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Undergraduate contests
Vojtěch Jarník IMC
2001 VJIMC
2001 VJIMC
Part of
Vojtěch Jarník IMC
Subcontests
(4)
Problem 1
2
Hide problems
x-y<=xy/25 if x,y in A, max |A|
Let
A
A
A
be a set of positive integers such that for any
x
,
y
∈
A
x,y\in A
x
,
y
∈
A
,
x
>
y
⟹
x
−
y
≥
x
y
25
.
x>y\implies x-y\ge\frac{xy}{25}.
x
>
y
⟹
x
−
y
≥
25
x
y
.
Find the maximal possible number of elements of the set
A
A
A
.
integers uniquely determined by pairwise sums
Let
n
≥
2
n\ge2
n
≥
2
be an integer and let
x
1
,
x
2
,
…
,
x
n
x_1,x_2,\ldots,x_n
x
1
,
x
2
,
…
,
x
n
be real numbers. Consider
N
=
(
n
2
)
N=\binom n2
N
=
(
2
n
)
sums
x
i
+
x
j
x_i+x_j
x
i
+
x
j
,
1
≤
i
<
j
≤
n
1\le i<j\le n
1
≤
i
<
j
≤
n
, and denote them by
y
1
,
y
2
,
…
,
y
N
y_1,y_2,\ldots,y_N
y
1
,
y
2
,
…
,
y
N
(in an arbitrary order). For which
n
n
n
are the numbers
x
1
,
x
2
,
…
,
x
n
x_1,x_2,\ldots,x_n
x
1
,
x
2
,
…
,
x
n
uniquely determined by the numbers
y
1
,
y
2
,
…
,
y
N
y_1,y_2,\ldots,y_N
y
1
,
y
2
,
…
,
y
N
?
Problem 3
2
Hide problems
ln2*ln3*...ln n < sqrt(n!)/n
Let
n
≥
2
n\ge2
n
≥
2
be a natural number. Prove that
∏
k
=
2
n
ln
k
<
n
!
n
.
\prod_{k=2}^n\ln k<\frac{\sqrt{n!}}n.
k
=
2
∏
n
ln
k
<
n
n
!
.
lim xf(x)=0 if f:R+->R+ and int^∞_0<∞
Let
f
:
(
0
,
+
∞
)
→
(
0
,
+
∞
)
f:(0,+\infty)\to(0,+\infty)
f
:
(
0
,
+
∞
)
→
(
0
,
+
∞
)
be a decreasing function which satisfies
∫
0
∞
f
(
x
)
d
x
<
+
∞
\int^\infty_0f(x)\text dx<+\infty
∫
0
∞
f
(
x
)
d
x
<
+
∞
. Prove that
lim
x
→
+
∞
x
f
(
x
)
=
0
\lim_{x\to+\infty}xf(x)=0
lim
x
→
+
∞
x
f
(
x
)
=
0
.
Problem 2
2
Hide problems
p^2|binomial sum of p choose k, k<2p/3
Prove that for any prime
p
≥
5
p\ge5
p
≥
5
, the number
∑
0
<
k
<
2
p
3
(
p
k
)
\sum_{0<k<\frac{2p}3}\binom pk
0
<
k
<
3
2
p
∑
(
k
p
)
is divisible by
p
2
p^2
p
2
.
f_(n+1)(x)=int^x_0 f_n(t)dt and f_n(1)=0 for all n
Let
f
:
[
0
,
1
]
→
R
f:[0,1]\to\mathbb R
f
:
[
0
,
1
]
→
R
be a continuous function. Define a sequence of functions
f
n
:
[
0
,
1
]
→
R
f_n:[0,1]\to\mathbb R
f
n
:
[
0
,
1
]
→
R
in the following way:
f
0
(
x
)
=
f
(
x
)
,
f
n
+
1
(
x
)
=
∫
0
x
f
n
(
t
)
d
t
,
n
=
0
,
1
,
2
,
…
.
f_0(x)=f(x),\qquad f_{n+1}(x)=\int^x_0f_n(t)\text dt,\qquad n=0,1,2,\ldots.
f
0
(
x
)
=
f
(
x
)
,
f
n
+
1
(
x
)
=
∫
0
x
f
n
(
t
)
d
t
,
n
=
0
,
1
,
2
,
…
.
Prove that if
f
n
(
1
)
=
0
f_n(1)=0
f
n
(
1
)
=
0
for all
n
n
n
, then
f
(
x
)
≡
0
f(x)\equiv0
f
(
x
)
≡
0
.
Problem 4
2
Hide problems
A+B subset A+C, A bounded, C closed, convex
Let
A
,
B
,
C
A,B,C
A
,
B
,
C
be nonempty sets in
R
n
\mathbb R^n
R
n
. Suppose that
A
A
A
is bounded,
C
C
C
is closed and convex, and
A
+
B
⊆
A
+
C
A+B\subseteq A+C
A
+
B
⊆
A
+
C
. Prove that
B
⊆
C
B\subseteq C
B
⊆
C
. Recall that
E
+
F
=
{
e
+
f
:
e
∈
E
,
f
∈
F
}
E+F=\{e+f:e\in E,f\in F\}
E
+
F
=
{
e
+
f
:
e
∈
E
,
f
∈
F
}
and
D
⊆
R
n
D\subseteq\mathbb R^n
D
⊆
R
n
is convex iff
t
x
+
(
1
−
t
)
y
∈
D
tx+(1-t)y\in D
t
x
+
(
1
−
t
)
y
∈
D
for all
x
,
y
∈
D
x,y\in D
x
,
y
∈
D
and any
t
∈
[
0
,
1
]
t\in[0,1]
t
∈
[
0
,
1
]
.
ring arithmetic, x^n=x for all x implies x,y^(n-1) commute
Let
R
R
R
be an associative non-commutative ring and let
n
>
2
n>2
n
>
2
be a fixed natural number. Assume that
x
n
=
x
x^n=x
x
n
=
x
for all
x
∈
R
x\in R
x
∈
R
. Prove that
x
y
n
−
1
=
y
n
−
1
x
xy^{n-1}=y^{n-1}x
x
y
n
−
1
=
y
n
−
1
x
holds for all
x
,
y
∈
R
x,y\in R
x
,
y
∈
R
.