document.write( "Question 1099833: The pattern forming the irrational number 0.543210123450054321000123450000....continues indefinitely. What is the 550th digit in this pattern?
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Algebra.Com's Answer #714371 by richwmiller(17219)\"\" \"About 
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I'd break the decimal place digits into groups:
\n" ); document.write( "543210 1234500 54321000 123450000 ... \r
\n" ); document.write( "\n" ); document.write( "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. \r
\n" ); document.write( "\n" ); document.write( "The total number of digits in the first n groups is
\n" ); document.write( "(5+1) + (5+2) + ... + (5+n) = 5n + n(n+1)/2 = (n^2 + 11n)/2 \r
\n" ); document.write( "\n" ); document.write( "You need that to be >= 550 to include the 550'th digit, so: \r
\n" ); document.write( "\n" ); document.write( "(n^2 + 11n)/2 >= 550
\n" ); document.write( "n^2 + 11n >= 1100
\n" ); document.write( "n^2 + 11n + (11/2)^2 >= 1100 + 121/4
\n" ); document.write( "(n + 11/2)^2 >= 1130.25
\n" ); document.write( "n >= sqrt(1130.25) - 5.5 ~~ 28.12 \r
\n" ); document.write( "\n" ); document.write( "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... \r
\n" ); document.write( "\n" ); document.write( "The previous 28 groups had 5*28 + (28)(29)/2 = 546 digits, so the 550'th digit is the 4th digit of group 29 \r
\n" ); document.write( "\n" ); document.write( "It's a 2, unless you count the 0 before the decimal point--in which case the answer would be 3 instead. \n" ); document.write( "
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