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VTRMC
2015 VTRMC
Problem 3
Problem 3
Part of
2015 VTRMC
Problems
(1)
2015! divides determinant of integer powers
Source: VTRMC 2015 P3
5/5/2021
Let
(
a
i
)
1
≤
i
≤
2015
(a_i)_{1\le i\le2015}
(
a
i
)
1
≤
i
≤
2015
be a sequence consisting of
2015
2015
2015
integers, and let
(
k
i
)
1
≤
i
≤
2015
(k_i)_{1\le i\le2015}
(
k
i
)
1
≤
i
≤
2015
be a sequence of
2015
2015
2015
positive integers (positive integer excludes
0
0
0
). Let
A
=
(
a
1
k
1
a
1
k
2
⋯
a
1
k
2015
a
2
k
1
a
2
k
2
⋯
a
2
k
2015
⋮
⋮
⋱
⋮
a
2015
k
1
a
2015
k
2
⋯
a
2015
k
2015
)
.
A=\begin{pmatrix}a_1^{k_1}&a_1^{k_2}&\cdots&a_1^{k_{2015}}\\a_2^{k_1}&a_2^{k_2}&\cdots&a_2^{k_{2015}}\\\vdots&\vdots&\ddots&\vdots\\a_{2015}^{k_1}&a_{2015}^{k_2}&\cdots&a_{2015}^{k_{2015}}\end{pmatrix}.
A
=
a
1
k
1
a
2
k
1
⋮
a
2015
k
1
a
1
k
2
a
2
k
2
⋮
a
2015
k
2
⋯
⋯
⋱
⋯
a
1
k
2015
a
2
k
2015
⋮
a
2015
k
2015
.
Prove that
2015
!
2015!
2015
!
divides
det
A
\det A
det
A
.
Matrices
linear algebra