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If a, b, and c are integers and abc ≠ 0, is a^2 – b^2

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Senior Manager
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If a, b, and c are integers and abc ≠ 0, is a^2 – b^2 [#permalink] New post 26 Mar 2013, 00:52
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73% (02:31) correct 27% (01:10) wrong based on 41 sessions
If a, b, and c are integers and abc ≠ 0, is a^2 – b^2 a multiple of 4?

(1) a = (c – 1)^2

(2) b = c^2 – 1

Need explanation
[Reveal] Spoiler: OA
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Re: If a, b, and c are integers and abc ≠ 0, is a^2 – b^2 [#permalink] New post 26 Mar 2013, 01:03
Expert's post
If a, b, and c are integers and abc ≠ 0, is a^2 – b^2 a multiple of 4?

(1) a = (c – 1)^2 --> a=c^2-2c+1. Not sufficient, since no info about b.

(2) b = c^2 – 1. Not sufficient, since no info about a.

(1)+(2) a^2-b^2 = (a-b)(a+b)=(c^2-2c+1-c^2+1)(c^2-2c+1+c^2-1)=(2-2c)(2c^2-2c)=4(1-c)(c^2-c) --> a^2-b^2 is a multiple of 4. Sufficient.

Answer: C.

Hope it's clear.
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Re: If a, b, and c are integers and abc ≠ 0, is a^2 – b^2 [#permalink] New post 26 Mar 2013, 01:15
mun23 wrote:
If a, b, and c are integers and abc ≠ 0, is a^2 – b^2 a multiple of 4?

(1) a = (c – 1)^2

(2) b = c^2 – 1

Need explanation


Hi mun23

(1) a = (c – 1)^2 ALONE

(1) is INSUFFICIENT because this statement do not give any information about b

(2) b = c^2 – 1 ALONE

(2) is INSUFFICIENT because this statement do not give any information about a

(1) + (2) a = (c – 1)^2 AND b = c^2 – 1

After combining (1) and (2) -->
a + b = (c-1)^2 + c^2 - 1 = c^2 - 2c + 1 + c^2 - 1 = 2c(c-1)
a - b = (c-1)^2 - c^2 + 1 = c^2 - 2c + 1 - c^2 +1 = -2c + 2 = 2(1-c)

a^2 - b^2 = (a+b)(a-b) = 4c(c-1)(1-c) = -4c (c-1)^2

Hence , a^2 - b^2 = 4 * Integer , So the answer is Yes --> (1) + (2) SUFFICIENT

Answer : C
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Re: If a, b, and c are integers and abc ≠ 0, is a^2 – b^2   [#permalink] 26 Mar 2013, 01:15
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