olymmath

CMO 1988 Problem 6

  • Day 2

Let nn ( n≥3n\ge 3 ) be a natural number. Denote by f(n)f(n) the least natural number by which nn is not divisible (e.g. f(12)=5f(12)=5 ). If f(n)≥3f(n)\ge 3 , we may have f(f(n))f(f(n)) in the same way. Similarly, if f(f(n))≥3f(f(n))\ge 3 , we may have f(f(f(n)))f(f(f(n))) , and so on. If f(f(…f⏟k times(n)… ))=2\underbrace{f(f(\dots f}_{k\text{ times}}(n)\dots ))=2 , we call kk the "length" of nn (also we denote by lnl_n the "length" of nn ). For arbitrary natural number nn ( n≥3n\ge 3 ), find lnl_n with proof.

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