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 $a+b=2+5=7$.