+1206 The first digit of a string of 2002 digits is a 1. Any two-digit number formed by consecutive digits within this string is divisible by 19 or 31. What is the largest possible last digit in this string?
+6248 2 digit multiples of 19 are (19, 38, 57, 76, 95)
2 digit multiples of 31 are (31, 62, 93)
1 -> 19
19 -> 193 or 195
193 -> 1931 or 1938
195 -> 1957
1931 -> 19319
1938 -> X
1957 -> 19576
19319 -> 193193 or 193195
19576 -> 195762
193193 -> 1931931 or 1931938
193195 -> 1931957
195762 -> X
1931931 -> 19319319
1931938 -> X
1931957 -> 19319576
continuing this way for a bit we see the following pattern
strings of (4+3k) digits will have the last digit in the set (1,7,8)
strings of (5+3k) digits will have the last digit in the set (6, 9)
strings of (6+3k) digits will have the last digit in the set (2, 3, 5)
2002 = 4 + 3(666)
so the possible last digits in the string will be 1, 7, or 8
8 is the largest of these
