MathDB
Problems
Contests
National and Regional Contests
Romania Contests
Romania - Local Contests
Grigore Moisil Intercounty
2005 Grigore Moisil Urziceni
2005 Grigore Moisil Urziceni
Part of
Grigore Moisil Intercounty
Subcontests
(3)
3
2
Hide problems
lnln limit
Let be a sequence
(
a
n
)
n
≥
1
\left( a_n \right)_{n\ge 1}
(
a
n
)
n
≥
1
with
a
1
>
0
a_1>0
a
1
>
0
and satisfying the equality
a
n
=
a
n
+
1
−
a
n
+
1
+
a
n
,
a_n=\sqrt{a_{n+1} -\sqrt{a_{n+1} +a_n}} ,
a
n
=
a
n
+
1
−
a
n
+
1
+
a
n
,
for all natural numbers
n
.
n.
n
.
a) Find a recurrence relation between two consecutive elements of
(
a
n
)
n
≥
1
.
\left( a_n \right)_{n\ge 1} .
(
a
n
)
n
≥
1
.
b) Prove that
lim
n
→
∞
ln
ln
a
n
n
=
ln
2.
\lim_{n\to\infty } \frac{\ln\ln a_n}{n} =\ln 2.
lim
n
→
∞
n
l
n
l
n
a
n
=
ln
2.
Rational product group under the operation (a,b)*(c,d) =(ac,ad+b)
Define the operation
(
a
,
b
)
∘
(
c
,
d
)
=
(
a
c
,
a
d
+
b
)
.
(a,b)\circ (c,d) =(ac,ad+b).
(
a
,
b
)
∘
(
c
,
d
)
=
(
a
c
,
a
d
+
b
)
.
a) Prove that
(
Q
∖
{
0
}
×
Q
,
∘
)
\left( \mathbb{Q}\setminus\{ 0\}\times\mathbb{Q} ,\circ \right)
(
Q
∖
{
0
}
×
Q
,
∘
)
is a group. b) Let
H
H
H
be an infinite subgroup of
(
Q
∖
{
0
}
×
Q
,
∘
)
\left( \mathbb{Q}\setminus\{ 0\}\times\mathbb{Q} ,\circ \right)
(
Q
∖
{
0
}
×
Q
,
∘
)
that is cyclic and doesn't contain any element of the form
(
1
,
q
)
,
(1,q) ,
(
1
,
q
)
,
where
q
q
q
is a nonzero rational. Show that there exist two rational numbers
a
,
b
a,b
a
,
b
such that
H
=
{
(
a
n
,
b
⋅
1
−
a
n
1
−
a
)
∣
n
∈
Z
}
H=\left\{ \left.\left( a^n, b\cdot\frac{1-a^n}{1-a} \right)\right| n\in\mathbb{Z} \right\}
H
=
{
(
a
n
,
b
⋅
1
−
a
1
−
a
n
)
n
∈
Z
}
2
3
Hide problems
Three nice limits
a) Prove that
lim
x
→
∞
x
⋅
∑
k
=
1
⌊
x
⌋
1
k
+
x
=
1.
\lim_{x\to\infty } \sqrt{x}\cdot\sum_{k=1}^{\lfloor \sqrt{x} \rfloor} \frac{1}{k+x}=1.
lim
x
→
∞
x
⋅
∑
k
=
1
⌊
x
⌋
k
+
x
1
=
1.
b) Show that
lim
x
→
∞
(
−
⌊
x
⌋
+
x
⋅
∑
k
=
1
⌊
x
⌋
1
k
+
x
)
=
−
1
2
\lim_{x\to\infty } \left( -\left\lfloor\sqrt{x}\right\rfloor +x\cdot\sum_{k=1}^{\lfloor \sqrt{x} \rfloor} \frac{1}{k+x} \right) =\frac{-1}{2}
lim
x
→
∞
(
−
⌊
x
⌋
+
x
⋅
∑
k
=
1
⌊
x
⌋
k
+
x
1
)
=
2
−
1
c) What about
lim
x
→
∞
(
−
x
+
x
⋅
∑
k
=
1
⌊
x
⌋
1
k
+
x
)
?
\lim_{x\to\infty } \left( -\sqrt{x} +x\cdot\sum_{k=1}^{\lfloor \sqrt{x} \rfloor} \frac{1}{k+x} \right) ?
lim
x
→
∞
(
−
x
+
x
⋅
∑
k
=
1
⌊
x
⌋
k
+
x
1
)
?
Characterization of geometric progressions of length 3 between 2004² and 2005²
Find all triples
(
x
,
y
,
z
)
(x,y,z)
(
x
,
y
,
z
)
of natural numbers that are in geometric progression and verify the inequalities
4016016
≤
x
<
y
<
z
≤
4020025.
4016016\le x<y<z\le 4020025.
4016016
≤
x
<
y
<
z
≤
4020025.
A classic result in antiderivative theory
Let be a function
f
:
R
⟶
R
≥
0
f:\mathbb{R}\longrightarrow\mathbb{R}_{\ge 0}
f
:
R
⟶
R
≥
0
that admits primitives and such that
lim
x
→
0
f
(
x
)
x
=
0.
\lim_{x\to 0 } \frac{f(x)}{x} =0.
lim
x
→
0
x
f
(
x
)
=
0.
Prove that the function
g
:
R
⟶
R
,
g:\mathbb{R}\longrightarrow\mathbb{R} ,
g
:
R
⟶
R
,
defined as
g
(
x
)
=
{
f
(
x
)
/
x
,
e
m
s
p
;
x
≠
0
0
,
e
m
s
p
;
x
=
0
,
g(x)=\left\{ \begin{matrix} f(x)/x ,&  x\neq 0\\ 0,&   x=0 \end{matrix} \right. ,
g
(
x
)
=
{
f
(
x
)
/
x
,
0
,
e
m
s
p
;
x
=
0
e
m
s
p
;
x
=
0
,
is primitivable.
1
2
Hide problems
Easy inequality
Prove that
5
x
+
6
x
≤
4
x
+
8
x
,
5^x+6^x\le 4^x+8^x,
5
x
+
6
x
≤
4
x
+
8
x
,
for any nonegative real numbers
x
.
x.
x
.
System of four variables
Find the nonnegative real numbers
a
,
b
,
c
,
d
a,b,c,d
a
,
b
,
c
,
d
that satisfy the following system:
{
a
3
+
2
a
b
c
+
b
c
d
−
6
=
a
a
2
b
+
b
2
c
+
a
b
d
+
b
d
2
=
b
a
2
b
+
a
2
c
+
b
c
2
+
c
d
2
=
c
d
3
+
a
b
2
+
a
b
c
+
b
c
d
−
6
=
d
\left\{ \begin{matrix} a^3+2abc+bcd-6&=&a \\a^2b+b^2c+abd+bd^2&=&b\\a^2b+a^2c+bc^2+cd^2&=&c\\d^3+ab^2+abc+bcd-6&=&d \end{matrix} \right.
⎩
⎨
⎧
a
3
+
2
ab
c
+
b
c
d
−
6
a
2
b
+
b
2
c
+
ab
d
+
b
d
2
a
2
b
+
a
2
c
+
b
c
2
+
c
d
2
d
3
+
a
b
2
+
ab
c
+
b
c
d
−
6
=
=
=
=
a
b
c
d