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if a and b are both positive integers ,is b(a+1)ba^b odd? [#permalink]
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04 May 2011, 23:56
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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, 13:56
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 bba is odd. From Statement 1 we learn that a is odd, so bba = b(1a) 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 bba is even.
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If a and b are both positive integers, is b^(a+1)  ba^b odd? [#permalink]
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10 Jul 2011, 05:42
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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
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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, 06:33
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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 5a8 odd 5*Integereven=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^EvenEven*(Odd)=eveneven=even If Integer=Odd Odd^EvenOdd*(Odd)=oddodd=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 Eveneven=even Sufficient. Ans: "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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10 Jul 2011, 11:55
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. 5a8 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^aa^b) is odd? let b=even=2 Expression: b(b^aa^b) 2(2^11^2)=2=even let b=odd=3 Expression : b(b^aa^b) 3(3^11^3)=6=even. Hence for b=even and b= odd, expression in question is even. b(b^aa^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^aa^b) 2(2^22^2)=0 = even let a=odd=1 Expression: b(b^aa^b) 2(2^11^2)=2=even Hence for a=even and a= odd, expression in question is even. b(b^aa^b) is odd? NO. Sufficient. OA 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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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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Re: If a and b are both positive integers, is b^(a+1)  ba^b odd? [#permalink]
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09 Aug 2011, 10:11
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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): 5a8 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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13 Aug 2011, 18:56
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We can factor b^(a+1)  ba^b and rephrase the question stem as "\(b(b^aa^b)\) odd?" \(b(b^aa^b)\) is odd only when both \(b\) and \((b^aa^b)\) are odd, since \(O * O = O\). Since \(b^a\) is odd if \(b\) is odd and \(a>0\), \((b^aa^b)\) 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.



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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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18 May 2013, 15:48
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): 5a8 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!



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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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18 May 2013, 21:00
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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.
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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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19 May 2013, 02:20
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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): 5a8 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.
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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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22 Dec 2014, 02:00
Using statement 1: 5a8 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^evenb*odd if b is odd>> we get oddodd=even if b is even >> we get eveneven=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) eveneven=even Suffieint So Ans 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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23 Dec 2015, 22:24
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), 5a8=odd, a=odd and b^3+3b^2+5b+7=odd > b=odd. Then, b^(a+1)ba^b=odd^evenodd(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 5a8=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 5051. > 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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Re: If a and b are both positive integers, is b^(a+1)  ba^b odd? [#permalink]
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14 Aug 2017, 22:34
\(b^{a+1}b*a^b.......b(b^aa^b)\).. 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 5a8 is odd... Possible only when 5a is odd, thus a is odd. Case 4 above. \(b(b^aa^b)\) If b is odd..Odd(oddodd)=odd*even=even If b is even... Even (evenodd)=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?
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