48 digits, 99 ... 9 (48 nines) Solution We notice that 1989 = 9·13·17. Also 10 = 1 mod 9, 10 16 = 1 mod 17 and 10 12 = 1 mod 13, so 10 48 = 1 mod 1989. Thus the number with 48 nines is certainly a solution. But can we do better? Note that 1989 = A necessary and sufficient condition for all the cyclical permutations to be divisible by 9 is simply that the sum of the digits should be divisible by 9. 2nd Irish 1989 (C) John Scholes jscholes@kalva.demon.co.uk 11 July 2003 Last updated/corrected 11 Jul 03
2nd Irish 1989 Problem A5 An n-digit number has the property that if you cyclically permute its digits it is always divisible by 1989. What is the smallest possible value of n? What is the smallest such number? [If we cyclically permute the digits of 3701 we get 7013, 137, 1370, and 3701.]