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Romania Team Selection Test
2002 Romania Team Selection Test
2002 Romania Team Selection Test
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Romania Team Selection Test
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3
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Choosing signs + or - to make the sum equal to 0
For any positive integer
n
n
n
, let
f
(
n
)
f(n)
f
(
n
)
be the number of possible choices of signs
+
or
−
+\ \text{or}\ -
+
or
−
in the algebraic expression
±
1
±
2
…
±
n
\pm 1\pm 2\ldots \pm n
±
1
±
2
…
±
n
, such that the obtained sum is zero. Show that
f
(
n
)
f(n)
f
(
n
)
satisfies the following conditions: a)
f
(
n
)
=
0
f(n)=0
f
(
n
)
=
0
for
n
=
1
(
m
o
d
4
)
n=1\pmod{4}
n
=
1
(
mod
4
)
or
n
=
2
(
m
o
d
4
)
n=2\pmod{4}
n
=
2
(
mod
4
)
. b)
2
n
2
−
1
≤
f
(
n
)
≤
2
n
−
2
⌊
n
2
⌋
+
1
2^{\frac{n}{2}-1}\le f(n)\le 2^n-2^{\lfloor\frac{n}{2}\rfloor+1}
2
2
n
−
1
≤
f
(
n
)
≤
2
n
−
2
⌊
2
n
⌋
+
1
, for
n
=
0
(
m
o
d
4
)
n=0\pmod{4}
n
=
0
(
mod
4
)
or
n
=
3
(
m
o
d
4
)
n=3\pmod{4}
n
=
3
(
mod
4
)
.Ioan Tomsecu
60% of participants can speak the same language
At an international conference there are four official languages. Any two participants can speak in one of these languages. Show that at least
60
%
60\%
60%
of the participants can speak the same language.Mihai Baluna
If x-y=2,3 or 5 then f(x) is not equal to f(y)
Let
f
:
Z
→
{
1
,
2
,
…
,
n
}
f:\mathbb{Z}\rightarrow\{ 1,2,\ldots ,n\}
f
:
Z
→
{
1
,
2
,
…
,
n
}
be a function such that
f
(
x
)
≠
f
(
y
)
f(x)\not= f(y)
f
(
x
)
=
f
(
y
)
, for all
x
,
y
∈
Z
x,y\in\mathbb{Z}
x
,
y
∈
Z
such that
∣
x
−
y
∣
∈
{
2
,
3
,
5
}
|x-y|\in\{2,3,5\}
∣
x
−
y
∣
∈
{
2
,
3
,
5
}
. Prove that
n
≥
4
n\ge 4
n
≥
4
.Ioan Tomescu
1
4
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3
5
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2
5
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