CMO 1988 Problem 6
- Day 2
Let ( ) be a natural number. Denote by the least natural number by which is not divisible (e.g. ). If , we may have in the same way. Similarly, if , we may have , and so on. If , we call the "length" of (also we denote by the "length" of ). For arbitrary natural number ( ), find with proof.
Dataset text AI-MO AoPS. Quoted with source.
Solutions
No solution is on olymmath yet. Official and human solutions come first when they arrive. AI solutions come after them, with a label.
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