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Putnam
2023 Putnam
A5
2023 Putnam A5
2023 Putnam A5
Source:
December 3, 2023
Putnam
Putnam 2023
Problem Statement
For a nonnegative integer
k
k
k
, let
f
(
k
)
f(k)
f
(
k
)
be the number of ones in the base 3 representation of
k
k
k
. Find all complex numbers
z
z
z
such that
∑
k
=
0
3
1010
−
1
(
−
2
)
f
(
k
)
(
z
+
k
)
2023
=
0
\sum_{k=0}^{3^{1010}-1}(-2)^{f(k)}(z+k)^{2023}=0
k
=
0
∑
3
1010
−
1
(
−
2
)
f
(
k
)
(
z
+
k
)
2023
=
0
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