O= Odd, E = even
(1) Equation 1 can also be written as 5a -8
For 5a - 8 to be odd, a has to be odd
now using a as an odd number in the given equation
b^(O+1) - b.O^b
if b is Odd
O^E - O.O^O
O - O.O
O - O
Even
if b is Even
E^E - E.O^E
E - E.O
Even
So Eq 1 is Sufficient
(2) For Equation 2 to be odd, (b^3 + 3b^2 + 5b) has to be odd, and therefore b has to Odd
now using b as an odd number in the given equation
O^(a+1) - O.a^O
if a is Odd
O^E - O.O^O
O - O.O
O - O
Even
if a is Even
O^O - O.E^O
O - E
Odd
So Eq 2 is Insufficient
Therefor answer should be probably A
Exponents + Odd and Evens
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Source: Beat The GMAT — Data Sufficiency |
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clock60
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hi guys
here i tend to disagree with above posters. and vote for D
as 1 st is elaborated, to the st 2
(2) b^3 + 3b^2 + 5b + 7 is odd
here i`ll try to prove that b is even, and if b is even the expression b*(b^a-a^b)will always be even regardless of the value of a, provided that a and b are integers
if b is even then
b^3-even, 3b^2-even, 5b-even, and 7-odd, rewrite the sum
(even+even+even)+odd=odd
but if b is odd
b^3-odd, 3b^2-odd. 5b-odd.7-odd
odd+odd+odd+odd=even but according to the st 2 it must be odd, so b can`t be odd, and as b is even the b*(b^a-a^b)=even(b^a-a^b)=even
here i tend to disagree with above posters. and vote for D
as 1 st is elaborated, to the st 2
(2) b^3 + 3b^2 + 5b + 7 is odd
here i`ll try to prove that b is even, and if b is even the expression b*(b^a-a^b)will always be even regardless of the value of a, provided that a and b are integers
if b is even then
b^3-even, 3b^2-even, 5b-even, and 7-odd, rewrite the sum
(even+even+even)+odd=odd
but if b is odd
b^3-odd, 3b^2-odd. 5b-odd.7-odd
odd+odd+odd+odd=even but according to the st 2 it must be odd, so b can`t be odd, and as b is even the b*(b^a-a^b)=even(b^a-a^b)=even
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Brent@GMATPrepNow
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For even/odd questions, it often helps to draw a table to explore all possibilities.tonebeeze 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
In the target question, (Is b^(a+1) - ba^b odd?), there are 4 cases to consider:
case a) a=even, b=even
case b) a=odd, b=even
case c) a=even, b=odd
case d) a=odd, b=odd
For each case, let's examine the output of b^(a+1) - ba^b
case a) a=even, b=even --> output is even
case b) a=odd, b=even --> output is even
case c) a=even, b=odd --> output is odd
case d) a=odd, b=odd --> output is even
So, for b^(a+1) - ba^b to be odd, it MUST be the case that a=even and b=odd
So, we can now rewrite our target question as "Is it the case that a is even and b is odd?
Now let's examine the statements.
Statement 1: a + (a + 4) + (a - 8) + (a + 6) + (a - 10) is odd
Simplify left-hand side to get: 5a - 8 is odd
Since 8 is even, we can conclude that 5a must be odd
For 5a to be odd, a MUST be odd
If a must be odd, we can answer the new target question (Is it the case that a is even and b is odd?) with certainty.
As such, statement 1 is SUFFICIENT
Statement 2: b^3 + 3b^2 + 5b + 7 is odd
It's hard to tell if this statement tells us anything about b.
Let's use a table to look at the two possible cases for b
case a) b is odd
case b) b is even
For each case, let's examine the output of b^3 + 3b^2 + 5b + 7
case a) b is odd --> output is even
case b) b is even --> output is odd
Since we are told that b^3 + 3b^2 + 5b + 7 is odd, we know that b MUST be even
If b must be even, we can answer the new target question with certainty.
As such, statement 2 is SUFFICIENT
So, the answer is D
Brent Hanneson - Creator of GMATPrepNow.com
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sourabh33
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sourabh33 wrote:O= Odd, E = even
(2) For Equation 2 to be odd, (b^3 + 3b^2 + 5b) has to be odd, and therefore b has to Odd
now using b as an odd number in the given equation
The answer should be D. I left 7 in (b^3 + 3b^2 + 5b + 7)
for (b^3 + 3b^2 + 5b + 7) to be odd...B has to be even...and so on
