If a, b, and c are integers and abc ≠ 0, is a^2 – b^2 : GMAT Data Sufficiency (DS)
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# If a, b, and c are integers and abc ≠ 0, is a^2 – b^2

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If a, b, and c are integers and abc ≠ 0, is a^2 – b^2 [#permalink]

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26 Mar 2013, 00:52
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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]

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26 Mar 2013, 01:03
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.

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]

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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

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Re: If a, b, and c are integers and abc ≠ 0, is a^2 – b^2 [#permalink]

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11 Nov 2014, 03:26
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If a, b, and c are integers and abc ≠ 0, is a2 – b2 a multiple of 4? [#permalink]

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11 Dec 2014, 15:27
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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
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Re: If a, b, and c are integers and abc ≠ 0, is a2 – b2 a multiple of 4? [#permalink]

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11 Dec 2014, 21:03
Question can be written as a^2 - b^2 or (a-b)*(a+b) is a multiple of 4
STAT1 alone is not sufficient as we don't know anything about b
STAT2 alone is not sufficient as we don't know anything about a
Taking both together

a = (c-1)^2 and b = c^2 - 1 or b = (c-1)*(c+1)

a+b = (c-1)^2 + (c-1)*(c+1) = (c-1)* (c-1 + c+1) = c*(c-1)
a-b = (c-1)^2 - (c-1)*(c+1) = (c-1)*(c-1 - (c+1)) = (c-1)*(-2) = -2*c*(c-1)
So, a^2 - b^2 = (a-b)*(a+b) = c*(c-1) * (-2*c*(c-1))
= -2*(c^2)*((c-1)^2)

If c is odd then c-1 will be even and c*c-1 will be a multiple of 2
else if c i even then again c*c-1) will be a multiple of 2

So, in both the cases c*(c-1) is a multiple of 2. So, -2*(c^2)*((c-1)^2) will be a multiple of 4 (Actually will be a multiple of 8 too)

Hope it helps!

viktorija 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

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Re: If a, b, and c are integers and abc ≠ 0, is a2 – b2 a multiple of 4? [#permalink]

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12 Dec 2014, 00:33
viktorija 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

st.1 alone is not sufficient as nothing is mentioned about b
st.2 alone is not sufficient as nothing is mentioned about a

st.1 and st.2

$$a= (c-1)^2$$ and $$b= (c^2-1)$$

$$a^2-b^2= (a-b)(a+b)$$
put the value of a and b in the above expression, we have

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

= $$((c^2+1-2c) - c^2 +1) ((c^2+1-2c) +c^2 - 1)$$

= $$(2-2C)(2C^2-2c)$$

=$$4c(1-c)(c-1)$$

as can be seen, the above expression is a multiple of 4. hence answer is C.
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Re: If a, b, and c are integers and abc ≠ 0, is a^2 – b^2 [#permalink]

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12 Dec 2014, 03:38
viktorija 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

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Re: If a, b, and c are integers and abc ≠ 0, is a^2 – b^2 [#permalink]

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08 Feb 2016, 17:38
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Re: If a, b, and c are integers and abc ≠ 0, is a^2 – b^2   [#permalink] 08 Feb 2016, 17:38
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