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# integers squared problem

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Intern
Joined: 28 Jun 2008
Posts: 45

Kudos [?]: 135 [0], given: 31

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09 Jun 2009, 20:43
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Hi guys was wondering if anyone could offer a good way to solve this problem:

(1001)*2 -(999)*2
________________
(101)*2-(99)*2

A. 10
B.20
C.40
D.80
E.100

Kudos [?]: 135 [0], given: 31

Current Student
Joined: 03 Aug 2006
Posts: 115

Kudos [?]: 305 [1], given: 3

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10 Jun 2009, 00:58
1
KUDOS
Lets recall

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

Now given

$$\frac{1001^2-999^2}{101^2-99^2}$$

$$= \frac{(1001+999)(1001-999)}{(101+99)(101-99)}$$

$$= \frac{(2000)(2)}{(200)(2)}$$

$$=\frac{2000}{200}$$

$$=10$$

Hence A

Kudos [?]: 305 [1], given: 3

Manager
Joined: 28 Jan 2004
Posts: 202

Kudos [?]: 28 [0], given: 4

Location: India

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10 Jun 2009, 15:43
same calculations as nookway.

Kudos [?]: 28 [0], given: 4

Manager
Joined: 04 Dec 2008
Posts: 104

Kudos [?]: 225 [0], given: 2

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13 Jun 2009, 07:30
Yup, mine is similar to nookway too.

Kudos [?]: 225 [0], given: 2

Re: integers squared problem   [#permalink] 13 Jun 2009, 07:30
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