1 − 0.000001 =

A. (1.01)(0.99)
B. (1.11)(0.99)
C. (1.001)(0.999)
D. (1.111)(0.999)
E. (1.0101)(0.0909)

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when you will do subtraction , you will get = 0.999999 or 999999 * \(10^{-6}\)
now you need to find factors of 999999 = 999 * 1001
(for example you need factors of 99 = 9 * 11 , 9999 = 99 * 101 and so on if 9 is repeated even times...)

so in answer choices option C , (1.001) * (0.999) is answer
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https://gmatclub.com/forum/if-n-0-is-4-n-1-divisible-by-5-1-n-is-the-integral-root-of-272101.html

GMAC official guides : https://gmatclub.com/forum/gmac-official-guides-the-master-directory-links-240610.html#p1854935

Give me kudos if you like it , it's totally harmless

Two ways..

1) algebraic..
\(1-0.000001=1^2-(0.001)^2=(1+0.001)(1-0.001)=1.001*0.999\)

2) choices
1-0.000001=0.999999
So the answer will have 6 decimal point.
A and B have 2+2=4 decimal points so eliminate
E has 4+4=8, so out
Left C and D
D is 1.111*0.999 the multiple cannot have all 9s as 11*99=1098 and also it will surely be >1.... eliminate

Left is C
1.001*0.999 will be <1 and will give all 9s
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1) Absolute modulus : http://gmatclub.com/forum/absolute-modulus-a-better-understanding-210849.html#p1622372
2)Combination of similar and dissimilar things : http://gmatclub.com/forum/topic215915.html
3) effects of arithmetic operations : https://gmatclub.com/forum/effects-of-arithmetic-operations-on-fractions-269413.html


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1 = 1²
0.000001 = 1/1,000,000 = 1/1,000² = (1/1,000)² = = (0.001)²

So, 1 and 0.000001 are SQUARES
This means 1 - 0.000001 is a difference of squares

We can factor a difference of squares as follows: x² - y² = (x + y)(x - y)

Likewise....
1 - 0.000001 = 1² - (0.001)²
= (1 + 0.001)(1 - 0.001)
= (1.001)(0.999)

Answer: C

Cheers,
Brent
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Brent Hanneson – Founder of gmatprepnow.com