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PEN I Problems
10
I 10
I 10
Source:
May 25, 2007
floor function
symmetry
Problem Statement
Show that for all primes
p
p
p
,
∑
k
=
1
p
−
1
⌊
k
3
p
⌋
=
(
p
+
1
)
(
p
−
1
)
(
p
−
2
)
4
.
\sum^{p-1}_{k=1}\left \lfloor \frac{k^{3}}{p}\right \rfloor =\frac{(p+1)(p-1)(p-2)}{4}.
k
=
1
∑
p
−
1
⌊
p
k
3
⌋
=
4
(
p
+
1
)
(
p
−
1
)
(
p
−
2
)
.
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