Question 1204618: The pattern forming the irrational number 0.12340432100123400043210000... continues indefinitely. What is the 1945th digit in this pattern?
Found 4 solutions by greenestamps, Edwin McCravy, mccravyedwin, ikleyn:
Answer by greenestamps(13195)   (Show Source):
Answer by Edwin McCravy(20054)  (Show Source):
You can put this solution on YOUR website!
I deleted this solution where I forgot to count the 0 before the decimal point
as the first digit. Below is the corrected version.
Edwin
Answer by mccravyedwin(405)  (Show Source):
You can put this solution on YOUR website!
[The first time I did this problem I forgot to count the 0 digit before the
decimal point as the 1st digit. Here is the correction. I'll delete the first
one.]
The 1's at the beginning of a 1234 appear at digits 2, 13, 28, 47.
Then I found the quadratic function that passes through (1,2), (2,13), (3,28),
by substituting those points in  , and finding a=2, b=5, and
c=-5. So the equation is:
 , where the yth term is a 1 at the beginning of a 1234.
So I set y=1945 and solved the equation:
 and got x as 30 exactly.
Then I substituted the whole part x=30 in
So the 1945th digit is a 1 at the beginning of a 1234.
Edwin
Answer by ikleyn(52754)  (Show Source):
You can put this solution on YOUR website! .
The pattern forming the irrational number 0.12340432100123400043210000... continues indefinitely.
What is the 1945th digit in this pattern?
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
The post does not instruct, from which digit to start counting;
So, in my solution I will start counting digits from the first digit after the decimal dot.
I look to the numerical string after the decimal dot.
What I see is this:
First I see a block of 4 digits 1234.
then I see 1 zero.
then I see a block of 4 digits 4321
then I see 2 zeros.
then I see a block of 4 digits 1234
then I see 3 zeros.
then I see a block of 4 digits 4321
then I see 4 zeros.
Then I see ellipses, but I assume that this pattern does continue,
so, regularly there is k-th block of 4 digits followed by k zeroes.
I presented it in the Table below
counter k 1 2 3 4 5 6 7
block of digits 4 4 4 4 4 4 4
1234 or 4321
block of zeros 1 2 3 4 5 6 7
The formula of the length N(k) of k blocks of 4 digits followed by k zeros is
N(k) = 4k + 
Then I write this equation
+ 4k = 1945
or
k^2 + k + 8k = 3890
k^2 + 9k - 3890 = 0.
From this equation, I want to determine how many described blocks are before the 1985-th digit
Therefore, I want to find the solution to this equation and to round it to the closest (lesser) integer number.
The root is 58.032, and I round it to 58.
So, before the 1945-th digit, I have 58 complete groups as described above, and N(58) = 1943.
The 59-th group starts from 1234;
so, the 1944-th digit is 1 (one), and 1945-th digit is 2. ANSWER
At the end, I remind, that I do not consider zero before the decimal dot and start counting digits
from the first digit after the decimal dot.
If you mean some other way of counting digits, you can modify the answer, based on what is written in my post.
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