The left most digit of an integer of length 2000 digits is 3. In this integer, any two consecutive digits must be divisible by 17 or 23 . The 2000th digit may be either ' $a$ ' or ' $b$ '. What is the value of $a+b$ ?
(A) 3
(B) 7
(C) 4
(D) 10
(E) 17
The left most digit of an integer of length 2000 digits is 3 . In this integer, any two consecutive digits must be divisible by 17 or 23 . The 2000th digit may be either ' $a$ ' or ' $b$ '. What is the value of $a+b$ ?
(A) 3
(B) 7
(C) 4
(D) 10
(E) 17

## Solution

We start by noting that the two-digit multiples of 17 are $17,34,51,68$, and 85 . Similarly we note that the two-digit multiples of 23 are $23,46,69$, and 92 . The first digit is 3 and since the only two-digit number in the two lists starting with 3 is 34 , the second digit is 4 . Similarly the third digit must be 6 . The fourth digit, however, can be either 8 or 9 . Let's consider this in two cases.

## Case 1

If the fourth digit is 8 , the number would be 3468517 and would stop here since there isn't a number in the two lists starting with 7.

## Case 2

If the fourth digit is 9 , the number would be $346923469234 \ldots$ and the five digits ' 34692 ' would continue repeating indefinitely as long as we choose 9 to follow 6 .

If we consider a 2000 digit number, its first 1995 digits must contain 399 groups of ' 34692 '. The last groups of five digits could be either 34692 or 34685 which means that the 2000th digit may be either 2 or 5 so that $a+b=2+5=7$.

ANSWER: (B)