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# 200^2 – 2(200)(199) + 199^2 =

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Joined: 01 Sep 2010
Posts: 3397
200^2 – 2(200)(199) + 199^2 =  [#permalink]

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Updated on: 18 Sep 2012, 06:15
4
00:00

Difficulty:

5% (low)

Question Stats:

98% (00:57) correct 3% (02:26) wrong based on 73 sessions

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200^2 – 2(200)(199) + 199^2 =

(A) –79,201
(B) –200
(C) 1
(D) 200
(E) 79,999

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Originally posted by carcass on 17 Sep 2012, 15:23.
Last edited by Bunuel on 18 Sep 2012, 06:15, edited 1 time in total.
Edited the question.
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Joined: 02 Sep 2009
Posts: 58340
Re: 200² – 2(200)(199) + 199² =  [#permalink]

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17 Sep 2012, 15:33
2
carcass wrote:
200² – 2(200)(199) + 199² =

(A) –79,201
(B) –200
(C) 1
(D) 200
(E) 79,999

$$a^2-2ab+b^2=(a-b)^2$$, therefore $$200^2-2(200)(199) + 199^2=(200-199)^2=1$$.

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Re: 200^2 − 2(200)(199) + 199^2 =  [#permalink]

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04 Mar 2017, 23:53
200^2 − 2(200)(199) + 199^2 =

(A) −79,201
(B) −200
(C) 1
(D) 200
(E) 79,999

$$200^2 − 2(200)(199) + 199^2 = ( 200 - 199 )^2$$

( 200 - 199 )^2 = 1^2

Thus, answer must be (C) 1

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Re: 200^2 − 2(200)(199) + 199^2 =  [#permalink]

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20 Oct 2018, 08:37
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Re: 200^2 − 2(200)(199) + 199^2 =   [#permalink] 20 Oct 2018, 08:37
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# 200^2 – 2(200)(199) + 199^2 =

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