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Putnam
1996 Putnam
5
Putnam 1996 A5
Putnam 1996 A5
Source:
June 4, 2014
Putnam
floor function
college contests
Problem Statement
Let
p
p
p
be a prime greater than
3
3
3
. Prove that
p
2
∣
∑
i
=
1
⌊
2
p
3
⌋
(
p
i
)
.
p^2\Big| \sum_{i=1}^{\left\lfloor\frac{2p}{3}\right\rfloor}\dbinom{p}{i}.
p
2
i
=
1
∑
⌊
3
2
p
⌋
(
i
p
)
.
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