Question 1197918
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Given number: 0.120210012000210000120000021...


Break up the decimal digits like so
120
2100
12000
210000
1200000
21...


A line break happens whenever we reach a nonzero digit
The nonzero digits are 1,2
The order of 1,2 swaps positions each time we get to a new row.
Also, the number of zeros increases.
The number of zeros in a given row is exactly equal to the row number.

<table border = "1" cellpadding = "5"><tr><td>Row</td><td>Term</td><td>Number of Digits</td><td>Partial Sum</td></tr><tr><td>1</td><td>120</td><td>3</td><td>3</td></tr><tr><td>2</td><td>2100</td><td>4</td><td>3+4 = 7</td></tr><tr><td>3</td><td>12000</td><td>5</td><td>7+5 = 12</td></tr><tr><td>4</td><td>210000</td><td>6</td><td>12+6 = 18</td></tr><tr><td>5</td><td>1200000</td><td>7</td><td>18+7 = 25</td></tr><tr><td>6</td><td>21...</td><td>8</td><td>25+8 = 33</td></tr></table>


The partial sums show us that we're adding consecutive integers
3+4 = 7
7+5 = 12 aka 3+4+5 = 12
12+6 = 18 aka 3+4+5+6 = 18
18+7 = 25 aka 3+4+5+6+7 = 25
and so on.


The partial sums give us a count how many decimal digits there are from row 1 to row n.


Recall that summing the integers from 1 to n is found through this formula n(n+1)/2
For example, if we wanted to sum from 1 to n = 5
1+2+3+4+5 = 15
n(n+1)/2 = 5*(5+1)/2 = 15
Or if we wanted to sum from 1 to n = 55
n(n+1)/2 = 55*(55+1)/2 = 1540


But we'll subtract off 1+2 = 3 since we're not starting at 1, but instead starting at 3.
The formula tailored to this problem is n(n+1)/2-3 which computes the sum 3+4+5+...+(n-1)+n.


If you were to solve n(n+1)/2-3 = 589, you'll get n = -34.91 and n = 33.91 approximately
This tells us there is no way to sum from 3 to n to get 589 exactly
Here are the two closest values.
n(n+1)/2-3 = 33(33+1)/2-3 = 558
n(n+1)/2-3 = 34(34+1)/2-3 = 592


Since 3+4+5+...+32+33 = 558, this means the first 33 rows consist of 558 digits (vast majority are 0's; rare sightings of 1's and 2's as well)
When going from slot 558 to slot 589, we move to the right 589-558 = 31 spaces.
The first two spaces are "12" or "21", the order doesn't really matter since we're skipping over these slots anyway.
The next 31-2 = 29 slots are zeros. 


This means that the 589th digit is a zero. 
This isn't much of a surprise considering zero is the most repeated value. Especially as you move further down the rows.



Answer: <font color=red>0</font>


Edit: The tutor @ikleyn has a good point. I somehow made a mistake with the indexing. 
Replace every n in n(n+1)/2-3 with n+2 to get (n+2)(n+3)/2-3 and that formula is equivalent to n(n+1)/2+2n that @ikleyn has written.
The shift from n to n+2 will correct the indexing error. My apologies for the mix up.
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