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Is (a^2 - b^2)/2 an odd integer?

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Is (a^2 - b^2)/2 an odd integer?  [#permalink]

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New post 25 Apr 2018, 23:57
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Is (a^2 - b^2)/2 an odd integer?

(1) a-b is odd

(2) a^2 + b^2 is odd

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Is (a^2 - b^2)/2 an odd integer?  [#permalink]

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New post 22 Aug 2018, 20:05
PP777 wrote:
Is (a^2 - b^2)/2 an odd integer?

(1) a-b is odd

(2) a^2 + b^2 is odd



Before starting such questions, always make a mental note that a and B need not be integer

So Is \(\frac{a^2-b^2}{2}=\frac{(a-b)(a+b)}{2}\) an odd integer.

1) a-b is odd.
Easy to fall for trap that since a-b is odd, a+B will be odd and hence O*O/2 will give answer NO
# But what if they are fractions say \(a= \frac{9}{2}\) and \(b=\frac{3}{2}\)
Then \(a-b=3\) and \(a+b=6\) thus \(\frac{3*6}{2} =9\) it is odd... YES
# And if a= 4 and b=1... \(\frac{(4-1)(4+1)}{2}=\frac{3*5}{2}=\frac{15}{2}\).. it is NO
Insufficient

2) a^2+b^2 is odd..
a=3 and b=2 so \(a^2+b^2= 9+4=13\) and \(a^2-b^2=9-4=5\), so 5/2 is a fraction, thus Ans NO
a=√\((\frac{9}{2})\) and b=√\((\frac{5}{2})\) so \(a^2+b^2= \frac{9}{2}+\frac{5}{2}= 7\), but \(a^2-b^2=\frac{9}{2}-\frac{5}{2}=2\) so Ans is \(\frac{2}{2}=1\)... So YES
Insufficient

Combined
Both a-b and a^2+b^2 cannot be integers if a and b are fractions, therefore a and b are integers and ANS is always NO

C
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Is (a^2 - b^2)/2 an odd integer?  [#permalink]

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New post 26 Apr 2018, 04:51
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PP777 wrote:
Is (a^2 - b^2)/2 an odd integer?

(1) a-b is odd

(2) a^2 + b^2 is odd


(1) a-b is odd. We are not given that a/b are integers. If we assume that a/b are integers, then a-b will be odd if one of a/b is odd and the other is even. So in this case a^2 - b^2 will also be odd, and thus not divisible by 2. This would make (a^2 - b^2)/2 a decimal, and thus not an integer. So the answer to the question asked would be NO.

But it is also possible that a/b are not integers, rather decimals. Eg, a=4.5 and b=1.5; this satisfies the first statement in that a-b = 3 (odd), and here a^2 - b^2 is 4.5^2 - 1.5^2 = 18. 18 divided by 2 is an odd integer (9), and so in this case the answer to the question asked would be YES.
This means statement 1 is not sufficient. Its also important to note that a-b will be odd either when both a/b are integers (one odd, one even) OR when they have some decimal values, but a/b cannot have irrational values.

(2) a^2 + b^2 is odd. If a/b are integers, then a^2 + b^2 will be odd if one of a/b is odd and other is even. And in that case a^2 - b^2 will also be odd, and thus not divisible by 2. This would make (a^2 - b^2)/2 a decimal, and thus not an integer. So the answer to the question asked would be NO.

But it is possible that a/b are not integers. Lets take a=(3)^1/2 and b=1; in this case a^2 - b^2 = 3-1 = 2; and here (a^2 - b^2)/2 = 2/2 = 1, odd integer. So in this case answer to the question asked is YES.
This means statement 2 is also not sufficient. Its also important to note that a^2 + b^2 will be odd either when both a/b are integers (one odd, one even) OR when at least one of them has irrational values, but a/b cannot have decimal values.

Combining the two statements, this formula comes to mind: the common info from both statements is that both a/b have to be integers, and one of them will be odd while the other will be even. In that case a^2 - b^2 is also odd, and thus not divisible by 2. So (a^2 - b^2)/2 cannot be an integer at all, thus the answer is a definite NO.

Hence C answer.
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Re: Is (a^2 - b^2)/2 an odd integer?  [#permalink]

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New post 22 Aug 2018, 18:20
Quote:
But it is possible that a/b are not integers. Lets take a=(3)^1/2 and b=1; in this case a^2 - b^2 = 3-1 = 2; and here (a^2 - b^2)/2 = 2/2 = 1, odd integer. So in this case answer to the question asked is YES.

i don't quite get it. statement 2 says a^2 + b^2 is odd, not a^2 - b^2. take a=(3)^1/2 and b=1,you'll get a^2 + b^2=4. it doesn't conform to statement 2.
please shed some light on it. thanks
Re: Is (a^2 - b^2)/2 an odd integer? &nbs [#permalink] 22 Aug 2018, 18:20
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