50th Putnam 1989 Problem A1 Which members of the sequence 101, 10101, 1010101, ... are prime? Solution
: 101. Fairly easy. Let k n represent the member of the sequence with n 1s. It is obvious that 101 divides k 2n . So we need only consider k 2n+1 . But k 2n+1 = 1 + 10 2 + 10 4 + ... + 10 4n = (10 4n+2 - 1)/99 = (10 2n+1 + 1)/11 (10 2n+1 - 1)/9. Each of these is integral: the first is 1 - 10 + 10 2 - ... + 10 2n , the second is 11...1 (2n+1 1s). 50th Putnam 1989 © John Scholes jscholes@kalva.demon.co.uk 3 Nov 1999