If a and b are both positive integers, is b^(a+1) - ba^b odd?

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.

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"

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

We can factor b^(a+1) - ba^b and rephrase the question stem as "\(b(b^a-a^b)\) odd?"

\(b(b^a-a^b)\) is odd only when both \(b\) and \((b^a-a^b)\) are odd, since \(O * O = O\). Since \(b^a\) is odd if \(b\) is odd and \(a>0\), \((b^a-a^b)\) is odd when \(a^b\) is even, since \(O - E = O\). \(a^b\) is even when \(a\) is even (and it's given that \(b>0\)).

We can rephrase once again: "Is both \(b\) odd and \(a\) even?"

Statement 1) \(5a - 8\) is odd, so \(5a\) is \(odd + even = odd\). \(a\) is thus odd since \(a = \frac{odd}{odd}\) cannot be even and it's given that \(a\) is an integer.

We are given that \(a\) is odd, so it cannot be that both \(b\) is odd and \(a\) is even. Sufficient.

Statement 2) \(b^3 + 3b^2 + 5b + 7\) is odd, so \(b^3 + 3b^2 + 5b\) is \(odd - odd = even\).

We can factor as \(b(b^2+3b+5) = even\) and test \(b\) as odd: \(O(O*O + 3(O) + 5) = O(O+O+O)=O\), so \(b\) cannot be odd. \(b\) must be even.

We are given that \(b\) is even, so it cannot be that both \(b\) is odd and \(a\) is even. Sufficient.

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!

All even numbers are divisible by 2 integrally. Thus 0 is EVEN.

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.

Using statement 1:
5a-8 is odd. so a is odd.

we know that odd^(positive integer) is odd and even ^(positive integer) is even.
So by evaluating the given expression we get b^even-b*odd
if b is odd>> we get odd-odd=even
if b is even >> we get even-even=even
Sufficient

Using Statement 2:
b^3 + 3b^2 + 5b + 7 is odd So b^3 + 3b^2 + 5b is even

so b is even. By evaluating expression we get even^(a+1)-even*(a^even)
even-even=even
Suffieint
So Ans D

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.

\(b^{a+1}-b*a^b.......b(b^a-a^b)\)..

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

Hi
Can we solve the part (2). of the question like following?
(2) b^3+3b^2+5^b+7 is odd.

Lets say b^3+3b^2+5^B +7 = any odd number (say9) {AS GIVEN}
thus, b^3+3b^2+5^B = 9-7 = 2 (EVEN)

now, ~9b has to be even which is only possible if B is even (if b is odd, 9b will be odd: 9*1=9, 9*2=18)
Thus the equation hold true only for even B
and we use even B to solve the main statement : b^(a+1) - bA^b

Simplifying the original expression: \(b^{a + 1}– b*(a^b)\) \(= b^a * b - b * a^b = b(b^a - a^b)\)

S1: \(a + (a + 4) + (a – 8 ) + (a + 6) + (a – 10)\) is odd [Sufficient]

This is only possible if \(a\) is odd. Now if we plug this information into \(b(b^a - a^b)\) we get an even output independent of whether \(b\) is even or odd

(1) \(b = odd\), then \(odd(odd^{odd} - odd^{odd}) = odd(odd - odd) = odd * even = even\)
(2) \(b = even\), then \(even(even^{odd} - odd^{even}) = even(even - odd) = even * odd = even\)

S2: \(b^3 + 3b^2 + 5b + 7\) is odd [Sufficient]

This is only possible if \(b\) is even. And if we plug in \(b\) is even in \(b(b^a - a^b)\) we get an even output independent of whether \(a\) is odd or even

Ans. D

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