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999,999^2-1 equals:

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999,999^2-1 equals: [#permalink]

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\(999,999^2-1\) equals:


(A) \((9^6)(11^6)\)

(B) \((10^6)(10^5 -2)\)

(C) \((10^6)(10^6 -2)\)

(D) \((10^5)^2\)

(E) \((10^6)^2\)


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[Reveal] Spoiler: OA

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Re: 999,999^2-1 equals: [#permalink]

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\((999,999)^2-1 = (999,999-1)(999,999+1) = 999,998 \times 1,000,000 = (10^6 - 2) \times 10^6\)

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Re: 999,999^2-1 equals: [#permalink]

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Bunuel wrote:
999,999^2-1 equals:

(A) (9^6)(11^6)

(B) (10^6)(10^5 -2)

(C) (10^6)(10^6 -2)

(D) (10^5)^2

(E) (10^6)^2


Kudos for a correct solution.


Just trying to find a different method (Not a better method essentially) :P

\(999,999^2-1 = (1000000-1)^2 - 1 = (10^6-1)^2 - 1 = (10^{12} +1 - 2*1*10^6) - 1 = (10^{12} - 2*10^6) = 10^6*(10^6-2)\)

Answer: option C
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Last edited by GMATinsight on 14 Sep 2015, 06:15, edited 1 time in total.

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Re: 999,999^2-1 equals: [#permalink]

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GMATinsight wrote:
Bunuel wrote:
999,999^2-1 equals:

(A) (9^6)(11^6)

(B) (10^6)(10^5 -2)

(C) (10^6)(10^6 -2)

(D) (10^5)^2

(E) (10^6)^2


Kudos for a correct solution.


Just trying to find a different method (Not a better method essentially) :P

\(999,999^2-1 = (1000000-1)^2 - 1 = (10^6-1)^2 - 1 = (10^{12} +1 - 2*1*10^6) - 1 = (10^{12} - 2*10^6) = 10^6*(10^6-2)\)

Answer: option E


GMATinsight, your answer is correct but the option is NOT E. It should be C.

Alternate method:

\(999999^2-1\) will have a unit's digit of 0 . Eliminate option A.

For, options B,C,D, look at it like this: using \(a^2-b^2=(a+b)(a-b)\)

\(999999^2-1 = (999999+1)(999999-1) = 10^6*(10^6-2)\), C is the correct answer. NOte here that maximum power of 10 in \(10^6*(10^6-2)\) is 11. Thus the answer is in order of 11.

Alternately, to eliminate options B,D,E look below:

Option B is in the order of \(10^6*(10^4)=10^{10}\). We need \(10^{11}\). Eliminate.

Option D is in the order of \(10^{10}\). We need \(10^{11}\). Eliminate.

Option E is in the order of \(10^{12}\). We need \(10^{11}\). Eliminate.

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Re: 999,999^2-1 equals: [#permalink]

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Engr2012 wrote:

GMATinsight, your answer is correct but the option is NOT E. It should be C.



Thank you for bringing in notice. :-)
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Re: 999,999^2-1 equals: [#permalink]

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999,999² - 1 = 999,999² - 1² = (999,999-1) (999,999+1) = 10^6 (999,998) = 10^6(10^6 - 2)
Answer: C
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Re: 999,999^2-1 equals: [#permalink]

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shailendra79s wrote:
999,999² - 1 = 999,999² - 1² = (999,999-1) (999,999+1) = 10^6 (999,998) = 10^6(10^6 - 2)
Answer: C







999,9992−1=(1000000−1)2−1=(106−1)2−1=(1012+1−2∗1∗106)−1=(1012−2∗106)=106∗(106−2)

Answer: option C
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Re: 999,999^2-1 equals: [#permalink]

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Consider this an algebraic expression of x2-1 .Break it down to (x-1)(x+1) .so it becomes (999,999+1)(999,999-1)
Reducing it further leads to (1000,000)(999,998) .This expression can be simplified further to be written as: 10^6(10^6 -2) .
Option C is the correct choice.
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Re: 999,999^2-1 equals: [#permalink]

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Bunuel wrote:
999,999^2-1 equals:

(A) (9^6)(11^6)

(B) (10^6)(10^5 -2)

(C) (10^6)(10^6 -2)

(D) (10^5)^2

(E) (10^6)^2


Kudos for a correct solution.


\(999,999^2-1\)

= \((10^6-1)^2 - 1\)

= \(((10^6)^2) + 1 - 2(10^6) -1\)

=\(((10^6)^2) - 2(10^6)\)

= \((10^6)(10^6 -2)\)

Answer:- C

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Re: 999,999^2-1 equals: [#permalink]

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New post 20 Sep 2015, 08:10
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Bunuel wrote:
999,999^2-1 equals:

(A) (9^6)(11^6)

(B) (10^6)(10^5 -2)

(C) (10^6)(10^6 -2)

(D) (10^5)^2

(E) (10^6)^2


Kudos for a correct solution.


VERITAS PREP OFFICIAL SOLUTION:

Here, the given information is in a quite-unusable form. You have no interest in squaring 999999. It’s a ridiculously large number, it will require an insanely long calculation, and most importantly it won’t look at all like the answer choices. The work would all be for naught, as ultimately you’d have to repackage the number to look like one of the expressions in A-E. Your mission here is not to calculate this number; as on many algebra-based problems, your mission is to repackage the given information as something much more useful. Two things should lead you to this decision:

1. The number itself is incalculable by hand, so it’s not a calculation-based problem
2. The answer choices are in algebraic form, so you’re looking for an expression and not a number

Seeing this, you should recognize that your job is to take the statement given and re-frame it as a restatement of the same fact…just a more useful form. Restatements can come in several ways:

* Factor common terms to turn addition/subtraction into multiplication
* Multiply a fraction by the same numerator and denominator (so it equals 1) to give it a new expression
* Take an equation and do the same thing to both sides to change the look of the expression

Here, we can use arguably the most powerful repackaging tool in all of algebra; the Difference of Squares rule. This rule states that \(x^2 -y^2 =(x -y)(x +y)\). And since we already have \(999999^2-1^2\) (remember: 1 is the same as 1^2), we can change it to \((999999 + 1)(999999-1)\). Why? because now we have 1000000 on the left, and that can be repackaged in exponent form as 10^6, which matches the notation in the answer choices. We needed something with an exponent, and transforming this expression as we did allowed us to do that. So we have:

\((10^6)(999998)\)

And it’s our job now to repackage the term on the right to match an answer choice. This can be done by stating 999998 as (10^6-2), giving us:

\((10^6)(10^6 -2)\) which is answer choice C.

What’s important to recognize here is that many algebra-based problems will hinge on your ability to take what you’re given and repackage it as something more useful. In this situation, know your assets. Tools like Difference of Squares can make repackaging algebra quite transformative. Seeing the answer choices as a guide is also instrumental to many of these problems as the choices give you a blueprint of what the algebra ultimately needs to look like. Repackaging is a huge component of the business world, as you can see with products like the Nike Free running shoes (want to run barefoot? We can make that happen – just buy these barefoot-running shoes!) and nearly every movie nowadays (remakes having replaced sequels as the easy-to-greenlight film category). Keep this principle in mind on the GMAT and you, too, can turn nothing into something, repackaging your 570 as a 750 in no time!
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Collection of Questions:
PS: 1. Tough and Tricky questions; 2. Hard questions; 3. Hard questions part 2; 4. Standard deviation; 5. Tough Problem Solving Questions With Solutions; 6. Probability and Combinations Questions With Solutions; 7 Tough and tricky exponents and roots questions; 8 12 Easy Pieces (or not?); 9 Bakers' Dozen; 10 Algebra set. ,11 Mixed Questions, 12 Fresh Meat

DS: 1. DS tough questions; 2. DS tough questions part 2; 3. DS tough questions part 3; 4. DS Standard deviation; 5. Inequalities; 6. 700+ GMAT Data Sufficiency Questions With Explanations; 7 Tough and tricky exponents and roots questions; 8 The Discreet Charm of the DS; 9 Devil's Dozen!!!; 10 Number Properties set., 11 New DS set.


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Re: 999,999^2-1 equals: [#permalink]

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New post 15 Oct 2017, 05:48
Bunuel wrote:
\(999,999^2-1\) equals:


(A) \((9^6)(11^6)\)

(B) \((10^6)(10^5 -2)\)

(C) \((10^6)(10^6 -2)\)

(D) \((10^5)^2\)

(E) \((10^6)^2\)


Kudos for a correct solution.


(999,9999 -1)(999,999 + 1) = 999,998 x 1,000,000 = (10^6 - 2) (10^6)
Thus, the answer is C

Kudos [?]: 5 [0], given: 2

Re: 999,999^2-1 equals:   [#permalink] 15 Oct 2017, 05:48
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