Weird Exponents

[guest612]

From GMATPREP CAT 2 #24:

(1001^2 - 999^2) / (101^2 - 99^2) = ?

A. 10
B. 20
C. 40
D. 80
E. 100

For some reason I thought I could eliminate D & E. And thought it would most likely be A or B but MAYBE C. I incorrectly guessed B. The correct answer is A. 10. How do you get to A. 10 from that division of exponents? Please help!

[StaceyKoprince]

Of course we wouldn't multiply this out, so there must be a pattern at work here.

The key is to know the difference between the sets of squares in the numerator and denominator without having to calculate the actual numbers. Because the numerator and denominator have two sets of very close numbers, we can use a nifty little trick here.

First, just to explain what we're about to do:
You can take any square and figure out the next higher square by doing the below:
given 12^2 = 144; I want to find 13^2
Take my starting point (144) and add a 12 (my starting base) and a 13 (the next base - the one I'm trying to find). 144 + 12 + 13 = 169... which also equals 13^2.
Let's say you're given 12^2 = 144 and you want to find 14^2. Add a 12 and 13 to get to 13^2 (which is 169) and then add a 13 and a 14 to get to 14^2: 169 + 13 + 14 = 196. So, every time, if I know the value of the lower square, I get to the next one by adding the starting base and the next consecutive base. If I want to jump two or three bases, I just have to do the process two or three times.

So, my numerator says 1001^2 - 999^2. 1001^2, therefore, equals 999^2 + 999 + 1000 + 1000 + 1001 = 999^2 + 4000. Since I subtract out the 999^2, I'm left with just 4000. (See below for step-by-step if you're not following this.)
1000^2 = 999^2 + 999+1000
1001^2 = 999^2 + 999+1000 + 1000 + 1001 = 999^2 + 4000
And, of course, I subtract out the 999^2 so I'm just left with 4000 on top.

Do the same for the denominator to get:
99^2 + 99 + 100 + 100 + 101 = 99^2 + 400, or simply 400 (once we subtract out the 99^2).

Now I've got 4000/400 or 10. Re: your educated guessing, yes, you might surmise that these sets of numbers are one order of magnitude apart (an order of magnitude is a power of 10), so the answer is likely to be closer to 10 or 20 than 80 or 100 (which is 2 orders of magnitude).

[guest612]

thank you. wow.

[Preet]

thank you. wow.

This sum can also be solved by a^2 - b^2 formula.

(1000 + 1)^2 - (1000 - 1)^2 ie. a^2 - b^2

= (1000 + 1 - 1000 + 1) (1000 + 1 + 1000 - 1) ie. (a - b) (a + b)

= 2*2000 ...and similar method for the denominator.

[StaceyKoprince]

Excellent, Preet - I love it!

[amsood]

Based on a^2 - b^2 formula, just keep it simple, it can be
(1001+999)(1001-999)/(101+99)(101-99) = 2000x2/200x2 = 10

[RonPurewal]

Based on a^2 - b^2 formula, just keep it simple, it can be
(1001+999)(1001-999)/(101+99)(101-99) = 2000x2/200x2 = 10

yes, this is definitely the best way to approach the problem. well done.

in general, it's best to try to apply the 'special' formulas (perfect squares of binomials, and, especially, differences of squares, as in this problem) as soon as possible, and in the simplest possible way. your solution succeeds admirably on both counts.

[Halo3]

I think I'm with guest612.... WOW - nice job

of the hundreds of thinks running through my head when i saw this question, neither of those methods were one of them