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If a and b are integers, is a^2 − b^2 even?
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24 Nov 2019, 23:37
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Competition Mode Question If a and b are integers, is \(a^2−b^2\) even? (1) \(a^2+2ab+b^2\) is odd. (2) \(a\) is odd.
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Re: If a and b are integers, is a^2 − b^2 even?
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25 Nov 2019, 00:51
(1) \(a^2+2ab+b^2\) is odd. > \((a + b)^2\) is odd > \(a + b\) is odd
Two cases are possible Case 1: a = even & b = odd > \(a^2  b^2\) = \(even^2  odd^2\) = \(even  odd = odd\) > A definite NO
Case 2: a = odd & b = even > \(a^2  b^2 = odd^2  even^2 = odd  even = odd\) > A definite NO > Sufficient
(2) \(a\) is odd. Nothing can be said about b > Insufficient
IMO Option A



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Re: If a and b are integers, is a^2 − b^2 even?
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25 Nov 2019, 01:06
#1 (a+b)^2 is odd so either of a or b is odd and even in that case a^2b^2 will not be even sufficient #2 a is odd , value of b not know insufficient IMO A
If a and b are integers, is a^2−b^2 even?
(1) a2+2ab+b2 is odd. (2) a is odd.



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Re: If a and b are integers, is a^2 − b^2 even?
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25 Nov 2019, 02:40
If a and b are integers, is a^2−b^2 even? \(a^2−b^2\) = even if o  o = e (both a and b are odd) e  e = e (both a and b are even) (1) \(a^2+2ab+b^2\) is odd. o + e + e = o (a  odd and b  even). Thus \(a^2−b^2\) = o  e = o NO e + e + o = o (a  even and b  odd). Thus \(a^2−b^2\) = e  o = o NO SUFFICIENT. (2) a is odd. nothing about b is given. INSUFFICIENT. Answer A.
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Re: If a and b are integers, is a^2 − b^2 even?
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25 Nov 2019, 03:48
Quote: If a and b are integers, is a^2−b^2 even?
(1) a^2+2ab+b^2 is odd (2) a is odd.
\((a,b)=integers\) \(a^2−b^2=(a+b)(ab)\) \(e+e=even…o+o=even…e+o=odd\) \(ee=even…eo=even…oo=odd\) (1) \(a^2+2ab+b^2\) is odd sufic\(a^2+2ab+b^2=(a+b)^2=odd…a+b=odd\) \((a,b)=(e,o):(a+b)(ab)=o*o=odd\) (2) a is odd. insufic\((a,b)=(o,e):(a+b)(ab)=o*o=odd\) \((a,b)=(o,o):(a+b)(ab)=e*e=even\) Ans (A)



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Re: If a and b are integers, is a^2 − b^2 even?
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25 Nov 2019, 04:18
for a2−b2 to be even......both a,b should be either odd or even
(1) a2+2ab+b2 is odd......................this is possible when one is odd and other is even.....from this we can say that a2−b2 cannot be even......SUFFICIENT
(2) a is odd......Clearly INSUFFICIENT
OA:A



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Re: If a and b are integers, is a^2 − b^2 even?
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25 Nov 2019, 08:39
Q: a2b2 is even?
The above expression will be even only when a & b are either odd or even as "eveneven" or "oddodd" is only even.
1) (a+b)2 is odd. Therefore a+b is odd, which is only possible when a & b are of different types i.e. odd and even. No. Sufficient 2) No info about b. insufficient. IMO A



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Re: If a and b are integers, is a^2 − b^2 even?
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25 Nov 2019, 12:00
We are to determine if a^2  b^2 is even. Now, a^2  b^2 = (a+b)(ab) For this to be true, then either (a+b) must be even or (ab) must be even. Better still, both a and b must be even or both a and b must be odd.
Statement 1: a^2 + 2ab + b^2 is odd. The only condition whereby statement 1 can be satisfied is either when a is odd and b is even or when a is even and b is odd. When a is odd and b is even, then a^2  b^2 is odd and not even. Similarly, when a is even and b is odd, then a^2  b^2 is odd and not even. Statement 1 is therefore sufficient on its own.
Statement 2: a is odd. This is insufficient. This is because when a is odd and b is odd as well, we have a^2  b^2 is even. However, if a is odd and b is even, then a^2  b^2 is odd. Since there is a possibility of a^2  b^2 to be either odd and even based on whether b is odd or even, statement 2 is not sufficient.
The answer is therefore A.



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Re: If a and b are integers, is a^2 − b^2 even?
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25 Nov 2019, 14:28
a, b —integers, —> Is \(a^{2} —b^{2}=( a—b)(a+b)\) even??
(Statement1): \((a+b)^{2}\) — odd —>( a+b) is odd In order (a+b) to be odd, one of either a or b must be odd integer. —> so, (a—b) — (even+odd= odd) or (odd+even= odd) —> Odd* Odd = Odd (always NO) Sufficient
(Statement2): a is odd. No info about what b is. b could be odd or even integer. —> Clearly insufficient
The answer is A.
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Re: If a and b are integers, is a^2 − b^2 even?
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25 Nov 2019, 20:35
If a and b are integers, is \(a^2−b^2\) even?
(1\() a^2+2ab+b^2\) is odd. (2) a is odd
\(a^2  b^2 \)will be even if both a & b are even or odd. (1) \((a+b)^2\) is odd. a+b is odd. so one of a & b is even and the other is odd. so\( a^2 b^2\) is odd. sufficient. (2) a is odd. no info about b. not sufficient. A is the answer.




Re: If a and b are integers, is a^2 − b^2 even?
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25 Nov 2019, 20:35






