A string of 2001 digits begins with a "6". Any number formed by two consecutive digits is divisible by either 17 or 23.

What is the last digit in this sequence?

What if the sequence had 2002 digits?

Legal pairs (multiples of 17 and 23 under 100):
17
23
34
46
51
68
69
85
92

Possibilities starting with 6:
68517 (uh-oh, nothing can go after that 7)
69234692346... repeating
So, the the first digit is 6, the 6th is, the 11th ... the 2001st is. The next (2002nd) is 9, of course.

Posted by Dave Turner on 2002-04-30 17:15:16