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# (999999)^2 - 1^2=

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Intern
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25 Aug 2003, 12:17
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(999999)^2 - 1^2=
Intern
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25 Aug 2003, 12:36
Nope!!!

the answers choices were something like-

10^10-2

10^5(10^4-2)

10^5(10^5 )-2

I don't remember the answer choices correctly

I was getting 10^6-2(or something like this), which wasn't there
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25 Aug 2003, 12:49
(10^6-1)^2-1^2=10^12-2*10^6=10^6*(10^6-2)
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25 Aug 2003, 12:53
999999^2 - 1^2 = (999999 + 1)(999999 - 1 ) , right ??
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Brainless

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25 Aug 2003, 13:06
yes... in fact:
10^6*(10^6-2)=10^6*999,998=(999999 + 1)(999999 - 1 )
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25 Aug 2003, 13:45
VERY NICE!!!

Let's also remember that trick if it were to be 999999^2 -1.

Even if the question is not written as 1^2, we can re-write the question to nail this one!!!
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Sept 3rd

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28 Aug 2003, 04:07
(999999)^2 can be calculated easily by doing it in a straightforward way

9*9 = 81
99*99 = 9801
999*999 = 998001
hence
999999*999999 = 999998000001
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Too much is not enough...

Intern
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28 Aug 2003, 22:20
bono wrote:
(999999)^2 can be calculated easily by doing it in a straightforward way

9*9 = 81
99*99 = 9801
999*999 = 998001
hence
999999*999999 = 999998000001

good.....but the answer choices do not require u to solve it this way.See my comments above.

The way javropu solved it is the right way.......
28 Aug 2003, 22:20
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