olymmath

Putnam 1997 B5

  • Session B

Prove that for n≥2n \geq 2, 22⋅⋅2⏟n terms≡22⋅⋅2⏟n−1 terms(modn).\underbrace{2^{2^{\cdot^{\cdot^{2}}}}}_{n \text{ terms}} \equiv \underbrace{2^{2^{\cdot^{\cdot^{2}}}}}_{n-1 \text{ terms}} \pmod{n}.

Dataset text Putnam 1995 to 2025. Quoted with source.

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