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Problems
Contests
National and Regional Contests
Netherlands Contests
Dutch Mathematical Olympiad
2022 Dutch Mathematical Olympiad
2022 Dutch Mathematical Olympiad
Part of
Dutch Mathematical Olympiad
Subcontests
(5)
3
1
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a sequence of fractions construction
Given a positive integer
c
c
c
, we construct a sequence of fractions
a
1
,
a
2
,
a
3
,
.
.
.
a_1, a_2, a_3,...
a
1
,
a
2
,
a
3
,
...
as follows:
∙
\bullet
∙
a
1
=
c
c
+
1
a_1 =\frac{c}{c+1}
a
1
=
c
+
1
c
∙
\bullet
∙
to get
a
n
a_n
a
n
, we take
a
n
−
1
a_{n-1}
a
n
−
1
(in its most simplified form, with both the numerator and denominator chosen to be positive) and we add
2
2
2
to the numerator and
3
3
3
to the denominator. Then we simplify the result again as much as possible, with positive numerator and denominator. For example, if we take
c
=
20
c = 20
c
=
20
, then
a
1
=
20
21
a_1 =\frac{20}{21}
a
1
=
21
20
and
a
2
=
22
24
=
11
12
a_2 =\frac{22}{24} = \frac{11}{12}
a
2
=
24
22
=
12
11
. Then we find that
a
3
=
13
15
a_3 =\frac{13}{15}
a
3
=
15
13
(which is already simplified) and
a
4
=
15
18
=
5
6
a_4 =\frac{15}{18} =\frac{5}{6}
a
4
=
18
15
=
6
5
. (a) Let
c
=
10
c = 10
c
=
10
, hence
a
1
=
10
11
a_1 =\frac{10}{11}
a
1
=
11
10
. Determine the largest
n
n
n
for which a simplification is needed in the construction of
a
n
a_n
a
n
. (b) Let
c
=
99
c = 99
c
=
99
, hence
a
1
=
99
100
a_1 =\frac{99}{100}
a
1
=
100
99
. Determine whether a simplification is needed somewhere in the sequence. (c) Find two values of
c
c
c
for which in the first step of the construction of
a
5
a_5
a
5
(before simplification) the numerator and denominator are divisible by
5
5
5
.
5
1
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3 blocks with the letter A, 3 with B and 3 with C
Kira has
3
3
3
blocks with the letter
A
A
A
,
3
3
3
blocks with the letter
B
B
B
, and
3
3
3
blocks with the letter
C
C
C
. She puts these
9
9
9
blocks in a sequence. She wants to have as many distinct distances between blocks with the same letter as possible. For example, in the sequence
A
B
C
A
A
B
C
B
C
ABCAABCBC
A
BC
AA
BCBC
the blocks with the letter A have distances
1
,
3
1, 3
1
,
3
, and
4
4
4
between one another, the blocks with the letter
B
B
B
have distances
2
,
4
2, 4
2
,
4
, and
6
6
6
between one another, and the blocks with the letter
C
C
C
have distances
2
,
4
2, 4
2
,
4
, and
6
6
6
between one another. Altogether, we got distances of
1
,
2
,
3
,
4
1, 2, 3, 4
1
,
2
,
3
,
4
, and
6
6
6
; these are
5
5
5
distinct distances. What is the maximum number of distinct distances that can occur?
4
1
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<BAC = <BCE, 2|AD| = |ED|
In triangle
A
B
C
ABC
A
BC
, the point
D
D
D
lies on segment
A
B
AB
A
B
such that
C
D
CD
C
D
is the angle bisector of angle
∠
C
\angle C
∠
C
. The perpendicular bisector of segment
C
D
CD
C
D
intersects the line
A
B
AB
A
B
in
E
E
E
. Suppose that
∣
B
E
∣
=
4
|BE| = 4
∣
BE
∣
=
4
and
∣
A
B
∣
=
5
|AB| = 5
∣
A
B
∣
=
5
. (a) Prove that
∠
B
A
C
=
∠
B
C
E
\angle BAC = \angle BCE
∠
B
A
C
=
∠
BCE
. (b) Prove that
2
∣
A
D
∣
=
∣
E
D
∣
2|AD| = |ED|
2∣
A
D
∣
=
∣
E
D
∣
.[asy] unitsize(1 cm);pair A, B, C, D, E;A = (0,0); B = (2,0); C = (1.8,1.8); D = extension(C, incenter(A,B,C), A, B); E = extension((C + D)/2, (C + D)/2 + rotate(90)*(C - D), A, B);draw((E + (0.5,0))--A--C--B); draw(C--D); draw(interp((C + D)/2,E,-0.3)--interp((C + D)/2,E,1.2));dot("
A
A
A
", A, SW); dot("
B
B
B
", B, S); dot("
C
C
C
", C, N); dot("
D
D
D
", D, S); dot("
E
E
E
", E, S); [/asy]
2
1
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all integers that can occur as the average of a centenary set
A set consisting of at least two distinct positive integers is called centenary if its greatest element is
100
100
100
. We will consider the average of all numbers in a centenary set, which we will call the average of the set. For example, the average of the centenary set
{
1
,
2
,
20
,
100
}
\{1, 2, 20, 100\}
{
1
,
2
,
20
,
100
}
is
123
4
\frac{123}{4}
4
123
and the average of the centenary set
{
74
,
90
,
100
}
\{74, 90, 100\}
{
74
,
90
,
100
}
is
88
88
88
. Determine all integers that can occur as the average of a centenary set.
1
1
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largest primary divisor number
A positive integer n is called primary divisor if for every positive divisor
d
d
d
of
n
n
n
at least one of the numbers
d
−
1
d - 1
d
−
1
and
d
+
1
d + 1
d
+
1
is prime. For example,
8
8
8
is divisor primary, because its positive divisors
1
1
1
,
2
2
2
,
4
4
4
, and
8
8
8
each differ by
1
1
1
from a prime number (
2
2
2
,
3
3
3
,
5
5
5
, and
7
7
7
, respectively), while
9
9
9
is not divisor primary, because the divisor
9
9
9
does not differ by
1
1
1
from a prime number (both
8
8
8
and
10
10
10
are composite). Determine the largest primary divisor number.