Even calculators have some LIMIT!

Algebra Level 3

$y=$ ${ 0.9999999999 }^{ 3 }$

what is the difference between the last 10 decimal digits and the first 10 decimal digits of $y$ ?

Example: If y=0.45^3=0.91125 so the difference between the last two decimal digits and first two decimal digits = 91-25=66 .

The answer is 2.

2 solutions

Otto Bretscher
Feb 20, 2016

$(1-10^{-10})^3=1-3\times10^{-10}+3\times 10^{-20}-10^{-30}$ . Thus the first ten digits are $10^{10}-3$ and the last ten digits are $10^{10}-1$ , with a difference of $\boxed{2}$

Really good solution.

nilav rudra - 5 years, 3 months ago

Nilav Rudra
Feb 19, 2016

we cannot calculate such problem by using calculators.so we need to find a pattern :

0.9^3=0.729 (difference of last digit and first digit is 2 )

0.99^3=0.970299 (difference of last 2 digits and first 2 digits is 2 )

0.999^3=0.997002999 (difference of last 3 digits and first 3 digits is 2 )

still not convinced...

0.9999^3=0.999700029999 (difference of last 4 digits and first 4 digits is 2 )

*similarly the difference of last 10 digits and first 10 digits of y=0.9999999999^3 is 2 *

Even calculators have some LIMIT!

The answer is 2.

2 solutions

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