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?

dolphinia Dec 30, 2022

#2**0 **

We can list out the first few digits.

1 is the first, followed by a 9 to make the first constraint true.

Then, the 9 has to have a following 3 because 93/31 = 3. (So far is 193____)

Then after the 3 there is an 8 or 5 to make the constraint true. (So far is 1938______ or 1935______)

If 8 then impossible because no 2 digit number that starts with 8 is divisible by 31 or 19. So 1935 is accurate, after the 5 there is a 7 because 57 is divisible by 19. (So far 19357______)

After a 7 there is a 6, because 76/19 = 4. No 2 digit number with 7s as tens digit is divisible by 31 so 31 is out of this digit part. (So far is 193576___).

After the 6 there is a 2, 62/31 = 2. No 2 digit number with 6 as tens digit is div by 19. (So far is 1935762____).

After the 2, there is nothing, so unfortunately we have to start over to step 3, where instead after the 9 there is a 5. (So far is 195_____)

After the 5 is a 7. (So far is 1957_____) After the 7 there is a 6 (1957) After 7 is a 6, then after 6 is a 2. So it loops back. So whoever made this problem forgot to realize there is no such string, since no 2 digit number that starts with a 2 or 8 is divisible by either 31 or 19 :|

proyaop Dec 30, 2022

#3**0 **

The only two digit numbers divisible by 19 or 31 are, 19, 38, 57, 76, 95 (those by 19), and 31, 62, 93, (by 31).

If the first digit is 1, the second digit has to be 9, (19 being the only possible continuation).

The sequence could then continue with either 95 or 93, forming 195 or 193 respectively.

Continuing in this way, the only possibles are,

195762, (19), (95), (57), (76), (62), the sequence terminating.

1938, (19), (93), (38), the sequence terminating.

1931931931.....(19), (93), (31), (19), (93), .....

This third sequence can continue indefinitely or it could be arranged to terminate by attaching either a 5 after a 9 or an 8 after a 3.

To reach 2002 digits, we have to go 193193193193...and, if we continue in this way, the 3 would land on the 2001th digit meaning the the 2002nd digit would be a 1.

However the sequence could be broken one cycle earlier, terminating with 1938 or 1957(62),

So 8 is the largest possible digit.

Guest Dec 31, 2022

#4**-1 **

I mam probably repeating what has been said but I will have a play too

any 2 ditits must be muliple of 19 or 31 so that is

19

38 none of the others start with an 8 so this one is scrapped

57

76

95 none of the others start with an 5 so this one is scrapped

31

62 none of the others start with an 2 so this one is scrapped

93

So this leaves possible adjacent digits of

19

31

57

76

93

since the first digit is 1 we have

193193193 etc

There are 2002 digits 2002mod 3 = 1

So the last digit is 1 becasue 1 is the first number of the repeating string.

Melody Jan 3, 2023