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Big Number Problems

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by vongdn » Sat Oct 02, 2010 10:33 pm
(1001^2 - 999^2) / (101^2 - 99^2) = ?

What is the best way to solve for this, I can't seem to find methods for stuff like this besides trying to work it all out by hand which I'm sure is not the right solution.
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Source: — Problem Solving |

by diebeatsthegmat » Sat Oct 02, 2010 10:38 pm
vongdn wrote:(1001^2 - 999^2) / (101^2 - 99^2) = ?

What is the best way to solve for this, I can't seem to find methods for stuff like this besides trying to work it all out by hand which I'm sure is not the right solution.
solve it with a^2-b^2=a+b)(a-b)
you will get the answer
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by GMATGuruNY » Sun Oct 03, 2010 3:49 am
vongdn wrote:(1001^2 - 999^2) / (101^2 - 99^2) = ?

What is the best way to solve for this, I can't seem to find methods for stuff like this besides trying to work it all out by hand which I'm sure is not the right solution.
Always be on the lookout for the following quadratic identities:

(x+y)^2 = x^2 + 2xy + y^2
(x-y)^2 = x^2 - 2xy + y^2
(x+y)(x-y) = x^2 - y^2


The problem above is testing the last identity in the list above, just in reverse order:

x^2 - y^2 = (x+y)(x-y)

Thus:

(1001^2 - 999^2) / (101^2 - 99^2)
= (1001+999)(1001-999)/(101+99)(101-99)
= (2000*2)/(200*2)
=10

Remember that the GMAT is NOT an arithmetic test. If you find yourself doing some long, involved arithmetic calculation, you're likely missing a "trick" that will allow you to solve more easily.
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by this_time_i_will » Sun Oct 03, 2010 7:01 am
[(1000+1)^2-(1000-1)^2]/(100+1)^2-(100-1)^2 = (2000+2000)/(200+200) = 10
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