are both positive integers, is b^(a + 1)– b*(a^b)   [#permalink] 14 Aug 2017, 23:34
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