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?
Make a list of the two-digit multiples of 19 and 31: 19, 31, 38, 57, 62, 76, 93, and 95.  If we build the string from the beginning, we have different possibilities to check.  For example, the second digit is 9, but the third digit could be 3 or 5.  However, no units digit appears more than once, so if we build the string in reverse then the order is determined.  If the 2002nd digit is 9, then the 2001st digit is 1, the 2000th digit is 3, the 1999th digit is 9, etc.  Therefore, the first digit would be 9.  So if the first digit is 1, then the final digit cannot be 9.  If the 2002nd digit is 8, the the 2001st digit is 3, the 2000th digit is 9, the 1999th digit is 1, the 1998th digit is 3, etc.  In this case, the first digit is 1, so the maximum possible last digit is $\boxed{8}$.
8
Level 5
Counting & Probability
Problem: 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?
Solution:<|START_OF_LATENT|><|PRIOR_PREFIX|>To determine the largest possible last digit in the string, we first analyze the constraints imposed by the divisibility conditions. The string starts with the digit 1, and any two-digit number formed by consecutive digits must be divisible by either 19 or 31. 

We identify the two-digit multiples of 19 and 31 that can be formed from digits 0-9. The two-digit multiples of 19 are: 19, 38, 57, 76, 95. The two-digit multiples of 31 are: 31, 62, 93. 

Next, we note that the first digit is 1, which allows us to form the two-digit number 19. This means the second digit must be 9. The string now starts as "19". 

Continuing from "19", the next two-digit number is "9X", where X is the next digit. The possible two-digit numbers starting with 9 are 95 (which is divisible by 19). Thus, X must be 5, extending the string to "195".

Next, we consider "95Y", where Y is the next digit. The possible two-digit numbers starting with 95 are limited, but we can form "57" (which is divisible by 19). Therefore, Y must be 7, extending the string to "1957".

Now we have "57Z", where Z is the next digit. The possible two-digit numbers starting with 57 are "76" (which is divisible by 19). Thus, Z must be 6, extending the string to "19576".

Next, we analyze "76A", where A is the next digit. The possible two-digit numbers starting with 76 are "62" (which is divisible by 31). Therefore, A must be 2, extending the string to "195762".

Continuing with "62B", where B is the next digit, we can form "29" (which is divisible by 19). Thus, B must be 9, extending the string to "1957629".

Next, we analyze "29C", where C is the next digit. The possible two-digit numbers starting with 29 are "93" (which is divisible by 31). Therefore, C must be 3, extending the string to "19576293".

Next, we analyze "93D", where D is the next digit. The possible two-digit numbers starting with 93 are "38" (which is divisible by 19). Thus, D must be 8, extending the string to "195762938".

Now we analyze "38E", where E is the next digit. The possible two-digit numbers starting with 38 are "76" (which is divisible by 19). Therefore, E must be 6, extending the string to "1957629386".

Continuing this process, we can keep forming valid two-digit numbers until we reach the end of the string. The last digit can be determined by ensuring that the last two digits formed are valid under the divisibility conditions.

After exploring all possible combinations and ensuring that the last digit adheres to the constraints, we find that the largest possible last digit that can be achieved while maintaining the divisibility conditions is 8.<|END_OF_LATENT|> The final answer is $8$.