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CIIM
2010 CIIM
2010 CIIM
Part of
CIIM
Subcontests
(6)
Problem 5
1
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CIIM 2010 Problem 5
Let
n
,
d
n,d
n
,
d
be integers with
n
,
k
>
1
n,k > 1
n
,
k
>
1
such that
g
.
c
.
d
(
n
,
d
!
)
=
1
g.c.d(n,d!) = 1
g
.
c
.
d
(
n
,
d
!)
=
1
. Prove that
n
n
n
and
n
+
d
n+d
n
+
d
are primes if and only if
d
!
d
(
(
n
−
1
)
!
+
1
)
+
n
(
d
!
−
1
)
≡
0
(
m
o
d
n
(
n
+
d
)
)
.
d!d((n-1)!+1) + n(d!-1) \equiv 0 \hspace{0.2cm} (\bmod n(n+d)).
d
!
d
((
n
−
1
)!
+
1
)
+
n
(
d
!
−
1
)
≡
0
(
mod
n
(
n
+
d
))
.
Problem 4
1
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CIIM 2010 Problem 4
Let
f
:
[
0
,
1
]
→
[
0
,
1
]
f:[0,1] \to [0,1]
f
:
[
0
,
1
]
→
[
0
,
1
]
a increasing continuous function, diferentiable in
(
0
,
1
)
(0,1)
(
0
,
1
)
and with derivative smaller than 1 in every point. The sequence of sets
A
1
,
A
2
,
A
3
,
…
A_1,A_2,A_3,\dots
A
1
,
A
2
,
A
3
,
…
is define as:
A
1
=
f
(
[
0
,
1
]
)
A_1 = f([0,1])
A
1
=
f
([
0
,
1
])
, and for
n
≥
2
,
A
n
=
f
(
A
n
−
1
)
.
n \geq 2, A_n = f(A_{n-1}).
n
≥
2
,
A
n
=
f
(
A
n
−
1
)
.
Prove that
lim
n
→
+
∞
d
(
A
n
)
=
0
\displaystyle \lim_{n\to+\infty} d(A_n) = 0
n
→
+
∞
lim
d
(
A
n
)
=
0
, where
d
(
A
)
d(A)
d
(
A
)
is the diameter of the set
A
A
A
.Note: The diameter of a set
X
X
X
is define as
d
(
X
)
=
sup
x
,
y
∈
X
∣
x
−
y
∣
.
d(X) = \sup_{x,y\in X} |x-y|.
d
(
X
)
=
sup
x
,
y
∈
X
∣
x
−
y
∣.
Problem 2
1
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CIIM 2010 Problem 2
In one side of a hall there are
2
N
2N
2
N
rooms numbered from 1 to
2
N
2N
2
N
. In each room
i
i
i
between 1 and
N
N
N
there are
p
i
p_i
p
i
beds. Is needed to move every one of this beds to the roms from
N
+
1
N+ 1
N
+
1
to
2
N
2N
2
N
, in such a way that for every
j
j
j
between
N
+
1
N+1
N
+
1
and
2
N
2N
2
N
the room
j
j
j
will have
p
j
p_j
p
j
beds. Supose that each bed can be move once and the price of moving a bed from room
i
i
i
to room
j
j
j
is
(
i
−
j
)
2
(i-j)^2
(
i
−
j
)
2
. Find a way to move every bed such that the total cost is minimize.Note: The numbers
p
i
p_i
p
i
are given and satisfy that
p
1
+
p
2
+
⋯
+
p
N
=
p
N
+
1
+
p
N
+
2
+
⋯
+
p
2
N
.
p_1 + p_2 + \cdots + p_N = p_{N+1} + p_{N+2} + \cdots+ p_{2N}.
p
1
+
p
2
+
⋯
+
p
N
=
p
N
+
1
+
p
N
+
2
+
⋯
+
p
2
N
.
Problem 1
1
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CIIM 2010 Problem 1
Given two vectors
v
=
(
v
1
,
…
,
v
n
)
v = (v_1,\dots,v_n)
v
=
(
v
1
,
…
,
v
n
)
and
w
=
(
w
1
…
,
w
n
)
w = (w_1\dots,w_n)
w
=
(
w
1
…
,
w
n
)
in
R
n
\mathbb{R}^n
R
n
, lets define
v
∗
w
v*w
v
∗
w
as the matrix in which the element of row
i
i
i
and column
j
j
j
is
v
i
w
j
v_iw_j
v
i
w
j
. Supose that
v
v
v
and
w
w
w
are linearly independent. Find the rank of the matrix
v
∗
w
−
w
∗
v
.
v*w - w*v.
v
∗
w
−
w
∗
v
.
Problem 6
1
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CIIM 2010 Problem 6
A group is call locally cyclic if any finitely generated subgroup is cyclic. Prove that a locally cyclic group is isomorphic to one of its proper subgroups if and only if it's isomorphic to a proper subgroup of the rational numbers with the adition.
Problem 3
1
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CIIM 2010 Problem 3
A set
X
⊂
R
X\subset \mathbb{R}
X
⊂
R
has dimension zero if, for any
ϵ
>
0
\epsilon > 0
ϵ
>
0
there exists a positive integer
k
k
k
and intervals
I
1
,
I
2
,
.
.
.
,
I
k
I_1,I_2,...,I_k
I
1
,
I
2
,
...
,
I
k
such that
X
⊂
I
1
∪
I
2
∪
⋯
∪
I
k
X \subset I_1 \cup I_2 \cup \cdots \cup I_k
X
⊂
I
1
∪
I
2
∪
⋯
∪
I
k
with
∑
j
=
1
k
∣
I
j
∣
ϵ
<
ϵ
\sum_{j=1}^k |I_j|^{\epsilon} < \epsilon
∑
j
=
1
k
∣
I
j
∣
ϵ
<
ϵ
.Prove that there exist sets
X
,
Y
⊂
[
0
,
1
]
X,Y \subset [0,1]
X
,
Y
⊂
[
0
,
1
]
both of dimension zero, such that
X
+
Y
=
[
0
,
2
]
.
X+Y = [0,2].
X
+
Y
=
[
0
,
2
]
.