MathDB
Problems
Contests
National and Regional Contests
Turkey Contests
National Olympiad First Round
2006 National Olympiad First Round
2006 National Olympiad First Round
Part of
National Olympiad First Round
Subcontests
(36)
36
1
Hide problems
Turkey NMO 2006 1st Round - P36 (Combinatorics)
In an exam with
n
n
n
problems where
n
n
n
is a positive integer, each problem was answered by at least one student. Each student answered an even number of problems. Any two students answered an even number of problems in common. What is the number of values that
n
n
n
cannot take?
<
s
p
a
n
c
l
a
s
s
=
′
l
a
t
e
x
−
b
o
l
d
′
>
(
A
)
<
/
s
p
a
n
>
3
<
s
p
a
n
c
l
a
s
s
=
′
l
a
t
e
x
−
b
o
l
d
′
>
(
B
)
<
/
s
p
a
n
>
4
<
s
p
a
n
c
l
a
s
s
=
′
l
a
t
e
x
−
b
o
l
d
′
>
(
C
)
<
/
s
p
a
n
>
5
<
s
p
a
n
c
l
a
s
s
=
′
l
a
t
e
x
−
b
o
l
d
′
>
(
D
)
<
/
s
p
a
n
>
Infinitely many
<
s
p
a
n
c
l
a
s
s
=
′
l
a
t
e
x
−
b
o
l
d
′
>
(
E
)
<
/
s
p
a
n
>
None of above
<span class='latex-bold'>(A)</span>\ 3 \qquad<span class='latex-bold'>(B)</span>\ 4 \qquad<span class='latex-bold'>(C)</span>\ 5 \qquad<span class='latex-bold'>(D)</span>\ \text{Infinitely many} \qquad<span class='latex-bold'>(E)</span>\ \text{None of above}
<
s
p
an
c
l
a
ss
=
′
l
a
t
e
x
−
b
o
l
d
′
>
(
A
)
<
/
s
p
an
>
3
<
s
p
an
c
l
a
ss
=
′
l
a
t
e
x
−
b
o
l
d
′
>
(
B
)
<
/
s
p
an
>
4
<
s
p
an
c
l
a
ss
=
′
l
a
t
e
x
−
b
o
l
d
′
>
(
C
)
<
/
s
p
an
>
5
<
s
p
an
c
l
a
ss
=
′
l
a
t
e
x
−
b
o
l
d
′
>
(
D
)
<
/
s
p
an
>
Infinitely many
<
s
p
an
c
l
a
ss
=
′
l
a
t
e
x
−
b
o
l
d
′
>
(
E
)
<
/
s
p
an
>
None of above
35
1
Hide problems
Turkey NMO 2006 1st Round - P35 (Algebra)
P
(
x
)
=
a
x
2
+
b
x
+
c
P(x)=ax^2+bx+c
P
(
x
)
=
a
x
2
+
b
x
+
c
has exactly
1
1
1
different real root where
a
,
b
,
c
a,b,c
a
,
b
,
c
are real numbers. If
P
(
P
(
P
(
x
)
)
)
P(P(P(x)))
P
(
P
(
P
(
x
)))
has exactly
3
3
3
different real roots, what is the minimum possible value of
a
b
c
abc
ab
c
?
<
s
p
a
n
c
l
a
s
s
=
′
l
a
t
e
x
−
b
o
l
d
′
>
(
A
)
<
/
s
p
a
n
>
−
3
<
s
p
a
n
c
l
a
s
s
=
′
l
a
t
e
x
−
b
o
l
d
′
>
(
B
)
<
/
s
p
a
n
>
−
2
<
s
p
a
n
c
l
a
s
s
=
′
l
a
t
e
x
−
b
o
l
d
′
>
(
C
)
<
/
s
p
a
n
>
2
3
<
s
p
a
n
c
l
a
s
s
=
′
l
a
t
e
x
−
b
o
l
d
′
>
(
D
)
<
/
s
p
a
n
>
3
3
<
s
p
a
n
c
l
a
s
s
=
′
l
a
t
e
x
−
b
o
l
d
′
>
(
E
)
<
/
s
p
a
n
>
None of above
<span class='latex-bold'>(A)</span>\ -3 \qquad<span class='latex-bold'>(B)</span>\ -2 \qquad<span class='latex-bold'>(C)</span>\ 2\sqrt 3 \qquad<span class='latex-bold'>(D)</span>\ 3\sqrt 3 \qquad<span class='latex-bold'>(E)</span>\ \text{None of above}
<
s
p
an
c
l
a
ss
=
′
l
a
t
e
x
−
b
o
l
d
′
>
(
A
)
<
/
s
p
an
>
−
3
<
s
p
an
c
l
a
ss
=
′
l
a
t
e
x
−
b
o
l
d
′
>
(
B
)
<
/
s
p
an
>
−
2
<
s
p
an
c
l
a
ss
=
′
l
a
t
e
x
−
b
o
l
d
′
>
(
C
)
<
/
s
p
an
>
2
3
<
s
p
an
c
l
a
ss
=
′
l
a
t
e
x
−
b
o
l
d
′
>
(
D
)
<
/
s
p
an
>
3
3
<
s
p
an
c
l
a
ss
=
′
l
a
t
e
x
−
b
o
l
d
′
>
(
E
)
<
/
s
p
an
>
None of above
34
1
Hide problems
Turkey NMO 2006 1st Round - P34 (Number Theory)
How many positive integers less than
1000
1000
1000
are there such that they cannot be written as sum of
2
2
2
or more successive positive integers?
<
s
p
a
n
c
l
a
s
s
=
′
l
a
t
e
x
−
b
o
l
d
′
>
(
A
)
<
/
s
p
a
n
>
6
<
s
p
a
n
c
l
a
s
s
=
′
l
a
t
e
x
−
b
o
l
d
′
>
(
B
)
<
/
s
p
a
n
>
10
<
s
p
a
n
c
l
a
s
s
=
′
l
a
t
e
x
−
b
o
l
d
′
>
(
C
)
<
/
s
p
a
n
>
26
<
s
p
a
n
c
l
a
s
s
=
′
l
a
t
e
x
−
b
o
l
d
′
>
(
D
)
<
/
s
p
a
n
>
68
<
s
p
a
n
c
l
a
s
s
=
′
l
a
t
e
x
−
b
o
l
d
′
>
(
E
)
<
/
s
p
a
n
>
72
<span class='latex-bold'>(A)</span>\ 6 \qquad<span class='latex-bold'>(B)</span>\ 10 \qquad<span class='latex-bold'>(C)</span>\ 26 \qquad<span class='latex-bold'>(D)</span>\ 68 \qquad<span class='latex-bold'>(E)</span>\ 72
<
s
p
an
c
l
a
ss
=
′
l
a
t
e
x
−
b
o
l
d
′
>
(
A
)
<
/
s
p
an
>
6
<
s
p
an
c
l
a
ss
=
′
l
a
t
e
x
−
b
o
l
d
′
>
(
B
)
<
/
s
p
an
>
10
<
s
p
an
c
l
a
ss
=
′
l
a
t
e
x
−
b
o
l
d
′
>
(
C
)
<
/
s
p
an
>
26
<
s
p
an
c
l
a
ss
=
′
l
a
t
e
x
−
b
o
l
d
′
>
(
D
)
<
/
s
p
an
>
68
<
s
p
an
c
l
a
ss
=
′
l
a
t
e
x
−
b
o
l
d
′
>
(
E
)
<
/
s
p
an
>
72
33
1
Hide problems
Turkey NMO 2006 1st Round - P33 (Geometry)
Let
A
B
C
D
ABCD
A
BC
D
be a convex quadrileteral such that
m
(
A
B
D
^
)
=
4
0
∘
m(\widehat{ABD})=40^\circ
m
(
A
B
D
)
=
4
0
∘
,
m
(
D
B
C
^
)
=
7
0
∘
m(\widehat{DBC})=70^\circ
m
(
D
BC
)
=
7
0
∘
,
m
(
B
D
A
^
)
=
8
0
∘
m(\widehat{BDA})=80^\circ
m
(
B
D
A
)
=
8
0
∘
, and
m
(
B
D
C
^
)
=
5
0
∘
m(\widehat{BDC})=50^\circ
m
(
B
D
C
)
=
5
0
∘
. What is
m
(
C
A
D
^
)
m(\widehat{CAD})
m
(
C
A
D
)
?
<
s
p
a
n
c
l
a
s
s
=
′
l
a
t
e
x
−
b
o
l
d
′
>
(
A
)
<
/
s
p
a
n
>
2
5
∘
<
s
p
a
n
c
l
a
s
s
=
′
l
a
t
e
x
−
b
o
l
d
′
>
(
B
)
<
/
s
p
a
n
>
3
0
∘
<
s
p
a
n
c
l
a
s
s
=
′
l
a
t
e
x
−
b
o
l
d
′
>
(
C
)
<
/
s
p
a
n
>
3
5
∘
<
s
p
a
n
c
l
a
s
s
=
′
l
a
t
e
x
−
b
o
l
d
′
>
(
D
)
<
/
s
p
a
n
>
3
8
∘
<
s
p
a
n
c
l
a
s
s
=
′
l
a
t
e
x
−
b
o
l
d
′
>
(
E
)
<
/
s
p
a
n
>
4
0
∘
<span class='latex-bold'>(A)</span>\ 25^\circ \qquad<span class='latex-bold'>(B)</span>\ 30^\circ \qquad<span class='latex-bold'>(C)</span>\ 35^\circ \qquad<span class='latex-bold'>(D)</span>\ 38^\circ \qquad<span class='latex-bold'>(E)</span>\ 40^\circ
<
s
p
an
c
l
a
ss
=
′
l
a
t
e
x
−
b
o
l
d
′
>
(
A
)
<
/
s
p
an
>
2
5
∘
<
s
p
an
c
l
a
ss
=
′
l
a
t
e
x
−
b
o
l
d
′
>
(
B
)
<
/
s
p
an
>
3
0
∘
<
s
p
an
c
l
a
ss
=
′
l
a
t
e
x
−
b
o
l
d
′
>
(
C
)
<
/
s
p
an
>
3
5
∘
<
s
p
an
c
l
a
ss
=
′
l
a
t
e
x
−
b
o
l
d
′
>
(
D
)
<
/
s
p
an
>
3
8
∘
<
s
p
an
c
l
a
ss
=
′
l
a
t
e
x
−
b
o
l
d
′
>
(
E
)
<
/
s
p
an
>
4
0
∘
32
1
Hide problems
Turkey NMO 2006 1st Round - P32 (Combinatorics)
What is the greatest integer
k
k
k
which makes the statement "When we take any
6
6
6
subsets with
5
5
5
elements of the set
{
1
,
2
,
…
,
9
}
\{1,2,\dots, 9\}
{
1
,
2
,
…
,
9
}
, there exist
k
k
k
of them having at least one common element." true?
<
s
p
a
n
c
l
a
s
s
=
′
l
a
t
e
x
−
b
o
l
d
′
>
(
A
)
<
/
s
p
a
n
>
1
<
s
p
a
n
c
l
a
s
s
=
′
l
a
t
e
x
−
b
o
l
d
′
>
(
B
)
<
/
s
p
a
n
>
2
<
s
p
a
n
c
l
a
s
s
=
′
l
a
t
e
x
−
b
o
l
d
′
>
(
C
)
<
/
s
p
a
n
>
3
<
s
p
a
n
c
l
a
s
s
=
′
l
a
t
e
x
−
b
o
l
d
′
>
(
D
)
<
/
s
p
a
n
>
4
<
s
p
a
n
c
l
a
s
s
=
′
l
a
t
e
x
−
b
o
l
d
′
>
(
E
)
<
/
s
p
a
n
>
5
<span class='latex-bold'>(A)</span>\ 1 \qquad<span class='latex-bold'>(B)</span>\ 2 \qquad<span class='latex-bold'>(C)</span>\ 3 \qquad<span class='latex-bold'>(D)</span>\ 4 \qquad<span class='latex-bold'>(E)</span>\ 5
<
s
p
an
c
l
a
ss
=
′
l
a
t
e
x
−
b
o
l
d
′
>
(
A
)
<
/
s
p
an
>
1
<
s
p
an
c
l
a
ss
=
′
l
a
t
e
x
−
b
o
l
d
′
>
(
B
)
<
/
s
p
an
>
2
<
s
p
an
c
l
a
ss
=
′
l
a
t
e
x
−
b
o
l
d
′
>
(
C
)
<
/
s
p
an
>
3
<
s
p
an
c
l
a
ss
=
′
l
a
t
e
x
−
b
o
l
d
′
>
(
D
)
<
/
s
p
an
>
4
<
s
p
an
c
l
a
ss
=
′
l
a
t
e
x
−
b
o
l
d
′
>
(
E
)
<
/
s
p
an
>
5
31
1
Hide problems
Turkey NMO 2006 1st Round - P31 (Algebra)
Let
P
(
x
)
=
x
3
+
a
x
2
+
b
x
+
c
P(x)=x^3+ax^2+bx+c
P
(
x
)
=
x
3
+
a
x
2
+
b
x
+
c
where
a
,
b
,
c
a,b,c
a
,
b
,
c
are positive real numbers. If
P
(
1
)
≥
2
P(1)\geq 2
P
(
1
)
≥
2
and
P
(
3
)
≤
31
P(3)\leq 31
P
(
3
)
≤
31
, how many of integers can
P
(
4
)
P(4)
P
(
4
)
take?
<
s
p
a
n
c
l
a
s
s
=
′
l
a
t
e
x
−
b
o
l
d
′
>
(
A
)
<
/
s
p
a
n
>
3
<
s
p
a
n
c
l
a
s
s
=
′
l
a
t
e
x
−
b
o
l
d
′
>
(
B
)
<
/
s
p
a
n
>
4
<
s
p
a
n
c
l
a
s
s
=
′
l
a
t
e
x
−
b
o
l
d
′
>
(
C
)
<
/
s
p
a
n
>
5
<
s
p
a
n
c
l
a
s
s
=
′
l
a
t
e
x
−
b
o
l
d
′
>
(
D
)
<
/
s
p
a
n
>
6
<
s
p
a
n
c
l
a
s
s
=
′
l
a
t
e
x
−
b
o
l
d
′
>
(
E
)
<
/
s
p
a
n
>
None of above
<span class='latex-bold'>(A)</span>\ 3 \qquad<span class='latex-bold'>(B)</span>\ 4 \qquad<span class='latex-bold'>(C)</span>\ 5 \qquad<span class='latex-bold'>(D)</span>\ 6 \qquad<span class='latex-bold'>(E)</span>\ \text{None of above}
<
s
p
an
c
l
a
ss
=
′
l
a
t
e
x
−
b
o
l
d
′
>
(
A
)
<
/
s
p
an
>
3
<
s
p
an
c
l
a
ss
=
′
l
a
t
e
x
−
b
o
l
d
′
>
(
B
)
<
/
s
p
an
>
4
<
s
p
an
c
l
a
ss
=
′
l
a
t
e
x
−
b
o
l
d
′
>
(
C
)
<
/
s
p
an
>
5
<
s
p
an
c
l
a
ss
=
′
l
a
t
e
x
−
b
o
l
d
′
>
(
D
)
<
/
s
p
an
>
6
<
s
p
an
c
l
a
ss
=
′
l
a
t
e
x
−
b
o
l
d
′
>
(
E
)
<
/
s
p
an
>
None of above
30
1
Hide problems
Turkey NMO 2006 1st Round - P30 (Number Theory)
How many integer triples
(
x
,
y
,
z
)
(x,y,z)
(
x
,
y
,
z
)
are there such that
x
−
y
z
2
≡
1
(
m
o
d
13
)
x
z
+
y
≡
4
(
m
o
d
13
)
\begin{array}{rcl} x - yz^2&\equiv & 1 \pmod {13} \\ xz+y&\equiv& 4 \pmod {13} \end{array}
x
−
y
z
2
x
z
+
y
≡
≡
1
(
mod
13
)
4
(
mod
13
)
where
0
≤
x
<
13
0\leq x < 13
0
≤
x
<
13
,
0
≤
y
<
13
0\leq y <13
0
≤
y
<
13
, and
0
≤
z
<
13
0\leq z< 13
0
≤
z
<
13
?
<
s
p
a
n
c
l
a
s
s
=
′
l
a
t
e
x
−
b
o
l
d
′
>
(
A
)
<
/
s
p
a
n
>
10
<
s
p
a
n
c
l
a
s
s
=
′
l
a
t
e
x
−
b
o
l
d
′
>
(
B
)
<
/
s
p
a
n
>
23
<
s
p
a
n
c
l
a
s
s
=
′
l
a
t
e
x
−
b
o
l
d
′
>
(
C
)
<
/
s
p
a
n
>
36
<
s
p
a
n
c
l
a
s
s
=
′
l
a
t
e
x
−
b
o
l
d
′
>
(
D
)
<
/
s
p
a
n
>
49
<
s
p
a
n
c
l
a
s
s
=
′
l
a
t
e
x
−
b
o
l
d
′
>
(
E
)
<
/
s
p
a
n
>
None of above
<span class='latex-bold'>(A)</span>\ 10 \qquad<span class='latex-bold'>(B)</span>\ 23 \qquad<span class='latex-bold'>(C)</span>\ 36 \qquad<span class='latex-bold'>(D)</span>\ 49 \qquad<span class='latex-bold'>(E)</span>\ \text{None of above}
<
s
p
an
c
l
a
ss
=
′
l
a
t
e
x
−
b
o
l
d
′
>
(
A
)
<
/
s
p
an
>
10
<
s
p
an
c
l
a
ss
=
′
l
a
t
e
x
−
b
o
l
d
′
>
(
B
)
<
/
s
p
an
>
23
<
s
p
an
c
l
a
ss
=
′
l
a
t
e
x
−
b
o
l
d
′
>
(
C
)
<
/
s
p
an
>
36
<
s
p
an
c
l
a
ss
=
′
l
a
t
e
x
−
b
o
l
d
′
>
(
D
)
<
/
s
p
an
>
49
<
s
p
an
c
l
a
ss
=
′
l
a
t
e
x
−
b
o
l
d
′
>
(
E
)
<
/
s
p
an
>
None of above
29
1
Hide problems
Turkey NMO 2006 1st Round - P29 (Geometry)
Let
I
I
I
be the center of incircle of
△
A
B
C
\triangle ABC
△
A
BC
, and
J
J
J
be the center of excircle tangent to
[
B
C
]
[BC]
[
BC
]
. If
m
(
B
^
)
=
4
5
∘
m(\widehat B) = 45^\circ
m
(
B
)
=
4
5
∘
,
m
(
A
^
)
=
12
0
∘
m(\widehat A) = 120^\circ
m
(
A
)
=
12
0
∘
, and
∣
I
J
∣
=
3
|IJ|=\sqrt 3
∣
I
J
∣
=
3
, then what is
∣
B
C
∣
|BC|
∣
BC
∣
?
<
s
p
a
n
c
l
a
s
s
=
′
l
a
t
e
x
−
b
o
l
d
′
>
(
A
)
<
/
s
p
a
n
>
3
2
<
s
p
a
n
c
l
a
s
s
=
′
l
a
t
e
x
−
b
o
l
d
′
>
(
B
)
<
/
s
p
a
n
>
3
2
<
s
p
a
n
c
l
a
s
s
=
′
l
a
t
e
x
−
b
o
l
d
′
>
(
C
)
<
/
s
p
a
n
>
3
4
<
s
p
a
n
c
l
a
s
s
=
′
l
a
t
e
x
−
b
o
l
d
′
>
(
D
)
<
/
s
p
a
n
>
6
2
<
s
p
a
n
c
l
a
s
s
=
′
l
a
t
e
x
−
b
o
l
d
′
>
(
E
)
<
/
s
p
a
n
>
3
−
1
<span class='latex-bold'>(A)</span>\ \frac 32 \qquad<span class='latex-bold'>(B)</span>\ \frac {\sqrt 3}2 \qquad<span class='latex-bold'>(C)</span>\ \frac 34 \qquad<span class='latex-bold'>(D)</span>\ \frac {\sqrt 6}2 \qquad<span class='latex-bold'>(E)</span>\ \sqrt3 - 1
<
s
p
an
c
l
a
ss
=
′
l
a
t
e
x
−
b
o
l
d
′
>
(
A
)
<
/
s
p
an
>
2
3
<
s
p
an
c
l
a
ss
=
′
l
a
t
e
x
−
b
o
l
d
′
>
(
B
)
<
/
s
p
an
>
2
3
<
s
p
an
c
l
a
ss
=
′
l
a
t
e
x
−
b
o
l
d
′
>
(
C
)
<
/
s
p
an
>
4
3
<
s
p
an
c
l
a
ss
=
′
l
a
t
e
x
−
b
o
l
d
′
>
(
D
)
<
/
s
p
an
>
2
6
<
s
p
an
c
l
a
ss
=
′
l
a
t
e
x
−
b
o
l
d
′
>
(
E
)
<
/
s
p
an
>
3
−
1
28
1
Hide problems
Turkey NMO 2006 1st Round - P28 (Combinatorics)
Ali who has
10
10
10
candies eats at least one candy a day. In how many different ways can he eat all candies (according to distribution among days)?
<
s
p
a
n
c
l
a
s
s
=
′
l
a
t
e
x
−
b
o
l
d
′
>
(
A
)
<
/
s
p
a
n
>
64
<
s
p
a
n
c
l
a
s
s
=
′
l
a
t
e
x
−
b
o
l
d
′
>
(
B
)
<
/
s
p
a
n
>
126
<
s
p
a
n
c
l
a
s
s
=
′
l
a
t
e
x
−
b
o
l
d
′
>
(
C
)
<
/
s
p
a
n
>
243
<
s
p
a
n
c
l
a
s
s
=
′
l
a
t
e
x
−
b
o
l
d
′
>
(
D
)
<
/
s
p
a
n
>
512
<
s
p
a
n
c
l
a
s
s
=
′
l
a
t
e
x
−
b
o
l
d
′
>
(
E
)
<
/
s
p
a
n
>
1025
<span class='latex-bold'>(A)</span>\ 64 \qquad<span class='latex-bold'>(B)</span>\ 126 \qquad<span class='latex-bold'>(C)</span>\ 243 \qquad<span class='latex-bold'>(D)</span>\ 512 \qquad<span class='latex-bold'>(E)</span>\ 1025
<
s
p
an
c
l
a
ss
=
′
l
a
t
e
x
−
b
o
l
d
′
>
(
A
)
<
/
s
p
an
>
64
<
s
p
an
c
l
a
ss
=
′
l
a
t
e
x
−
b
o
l
d
′
>
(
B
)
<
/
s
p
an
>
126
<
s
p
an
c
l
a
ss
=
′
l
a
t
e
x
−
b
o
l
d
′
>
(
C
)
<
/
s
p
an
>
243
<
s
p
an
c
l
a
ss
=
′
l
a
t
e
x
−
b
o
l
d
′
>
(
D
)
<
/
s
p
an
>
512
<
s
p
an
c
l
a
ss
=
′
l
a
t
e
x
−
b
o
l
d
′
>
(
E
)
<
/
s
p
an
>
1025
27
1
Hide problems
Turkey NMO 2006 1st Round - P27 (Algebra)
If
x
,
y
,
z
x,y,z
x
,
y
,
z
are positive real numbers such that
x
y
+
y
z
+
z
x
=
5
xy+yz+zx=5
x
y
+
yz
+
z
x
=
5
,
x
2
+
y
2
+
z
2
−
x
y
z
x^2+y^2+z^2-xyz
x
2
+
y
2
+
z
2
−
x
yz
cannot be
‾
\underline{\hspace{1cm}}
.
<
s
p
a
n
c
l
a
s
s
=
′
l
a
t
e
x
−
b
o
l
d
′
>
(
A
)
<
/
s
p
a
n
>
3
<
s
p
a
n
c
l
a
s
s
=
′
l
a
t
e
x
−
b
o
l
d
′
>
(
B
)
<
/
s
p
a
n
>
4
<
s
p
a
n
c
l
a
s
s
=
′
l
a
t
e
x
−
b
o
l
d
′
>
(
C
)
<
/
s
p
a
n
>
5
<
s
p
a
n
c
l
a
s
s
=
′
l
a
t
e
x
−
b
o
l
d
′
>
(
D
)
<
/
s
p
a
n
>
3
3
<
s
p
a
n
c
l
a
s
s
=
′
l
a
t
e
x
−
b
o
l
d
′
>
(
E
)
<
/
s
p
a
n
>
None of above
<span class='latex-bold'>(A)</span>\ 3 \qquad<span class='latex-bold'>(B)</span>\ 4 \qquad<span class='latex-bold'>(C)</span>\ 5 \qquad<span class='latex-bold'>(D)</span>\ 3\sqrt 3 \qquad<span class='latex-bold'>(E)</span>\ \text{None of above}
<
s
p
an
c
l
a
ss
=
′
l
a
t
e
x
−
b
o
l
d
′
>
(
A
)
<
/
s
p
an
>
3
<
s
p
an
c
l
a
ss
=
′
l
a
t
e
x
−
b
o
l
d
′
>
(
B
)
<
/
s
p
an
>
4
<
s
p
an
c
l
a
ss
=
′
l
a
t
e
x
−
b
o
l
d
′
>
(
C
)
<
/
s
p
an
>
5
<
s
p
an
c
l
a
ss
=
′
l
a
t
e
x
−
b
o
l
d
′
>
(
D
)
<
/
s
p
an
>
3
3
<
s
p
an
c
l
a
ss
=
′
l
a
t
e
x
−
b
o
l
d
′
>
(
E
)
<
/
s
p
an
>
None of above
26
1
Hide problems
Turkey NMO 2006 1st Round - P26 (Number Theory)
For how many primes
p
p
p
, there exists an integr
m
m
m
such that
m
3
+
3
m
−
2
≡
0
(
m
o
d
p
)
m^3+3m-2 \equiv 0 \pmod p
m
3
+
3
m
−
2
≡
0
(
mod
p
)
and
m
2
+
4
m
+
5
≡
0
(
m
o
d
p
)
m^2+4m+5\equiv 0 \pmod p
m
2
+
4
m
+
5
≡
0
(
mod
p
)
?
<
s
p
a
n
c
l
a
s
s
=
′
l
a
t
e
x
−
b
o
l
d
′
>
(
A
)
<
/
s
p
a
n
>
1
<
s
p
a
n
c
l
a
s
s
=
′
l
a
t
e
x
−
b
o
l
d
′
>
(
B
)
<
/
s
p
a
n
>
2
<
s
p
a
n
c
l
a
s
s
=
′
l
a
t
e
x
−
b
o
l
d
′
>
(
C
)
<
/
s
p
a
n
>
3
<
s
p
a
n
c
l
a
s
s
=
′
l
a
t
e
x
−
b
o
l
d
′
>
(
D
)
<
/
s
p
a
n
>
4
<
s
p
a
n
c
l
a
s
s
=
′
l
a
t
e
x
−
b
o
l
d
′
>
(
E
)
<
/
s
p
a
n
>
Infinitely many
<span class='latex-bold'>(A)</span>\ 1 \qquad<span class='latex-bold'>(B)</span>\ 2 \qquad<span class='latex-bold'>(C)</span>\ 3 \qquad<span class='latex-bold'>(D)</span>\ 4 \qquad<span class='latex-bold'>(E)</span>\ \text{Infinitely many}
<
s
p
an
c
l
a
ss
=
′
l
a
t
e
x
−
b
o
l
d
′
>
(
A
)
<
/
s
p
an
>
1
<
s
p
an
c
l
a
ss
=
′
l
a
t
e
x
−
b
o
l
d
′
>
(
B
)
<
/
s
p
an
>
2
<
s
p
an
c
l
a
ss
=
′
l
a
t
e
x
−
b
o
l
d
′
>
(
C
)
<
/
s
p
an
>
3
<
s
p
an
c
l
a
ss
=
′
l
a
t
e
x
−
b
o
l
d
′
>
(
D
)
<
/
s
p
an
>
4
<
s
p
an
c
l
a
ss
=
′
l
a
t
e
x
−
b
o
l
d
′
>
(
E
)
<
/
s
p
an
>
Infinitely many
25
1
Hide problems
Turkey NMO 2006 1st Round - P25 (Geometry)
Let
E
E
E
be the midpoint of the side
[
B
C
]
[BC]
[
BC
]
of
△
A
B
C
\triangle ABC
△
A
BC
with
∣
A
B
∣
=
7
|AB|=7
∣
A
B
∣
=
7
,
∣
B
C
∣
=
6
|BC|=6
∣
BC
∣
=
6
, and
∣
A
C
∣
=
5
|AC|=5
∣
A
C
∣
=
5
. The line, which passes through
E
E
E
and is perpendicular to the angle bisector of
∠
A
\angle A
∠
A
, intersects
A
B
AB
A
B
at
D
D
D
. What is
∣
A
D
∣
|AD|
∣
A
D
∣
?
<
s
p
a
n
c
l
a
s
s
=
′
l
a
t
e
x
−
b
o
l
d
′
>
(
A
)
<
/
s
p
a
n
>
5
<
s
p
a
n
c
l
a
s
s
=
′
l
a
t
e
x
−
b
o
l
d
′
>
(
B
)
<
/
s
p
a
n
>
6
<
s
p
a
n
c
l
a
s
s
=
′
l
a
t
e
x
−
b
o
l
d
′
>
(
C
)
<
/
s
p
a
n
>
9
2
<
s
p
a
n
c
l
a
s
s
=
′
l
a
t
e
x
−
b
o
l
d
′
>
(
D
)
<
/
s
p
a
n
>
3
2
<
s
p
a
n
c
l
a
s
s
=
′
l
a
t
e
x
−
b
o
l
d
′
>
(
E
)
<
/
s
p
a
n
>
None of above
<span class='latex-bold'>(A)</span>\ 5 \qquad<span class='latex-bold'>(B)</span>\ 6 \qquad<span class='latex-bold'>(C)</span>\ \frac 92 \qquad<span class='latex-bold'>(D)</span>\ 3\sqrt 2 \qquad<span class='latex-bold'>(E)</span>\ \text{None of above}
<
s
p
an
c
l
a
ss
=
′
l
a
t
e
x
−
b
o
l
d
′
>
(
A
)
<
/
s
p
an
>
5
<
s
p
an
c
l
a
ss
=
′
l
a
t
e
x
−
b
o
l
d
′
>
(
B
)
<
/
s
p
an
>
6
<
s
p
an
c
l
a
ss
=
′
l
a
t
e
x
−
b
o
l
d
′
>
(
C
)
<
/
s
p
an
>
2
9
<
s
p
an
c
l
a
ss
=
′
l
a
t
e
x
−
b
o
l
d
′
>
(
D
)
<
/
s
p
an
>
3
2
<
s
p
an
c
l
a
ss
=
′
l
a
t
e
x
−
b
o
l
d
′
>
(
E
)
<
/
s
p
an
>
None of above
24
1
Hide problems
Turkey NMO 2006 1st Round - P24 (Combinatorics)
In a handball tournament with
n
n
n
teams, each team played against other teams exactly once. In each game, the winner got
2
2
2
points, the loser got
0
0
0
point, and each team got
1
1
1
point if there was a tie. After the tournament ended, each team had different score than the others, and the last team defeated the first three teams. What is the least possible value of
n
n
n
?
<
s
p
a
n
c
l
a
s
s
=
′
l
a
t
e
x
−
b
o
l
d
′
>
(
A
)
<
/
s
p
a
n
>
8
<
s
p
a
n
c
l
a
s
s
=
′
l
a
t
e
x
−
b
o
l
d
′
>
(
B
)
<
/
s
p
a
n
>
9
<
s
p
a
n
c
l
a
s
s
=
′
l
a
t
e
x
−
b
o
l
d
′
>
(
C
)
<
/
s
p
a
n
>
10
<
s
p
a
n
c
l
a
s
s
=
′
l
a
t
e
x
−
b
o
l
d
′
>
(
D
)
<
/
s
p
a
n
>
12
<
s
p
a
n
c
l
a
s
s
=
′
l
a
t
e
x
−
b
o
l
d
′
>
(
E
)
<
/
s
p
a
n
>
None of above
<span class='latex-bold'>(A)</span>\ 8 \qquad<span class='latex-bold'>(B)</span>\ 9 \qquad<span class='latex-bold'>(C)</span>\ 10 \qquad<span class='latex-bold'>(D)</span>\ 12 \qquad<span class='latex-bold'>(E)</span>\ \text{None of above}
<
s
p
an
c
l
a
ss
=
′
l
a
t
e
x
−
b
o
l
d
′
>
(
A
)
<
/
s
p
an
>
8
<
s
p
an
c
l
a
ss
=
′
l
a
t
e
x
−
b
o
l
d
′
>
(
B
)
<
/
s
p
an
>
9
<
s
p
an
c
l
a
ss
=
′
l
a
t
e
x
−
b
o
l
d
′
>
(
C
)
<
/
s
p
an
>
10
<
s
p
an
c
l
a
ss
=
′
l
a
t
e
x
−
b
o
l
d
′
>
(
D
)
<
/
s
p
an
>
12
<
s
p
an
c
l
a
ss
=
′
l
a
t
e
x
−
b
o
l
d
′
>
(
E
)
<
/
s
p
an
>
None of above
23
1
Hide problems
Turkey NMO 2006 1st Round - P23 (Algebra)
What is the greatest integer not exceeding the number
(
1
+
2
+
3
+
4
2
+
3
+
6
+
8
+
4
)
10
\left( 1 + \frac{\sqrt 2 + \sqrt 3 + \sqrt 4}{\sqrt 2 + \sqrt 3 + \sqrt 6 + \sqrt 8 + 4}\right)^{10}
(
1
+
2
+
3
+
6
+
8
+
4
2
+
3
+
4
)
10
?
<
s
p
a
n
c
l
a
s
s
=
′
l
a
t
e
x
−
b
o
l
d
′
>
(
A
)
<
/
s
p
a
n
>
2
<
s
p
a
n
c
l
a
s
s
=
′
l
a
t
e
x
−
b
o
l
d
′
>
(
B
)
<
/
s
p
a
n
>
10
<
s
p
a
n
c
l
a
s
s
=
′
l
a
t
e
x
−
b
o
l
d
′
>
(
C
)
<
/
s
p
a
n
>
21
<
s
p
a
n
c
l
a
s
s
=
′
l
a
t
e
x
−
b
o
l
d
′
>
(
D
)
<
/
s
p
a
n
>
32
<
s
p
a
n
c
l
a
s
s
=
′
l
a
t
e
x
−
b
o
l
d
′
>
(
E
)
<
/
s
p
a
n
>
36
<span class='latex-bold'>(A)</span>\ 2 \qquad<span class='latex-bold'>(B)</span>\ 10 \qquad<span class='latex-bold'>(C)</span>\ 21 \qquad<span class='latex-bold'>(D)</span>\ 32 \qquad<span class='latex-bold'>(E)</span>\ 36
<
s
p
an
c
l
a
ss
=
′
l
a
t
e
x
−
b
o
l
d
′
>
(
A
)
<
/
s
p
an
>
2
<
s
p
an
c
l
a
ss
=
′
l
a
t
e
x
−
b
o
l
d
′
>
(
B
)
<
/
s
p
an
>
10
<
s
p
an
c
l
a
ss
=
′
l
a
t
e
x
−
b
o
l
d
′
>
(
C
)
<
/
s
p
an
>
21
<
s
p
an
c
l
a
ss
=
′
l
a
t
e
x
−
b
o
l
d
′
>
(
D
)
<
/
s
p
an
>
32
<
s
p
an
c
l
a
ss
=
′
l
a
t
e
x
−
b
o
l
d
′
>
(
E
)
<
/
s
p
an
>
36
22
1
Hide problems
Turkey NMO 2006 1st Round - P22 (Number Theory)
How many integer pairs
(
x
,
y
)
(x,y)
(
x
,
y
)
are there such that 0\leq x < 165, 0\leq y < 165 \text{ and } y^2\equiv x^3+x \pmod {165}?
<
s
p
a
n
c
l
a
s
s
=
′
l
a
t
e
x
−
b
o
l
d
′
>
(
A
)
<
/
s
p
a
n
>
80
<
s
p
a
n
c
l
a
s
s
=
′
l
a
t
e
x
−
b
o
l
d
′
>
(
B
)
<
/
s
p
a
n
>
99
<
s
p
a
n
c
l
a
s
s
=
′
l
a
t
e
x
−
b
o
l
d
′
>
(
C
)
<
/
s
p
a
n
>
120
<
s
p
a
n
c
l
a
s
s
=
′
l
a
t
e
x
−
b
o
l
d
′
>
(
D
)
<
/
s
p
a
n
>
315
<
s
p
a
n
c
l
a
s
s
=
′
l
a
t
e
x
−
b
o
l
d
′
>
(
E
)
<
/
s
p
a
n
>
None of above
<span class='latex-bold'>(A)</span>\ 80 \qquad<span class='latex-bold'>(B)</span>\ 99 \qquad<span class='latex-bold'>(C)</span>\ 120 \qquad<span class='latex-bold'>(D)</span>\ 315 \qquad<span class='latex-bold'>(E)</span>\ \text{None of above}
<
s
p
an
c
l
a
ss
=
′
l
a
t
e
x
−
b
o
l
d
′
>
(
A
)
<
/
s
p
an
>
80
<
s
p
an
c
l
a
ss
=
′
l
a
t
e
x
−
b
o
l
d
′
>
(
B
)
<
/
s
p
an
>
99
<
s
p
an
c
l
a
ss
=
′
l
a
t
e
x
−
b
o
l
d
′
>
(
C
)
<
/
s
p
an
>
120
<
s
p
an
c
l
a
ss
=
′
l
a
t
e
x
−
b
o
l
d
′
>
(
D
)
<
/
s
p
an
>
315
<
s
p
an
c
l
a
ss
=
′
l
a
t
e
x
−
b
o
l
d
′
>
(
E
)
<
/
s
p
an
>
None of above
21
1
Hide problems
Turkey NMO 2006 1st Round - P21 (Geometry)
Let
A
B
C
ABC
A
BC
be a triangle with
m
(
A
^
)
=
7
0
∘
m(\widehat A) = 70^\circ
m
(
A
)
=
7
0
∘
and the incenter
I
I
I
. If
∣
B
C
∣
=
∣
A
C
∣
+
∣
A
I
∣
|BC|=|AC|+|AI|
∣
BC
∣
=
∣
A
C
∣
+
∣
A
I
∣
, then what is
m
(
B
^
)
m(\widehat B)
m
(
B
)
?
<
s
p
a
n
c
l
a
s
s
=
′
l
a
t
e
x
−
b
o
l
d
′
>
(
A
)
<
/
s
p
a
n
>
3
5
∘
<
s
p
a
n
c
l
a
s
s
=
′
l
a
t
e
x
−
b
o
l
d
′
>
(
B
)
<
/
s
p
a
n
>
3
6
∘
<
s
p
a
n
c
l
a
s
s
=
′
l
a
t
e
x
−
b
o
l
d
′
>
(
C
)
<
/
s
p
a
n
>
4
2
∘
<
s
p
a
n
c
l
a
s
s
=
′
l
a
t
e
x
−
b
o
l
d
′
>
(
D
)
<
/
s
p
a
n
>
4
5
∘
<
s
p
a
n
c
l
a
s
s
=
′
l
a
t
e
x
−
b
o
l
d
′
>
(
E
)
<
/
s
p
a
n
>
None of above
<span class='latex-bold'>(A)</span>\ 35^\circ \qquad<span class='latex-bold'>(B)</span>\ 36^\circ \qquad<span class='latex-bold'>(C)</span>\ 42^\circ \qquad<span class='latex-bold'>(D)</span>\ 45^\circ \qquad<span class='latex-bold'>(E)</span>\ \text{None of above}
<
s
p
an
c
l
a
ss
=
′
l
a
t
e
x
−
b
o
l
d
′
>
(
A
)
<
/
s
p
an
>
3
5
∘
<
s
p
an
c
l
a
ss
=
′
l
a
t
e
x
−
b
o
l
d
′
>
(
B
)
<
/
s
p
an
>
3
6
∘
<
s
p
an
c
l
a
ss
=
′
l
a
t
e
x
−
b
o
l
d
′
>
(
C
)
<
/
s
p
an
>
4
2
∘
<
s
p
an
c
l
a
ss
=
′
l
a
t
e
x
−
b
o
l
d
′
>
(
D
)
<
/
s
p
an
>
4
5
∘
<
s
p
an
c
l
a
ss
=
′
l
a
t
e
x
−
b
o
l
d
′
>
(
E
)
<
/
s
p
an
>
None of above
20
1
Hide problems
Turkey NMO 2006 1st Round - P20 (Combinatorics)
The integer
k
k
k
is a good number, if we can divide a square into
k
k
k
squares. How many good numbers not greater than
2006
2006
2006
are there?
<
s
p
a
n
c
l
a
s
s
=
′
l
a
t
e
x
−
b
o
l
d
′
>
(
A
)
<
/
s
p
a
n
>
1003
<
s
p
a
n
c
l
a
s
s
=
′
l
a
t
e
x
−
b
o
l
d
′
>
(
B
)
<
/
s
p
a
n
>
1026
<
s
p
a
n
c
l
a
s
s
=
′
l
a
t
e
x
−
b
o
l
d
′
>
(
C
)
<
/
s
p
a
n
>
2000
<
s
p
a
n
c
l
a
s
s
=
′
l
a
t
e
x
−
b
o
l
d
′
>
(
D
)
<
/
s
p
a
n
>
2003
<
s
p
a
n
c
l
a
s
s
=
′
l
a
t
e
x
−
b
o
l
d
′
>
(
E
)
<
/
s
p
a
n
>
2004
<span class='latex-bold'>(A)</span>\ 1003 \qquad<span class='latex-bold'>(B)</span>\ 1026 \qquad<span class='latex-bold'>(C)</span>\ 2000 \qquad<span class='latex-bold'>(D)</span>\ 2003 \qquad<span class='latex-bold'>(E)</span>\ 2004
<
s
p
an
c
l
a
ss
=
′
l
a
t
e
x
−
b
o
l
d
′
>
(
A
)
<
/
s
p
an
>
1003
<
s
p
an
c
l
a
ss
=
′
l
a
t
e
x
−
b
o
l
d
′
>
(
B
)
<
/
s
p
an
>
1026
<
s
p
an
c
l
a
ss
=
′
l
a
t
e
x
−
b
o
l
d
′
>
(
C
)
<
/
s
p
an
>
2000
<
s
p
an
c
l
a
ss
=
′
l
a
t
e
x
−
b
o
l
d
′
>
(
D
)
<
/
s
p
an
>
2003
<
s
p
an
c
l
a
ss
=
′
l
a
t
e
x
−
b
o
l
d
′
>
(
E
)
<
/
s
p
an
>
2004
19
1
Hide problems
Turkey NMO 2006 1st Round - P19 (Algebra)
How many real triples
(
x
,
y
,
z
)
(x,y,z)
(
x
,
y
,
z
)
are there such that
x
4
+
y
4
+
z
4
+
1
=
4
x
y
z
x^4+y^4+z^4+1 = 4xyz
x
4
+
y
4
+
z
4
+
1
=
4
x
yz
?
<
s
p
a
n
c
l
a
s
s
=
′
l
a
t
e
x
−
b
o
l
d
′
>
(
A
)
<
/
s
p
a
n
>
0
<
s
p
a
n
c
l
a
s
s
=
′
l
a
t
e
x
−
b
o
l
d
′
>
(
B
)
<
/
s
p
a
n
>
4
<
s
p
a
n
c
l
a
s
s
=
′
l
a
t
e
x
−
b
o
l
d
′
>
(
C
)
<
/
s
p
a
n
>
6
<
s
p
a
n
c
l
a
s
s
=
′
l
a
t
e
x
−
b
o
l
d
′
>
(
D
)
<
/
s
p
a
n
>
10
<
s
p
a
n
c
l
a
s
s
=
′
l
a
t
e
x
−
b
o
l
d
′
>
(
E
)
<
/
s
p
a
n
>
Infinitely many
<span class='latex-bold'>(A)</span>\ 0 \qquad<span class='latex-bold'>(B)</span>\ 4 \qquad<span class='latex-bold'>(C)</span>\ 6 \qquad<span class='latex-bold'>(D)</span>\ 10 \qquad<span class='latex-bold'>(E)</span>\ \text{Infinitely many}
<
s
p
an
c
l
a
ss
=
′
l
a
t
e
x
−
b
o
l
d
′
>
(
A
)
<
/
s
p
an
>
0
<
s
p
an
c
l
a
ss
=
′
l
a
t
e
x
−
b
o
l
d
′
>
(
B
)
<
/
s
p
an
>
4
<
s
p
an
c
l
a
ss
=
′
l
a
t
e
x
−
b
o
l
d
′
>
(
C
)
<
/
s
p
an
>
6
<
s
p
an
c
l
a
ss
=
′
l
a
t
e
x
−
b
o
l
d
′
>
(
D
)
<
/
s
p
an
>
10
<
s
p
an
c
l
a
ss
=
′
l
a
t
e
x
−
b
o
l
d
′
>
(
E
)
<
/
s
p
an
>
Infinitely many
18
1
Hide problems
Turkey NMO 2006 1st Round - P18 (Number Theory)
What is the least positive integer
k
k
k
satisfying that
n
+
k
∈
S
n+k\in S
n
+
k
∈
S
for every
n
∈
S
n\in S
n
∈
S
where
S
=
{
n
:
n
3
n
+
(
2
n
+
1
)
5
n
≡
0
(
m
o
d
7
)
}
S=\{n : n3^n + (2n+1)5^n \equiv 0 \pmod 7\}
S
=
{
n
:
n
3
n
+
(
2
n
+
1
)
5
n
≡
0
(
mod
7
)}
?
<
s
p
a
n
c
l
a
s
s
=
′
l
a
t
e
x
−
b
o
l
d
′
>
(
A
)
<
/
s
p
a
n
>
6
<
s
p
a
n
c
l
a
s
s
=
′
l
a
t
e
x
−
b
o
l
d
′
>
(
B
)
<
/
s
p
a
n
>
7
<
s
p
a
n
c
l
a
s
s
=
′
l
a
t
e
x
−
b
o
l
d
′
>
(
C
)
<
/
s
p
a
n
>
14
<
s
p
a
n
c
l
a
s
s
=
′
l
a
t
e
x
−
b
o
l
d
′
>
(
D
)
<
/
s
p
a
n
>
21
<
s
p
a
n
c
l
a
s
s
=
′
l
a
t
e
x
−
b
o
l
d
′
>
(
E
)
<
/
s
p
a
n
>
42
<span class='latex-bold'>(A)</span>\ 6 \qquad<span class='latex-bold'>(B)</span>\ 7 \qquad<span class='latex-bold'>(C)</span>\ 14 \qquad<span class='latex-bold'>(D)</span>\ 21 \qquad<span class='latex-bold'>(E)</span>\ 42
<
s
p
an
c
l
a
ss
=
′
l
a
t
e
x
−
b
o
l
d
′
>
(
A
)
<
/
s
p
an
>
6
<
s
p
an
c
l
a
ss
=
′
l
a
t
e
x
−
b
o
l
d
′
>
(
B
)
<
/
s
p
an
>
7
<
s
p
an
c
l
a
ss
=
′
l
a
t
e
x
−
b
o
l
d
′
>
(
C
)
<
/
s
p
an
>
14
<
s
p
an
c
l
a
ss
=
′
l
a
t
e
x
−
b
o
l
d
′
>
(
D
)
<
/
s
p
an
>
21
<
s
p
an
c
l
a
ss
=
′
l
a
t
e
x
−
b
o
l
d
′
>
(
E
)
<
/
s
p
an
>
42
17
1
Hide problems
Turkey NMO 2006 1st Round - P17 (Geometry)
Let
D
D
D
be a point on the side
[
B
C
]
[BC]
[
BC
]
of
△
A
B
C
\triangle ABC
△
A
BC
such that
∣
B
D
∣
=
2
|BD|=2
∣
B
D
∣
=
2
and
∣
D
C
∣
=
6
|DC|=6
∣
D
C
∣
=
6
. If
∣
A
B
∣
=
4
|AB|=4
∣
A
B
∣
=
4
and
m
(
A
C
B
^
)
=
2
0
∘
m(\widehat{ACB})=20^\circ
m
(
A
CB
)
=
2
0
∘
, then what is
m
(
B
A
D
^
)
m(\widehat {BAD})
m
(
B
A
D
)
?
<
s
p
a
n
c
l
a
s
s
=
′
l
a
t
e
x
−
b
o
l
d
′
>
(
A
)
<
/
s
p
a
n
>
1
0
∘
<
s
p
a
n
c
l
a
s
s
=
′
l
a
t
e
x
−
b
o
l
d
′
>
(
B
)
<
/
s
p
a
n
>
1
8
∘
<
s
p
a
n
c
l
a
s
s
=
′
l
a
t
e
x
−
b
o
l
d
′
>
(
C
)
<
/
s
p
a
n
>
2
0
∘
<
s
p
a
n
c
l
a
s
s
=
′
l
a
t
e
x
−
b
o
l
d
′
>
(
D
)
<
/
s
p
a
n
>
2
2
∘
<
s
p
a
n
c
l
a
s
s
=
′
l
a
t
e
x
−
b
o
l
d
′
>
(
E
)
<
/
s
p
a
n
>
2
5
∘
<span class='latex-bold'>(A)</span>\ 10^\circ \qquad<span class='latex-bold'>(B)</span>\ 18^\circ \qquad<span class='latex-bold'>(C)</span>\ 20^\circ \qquad<span class='latex-bold'>(D)</span>\ 22^\circ \qquad<span class='latex-bold'>(E)</span>\ 25^\circ
<
s
p
an
c
l
a
ss
=
′
l
a
t
e
x
−
b
o
l
d
′
>
(
A
)
<
/
s
p
an
>
1
0
∘
<
s
p
an
c
l
a
ss
=
′
l
a
t
e
x
−
b
o
l
d
′
>
(
B
)
<
/
s
p
an
>
1
8
∘
<
s
p
an
c
l
a
ss
=
′
l
a
t
e
x
−
b
o
l
d
′
>
(
C
)
<
/
s
p
an
>
2
0
∘
<
s
p
an
c
l
a
ss
=
′
l
a
t
e
x
−
b
o
l
d
′
>
(
D
)
<
/
s
p
an
>
2
2
∘
<
s
p
an
c
l
a
ss
=
′
l
a
t
e
x
−
b
o
l
d
′
>
(
E
)
<
/
s
p
an
>
2
5
∘
16
1
Hide problems
Turkey NMO 2006 1st Round - P16 (Combinatorics)
How many positive integer tuples
(
x
1
,
x
2
,
…
,
x
13
)
(x_1,x_2,\dots, x_{13})
(
x
1
,
x
2
,
…
,
x
13
)
are there satisfying the inequality
x
1
+
x
2
+
⋯
+
x
13
≤
2006
x_1+x_2+\dots + x_{13}\leq 2006
x
1
+
x
2
+
⋯
+
x
13
≤
2006
?
<
s
p
a
n
c
l
a
s
s
=
′
l
a
t
e
x
−
b
o
l
d
′
>
(
A
)
<
/
s
p
a
n
>
2006
!
13
!
1993
!
<
s
p
a
n
c
l
a
s
s
=
′
l
a
t
e
x
−
b
o
l
d
′
>
(
B
)
<
/
s
p
a
n
>
2006
!
14
!
1992
!
<
s
p
a
n
c
l
a
s
s
=
′
l
a
t
e
x
−
b
o
l
d
′
>
(
C
)
<
/
s
p
a
n
>
1993
!
12
!
1981
!
<
s
p
a
n
c
l
a
s
s
=
′
l
a
t
e
x
−
b
o
l
d
′
>
(
D
)
<
/
s
p
a
n
>
1993
!
13
!
1980
!
<
s
p
a
n
c
l
a
s
s
=
′
l
a
t
e
x
−
b
o
l
d
′
>
(
E
)
<
/
s
p
a
n
>
None of above
<span class='latex-bold'>(A)</span>\ \frac{2006!}{13!1993!} \qquad<span class='latex-bold'>(B)</span>\ \frac{2006!}{14!1992!} \qquad<span class='latex-bold'>(C)</span>\ \frac{1993!}{12!1981!} \qquad<span class='latex-bold'>(D)</span>\ \frac{1993!}{13!1980!} \qquad<span class='latex-bold'>(E)</span>\ \text{None of above}
<
s
p
an
c
l
a
ss
=
′
l
a
t
e
x
−
b
o
l
d
′
>
(
A
)
<
/
s
p
an
>
13
!
1993
!
2006
!
<
s
p
an
c
l
a
ss
=
′
l
a
t
e
x
−
b
o
l
d
′
>
(
B
)
<
/
s
p
an
>
14
!
1992
!
2006
!
<
s
p
an
c
l
a
ss
=
′
l
a
t
e
x
−
b
o
l
d
′
>
(
C
)
<
/
s
p
an
>
12
!
1981
!
1993
!
<
s
p
an
c
l
a
ss
=
′
l
a
t
e
x
−
b
o
l
d
′
>
(
D
)
<
/
s
p
an
>
13
!
1980
!
1993
!
<
s
p
an
c
l
a
ss
=
′
l
a
t
e
x
−
b
o
l
d
′
>
(
E
)
<
/
s
p
an
>
None of above
15
1
Hide problems
Turkey NMO 2006 1st Round - P15 (Algebra)
How many different real roots does the equation
x
2
−
5
x
−
4
x
+
13
=
0
x^2-5x-4\sqrt x + 13 = 0
x
2
−
5
x
−
4
x
+
13
=
0
have?
<
s
p
a
n
c
l
a
s
s
=
′
l
a
t
e
x
−
b
o
l
d
′
>
(
A
)
<
/
s
p
a
n
>
0
<
s
p
a
n
c
l
a
s
s
=
′
l
a
t
e
x
−
b
o
l
d
′
>
(
B
)
<
/
s
p
a
n
>
1
<
s
p
a
n
c
l
a
s
s
=
′
l
a
t
e
x
−
b
o
l
d
′
>
(
C
)
<
/
s
p
a
n
>
2
<
s
p
a
n
c
l
a
s
s
=
′
l
a
t
e
x
−
b
o
l
d
′
>
(
D
)
<
/
s
p
a
n
>
3
<
s
p
a
n
c
l
a
s
s
=
′
l
a
t
e
x
−
b
o
l
d
′
>
(
E
)
<
/
s
p
a
n
>
4
<span class='latex-bold'>(A)</span>\ 0 \qquad<span class='latex-bold'>(B)</span>\ 1 \qquad<span class='latex-bold'>(C)</span>\ 2 \qquad<span class='latex-bold'>(D)</span>\ 3 \qquad<span class='latex-bold'>(E)</span>\ 4
<
s
p
an
c
l
a
ss
=
′
l
a
t
e
x
−
b
o
l
d
′
>
(
A
)
<
/
s
p
an
>
0
<
s
p
an
c
l
a
ss
=
′
l
a
t
e
x
−
b
o
l
d
′
>
(
B
)
<
/
s
p
an
>
1
<
s
p
an
c
l
a
ss
=
′
l
a
t
e
x
−
b
o
l
d
′
>
(
C
)
<
/
s
p
an
>
2
<
s
p
an
c
l
a
ss
=
′
l
a
t
e
x
−
b
o
l
d
′
>
(
D
)
<
/
s
p
an
>
3
<
s
p
an
c
l
a
ss
=
′
l
a
t
e
x
−
b
o
l
d
′
>
(
E
)
<
/
s
p
an
>
4
14
1
Hide problems
Turkey NMO 2006 1st Round - P14 (Number Theory)
How many four digit perfect square numbers are there in the form
A
A
B
B
AABB
AA
BB
where
A
,
B
∈
{
1
,
2
,
…
,
9
}
A,B \in \{1,2,\dots, 9\}
A
,
B
∈
{
1
,
2
,
…
,
9
}
?
<
s
p
a
n
c
l
a
s
s
=
′
l
a
t
e
x
−
b
o
l
d
′
>
(
A
)
<
/
s
p
a
n
>
3
<
s
p
a
n
c
l
a
s
s
=
′
l
a
t
e
x
−
b
o
l
d
′
>
(
B
)
<
/
s
p
a
n
>
2
<
s
p
a
n
c
l
a
s
s
=
′
l
a
t
e
x
−
b
o
l
d
′
>
(
C
)
<
/
s
p
a
n
>
1
<
s
p
a
n
c
l
a
s
s
=
′
l
a
t
e
x
−
b
o
l
d
′
>
(
D
)
<
/
s
p
a
n
>
0
<
s
p
a
n
c
l
a
s
s
=
′
l
a
t
e
x
−
b
o
l
d
′
>
(
E
)
<
/
s
p
a
n
>
None of above
<span class='latex-bold'>(A)</span>\ 3 \qquad<span class='latex-bold'>(B)</span>\ 2 \qquad<span class='latex-bold'>(C)</span>\ 1 \qquad<span class='latex-bold'>(D)</span>\ 0 \qquad<span class='latex-bold'>(E)</span>\ \text{None of above}
<
s
p
an
c
l
a
ss
=
′
l
a
t
e
x
−
b
o
l
d
′
>
(
A
)
<
/
s
p
an
>
3
<
s
p
an
c
l
a
ss
=
′
l
a
t
e
x
−
b
o
l
d
′
>
(
B
)
<
/
s
p
an
>
2
<
s
p
an
c
l
a
ss
=
′
l
a
t
e
x
−
b
o
l
d
′
>
(
C
)
<
/
s
p
an
>
1
<
s
p
an
c
l
a
ss
=
′
l
a
t
e
x
−
b
o
l
d
′
>
(
D
)
<
/
s
p
an
>
0
<
s
p
an
c
l
a
ss
=
′
l
a
t
e
x
−
b
o
l
d
′
>
(
E
)
<
/
s
p
an
>
None of above
13
1
Hide problems
Turkey NMO 2006 1st Round - P13 (Geometry)
Let
D
D
D
be a point on the side
[
A
B
]
[AB]
[
A
B
]
of the isosceles triangle
A
B
C
ABC
A
BC
such that
∣
A
B
∣
=
∣
A
C
∣
|AB|=|AC|
∣
A
B
∣
=
∣
A
C
∣
. The parallel line to
B
C
BC
BC
passing through
D
D
D
intersects
A
C
AC
A
C
at
E
E
E
. If
m
(
A
^
)
=
2
0
∘
m(\widehat A) = 20^\circ
m
(
A
)
=
2
0
∘
,
∣
D
E
∣
=
1
|DE|=1
∣
D
E
∣
=
1
,
∣
B
C
∣
=
a
|BC|=a
∣
BC
∣
=
a
, and
∣
B
E
∣
=
a
+
1
|BE|=a+1
∣
BE
∣
=
a
+
1
, then which of the followings is equal to
∣
A
B
∣
|AB|
∣
A
B
∣
?
<
s
p
a
n
c
l
a
s
s
=
′
l
a
t
e
x
−
b
o
l
d
′
>
(
A
)
<
/
s
p
a
n
>
2
a
<
s
p
a
n
c
l
a
s
s
=
′
l
a
t
e
x
−
b
o
l
d
′
>
(
B
)
<
/
s
p
a
n
>
a
2
−
a
<
s
p
a
n
c
l
a
s
s
=
′
l
a
t
e
x
−
b
o
l
d
′
>
(
C
)
<
/
s
p
a
n
>
a
2
+
1
<
s
p
a
n
c
l
a
s
s
=
′
l
a
t
e
x
−
b
o
l
d
′
>
(
D
)
<
/
s
p
a
n
>
(
a
+
1
)
2
<
s
p
a
n
c
l
a
s
s
=
′
l
a
t
e
x
−
b
o
l
d
′
>
(
E
)
<
/
s
p
a
n
>
a
2
+
a
<span class='latex-bold'>(A)</span>\ 2a \qquad<span class='latex-bold'>(B)</span>\ a^2-a \qquad<span class='latex-bold'>(C)</span>\ a^2+1 \qquad<span class='latex-bold'>(D)</span>\ (a+1)^2 \qquad<span class='latex-bold'>(E)</span>\ a^2+a
<
s
p
an
c
l
a
ss
=
′
l
a
t
e
x
−
b
o
l
d
′
>
(
A
)
<
/
s
p
an
>
2
a
<
s
p
an
c
l
a
ss
=
′
l
a
t
e
x
−
b
o
l
d
′
>
(
B
)
<
/
s
p
an
>
a
2
−
a
<
s
p
an
c
l
a
ss
=
′
l
a
t
e
x
−
b
o
l
d
′
>
(
C
)
<
/
s
p
an
>
a
2
+
1
<
s
p
an
c
l
a
ss
=
′
l
a
t
e
x
−
b
o
l
d
′
>
(
D
)
<
/
s
p
an
>
(
a
+
1
)
2
<
s
p
an
c
l
a
ss
=
′
l
a
t
e
x
−
b
o
l
d
′
>
(
E
)
<
/
s
p
an
>
a
2
+
a
12
1
Hide problems
Turkey NMO 2006 1st Round - P12 (Combinatorics)
In how many different ways can the set
{
1
,
2
,
…
,
2006
}
\{1,2,\dots, 2006\}
{
1
,
2
,
…
,
2006
}
be divided into three non-empty sets such that no set contains two successive numbers?
<
s
p
a
n
c
l
a
s
s
=
′
l
a
t
e
x
−
b
o
l
d
′
>
(
A
)
<
/
s
p
a
n
>
3
2006
−
3
⋅
2
2006
+
1
<
s
p
a
n
c
l
a
s
s
=
′
l
a
t
e
x
−
b
o
l
d
′
>
(
B
)
<
/
s
p
a
n
>
2
2005
−
2
<
s
p
a
n
c
l
a
s
s
=
′
l
a
t
e
x
−
b
o
l
d
′
>
(
C
)
<
/
s
p
a
n
>
3
2004
<
s
p
a
n
c
l
a
s
s
=
′
l
a
t
e
x
−
b
o
l
d
′
>
(
D
)
<
/
s
p
a
n
>
3
2005
−
1
<
s
p
a
n
c
l
a
s
s
=
′
l
a
t
e
x
−
b
o
l
d
′
>
(
E
)
<
/
s
p
a
n
>
None of above
<span class='latex-bold'>(A)</span>\ 3^{2006}-3\cdot 2^{2006}+1 \qquad<span class='latex-bold'>(B)</span>\ 2^{2005}-2 \qquad<span class='latex-bold'>(C)</span>\ 3^{2004} \qquad<span class='latex-bold'>(D)</span>\ 3^{2005}-1 \qquad<span class='latex-bold'>(E)</span>\ \text{None of above}
<
s
p
an
c
l
a
ss
=
′
l
a
t
e
x
−
b
o
l
d
′
>
(
A
)
<
/
s
p
an
>
3
2006
−
3
⋅
2
2006
+
1
<
s
p
an
c
l
a
ss
=
′
l
a
t
e
x
−
b
o
l
d
′
>
(
B
)
<
/
s
p
an
>
2
2005
−
2
<
s
p
an
c
l
a
ss
=
′
l
a
t
e
x
−
b
o
l
d
′
>
(
C
)
<
/
s
p
an
>
3
2004
<
s
p
an
c
l
a
ss
=
′
l
a
t
e
x
−
b
o
l
d
′
>
(
D
)
<
/
s
p
an
>
3
2005
−
1
<
s
p
an
c
l
a
ss
=
′
l
a
t
e
x
−
b
o
l
d
′
>
(
E
)
<
/
s
p
an
>
None of above
11
1
Hide problems
Turkey NMO 2006 1st Round - P11 (Algebra)
What is the sum of the real roots of the equation
4
x
4
−
3
x
2
+
7
x
−
3
=
0
4x^4-3x^2+7x-3=0
4
x
4
−
3
x
2
+
7
x
−
3
=
0
?
<
s
p
a
n
c
l
a
s
s
=
′
l
a
t
e
x
−
b
o
l
d
′
>
(
A
)
<
/
s
p
a
n
>
−
1
<
s
p
a
n
c
l
a
s
s
=
′
l
a
t
e
x
−
b
o
l
d
′
>
(
B
)
<
/
s
p
a
n
>
−
2
<
s
p
a
n
c
l
a
s
s
=
′
l
a
t
e
x
−
b
o
l
d
′
>
(
C
)
<
/
s
p
a
n
>
−
3
<
s
p
a
n
c
l
a
s
s
=
′
l
a
t
e
x
−
b
o
l
d
′
>
(
D
)
<
/
s
p
a
n
>
−
4
<
s
p
a
n
c
l
a
s
s
=
′
l
a
t
e
x
−
b
o
l
d
′
>
(
E
)
<
/
s
p
a
n
>
None of above
<span class='latex-bold'>(A)</span>\ -1 \qquad<span class='latex-bold'>(B)</span>\ -2 \qquad<span class='latex-bold'>(C)</span>\ -3 \qquad<span class='latex-bold'>(D)</span>\ -4 \qquad<span class='latex-bold'>(E)</span>\ \text {None of above}
<
s
p
an
c
l
a
ss
=
′
l
a
t
e
x
−
b
o
l
d
′
>
(
A
)
<
/
s
p
an
>
−
1
<
s
p
an
c
l
a
ss
=
′
l
a
t
e
x
−
b
o
l
d
′
>
(
B
)
<
/
s
p
an
>
−
2
<
s
p
an
c
l
a
ss
=
′
l
a
t
e
x
−
b
o
l
d
′
>
(
C
)
<
/
s
p
an
>
−
3
<
s
p
an
c
l
a
ss
=
′
l
a
t
e
x
−
b
o
l
d
′
>
(
D
)
<
/
s
p
an
>
−
4
<
s
p
an
c
l
a
ss
=
′
l
a
t
e
x
−
b
o
l
d
′
>
(
E
)
<
/
s
p
an
>
None of above
10
1
Hide problems
Turkey NMO 2006 1st Round - P10 (Number Theory)
What is the larget integer
n
n
n
such that
5
n
5^n
5
n
divides
2006
!
(
1003
!
)
2
\frac {2006!}{(1003!)^2}
(
1003
!
)
2
2006
!
?
<
s
p
a
n
c
l
a
s
s
=
′
l
a
t
e
x
−
b
o
l
d
′
>
(
A
)
<
/
s
p
a
n
>
0
<
s
p
a
n
c
l
a
s
s
=
′
l
a
t
e
x
−
b
o
l
d
′
>
(
B
)
<
/
s
p
a
n
>
1
<
s
p
a
n
c
l
a
s
s
=
′
l
a
t
e
x
−
b
o
l
d
′
>
(
C
)
<
/
s
p
a
n
>
2
<
s
p
a
n
c
l
a
s
s
=
′
l
a
t
e
x
−
b
o
l
d
′
>
(
D
)
<
/
s
p
a
n
>
3
<
s
p
a
n
c
l
a
s
s
=
′
l
a
t
e
x
−
b
o
l
d
′
>
(
E
)
<
/
s
p
a
n
>
500
<span class='latex-bold'>(A)</span>\ 0 \qquad<span class='latex-bold'>(B)</span>\ 1 \qquad<span class='latex-bold'>(C)</span>\ 2 \qquad<span class='latex-bold'>(D)</span>\ 3 \qquad<span class='latex-bold'>(E)</span>\ 500
<
s
p
an
c
l
a
ss
=
′
l
a
t
e
x
−
b
o
l
d
′
>
(
A
)
<
/
s
p
an
>
0
<
s
p
an
c
l
a
ss
=
′
l
a
t
e
x
−
b
o
l
d
′
>
(
B
)
<
/
s
p
an
>
1
<
s
p
an
c
l
a
ss
=
′
l
a
t
e
x
−
b
o
l
d
′
>
(
C
)
<
/
s
p
an
>
2
<
s
p
an
c
l
a
ss
=
′
l
a
t
e
x
−
b
o
l
d
′
>
(
D
)
<
/
s
p
an
>
3
<
s
p
an
c
l
a
ss
=
′
l
a
t
e
x
−
b
o
l
d
′
>
(
E
)
<
/
s
p
an
>
500
9
1
Hide problems
Turkey NMO 2006 1st Round - P09 (Geometry)
A
B
C
ABC
A
BC
is a triangle with
∣
A
B
∣
=
6
|AB|=6
∣
A
B
∣
=
6
,
∣
B
C
∣
=
7
|BC|=7
∣
BC
∣
=
7
, and
∣
A
C
∣
=
8
|AC|=8
∣
A
C
∣
=
8
. Let the angle bisector of
∠
A
\angle A
∠
A
intersect
B
C
BC
BC
at
D
D
D
. If
E
E
E
is a point on
[
A
C
]
[AC]
[
A
C
]
such that
∣
C
E
∣
=
2
|CE|=2
∣
CE
∣
=
2
, what is
∣
D
E
∣
|DE|
∣
D
E
∣
?
<
s
p
a
n
c
l
a
s
s
=
′
l
a
t
e
x
−
b
o
l
d
′
>
(
A
)
<
/
s
p
a
n
>
3
<
s
p
a
n
c
l
a
s
s
=
′
l
a
t
e
x
−
b
o
l
d
′
>
(
B
)
<
/
s
p
a
n
>
17
5
<
s
p
a
n
c
l
a
s
s
=
′
l
a
t
e
x
−
b
o
l
d
′
>
(
C
)
<
/
s
p
a
n
>
7
2
<
s
p
a
n
c
l
a
s
s
=
′
l
a
t
e
x
−
b
o
l
d
′
>
(
D
)
<
/
s
p
a
n
>
2
3
<
s
p
a
n
c
l
a
s
s
=
′
l
a
t
e
x
−
b
o
l
d
′
>
(
E
)
<
/
s
p
a
n
>
3
2
<span class='latex-bold'>(A)</span>\ 3 \qquad<span class='latex-bold'>(B)</span>\ \frac {17}5 \qquad<span class='latex-bold'>(C)</span>\ \frac 72 \qquad<span class='latex-bold'>(D)</span>\ 2\sqrt 3 \qquad<span class='latex-bold'>(E)</span>\ 3\sqrt 2
<
s
p
an
c
l
a
ss
=
′
l
a
t
e
x
−
b
o
l
d
′
>
(
A
)
<
/
s
p
an
>
3
<
s
p
an
c
l
a
ss
=
′
l
a
t
e
x
−
b
o
l
d
′
>
(
B
)
<
/
s
p
an
>
5
17
<
s
p
an
c
l
a
ss
=
′
l
a
t
e
x
−
b
o
l
d
′
>
(
C
)
<
/
s
p
an
>
2
7
<
s
p
an
c
l
a
ss
=
′
l
a
t
e
x
−
b
o
l
d
′
>
(
D
)
<
/
s
p
an
>
2
3
<
s
p
an
c
l
a
ss
=
′
l
a
t
e
x
−
b
o
l
d
′
>
(
E
)
<
/
s
p
an
>
3
2
8
1
Hide problems
Turkey NMO 2006 1st Round - P08 (Combinatorics)
Let
d
1
d_1
d
1
and
d
2
d_2
d
2
be parallel lines in the plane. We are marking
11
11
11
black points on
d
1
d_1
d
1
, and
16
16
16
white points on
d
2
d_2
d
2
. We are drawig the segments connecting black points with white points. What is the maximum number of points of intersection of these segments that lies on between the parallel lines (excluding the intersection points on the lines) ?
<
s
p
a
n
c
l
a
s
s
=
′
l
a
t
e
x
−
b
o
l
d
′
>
(
A
)
<
/
s
p
a
n
>
5600
<
s
p
a
n
c
l
a
s
s
=
′
l
a
t
e
x
−
b
o
l
d
′
>
(
B
)
<
/
s
p
a
n
>
5650
<
s
p
a
n
c
l
a
s
s
=
′
l
a
t
e
x
−
b
o
l
d
′
>
(
C
)
<
/
s
p
a
n
>
6500
<
s
p
a
n
c
l
a
s
s
=
′
l
a
t
e
x
−
b
o
l
d
′
>
(
D
)
<
/
s
p
a
n
>
6560
<
s
p
a
n
c
l
a
s
s
=
′
l
a
t
e
x
−
b
o
l
d
′
>
(
E
)
<
/
s
p
a
n
>
6600
<span class='latex-bold'>(A)</span>\ 5600 \qquad<span class='latex-bold'>(B)</span>\ 5650 \qquad<span class='latex-bold'>(C)</span>\ 6500 \qquad<span class='latex-bold'>(D)</span>\ 6560 \qquad<span class='latex-bold'>(E)</span>\ 6600
<
s
p
an
c
l
a
ss
=
′
l
a
t
e
x
−
b
o
l
d
′
>
(
A
)
<
/
s
p
an
>
5600
<
s
p
an
c
l
a
ss
=
′
l
a
t
e
x
−
b
o
l
d
′
>
(
B
)
<
/
s
p
an
>
5650
<
s
p
an
c
l
a
ss
=
′
l
a
t
e
x
−
b
o
l
d
′
>
(
C
)
<
/
s
p
an
>
6500
<
s
p
an
c
l
a
ss
=
′
l
a
t
e
x
−
b
o
l
d
′
>
(
D
)
<
/
s
p
an
>
6560
<
s
p
an
c
l
a
ss
=
′
l
a
t
e
x
−
b
o
l
d
′
>
(
E
)
<
/
s
p
an
>
6600
7
1
Hide problems
Turkey NMO 2006 1st Round - P07 (Algebra)
How many positive integers are there such that
⌊
m
11
⌋
=
⌊
m
10
⌋
\left \lfloor \frac m{11} \right \rfloor = \left \lfloor \frac m{10} \right \rfloor
⌊
11
m
⌋
=
⌊
10
m
⌋
? (
⌊
x
⌋
\left \lfloor x \right \rfloor
⌊
x
⌋
denotes the greatest integer not exceeding
x
x
x
.)
<
s
p
a
n
c
l
a
s
s
=
′
l
a
t
e
x
−
b
o
l
d
′
>
(
A
)
<
/
s
p
a
n
>
44
<
s
p
a
n
c
l
a
s
s
=
′
l
a
t
e
x
−
b
o
l
d
′
>
(
B
)
<
/
s
p
a
n
>
48
<
s
p
a
n
c
l
a
s
s
=
′
l
a
t
e
x
−
b
o
l
d
′
>
(
C
)
<
/
s
p
a
n
>
52
<
s
p
a
n
c
l
a
s
s
=
′
l
a
t
e
x
−
b
o
l
d
′
>
(
D
)
<
/
s
p
a
n
>
54
<
s
p
a
n
c
l
a
s
s
=
′
l
a
t
e
x
−
b
o
l
d
′
>
(
E
)
<
/
s
p
a
n
>
56
<span class='latex-bold'>(A)</span>\ 44 \qquad<span class='latex-bold'>(B)</span>\ 48 \qquad<span class='latex-bold'>(C)</span>\ 52 \qquad<span class='latex-bold'>(D)</span>\ 54 \qquad<span class='latex-bold'>(E)</span>\ 56
<
s
p
an
c
l
a
ss
=
′
l
a
t
e
x
−
b
o
l
d
′
>
(
A
)
<
/
s
p
an
>
44
<
s
p
an
c
l
a
ss
=
′
l
a
t
e
x
−
b
o
l
d
′
>
(
B
)
<
/
s
p
an
>
48
<
s
p
an
c
l
a
ss
=
′
l
a
t
e
x
−
b
o
l
d
′
>
(
C
)
<
/
s
p
an
>
52
<
s
p
an
c
l
a
ss
=
′
l
a
t
e
x
−
b
o
l
d
′
>
(
D
)
<
/
s
p
an
>
54
<
s
p
an
c
l
a
ss
=
′
l
a
t
e
x
−
b
o
l
d
′
>
(
E
)
<
/
s
p
an
>
56
6
1
Hide problems
Turkey NMO 2006 1st Round - P06 (Number Theory)
What is the sum of
3
+
3
2
+
3
2
2
+
3
2
3
+
⋯
+
3
2
2006
3+3^2+3^{2^2} + 3^{2^3} + \dots + 3^{2^{2006}}
3
+
3
2
+
3
2
2
+
3
2
3
+
⋯
+
3
2
2006
in
m
o
d
11
\mod 11
mod
11
?
<
s
p
a
n
c
l
a
s
s
=
′
l
a
t
e
x
−
b
o
l
d
′
>
(
A
)
<
/
s
p
a
n
>
0
<
s
p
a
n
c
l
a
s
s
=
′
l
a
t
e
x
−
b
o
l
d
′
>
(
B
)
<
/
s
p
a
n
>
1
<
s
p
a
n
c
l
a
s
s
=
′
l
a
t
e
x
−
b
o
l
d
′
>
(
C
)
<
/
s
p
a
n
>
2
<
s
p
a
n
c
l
a
s
s
=
′
l
a
t
e
x
−
b
o
l
d
′
>
(
D
)
<
/
s
p
a
n
>
5
<
s
p
a
n
c
l
a
s
s
=
′
l
a
t
e
x
−
b
o
l
d
′
>
(
E
)
<
/
s
p
a
n
>
10
<span class='latex-bold'>(A)</span>\ 0 \qquad<span class='latex-bold'>(B)</span>\ 1 \qquad<span class='latex-bold'>(C)</span>\ 2 \qquad<span class='latex-bold'>(D)</span>\ 5 \qquad<span class='latex-bold'>(E)</span>\ 10
<
s
p
an
c
l
a
ss
=
′
l
a
t
e
x
−
b
o
l
d
′
>
(
A
)
<
/
s
p
an
>
0
<
s
p
an
c
l
a
ss
=
′
l
a
t
e
x
−
b
o
l
d
′
>
(
B
)
<
/
s
p
an
>
1
<
s
p
an
c
l
a
ss
=
′
l
a
t
e
x
−
b
o
l
d
′
>
(
C
)
<
/
s
p
an
>
2
<
s
p
an
c
l
a
ss
=
′
l
a
t
e
x
−
b
o
l
d
′
>
(
D
)
<
/
s
p
an
>
5
<
s
p
an
c
l
a
ss
=
′
l
a
t
e
x
−
b
o
l
d
′
>
(
E
)
<
/
s
p
an
>
10
5
1
Hide problems
Turkey NMO 2006 1st Round - P05 (Geometry)
Let
D
D
D
be a point on the side
[
B
C
]
[BC]
[
BC
]
of
△
A
B
C
\triangle ABC
△
A
BC
such that
∣
A
B
∣
+
∣
B
D
∣
=
∣
A
C
∣
|AB|+|BD|=|AC|
∣
A
B
∣
+
∣
B
D
∣
=
∣
A
C
∣
and
m
(
B
A
D
^
)
=
m
(
D
A
C
^
)
=
3
0
∘
m(\widehat{BAD})=m(\widehat{DAC})=30^\circ
m
(
B
A
D
)
=
m
(
D
A
C
)
=
3
0
∘
. What is
m
(
A
C
B
^
)
m(\widehat{ACB})
m
(
A
CB
)
?
<
s
p
a
n
c
l
a
s
s
=
′
l
a
t
e
x
−
b
o
l
d
′
>
(
A
)
<
/
s
p
a
n
>
3
0
∘
<
s
p
a
n
c
l
a
s
s
=
′
l
a
t
e
x
−
b
o
l
d
′
>
(
B
)
<
/
s
p
a
n
>
4
0
∘
<
s
p
a
n
c
l
a
s
s
=
′
l
a
t
e
x
−
b
o
l
d
′
>
(
C
)
<
/
s
p
a
n
>
4
5
∘
<
s
p
a
n
c
l
a
s
s
=
′
l
a
t
e
x
−
b
o
l
d
′
>
(
D
)
<
/
s
p
a
n
>
4
8
∘
<
s
p
a
n
c
l
a
s
s
=
′
l
a
t
e
x
−
b
o
l
d
′
>
(
E
)
<
/
s
p
a
n
>
5
0
∘
<span class='latex-bold'>(A)</span>\ 30^\circ \qquad<span class='latex-bold'>(B)</span>\ 40^\circ \qquad<span class='latex-bold'>(C)</span>\ 45^\circ \qquad<span class='latex-bold'>(D)</span>\ 48^\circ \qquad<span class='latex-bold'>(E)</span>\ 50^\circ
<
s
p
an
c
l
a
ss
=
′
l
a
t
e
x
−
b
o
l
d
′
>
(
A
)
<
/
s
p
an
>
3
0
∘
<
s
p
an
c
l
a
ss
=
′
l
a
t
e
x
−
b
o
l
d
′
>
(
B
)
<
/
s
p
an
>
4
0
∘
<
s
p
an
c
l
a
ss
=
′
l
a
t
e
x
−
b
o
l
d
′
>
(
C
)
<
/
s
p
an
>
4
5
∘
<
s
p
an
c
l
a
ss
=
′
l
a
t
e
x
−
b
o
l
d
′
>
(
D
)
<
/
s
p
an
>
4
8
∘
<
s
p
an
c
l
a
ss
=
′
l
a
t
e
x
−
b
o
l
d
′
>
(
E
)
<
/
s
p
an
>
5
0
∘
4
1
Hide problems
Turkey NMO 2006 1st Round - P04 (Combinatorics)
There are
27
27
27
unit cubes. We are marking one point on each of the two opposing faces, two points on each of the other two opposing faces, and three points on each of the remaining two opposing faces of each cube. We are constructing a
3
×
3
×
3
3\times 3 \times 3
3
×
3
×
3
cube with these
27
27
27
cubes. What is the least number of marked points on the faces of the new cube?
<
s
p
a
n
c
l
a
s
s
=
′
l
a
t
e
x
−
b
o
l
d
′
>
(
A
)
<
/
s
p
a
n
>
54
<
s
p
a
n
c
l
a
s
s
=
′
l
a
t
e
x
−
b
o
l
d
′
>
(
B
)
<
/
s
p
a
n
>
60
<
s
p
a
n
c
l
a
s
s
=
′
l
a
t
e
x
−
b
o
l
d
′
>
(
C
)
<
/
s
p
a
n
>
72
<
s
p
a
n
c
l
a
s
s
=
′
l
a
t
e
x
−
b
o
l
d
′
>
(
D
)
<
/
s
p
a
n
>
90
<
s
p
a
n
c
l
a
s
s
=
′
l
a
t
e
x
−
b
o
l
d
′
>
(
E
)
<
/
s
p
a
n
>
96
<span class='latex-bold'>(A)</span>\ 54 \qquad<span class='latex-bold'>(B)</span>\ 60 \qquad<span class='latex-bold'>(C)</span>\ 72 \qquad<span class='latex-bold'>(D)</span>\ 90 \qquad<span class='latex-bold'>(E)</span>\ 96
<
s
p
an
c
l
a
ss
=
′
l
a
t
e
x
−
b
o
l
d
′
>
(
A
)
<
/
s
p
an
>
54
<
s
p
an
c
l
a
ss
=
′
l
a
t
e
x
−
b
o
l
d
′
>
(
B
)
<
/
s
p
an
>
60
<
s
p
an
c
l
a
ss
=
′
l
a
t
e
x
−
b
o
l
d
′
>
(
C
)
<
/
s
p
an
>
72
<
s
p
an
c
l
a
ss
=
′
l
a
t
e
x
−
b
o
l
d
′
>
(
D
)
<
/
s
p
an
>
90
<
s
p
an
c
l
a
ss
=
′
l
a
t
e
x
−
b
o
l
d
′
>
(
E
)
<
/
s
p
an
>
96
3
1
Hide problems
Turkey NMO 2006 1st Round - P03 (Algebra)
a
1
=
−
1
a_1=-1
a
1
=
−
1
,
a
2
=
2
a_2=2
a
2
=
2
, and
a
n
=
a
n
−
1
a
n
−
2
a_n=\frac {a_{n-1}}{a_{n-2}}
a
n
=
a
n
−
2
a
n
−
1
for
n
≥
3
n\geq 3
n
≥
3
. What is
a
2006
a_{2006}
a
2006
?
<
s
p
a
n
c
l
a
s
s
=
′
l
a
t
e
x
−
b
o
l
d
′
>
(
A
)
<
/
s
p
a
n
>
−
2
<
s
p
a
n
c
l
a
s
s
=
′
l
a
t
e
x
−
b
o
l
d
′
>
(
B
)
<
/
s
p
a
n
>
−
1
<
s
p
a
n
c
l
a
s
s
=
′
l
a
t
e
x
−
b
o
l
d
′
>
(
C
)
<
/
s
p
a
n
>
−
1
2
<
s
p
a
n
c
l
a
s
s
=
′
l
a
t
e
x
−
b
o
l
d
′
>
(
D
)
<
/
s
p
a
n
>
1
2
<
s
p
a
n
c
l
a
s
s
=
′
l
a
t
e
x
−
b
o
l
d
′
>
(
E
)
<
/
s
p
a
n
>
2
<span class='latex-bold'>(A)</span>\ -2 \qquad<span class='latex-bold'>(B)</span>\ -1 \qquad<span class='latex-bold'>(C)</span>\ -\frac 12 \qquad<span class='latex-bold'>(D)</span>\ \frac 12 \qquad<span class='latex-bold'>(E)</span>\ 2
<
s
p
an
c
l
a
ss
=
′
l
a
t
e
x
−
b
o
l
d
′
>
(
A
)
<
/
s
p
an
>
−
2
<
s
p
an
c
l
a
ss
=
′
l
a
t
e
x
−
b
o
l
d
′
>
(
B
)
<
/
s
p
an
>
−
1
<
s
p
an
c
l
a
ss
=
′
l
a
t
e
x
−
b
o
l
d
′
>
(
C
)
<
/
s
p
an
>
−
2
1
<
s
p
an
c
l
a
ss
=
′
l
a
t
e
x
−
b
o
l
d
′
>
(
D
)
<
/
s
p
an
>
2
1
<
s
p
an
c
l
a
ss
=
′
l
a
t
e
x
−
b
o
l
d
′
>
(
E
)
<
/
s
p
an
>
2
2
1
Hide problems
Turkey NMO 2006 1st Round - P02 (Number Theory)
If
p
p
p
and
p
2
+
2
p^2+2
p
2
+
2
are prime numbers, at most how many prime divisors can
p
3
+
3
p^3+3
p
3
+
3
have?
<
s
p
a
n
c
l
a
s
s
=
′
l
a
t
e
x
−
b
o
l
d
′
>
(
A
)
<
/
s
p
a
n
>
1
<
s
p
a
n
c
l
a
s
s
=
′
l
a
t
e
x
−
b
o
l
d
′
>
(
B
)
<
/
s
p
a
n
>
2
<
s
p
a
n
c
l
a
s
s
=
′
l
a
t
e
x
−
b
o
l
d
′
>
(
C
)
<
/
s
p
a
n
>
3
<
s
p
a
n
c
l
a
s
s
=
′
l
a
t
e
x
−
b
o
l
d
′
>
(
D
)
<
/
s
p
a
n
>
4
<
s
p
a
n
c
l
a
s
s
=
′
l
a
t
e
x
−
b
o
l
d
′
>
(
E
)
<
/
s
p
a
n
>
5
<span class='latex-bold'>(A)</span>\ 1 \qquad<span class='latex-bold'>(B)</span>\ 2 \qquad<span class='latex-bold'>(C)</span>\ 3 \qquad<span class='latex-bold'>(D)</span>\ 4 \qquad<span class='latex-bold'>(E)</span>\ 5
<
s
p
an
c
l
a
ss
=
′
l
a
t
e
x
−
b
o
l
d
′
>
(
A
)
<
/
s
p
an
>
1
<
s
p
an
c
l
a
ss
=
′
l
a
t
e
x
−
b
o
l
d
′
>
(
B
)
<
/
s
p
an
>
2
<
s
p
an
c
l
a
ss
=
′
l
a
t
e
x
−
b
o
l
d
′
>
(
C
)
<
/
s
p
an
>
3
<
s
p
an
c
l
a
ss
=
′
l
a
t
e
x
−
b
o
l
d
′
>
(
D
)
<
/
s
p
an
>
4
<
s
p
an
c
l
a
ss
=
′
l
a
t
e
x
−
b
o
l
d
′
>
(
E
)
<
/
s
p
an
>
5
1
1
Hide problems
Turkey NMO 2006 1st Round - P01 (Geometry)
Let
A
B
C
ABC
A
BC
be an equilateral triangle.
D
D
D
and
E
E
E
are midpoints of
[
A
B
]
[AB]
[
A
B
]
and
[
A
C
]
[AC]
[
A
C
]
. The ray
[
D
E
[DE
[
D
E
cuts the circumcircle of
△
A
B
C
\triangle ABC
△
A
BC
at
F
F
F
. What is
∣
D
E
∣
∣
D
F
∣
\frac {|DE|}{|DF|}
∣
D
F
∣
∣
D
E
∣
?
<
s
p
a
n
c
l
a
s
s
=
′
l
a
t
e
x
−
b
o
l
d
′
>
(
A
)
<
/
s
p
a
n
>
1
2
<
s
p
a
n
c
l
a
s
s
=
′
l
a
t
e
x
−
b
o
l
d
′
>
(
B
)
<
/
s
p
a
n
>
3
3
<
s
p
a
n
c
l
a
s
s
=
′
l
a
t
e
x
−
b
o
l
d
′
>
(
C
)
<
/
s
p
a
n
>
2
3
(
3
−
1
)
<
s
p
a
n
c
l
a
s
s
=
′
l
a
t
e
x
−
b
o
l
d
′
>
(
D
)
<
/
s
p
a
n
>
2
3
<
s
p
a
n
c
l
a
s
s
=
′
l
a
t
e
x
−
b
o
l
d
′
>
(
E
)
<
/
s
p
a
n
>
5
−
1
2
<span class='latex-bold'>(A)</span>\ \frac 12 \qquad<span class='latex-bold'>(B)</span>\ \frac {\sqrt 3}3 \qquad<span class='latex-bold'>(C)</span>\ \frac 23(\sqrt 3 - 1) \qquad<span class='latex-bold'>(D)</span>\ \frac 23 \qquad<span class='latex-bold'>(E)</span>\ \frac {\sqrt 5 - 1}2
<
s
p
an
c
l
a
ss
=
′
l
a
t
e
x
−
b
o
l
d
′
>
(
A
)
<
/
s
p
an
>
2
1
<
s
p
an
c
l
a
ss
=
′
l
a
t
e
x
−
b
o
l
d
′
>
(
B
)
<
/
s
p
an
>
3
3
<
s
p
an
c
l
a
ss
=
′
l
a
t
e
x
−
b
o
l
d
′
>
(
C
)
<
/
s
p
an
>
3
2
(
3
−
1
)
<
s
p
an
c
l
a
ss
=
′
l
a
t
e
x
−
b
o
l
d
′
>
(
D
)
<
/
s
p
an
>
3
2
<
s
p
an
c
l
a
ss
=
′
l
a
t
e
x
−
b
o
l
d
′
>
(
E
)
<
/
s
p
an
>
2
5
−
1