Using statement (1):
5a-8 is odd
=> a is odd

Now evaluating b^(a+1) - ba^b, we get b^even -b*odd
If b is even, the expression is even^even - even = even
If b is odd, the expression is odd^even - odd = even
Therefore in either case, we can say that the expression is not odd. Sufficient.

Using statement 2:
b^3 + 3b^2 + 5b + 7 is odd
=> b^3 + 3b^2 + 5b is even
=> b is even (because if b is odd, the expression would be odd + odd + odd = odd)

Now evaluating b^(a+1) - ba^b
we get even^(a+1) - even*a^odd
= even - even*a^odd
= even
Therefore statement (2) is sufficient.

The answer should therefore be (D).
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The answer is D here; each statement alone guarantees that the expression in the question is *even* so we know the answer is 'no'.

A bit of theory first: when you are dealing with positive integers b and x, then when b^x is odd, b is odd, and when b^x is even, then b is even. That is, the exponent never makes any difference at all - we only care about the base. So as long as we know we're dealing with positive integers, we can always ignore exponents in an even/odd question, because they change nothing.

So if a question asks if b^(a+1)-b(a^b) is odd, it's really just asking if b-ba is odd. From Statement 1 we learn that a is odd, so b-ba = b(1-a) will be even. In Statement 2, again you can just ignore all the exponents; Statement 2 then tells you that b+3b+5b+7 is odd, so 9b+7 is odd and b is even, and thus b-ba is even.
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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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Hi guys,

I have a question about statement 1

Simplifying main stem: b^(a+1) - b(a^b) gives us b[(b^a) - (a^b)]

From statement 1 we get that a has to be odd
Then to see if statement 1 is S, we try odd/even values for b and see that all of them give even answers, except when a=3 b=3, which yield and answer of 0, which isnt considered neither even or odd.
Am I not understanding something here, which prevents the use of these values?

Thanks in advance!

Zero is an even integer.

An even number is an integer that is "evenly divisible" by 2, i.e., divisible by 2 without a remainder and as zero is evenly divisible by 2 then it must be even (in fact zero is divisible by every integer except zero itself).

Hope it's clear.
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Forget conventional ways of solving math questions. In DS, Variable approach is the easiest and quickest way to find the answer without actually solving the problem. Remember equal number of variables and independent equations ensures a solution.

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 original condition, there are 2 variables(a,b), which should match with the number of equations. So you need 2 equations. For 1) 1 equation, for 2) 1 equation, which is likely to make C the answer. In 1)&2), 5a-8=odd, a=odd and b^3+3b^2+5b+7=odd -> b=odd. Then, b^(a+1)-ba^b=odd^even-odd(odd)^odd=even, which is no and sufficient. However, since this is an integers questions which is one of the key questions, apply the mistake type 4(A). In 1), when 5a-8=odd, a=odd and b=odd is even, which is b^(a+1) - ba^b=even. So it is no and sufficient. In 2), when b=odd, a=odd, which is even. Then it becomes b^(a+1) - ba^b=even, which is no and sufficient. Therefore the answer is D. You should get this type of question right in order to score 50-51.


-> For cases where we need 2 more equations, such as original conditions with “2 variables”, or “3 variables and 1 equation”, or “4 variables and 2 equations”, we have 1 equation each in both 1) and 2). Therefore, there is 70% chance that C is the answer, while E has 25% chance. These two are the majority. In case of common mistake type 3,4, the answer may be from A, B or D but there is only 5% chance. Since C is most likely to be the answer using 1) and 2) separately according to DS definition (It saves us time). Obviously there may be cases where the answer is A, B, D or E.
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