olymmath

Putnam 2006 A3

  • Session A

Let 1,2,3,…,2005,2006,2007,2009,2012,2016,…1, 2, 3, \dots, 2005, 2006, 2007, 2009, 2012, 2016, \dots be a sequence defined by xk=kx_k = k for k=1,2,…,2006k=1, 2, \dots, 2006 and xk+1=xk+xk−2005x_{k+1} = x_k + x_{k-2005} for k≥2006k \geq 2006. Show that the sequence has 2005 consecutive terms each divisible by 2006.

Curated text Kedlaya Putnam archive. Quoted with source.

Solutions

Human solution Solution 1 by Kiran Kedlaya and Lenny Ng

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