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(99999)^2-1^2

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(99999)^2-1^2 [#permalink] New post 28 Aug 2009, 05:50
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A
B
C
D
E

Difficulty:

  35% (medium)

Question Stats:

73% (02:01) correct 28% (01:09) wrong based on 80 sessions
\(99999^2-1^2\) =

A) \(10^5\)
B) \(10^{10}\)
C) \((10^5)(10^5-2)\)
D) \((10^5)(10-2)\)
E) \((10^{10})(10^{10}-2)\)
[Reveal] Spoiler: OA

Last edited by Bunuel on 17 Jan 2015, 15:42, edited 1 time in total.
Edited the OA.
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Re: (99999)^2-1^2 [#permalink] New post 28 Aug 2009, 05:56
I did it in my head, so C? anyone else
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Re: (99999)^2-1^2 [#permalink] New post 28 Aug 2009, 06:14
I am trying to do it, but can not find the answer I got among answers:

9999^2-1^2

(9999+1)(9999-1)
Thus getting 10 0000 (9998).
We can rearrange that as 10^5(10^-2). But there is no such answer in there though
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Re: (99999)^2-1^2 [#permalink] New post 28 Aug 2009, 06:55
it is C!!!!! thanks your explanation I got to the right path

her it comes

99999^2-1^2

(100000–1)^2–(1^2)=

(10^5)^2 – 2(10^5)+1–1=

(10^5)(10^2–2)
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Re: (99999)^2-1^2 [#permalink] New post 28 Aug 2009, 07:30
TheRob wrote:
it is C!!!!! thanks your explanation I got to the right path

her it comes

99999^2-1^2

(100000–1)^2–(1^2)=

(10^5)^2 – 2(10^5)+1–1=

(10^5)(10^2–2)


Can you pls elaborate how you moved from
(10^5)^2 – 2(10^5)+1–1
to
(10^5)(10^2–2)
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Re: (99999)^2-1^2 [#permalink] New post 28 Aug 2009, 07:59
defoue wrote:
TheRob wrote:
it is C!!!!! thanks your explanation I got to the right path

her it comes

99999^2-1^2

(100000–1)^2–(1^2)=

(10^5)^2 – 2(10^5)+1–1=

(10^5)(10^2–2)


Can you pls elaborate how you moved from
(10^5)^2 – 2(10^5)+1–1
to
(10^5)(10^2–2)



It should be \((10^5)^2-2\times 10^5=10^5\times (10^5-2)\)...I believe it is a mistake...
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Re: (99999)^2-1^2 [#permalink] New post 28 Aug 2009, 08:08
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This question can be solved the same way as what is \(99^2\)?

Personally I can't do 99x99 in my head. But I can easily do (99x100) - 99 which would give the same result.

Here we do the same thing 99,999x 100,000 - 99,999

= 9,999,900,000 - 99,999 = 9,999,800,001 - now subtract \(1^2\)

All that remains is to figure out the exponents. \(10^5\) takes care of the 0's and \(10^5-2\) should take care of the remainder. Unfortunately that doesn't look like any answer given. Are you sure answer choices are correct?

Answer should be \((10^5)(10^5-2)\)
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Re: (99999)^2-1^2 [#permalink] New post 28 Aug 2009, 09:31
Can you pls elaborate how you moved from
(10^5)^2 – 2(10^5)+1–1
to
(10^5)(10^2–2)



well I factorized 10^5 from the terms

(10^5)(10^2 -2 ) + 1 - 1
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Re: (99999)^2-1^2 [#permalink] New post 28 Aug 2009, 10:11
Answer is 10^5(10^5 - 2)
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Re: (99999)^2-1^2 [#permalink] New post 28 Aug 2009, 10:27
TheRob wrote:
Can you pls elaborate how you moved from
(10^5)^2 – 2(10^5)+1–1
to
(10^5)(10^2–2)



well I factorized 10^5 from the terms

(10^5)(10^2 -2 ) + 1 - 1


Your factorization is not correct, please see my previous post regarding it.
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Re: (99999)^2-1^2 [#permalink] New post 28 Aug 2009, 12:24
Answer should be 10^5(10^5 - 2)..

10^5 * 10^2 = 10^7 (not 10^10)

10^5 * 10^5 = 10^10
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Re: (99999)^2-1^2 [#permalink] New post 29 Aug 2009, 00:08
TheRob wrote:
\(99999^2-1^2\)


\(99999^2-1^2\)
=(99999+1)(99999-1)
=100000*99998
=9999800000
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Re: (99999)^2-1^2 [#permalink] New post 31 Aug 2009, 05:23
It is true my factorization is wrong

but please look at this

\(99999^2-1^2\)

(99999-1)(99999+1)

(100000-2)(100000)

(10^5 - 2) (10^5)


What do you think?
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Re: (99999)^2-1^2 [#permalink] New post 13 Jan 2015, 06:15
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Re: (99999)^2-1^2 [#permalink] New post 17 Jan 2015, 11:18
TheRob - can u please edit answer to 10^5(10^5 -1)
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Re: (99999)^2-1^2 [#permalink] New post 17 Jan 2015, 11:31
Expert's post
Hi vedavyas9,

It looks like these posts are from over 5 YEARS ago, so I'm not sure if anyone from this thread is still around. The correct answer IS 10^5(10^5 - 2). Based on the "order" of the answer choices, it seems possible that Answer D just wasn't properly copied (or the original source material had a typo in it) - if the extra "power of 5" were written in, then the correct answer would be D. Maybe Bunuel can edit this?

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Re: (99999)^2-1^2 [#permalink] New post 17 Jan 2015, 11:43
Please Check the options provided. There seems to be some typo

\(99999^2 - 1^2\)
= (99999+1)(99999-1)
(Using the formula \(a^2 - b^2 = (a+b)(a-b)\))
=\((100000)(99998)\)
=\((100000)(100000-2)\)
=\((10^5)(10^5 - 2)\)
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Re: (99999)^2-1^2 [#permalink] New post 21 Jan 2015, 02:45
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\(99999^2 - 1^2\)

\(= (10^5 - 1)^2 - 1\)

\(= 10^{10} - 2*10^5 - 1 + 1\)

\(= (10^5)(10^5-2)\)

Answer = C
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(99999)^2-1^2 [#permalink] New post 28 Feb 2015, 04:59
I did it sort of differently, but looks simpler to me...:

99999^2 - 1 = (3^2 * 11111) - 1 = 99999 - 1 = 99998

From the answer choices we eliminate A and B right away, and move to C which is:
10^5 - 2 = 100000 - 2 = 99998.

Right...??
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(99999)^2-1^2 [#permalink] New post 28 Feb 2015, 05:03
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pacifist85 wrote:
I did it sort of differently, but looks simpler to me...:

99999^2 - 1 = (3^2 * 11111) - 1 = 99999 - 1 = 99998

From the answer choices we eliminate A and B right away, and move to C which is:
10^5 - 2 = 100000 - 2 = 99998.

Right...??


hi pacifist,
you have missed out the power 2 of 99999..
and C is 10^5(10^5-2)
(99999)^2-1^2   [#permalink] 28 Feb 2015, 05:03

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