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If a is a negative integer, is |a|+|b| an even integer?

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If a is a negative integer, is |a|+|b| an even integer?  [#permalink]

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New post Updated on: 10 Jan 2014, 11:33
9
00:00
A
B
C
D
E

Difficulty:

  95% (hard)

Question Stats:

46% (01:33) correct 54% (01:24) wrong based on 228 sessions

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If a is a negative integer, is |a|+|b| an even integer?

(1) x^a·x^b=1

(2) a≠−1

*****SPOILER: IF you want to attempt this problem yourself do not read below here*****


I answered A because x^a times x^b = 1 means that x^a = 1/x^b or that a = -b. Therefore, a and b are the same number with opposite signs, thus |a| + |b| must be even because odd+odd=even and even+even=even. If a and b are same number with opposite signs, they cannot be one odd and one even. therefore 1 sufficient

Obviously 2 isnt sufficient without any info on B.

But the official answer states this:

Answer A is incorrect because:
Incorrect.Stat. (1): multiply the powers with the same base of x: xa·xb = xa+b = 1. This seems to indicate that a+b=0, since x0=1. Since zero is even, then answer is "Yes". But is it always "Yes", for any number? If x=1, then a+b could be anything (for instance, 13 also equals 1.) Therefore, if x=1 the answer could be "No". That's a "Maybe", so Stat.(1)->IS->BCE.

Answer E is correct because:
Correct.Stat. (1): multiply the powers with the same base of x: xa·xb = xa+b = 1. This seems to indicate that a+b=0, since x0=1. Since zero is even, then answer is "Yes". But is it always "Yes", for any number? If x=1, then a+b could be anything (for instance, 13 also equals 1.) Therefore, if x=1 the answer could be "No". That's a "Maybe", so Stat.(1)->IS->BCE.Stat. (2) alone tells you nothing about b. If a=-2 and b=0, then |a|+|b| = |-2| + |0| = 2, which is even, yielding an answer of "Yes". However, if a=-2 and b=1, then |a|+|b| = |-2| + |1| = 3, which is odd, yielding an answer of "No". No definite answer, so Stat.(2)->IS->CE.Stat. (1+2) combined still allow both cases - either |a|+|b|=0 and the answer is "Yes", or x=1, which then allows |a|+|b| to be an odd number, yielding a "No" answer. Still no definite answer, so Stat.(1+2)->IS->E.


Please help explain and tell me if this is a mistake, thank you!

Originally posted by omgmat on 10 Jan 2014, 11:06.
Last edited by Bunuel on 10 Jan 2014, 11:33, edited 1 time in total.
Renamed the topic and edited the question.
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Re: If a is a negative integer, is |a|+|b| an even integer?  [#permalink]

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New post 10 Jan 2014, 11:48
Hi,

Chose E because we do not know the value of X and B and the result changes according to them. (could be negative or positive)

Hope it helps!
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Re: If a is a negative integer, is |a|+|b| an even integer?  [#permalink]

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New post 10 Jan 2014, 12:38
1
Paris75 wrote:
Hi,

Chose E because we do not know the value of X and B and the result changes according to them. (could be negative or positive)

Hope it helps!



Yes, but you do not need to know the value of X or B to have sufficient information that the sum of the absolute values of a and b will be even.

Like I said in the original post, if a = -b then it doesn't matter what values they are it must be even when you sum them (and their abs values).

for example: a = -1, b = 1 then answer is 2 (even)
a = -2, b = 2, then answer is 4 (even)

as long as a and b are the same number, the result will be even. and we know that they must be the same number because x^a = 1/x^b

I'm fairly convinced that Economist has made an error, if I have made a mistake someone please correct me.
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Re: If a is a negative integer, is |a|+|b| an even integer?  [#permalink]

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New post 11 Jan 2014, 02:46
1
In
(1) x^a·x^b=1

There are two cases
a)
x is NOT equal to 1
Your solution is valid and the statement(1) seems sufficient

b)
x is equal to 1
a and b can take any values
Your case is not totally valid (not valid on all values)
statement(1) seems insufficient
Example:
x=1;a=5;b=6
x=1;a=5;b=5
(though your case is still valid x=1;a=5;b=-5)

Final result: statement(1) insufficient
Out of infinite cases sufficient we just need one case, giving a different result, to say the statement is insufficient
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Re: If a is a negative integer, is |a|+|b| an even integer?  [#permalink]

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New post 11 Jan 2014, 03:01
1
[quote="omgmat"]If a is a negative integer, is |a|+|b| an even integer?

(1) x^a·x^b=1

(2) a≠−1

Statement I is insufficient

X^(a+b) = 1

x = 1, a = -2, b = +2 (YES |a| + |b| is an even integer)
x = 1, a = -3, b = 2 (NO |a| + |b| is not an even integer)

Since we have a yes and a no hence statement is insufficient

Statement II and combining is insufficient

a is not equal to -1 in our cases above hence combining is not even required.

Answer is E
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Re: If a is a negative integer, is |a|+|b| an even integer?  [#permalink]

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New post 12 Jan 2014, 00:13
sahildrift wrote:
In
(1) x^a·x^b=1

There are two cases
a)
x is NOT equal to 1
Your solution is valid and the statement(1) seems sufficient

b)
x is equal to 1
a and b can take any values
Your case is not totally valid (not valid on all values)
statement(1) seems insufficient
Example:
x=1;a=5;b=6
x=1;a=5;b=5
(though your case is still valid x=1;a=5;b=-5)

Final result: statement(1) insufficient
Out of infinite cases sufficient we just need one case, giving a different result, to say the statement is insufficient


Awesome, thanks for your help
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Re: If a is a negative integer, is |a|+|b| an even integer?  [#permalink]

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New post 13 Dec 2015, 17:29
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 is a negative integer, is |a|+|b| an even integer?

(1) x^a•x^b=1

(2) a≠−1

-> In the original condition, there are 3 variables(a,b,x) and they should match with the number of equations, which requires 3 equations. For 1) 1 equation, for 2) 1 equation, which is likely to make E the answer. In 1) & 2), x=1 a=b=-2 -> yes and x=1 a=-2, b=-3 -> no, which is not sufficient. Therefore the answer is E.

-> For cases where we need 3 more equations, such as original conditions with “3 variables”, or “4 variables and 1 equation”, or “5 variables and 2 equations”, we have 1 equation each in both 1) and 2). Therefore, there is 80% chance that E is the answer (especially about 90% of 2 by 2 questions where there are more than 3 variables), while C has 15% 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 E 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, C or D.
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Re: If a is a negative integer, is |a|+|b| an even integer?  [#permalink]

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New post 24 Dec 2015, 21:42
sahildrift wrote:
In
(1) x^a·x^b=1

There are two cases
a)
x is NOT equal to 1
Your solution is valid and the statement(1) seems sufficient

b)
x is equal to 1
a and b can take any values
Your case is not totally valid (not valid on all values)
statement(1) seems insufficient
Example:
x=1;a=5;b=6
x=1;a=5;b=5
(though your case is still valid x=1;a=5;b=-5)

Final result: statement(1) insufficient
Out of infinite cases sufficient we just need one case, giving a different result, to say the statement is insufficient

sahildrift
Why we cannot write statement 1 as
x^a.x^b = 1
X^a+b = x^0
a+b = 0
a = -b
Then, lal + lbl has to be even and hence this statement should be sufficient?
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If a is a negative integer, is |a|+|b| an even integer?  [#permalink]

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New post 24 Dec 2015, 22:00
Sum of two numbers are even when both are odd or both are even.

None of the statements give an idea of the same, individually or together

Hence, E
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Re: If a is a negative integer, is |a|+|b| an even integer?  [#permalink]

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Re: If a is a negative integer, is |a|+|b| an even integer? &nbs [#permalink] 04 Oct 2018, 06:16
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