MathDB
Problems
Contests
National and Regional Contests
China Contests
China Team Selection Test
1993 China Team Selection Test
1
find the value of f(p)
find the value of f(p)
Source: China TST 1993, problem 1
June 27, 2005
modular arithmetic
number theory unsolved
number theory
Problem Statement
For all primes
p
≥
3
,
p \geq 3,
p
≥
3
,
define
F
(
p
)
=
∑
k
=
1
p
−
1
2
k
120
F(p) = \sum^{\frac{p-1}{2}}_{k=1}k^{120}
F
(
p
)
=
∑
k
=
1
2
p
−
1
k
120
and
f
(
p
)
=
1
2
−
{
F
(
p
)
p
}
f(p) = \frac{1}{2} - \left\{ \frac{F(p)}{p} \right\}
f
(
p
)
=
2
1
−
{
p
F
(
p
)
}
, where
{
x
}
=
x
−
[
x
]
,
\{x\} = x - [x],
{
x
}
=
x
−
[
x
]
,
find the value of
f
(
p
)
.
f(p).
f
(
p
)
.
Back to Problems
View on AoPS