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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



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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 = (c1)^2 + c^2  1 = c^2  2c + 1 + c^2  1 = 2c(c1) a  b = (c1)^2  c^2 + 1 = c^2  2c + 1  c^2 +1 = 2c + 2 = 2(1c) a^2  b^2 = (a+b)(ab) = 4c(c1)(1c) = 4c (c1)^2 Hence , a^2  b^2 = 4 * Integer , So the answer is Yes > (1) + (2) SUFFICIENT Answer : C
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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 (ab)*(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 = (c1)^2 and b = c^2  1 or b = (c1)*(c+1) a+b = (c1)^2 + (c1)*(c+1) = (c1)* (c1 + c+1) = c*(c1) ab = (c1)^2  (c1)*(c+1) = (c1)*(c1  (c+1)) = (c1)*(2) = 2*c*(c1) So, a^2  b^2 = (ab)*(a+b) = c*(c1) * (2*c*(c1)) = 2*(c^2)*((c1)^2) If c is odd then c1 will be even and c*c1 will be a multiple of 2 else if c i even then again c*c1) will be a multiple of 2 So, in both the cases c*(c1) is a multiple of 2. So, 2*(c^2)*((c1)^2) will be a multiple of 4 (Actually will be a multiple of 8 too) So, Answer will be C 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 \(a= (c1)^2\) and \(b= (c^21)\) \(a^2b^2= (ab)(a+b)\) put the value of a and b in the above expression, we have \(a^2b^2 = ((c1)^2  (c^21))((c1)^2 + (c^2  1))\) = \(((c^2+12c)  c^2 +1) ((c^2+12c) +c^2  1)\) = \((22C)(2C^22c)\) =\(4c(1c)(c1)\) 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



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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^2b^2=4c(c1)^2 Sent from my SMJ210F using GMAT Club Forum mobile app



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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 amanvermagmatQuote: 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^22c+1.
(2) b = c^2 – 1.
How about this approach? 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
adkikani wrote: Bunuel niks18 chetan2u amanvermagmatQuote: 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^22c+1.
(2) b = c^2 – 1.
How about this approach? 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 = 259 = 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
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