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Problems
Contests
Undergraduate contests
Brazil Undergrad MO
2019 Brazil Undergrad MO
2019 Brazil Undergrad MO
Part of
Brazil Undergrad MO
Subcontests
(5)
Problem 5
1
Hide problems
Small number with big primes and big exponents
Let
M
,
k
>
0
M, k>0
M
,
k
>
0
integers.Let
X
(
M
,
k
)
X(M,k)
X
(
M
,
k
)
the (infinite) set of all integers that can be factored as
p
1
e
1
⋅
p
2
e
2
⋅
…
⋅
p
r
e
r
{p_1}^{e_1} \cdot {p_2}^{e_2} \cdot \ldots \cdot {p_r}^{e_r}
p
1
e
1
⋅
p
2
e
2
⋅
…
⋅
p
r
e
r
where each
p
i
p_i
p
i
is not smaller than
M
M
M
and also each
e
i
e_i
e
i
is not smaller than
k
k
k
. Let
Z
(
M
,
k
,
n
)
Z(M,k,n)
Z
(
M
,
k
,
n
)
the number of elements of
X
(
M
,
k
)
X(M,k)
X
(
M
,
k
)
not bigger than
n
n
n
.Show that there are positive reals
c
(
M
,
k
)
c(M,k)
c
(
M
,
k
)
and
β
(
M
,
k
)
\beta(M,k)
β
(
M
,
k
)
such that
lim
n
→
∞
Z
(
M
,
k
,
n
)
n
β
(
M
,
k
)
=
c
(
M
,
k
)
\lim_{n \rightarrow \infty}{\frac{Z(M,k,n)}{n^{\beta(M,k)}}} = c(M,k)
n
→
∞
lim
n
β
(
M
,
k
)
Z
(
M
,
k
,
n
)
=
c
(
M
,
k
)
and find
β
(
M
,
k
)
\beta(M,k)
β
(
M
,
k
)
6
1
Hide problems
Equal to the number of permutations
In a hidden friend, suppose no one takes oneself. We say that the hidden friend has "marmalade" if there are two people
A
A
A
and
B
B
B
such that A took
B
B
B
and
B
B
B
took
A
A
A
. For each positive integer n, let
f
(
n
)
f (n)
f
(
n
)
be the number of hidden friends with n people where there is no “marmalade”, i.e.
f
(
n
)
f (n)
f
(
n
)
is equal to the number of permutations
σ
\sigma
σ
of {
1
,
2
,
.
.
.
,
n
1, 2,. . . , n
1
,
2
,
...
,
n
} such that:*
σ
(
i
)
≠
i
\sigma (i) \neq i
σ
(
i
)
=
i
for all
i
=
1
,
2
,
.
.
.
,
n
i=1,2,...,n
i
=
1
,
2
,
...
,
n
* there are no
1
≤
i
<
j
≤
n
1 \leq i <j \leq n
1
≤
i
<
j
≤
n
such that
σ
(
i
)
=
j
\sigma (i) = j
σ
(
i
)
=
j
and
σ
(
j
)
=
i
.
\sigma (j) = i.
σ
(
j
)
=
i
.
Determine the limit
lim
n
→
+
∞
f
(
n
)
n
!
\lim_{n \to + \infty} \frac{f(n)}{n!}
lim
n
→
+
∞
n
!
f
(
n
)
3
1
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System of equations
Let
a
,
b
,
c
a,b,c
a
,
b
,
c
be constants and
a
,
b
,
c
a,b,c
a
,
b
,
c
are positive real numbers. Prove that the equations
2
x
+
y
+
z
=
c
2
+
z
2
+
c
2
+
y
2
2x+y+z=\sqrt{c^2+z^2}+\sqrt{c^2+y^2}
2
x
+
y
+
z
=
c
2
+
z
2
+
c
2
+
y
2
x
+
2
y
+
z
=
b
2
+
x
2
+
b
2
+
z
2
x+2y+z=\sqrt{b^2+x^2}+\sqrt{b^2+z^2}
x
+
2
y
+
z
=
b
2
+
x
2
+
b
2
+
z
2
x
+
y
+
2
z
=
a
2
+
x
2
+
a
2
+
y
2
x+y+2z=\sqrt{a^2+x^2}+\sqrt{a^2+y^2}
x
+
y
+
2
z
=
a
2
+
x
2
+
a
2
+
y
2
have exactly one real solution
(
x
,
y
,
z
)
(x,y,z)
(
x
,
y
,
z
)
with
x
,
y
,
z
≥
0
x,y,z \geq 0
x
,
y
,
z
≥
0
.
4
1
Hide problems
Normal function.
Find all functions
f
:
R
→
R
f:\mathbb{R}\rightarrow \mathbb{R}
f
:
R
→
R
such that for any
(
x
,
y
)
(x, y)
(
x
,
y
)
real numbers we have
f
(
x
f
(
y
)
+
f
(
x
)
)
+
f
(
y
2
)
=
f
(
x
)
+
y
f
(
x
+
y
)
f(xf(y)+f(x))+f(y^2)=f(x)+yf(x+y)
f
(
x
f
(
y
)
+
f
(
x
))
+
f
(
y
2
)
=
f
(
x
)
+
y
f
(
x
+
y
)
1
1
Hide problems
Existence of an array c
Let
I
I
I
and
0
0
0
be the square identity and null matrices, both of size
2019
2019
2019
. There is a square matrix
A
A
A
with rational entries and size
2019
2019
2019
such that: a)
A
3
+
6
A
2
−
2
I
=
0
A ^ 3 + 6A ^ 2-2I = 0
A
3
+
6
A
2
−
2
I
=
0
? b)
A
4
+
6
A
3
−
2
I
=
0
A ^ 4 + 6A ^ 3-2I = 0
A
4
+
6
A
3
−
2
I
=
0
?