A string of 2015 digits starts with a "4". Any number formed by two consecutive digits is divisible by either 19 or 23.
What is the last digit in this sequence?
What if the sequence had 2016 digits?
The following table shows possible preceding digits followed by the allowable following digit or digits:
1 9
2 3
3 8
4 6
5 7
6 9
7 6
8
9 2 5
As we start with a 4, the first three digits are 469. At that point, we could go to 238, but that's a dead end, so we have to go to 5769. We always have to take the 5 path rather than the 2 path when we get to 9, so the start looks like 469576957695769....
2015 is congruent to 3 mod 4, so the 2015th digit is 9; and the 2016th digit of such a 2016-digit number is either 2 or 5 as we need not worry about continuation in that instance.
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Posted by Charlie
on 2014-09-04 11:22:35 |
solution
