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Nov 9th, 2002, 03:37 PM
#5
Fanatic Member
I will define my own blocks:
block 1:
1234
block 2:
01234
block 3:
001234
...
block n:
000...0001234
(n-1) 0's
therefore the first block has 4 digits
2nd has 5 digits
...
nth block has n+3 digits
after exactly nth block, the total number of digits would be:
(4+n+3)n/2
or (n^2+7n)/2
I will now find the number of complete blocks such that the # of digits does not exceed 2550:
(n^2+7n)/2<=2550
gives me:
n^2+7n<= 5100
n=69....
so when n=69
(n^2+7n)/2-2550>4
so the digit is probably a 0
Massey RuleZ! ^-^__  Cheers!  __^-^ Massey RuleZ!
Did you know that...
The probability that a random rational number has an even denominator is 1/3 (Salamin and Gosper 1972)? This result is independently verified by me (2002)!
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