MathDB
Problems
Contests
National and Regional Contests
Singapore Contests
Singapore MO Open
2006 Singapore MO Open
2006 Singapore MO Open
Part of
Singapore MO Open
Subcontests
(5)
4
1
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smo open 2nd round 2006 q4
Let
n
n
n
be positive integer. Let
S
1
,
S
2
,
⋯
,
S
k
S_1,S_2,\cdots,S_k
S
1
,
S
2
,
⋯
,
S
k
be a collection of
2
n
2n
2
n
-element subsets of
{
1
,
2
,
3
,
4
,
.
.
.
,
4
n
−
1
,
4
n
}
\{1,2,3,4,...,4n-1,4n\}
{
1
,
2
,
3
,
4
,
...
,
4
n
−
1
,
4
n
}
so that
S
i
∩
S
j
S_{i}\cap S_{j}
S
i
∩
S
j
contains at most
n
n
n
elements for all
1
≤
i
<
j
≤
k
1\leq i<j\leq k
1
≤
i
<
j
≤
k
. Show that
k
≤
6
(
n
+
1
)
/
2
k\leq 6^{(n+1)/2}
k
≤
6
(
n
+
1
)
/2
1
1
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Angle in triangle
In the triangle
A
B
C
,
∠
A
=
π
3
,
D
,
M
ABC,\angle A=\frac{\pi}{3},D,M
A
BC
,
∠
A
=
3
π
,
D
,
M
are points on the line
A
C
AC
A
C
and
E
,
N
E,N
E
,
N
are points on the line
A
B
AB
A
B
such that
D
N
DN
D
N
and
E
M
EM
EM
are the perpendicular bisectors of
A
C
AC
A
C
and
A
B
AB
A
B
respectively. Let
L
L
L
be the midpoint of
M
N
MN
MN
. Prove that
∠
E
D
L
=
∠
E
L
D
\angle EDL=\angle ELD
∠
E
D
L
=
∠
E
L
D
5
1
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Divides
Let
a
,
b
,
n
a,b,n
a
,
b
,
n
be positive integers. Prove that
n
!
n!
n
!
divides
b
n
−
1
a
(
a
+
b
)
(
a
+
2
b
)
.
.
.
(
a
+
(
n
−
1
)
b
)
b^{n-1}a(a+b)(a+2b)...(a+(n-1)b)
b
n
−
1
a
(
a
+
b
)
(
a
+
2
b
)
...
(
a
+
(
n
−
1
)
b
)
3
1
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Sequence of Prime
Consider the sequence
p
1
,
p
2
,
.
.
.
p_{1},p_{2},...
p
1
,
p
2
,
...
of primes such that for each
i
≥
2
i\geq2
i
≥
2
, either
p
i
=
2
p
i
−
1
−
1
p_{i}=2p_{i-1}-1
p
i
=
2
p
i
−
1
−
1
or
p
i
=
2
p
i
−
1
+
1
p_{i}=2p_{i-1}+1
p
i
=
2
p
i
−
1
+
1
. Show that any such sequence has a finite number of terms.
2
1
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Representation of 1
Show that any representation of 1 as the sum of distinct reciprocals of numbers drawn from the arithmetic progression
{
2
,
5
,
8
,
11
,
.
.
.
}
\{2,5,8,11,...\}
{
2
,
5
,
8
,
11
,
...
}
such as given in the following example must have at least eight terms:
1
=
1
2
+
1
5
+
1
8
+
1
11
+
1
20
+
1
41
+
1
110
+
1
1640
1=\frac{1}{2}+\frac{1}{5}+\frac{1}{8}+\frac{1}{11}+\frac{1}{20}+\frac{1}{41}+\frac{1}{110}+\frac{1}{1640}
1
=
2
1
+
5
1
+
8
1
+
11
1
+
20
1
+
41
1
+
110
1
+
1640
1