Question 1099833: The pattern forming the irrational number 0.543210123450054321000123450000....continues indefinitely. What is the 550th digit in this pattern?
Answer by richwmiller(17219)   (Show Source):
You can put this solution on YOUR website! I'd break the decimal place digits into groups:
543210 1234500 54321000 123450000 ...
and number them 1, 2, 3,... so that group n has 5 digits (12345 if n is even, 54321 if n is odd) followed by n zeroes.
The total number of digits in the first n groups is
(5+1) + (5+2) + ... + (5+n) = 5n + n(n+1)/2 = (n^2 + 11n)/2
You need that to be >= 550 to include the 550'th digit, so:
(n^2 + 11n)/2 >= 550
n^2 + 11n >= 1100
n^2 + 11n + (11/2)^2 >= 1100 + 121/4
(n + 11/2)^2 >= 1130.25
n >= sqrt(1130.25) - 5.5 ~~ 28.12
The smallest integer n that works is n=29, so digit 550 is in the 29th group. That's odd so the group is 543210000000...
The previous 28 groups had 5*28 + (28)(29)/2 = 546 digits, so the 550'th digit is the 4th digit of group 29
It's a 2, unless you count the 0 before the decimal point--in which case the answer would be 3 instead.
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