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5
2024 Alg/NT Problem 5
2024 Alg/NT Problem 5
Source:
April 14, 2024
algebra
Problem Statement
Let
f
(
x
)
=
(
x
+
1
)
6
+
(
x
−
1
)
5
+
(
x
+
1
)
4
+
(
x
−
1
)
3
+
(
x
+
1
)
2
+
(
x
−
1
)
1
+
1.
f(x)=(x+1)^{6}+(x-1)^{5}+(x+1)^{4}+(x-1)^3+(x+1)^2+(x-1)^1+1.
f
(
x
)
=
(
x
+
1
)
6
+
(
x
−
1
)
5
+
(
x
+
1
)
4
+
(
x
−
1
)
3
+
(
x
+
1
)
2
+
(
x
−
1
)
1
+
1.
Find the remainder when
∑
j
=
−
126
126
j
f
(
j
)
\sum_{j=-126}^{126}jf(j)
∑
j
=
−
126
126
j
f
(
j
)
is divided by 1000.Proposed by Hari Desikan
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