olymmath

Putnam 1996 A5

  • Session A

If pp is a prime number greater than 3 and k=⌊2p/3⌋k = \lfloor 2p/3 \rfloor, prove that the sum (p1)+(p2)+⋯+(pk)\binom{p}{1} + \binom{p}{2} + \cdots + \binom{p}{k} of binomial coefficients is divisible by p2p^2.

Dataset text Putnam 1995 to 2025. Quoted with source.

Solutions

Human solution Solution 1 by Manjul Bhargava and Kiran Kedlaya

The terms of the source let olymmath link to this solution but not copy it. Read the solution (Kedlaya Putnam archive).

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