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 requyired 2001st digit is 6.

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