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1990 Nordic
1
1
Part of
1990 Nordic
Problems
(1)
sum from k=1 to (n-1)^p, of k^m
Source: Nordic Mathematical Contest 1990 #1
10/5/2017
Let
m
,
n
,
m, n,
m
,
n
,
and
p
p
p
be odd positive integers. Prove that the number
∑
k
=
1
(
n
−
1
)
p
k
m
\sum\limits_{k=1}^{{{(n-1)}^{p}}}{{{k}^{m}}}
k
=
1
∑
(
n
−
1
)
p
k
m
is divisible by
n
n
n
number theory
Sum of powers