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?

(In reply to answer by K Sengupta)

At the outset, we observe that the two digit multiples of 23 are 23, 46, 69 and 92;  while the two digit multiples of 17 are 17, 34, 51, 68 and 85. Thus, in conformity with the given conditions, the possible sequences are:

(i) 6923469234...... with the string '69234' repeated indefinitely, so that the respective (5p+1)st digit and the (5p+2)nd digit will be 6 and 9, for all nonnegative integers p.

(ii) The sequence "68517" consisting of only 5 terms. This is a contradiction, since 5< 2001

(iii) The first through 2000 digits consist of the string '69234' repeated precisely 400 times with the sequence terminating with the string '68517'.

In terms of (i), substituting p=400, the respective 200st and the 2002nd digits are 6 and 9.

In terms of (iii), the respective 2001st and 2002nd digits are 6 and 8.

Consequently,

(i) The required 2001st digit is 6.

(ii) The 2002nd digit is either 9, or 8. 

Edited on June 12, 2022, 10:17 am

Posted by K Sengupta on 2008-07-21 04:27:44