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Contests
National and Regional Contests
Switzerland Contests
Switzerland - Final Round
2011 Switzerland - Final Round
2011 Switzerland - Final Round
Part of
Switzerland - Final Round
Subcontests
(6)
10
1
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Bugs on a chessboard - Switzerland 2011
On each square of an
n
×
n
n\times n
n
×
n
-chessboard, there are two bugs. In a move, each bug moves to a (vertically of horizontally) adjacent square. Bugs from the same square always move to different squares. Determine the maximal number of free squares that can occur after one move.(Swiss Mathematical Olympiad 2011, Final round, problem 10)
9
1
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Divisors with special properties - Switzerland 2011
For any positive integer
n
n
n
let
f
(
n
)
f(n)
f
(
n
)
be the number of divisors of
n
n
n
ending with
1
1
1
or
9
9
9
in base
10
10
10
and let
g
(
n
)
g(n)
g
(
n
)
be the number of divisors of
n
n
n
ending with digit
3
3
3
or
7
7
7
in base
10
10
10
. Prove that
f
(
n
)
⩾
g
(
n
)
f(n)\geqslant g(n)
f
(
n
)
⩾
g
(
n
)
for all nonnegative integers
n
n
n
.(Swiss Mathematical Olympiad 2011, Final round, problem 9)
7
1
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Equation in Z - Switzerland 2011
For a given rational number
r
r
r
, find all integers
z
z
z
such that 2^z + 2 = r^2\mbox{.}(Swiss Mathematical Olympiad 2011, Final round, problem 7)
6
1
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Four-Variables Inequality - Switzerland 2011
Let
a
,
b
,
c
,
d
a, b, c, d
a
,
b
,
c
,
d
be positive real numbers satisfying
a
+
b
+
c
+
d
=
1
a+b+c+d =1
a
+
b
+
c
+
d
=
1
. Show that \frac{2}{(a+b)(c+d)} \leq \frac{1}{\sqrt{ab}}+ \frac{1}{\sqrt{cd}}\mbox{.}(Swiss Mathematical Olympiad 2011, Final round, problem 6)
4
1
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Functional equation with abcd=1 - Switzerland 2011
Find all functions
f
:
R
+
→
R
+
f:\mathbb{R}^+\to\mathbb{R}^+
f
:
R
+
→
R
+
such that for any real numbers
a
,
b
,
c
,
d
>
0
a, b, c, d >0
a
,
b
,
c
,
d
>
0
satisfying
a
b
c
d
=
1
abcd=1
ab
c
d
=
1
,
(
f
(
a
)
+
f
(
b
)
)
(
f
(
c
)
+
f
(
d
)
)
=
(
a
+
b
)
(
c
+
d
)
(f(a)+f(b))(f(c)+f(d))=(a+b)(c+d)
(
f
(
a
)
+
f
(
b
))
(
f
(
c
)
+
f
(
d
))
=
(
a
+
b
)
(
c
+
d
)
holds true.(Swiss Mathematical Olympiad 2011, Final round, problem 4)
1
1
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Clinking glasses - Switzerland 2011
At a party, there are
2011
2011
2011
people with a glass of fruit juice each sitting around a circular table. Once a second, they clink glasses obeying the following two rules: (a) They do not clink glasses crosswise. (b) At each point of time, everyone can clink glasses with at most one other person. How many seconds pass at least until everyone clinked glasses with everybody else?(Swiss Mathematical Olympiad 2011, Final round, problem 1)