50th Putnam 1989

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Problem A1

Which members of the sequence 101, 10101, 1010101, ... are prime?

 

Solution

Answer: 101.

Fairly easy.

Let kn represent the member of the sequence with n 1s. It is obvious that 101 divides k2n. So we need only consider k2n+1.

But k2n+1 = 1 + 102 + 104 + ... + 104n = (104n+2 - 1)/99 = (102n+1 + 1)/11 (102n+1 - 1)/9. Each of these is integral: the first is 1 - 10 + 102 - ... + 102n, the second is 11...1 (2n+1 1s).

 


 

50th Putnam 1989

© John Scholes
jscholes@kalva.demon.co.uk
3 Nov 1999