MathDB
Problems
Contests
National and Regional Contests
Sweden Contests
Swedish Mathematical Competition
1991 Swedish Mathematical Competition
1991 Swedish Mathematical Competition
Part of
Swedish Mathematical Competition
Subcontests
(6)
3
1
Hide problems
x_0 = 0, x_{k+1} = [(n - \sum_0^k x_i)/2
The sequence
x
0
,
x
1
,
x
2
,
.
.
.
x_0, x_1, x_2, ...
x
0
,
x
1
,
x
2
,
...
is defined by
x
0
=
0
x_0 = 0
x
0
=
0
,
x
k
+
1
=
[
(
n
−
∑
0
k
x
i
)
/
2
]
x_{k+1} = [(n - \sum_0^k x_i)/2]
x
k
+
1
=
[(
n
−
∑
0
k
x
i
)
/2
]
. Show that
x
k
=
0
x_k = 0
x
k
=
0
for all sufficiently large
k
k
k
and that the sum of the non-zero terms
x
k
x_k
x
k
is
n
−
1
n-1
n
−
1
.
2
1
Hide problems
y - \sqrt{y} < x - 1/4 < y + \sqrt{y} if x - \sqrt{x} < y - 1/4 <= x + \sqrt{x}
x
,
y
x, y
x
,
y
are positive reals such that
x
−
x
≤
y
−
1
/
4
≤
x
+
x
x - \sqrt{x} \le y - 1/4 \le x + \sqrt{x}
x
−
x
≤
y
−
1/4
≤
x
+
x
. Show that
y
−
y
≤
x
−
1
/
4
≤
y
+
y
y - \sqrt{y} \le x - 1/4 \le y + \sqrt{y}
y
−
y
≤
x
−
1/4
≤
y
+
y
.
4
1
Hide problems
moves on a permutation of 1, 2, ..., 8
x
1
,
x
2
,
.
.
.
,
x
8
x_1, x_2, ... , x_8
x
1
,
x
2
,
...
,
x
8
is a permutation of
1
,
2
,
.
.
.
,
8
1, 2, ..., 8
1
,
2
,
...
,
8
. A move is to take
x
3
x_3
x
3
or
x
8
x_8
x
8
and place it at the start to from a new sequence. Show that by a sequence of moves we can always arrive at
1
,
2
,
.
.
.
,
8
1, 2, ..., 8
1
,
2
,
...
,
8
.
6
1
Hide problems
exists equilateral with area <=1/4, inscribed in each triangle
Given any triangle, show that we can always pick a point on each side so that the three points form an equilateral triangle with area at most one quarter of the original triangle.
5
1
Hide problems
odd positive integers n such that in binary n has more 1s than n^2
Show that there are infinitely many odd positive integers
n
n
n
such that in binary
n
n
n
has more
1
1
1
s than
n
2
n^2
n
2
.
1
1
Hide problems
1/m + 1/n - 1/(mn) = 2/5
Find all positive integers
m
,
n
m, n
m
,
n
such that
1
m
+
1
n
−
1
m
n
=
2
5
\frac{1}{m} + \frac{1}{n} - \frac{1}{mn} =\frac{2}{5}
m
1
+
n
1
−
mn
1
=
5
2
.