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N2.
2023 Fall TJ Proof TST, P6
Number Theory
3
For every positive integer
n
n
n
, there exists exactly one function
f
n
:
{
1
,
2
,
…
,
n
}
→
R
f_n\colon\{1,2,\dots,n\}\to\mathbb{R}
f
n
:
{
1
,
2
,
…
,
n
}
→
R
such that for all positive integers
k
≤
n
k\le n
k
≤
n
,
f
n
(
k
)
+
f
n
(
2
k
)
+
⋯
+
f
n
(
k
⌊
n
k
⌋
)
=
1
k
⌊
n
k
⌋
.
f_n(k)+f_n(2k)+\dots+f_n\left(k\left\lfloor\frac nk\right\rfloor\right) = \frac1k\left\lfloor\dfrac nk\right\rfloor.
f
n
(
k
)
+
f
n
(
2
k
)
+
⋯
+
f
n
(
k
⌊
k
n
⌋
)
=
k
1
⌊
k
n
⌋
.
Find
f
1000
(
1
)
−
f
999
(
1
)
f_{1000}(1)-f_{999}(1)
f
1000
(
1
)
−
f
999
(
1
)
Written by Calvin Wang
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Oron Wang
12/28/2024
JAT.
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