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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, 01:52
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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
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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, 02:03
1
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, 02:15
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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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, 16:27
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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, 22:03
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, 01:33
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, 04:38
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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03 Aug 2017, 08:55
Ans : C
a^2-b^2=-4c(c-1)^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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13 Apr 2018, 17:06
Bunuel niks18 chetan2u amanvermagmat

Quote:
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.

(2) b = c^2 – 1.

If I simplify question stem to best possible, what am I asked?
Is a / b even?

Why?
a. A multiple pf 4 is always even
b. Squaring of odd / even is always odd /even
c. Even - Even = Even.
or Odd - Odd = Even

I did above in my head.
Essentially, either of statements is clearly insuff
since I need to know both a and b.

What happens when I combine and simplify $$a^2$$ - $$b^2$$
I get -2c + 1.

Even if I do not know value of c, I know -2 will always be even
Even + Odd = Odd. So, we do get an unique ans: NO
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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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14 Apr 2018, 11:07
Bunuel niks18 chetan2u amanvermagmat

Quote:
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.

(2) b = c^2 – 1.

If I simplify question stem to best possible, what am I asked?
Is a / b even?

Why?
a. A multiple pf 4 is always even
b. Squaring of odd / even is always odd /even
c. Even - Even = Even.
or Odd - Odd = Even

I did above in my head.
Essentially, either of statements is clearly insuff
since I need to know both a and b.

What happens when I combine and simplify $$a^2$$ - $$b^2$$
I get -2c + 1.

Even if I do not know value of c, I know -2 will always be even
Even + Odd = Odd. So, we do get an unique ans: NO

Hello

Its NOT necessary for a & b to be even for this to be true. a & b could be both odd also. Eg, a=5, b=3, here a^2 - b^2 = 25-9 = 16, which is a multiple of 4.
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If a, b, and c are integers and abc ≠ 0, is a^2 – b^2  [#permalink]

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22 Apr 2018, 09:54
The math on the previous posts is clear, so I can't add anything to that. That's a surefire way of getting the problem correct.

I took a different approach in plugging in numbers to the prompts. It was obvious right away that 1 and 2 alone are insufficient because they don't give full details about the variables.

After combining, start picking smart numbers for C and work out what A and B would be. I made C = 2, 3, and 4, and then worked out what that would make A and B equal to. This gave me a good enough sample size to test divisibility of 4 and to feel comfortable enough in picking answer choice C.

Is this a perfect approach? No, but I got to the right answer in well under 2 minutes with a high probability of being correct.
If a, b, and c are integers and abc ≠ 0, is a^2 – b^2 &nbs [#permalink] 22 Apr 2018, 09:54
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