50th Putnam 1989

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

Solution

Answer: 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 + 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