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(99999)^2-1^2

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(99999)^2-1^2  [#permalink]

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New post Updated on: 17 Jan 2015, 16:42
4
00:00
A
B
C
D
E

Difficulty:

  15% (low)

Question Stats:

76% (01:04) correct 24% (01:10) wrong based on 200 sessions

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\(99999^2-1^2\) =

A) \(10^5\)
B) \(10^{10}\)
C) \((10^5)(10^5-2)\)
D) \((10^5)(10-2)\)
E) \((10^{10})(10^{10}-2)\)

Originally posted by TheRob on 28 Aug 2009, 06:50.
Last edited by Bunuel on 17 Jan 2015, 16:42, edited 1 time in total.
Edited the OA.
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Re: (99999)^2-1^2  [#permalink]

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New post 28 Aug 2009, 07:14
1
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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New post 28 Aug 2009, 07:55
1
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]

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New post 28 Aug 2009, 08:59
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]

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New post 28 Aug 2009, 09:08
2
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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New post 28 Aug 2009, 10:31
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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New post 28 Aug 2009, 11:11
Answer is 10^5(10^5 - 2)
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New post 28 Aug 2009, 11:27
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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New post 28 Aug 2009, 13:24
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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New post 29 Aug 2009, 01:08
TheRob wrote:
\(99999^2-1^2\)


\(99999^2-1^2\)
=(99999+1)(99999-1)
=100000*99998
=9999800000
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New post 31 Aug 2009, 06:23
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]

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New post 17 Jan 2015, 12:18
TheRob - can u please edit answer to 10^5(10^5 -1)
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Re: (99999)^2-1^2  [#permalink]

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New post 17 Jan 2015, 12:31
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]

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New post 17 Jan 2015, 12:43
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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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New post 21 Jan 2015, 03:45
5
\(99999^2 - 1^2\)

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

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

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

Answer = C
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New post 28 Feb 2015, 05:59
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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New post 28 Feb 2015, 06:03
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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)
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New post 28 Feb 2015, 06:05
chetan2u wrote:
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)



Yes, I was just realizing exactly that! Also, I very conveniently decided to ignore the first part of answer C. Which worked like charm! Hahahahaha!

Thank you though!
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New post 22 Feb 2017, 23:34
TheRob wrote:
\(99999^2-1^2\) =

A) \(10^5\)
B) \(10^{10}\)
C) \((10^5)(10^5-2)\)
D) \((10^5)(10-2)\)
E) \((10^{10})(10^{10}-2)\)


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

\(99999^2-1^2\)

= (99999-1)(99999+1)

= (100000-2)(100000)

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

Hence Option C is correct
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New post 19 Sep 2018, 23:30
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Re: (99999)^2-1^2   [#permalink] 19 Sep 2018, 23:30
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