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

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Joined: 02 Sep 2009
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if a and b are both positive integers ,is b(a+1)-ba^b odd?  [#permalink]

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05 May 2011, 00:56
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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

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Re: If a and b are both positive integers, is b^(a+1) - ba^b odd?  [#permalink]

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09 Aug 2011, 11:11
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8
Quote:
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

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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Re: If a and b are both positive integers, is b^(a+1) - ba^b odd?  [#permalink]

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09 Aug 2011, 05:20
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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
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If a and b are both positive integers, is b^(a+1) - ba^b odd?  [#permalink]

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05 May 2011, 14:56
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Bunuel 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

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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Re: If a and b are both positive integers, is b^(a+1) - ba^b odd?  [#permalink]

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10 Jul 2011, 07:33
3
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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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Schools: Simon '16 (M$) Re: If a and b are both positive integers, is b^(a+1) - ba^b odd? [#permalink] ### Show Tags 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 _________________ My dad once said to me: Son, nothing succeeds like success. Intern Joined: 17 May 2009 Posts: 27 Location: United States GMAT 1: 770 Q51 V44 GPA: 3.62 WE: Corporate Finance (Manufacturing) Re: If a and b are both positive integers, is b^(a+1) - ba^b odd? [#permalink] ### Show Tags 13 Aug 2011, 19:56 1 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. Intern Joined: 11 Feb 2013 Posts: 1 Re: If a and b are both positive integers, is b^(a+1) - ba^b odd? [#permalink] ### Show Tags 18 May 2013, 16:48 GyanOne wrote: Quote: 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 Using statement (1): 5a-8 is odd => a is odd Now evaluating b^(a+1) - b(a^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). 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! Verbal Forum Moderator Joined: 10 Oct 2012 Posts: 590 Re: If a and b are both positive integers, is b^(a+1) - ba^b odd? [#permalink] ### Show Tags 18 May 2013, 22:00 1 Quote: 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 an answer of 0, which isnt considered neither even or odd. All even numbers are divisible by 2 integrally. Thus 0 is EVEN. _________________ Math Expert Joined: 02 Sep 2009 Posts: 58434 Re: If a and b are both positive integers, is b^(a+1) - ba^b odd? [#permalink] ### Show Tags 19 May 2013, 03:20 1 1 retropecexy82 wrote: GyanOne wrote: Quote: 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 Using statement (1): 5a-8 is odd => a is odd Now evaluating b^(a+1) - b(a^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). 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. _________________ Manager Joined: 14 Dec 2014 Posts: 51 Location: India Concentration: Technology, Finance GPA: 3.87 WE: Programming (Computer Software) Re: If a and b are both positive integers, is b^(a+1) - ba^b odd? [#permalink] ### Show Tags 22 Dec 2014, 03:00 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 _________________ If you like my posts appreciate them with Kudos Cheers!! Math Revolution GMAT Instructor Joined: 16 Aug 2015 Posts: 8017 GMAT 1: 760 Q51 V42 GPA: 3.82 Re: If a and b are both positive integers, is b^(a+1) - ba^b odd? [#permalink] ### Show Tags 23 Dec 2015, 23:24 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. _________________ MathRevolution: Finish GMAT Quant Section with 10 minutes to spare The one-and-only World’s First Variable Approach for DS and IVY Approach for PS with ease, speed and accuracy. "Only$79 for 1 month Online Course"
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Re: If a and b are both positive integers, is b^(a+1) - ba^b odd?  [#permalink]

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14 Aug 2017, 23:34
1
$$b^{a+1}-b*a^b.......b(b^a-a^b)$$..

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) - ba^b odd?  [#permalink]

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05 Sep 2019, 23:22
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Re: If a and b are both positive integers, is b^(a+1) - ba^b odd?   [#permalink] 05 Sep 2019, 23:22
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