Problem E and Solution

Two Digits at a Time

The number \(34\,692\) contains \(5\) digits, and so we say its digit length is \(5\). The first digit of \(34\,692\) is \(3\). Also, the two-digit integers formed by choosing any pair of consecutive digits, that is, \(34\), \(46\), \(69\), and \(92\), are all divisible by either \(17\) or \(23\).

An integer with digit length \(2022\) has first digit \(3\). This integer also has the property that the two-digit integers formed by choosing any pair of consecutive digits are all divisible by either \(17\) or \(23\).

List all of the possibilities for the last three digits of this integer.

We first list out all two-digit multiples of \(17\) and \(23\).

The two-digit multiples of \(17\) are \(17\), \(34\), \(51\), \(68\), and \(85\).

The two-digit multiples of \(23\) are \(23\), \(46\), \(69\), and \(92\).

Since the integer starts with a \(3\) and the only two-digit number in the two lists that starts with a \(3\) is \(34\), the second digit must be \(4\). Similarly, the third digit is \(6\), since the only two-digit number in the two lists starting with a \(4\) is \(46\). However, the fourth digit can be \(8\) or \(9\), since there are two two-digit numbers, \(68\) and \(69\), in the two lists that start with a \(6\). Now we need to consider two cases.

Case 1: The fourth digit is an \(8\).

Since the only two-digit number in the two lists starting with an \(8\) is \(85\), the fifth digit must then be \(5\). Similarly, the sixth digit is \(1\), and the seventh digit is \(7\). We must stop here, since there is no two-digit number in either list that starts with a \(7\). Therefore, the fourth digit must not be \(8\).Case 2: The fourth digit is \(9\).

Since the only two-digit number in the two lists starting with a \(9\) is \(92\), the fifth digit must then be \(2\). Similarly, the sixth digit is \(3\). We can now repeat our argument from the beginning. The digit after the \(3\) must be a \(4\), then a \(6\), then a \(9\), then a \(2\). The five digits `\(34692\)’ will continue to repeat as long as a \(9\) follows the \(6\).

Since \(2022\div 5 = 404.4\), the number of length \(2022\) will consist of \(404\) blocks of five digits followed by two more digits. As we saw above, the first \(403\) blocks of digits must each be the five digits \(34692\). However, the last seven digits could be \(3469234\) or \(3468517\).

Therefore, there are two possibilities for the last three digits. The last three digits could be \(234\) or \(517\).