The four digit number \(2652\) is such that any two consecutive digits from it make a multiple of \(13\). Another number \(N\) has this same property, is \(100\) digits long and begins in a \(9\). What is the last digit of \(N\)?

2

3

6

9

The two-digit multiples of \(13\) are \[13,\ 26,\ 39,\ 52,\ 65,\ 78,\ 91.\]

\(N\) starts with a \(9\), so the first two digits must be \(91\). Then the second and third must be \(13\), and the third and fourth \(39\).

Then the sequence repeats, so \(N\) is \(913913913...913\) up to the \(99\)th digit, with \(9\) as its \(100\)th digit.

Hence the answer is (d).