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# If a and b are integers, is a^2 − b^2 even?

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If a and b are integers, is a^2 − b^2 even?  [#permalink]

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25 Nov 2019, 00:37
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65% (01:17) correct 35% (01:48) wrong based on 53 sessions

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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?  [#permalink]

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25 Nov 2019, 01:51
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(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?  [#permalink]

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25 Nov 2019, 02:06
1
#1
(a+b)^2 is odd
so either of a or b is odd and even
in that case a^2-b^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?  [#permalink]

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25 Nov 2019, 03:40
1
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.

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Re: If a and b are integers, is a^2 − b^2 even?  [#permalink]

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25 Nov 2019, 04:48
1
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.

$$(a,b)=integers$$
$$a^2−b^2=(a+b)(a-b)$$
$$e+-e=even…o+-o=even…e+-o=odd$$
$$ee=even…eo=even…oo=odd$$

(1) $$a^2+2ab+b^2$$ is odd sufic

$$a^2+2ab+b^2=(a+b)^2=odd…a+b=odd$$
$$(a,b)=(e,o):(a+b)(a-b)=o*o=odd$$

(2) a is odd. insufic

$$(a,b)=(o,e):(a+b)(a-b)=o*o=odd$$
$$(a,b)=(o,o):(a+b)(a-b)=e*e=even$$

Ans (A)
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Re: If a and b are integers, is a^2 − b^2 even?  [#permalink]

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25 Nov 2019, 05:18
1
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

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Re: If a and b are integers, is a^2 − b^2 even?  [#permalink]

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25 Nov 2019, 09:39
1
Q: a2-b2 is even?

The above expression will be even only when a & b are either odd or even as "even-even" or "odd-odd" 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?  [#permalink]

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25 Nov 2019, 13:00
1
We are to determine if a^2 - b^2 is even.
Now, a^2 - b^2 = (a+b)(a-b)
For this to be true, then either (a+b) must be even or (a-b) 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?  [#permalink]

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25 Nov 2019, 15:28
1
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?  [#permalink]

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25 Nov 2019, 21:35
1
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?   [#permalink] 25 Nov 2019, 21:35
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# If a and b are integers, is a^2 − b^2 even?

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