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Problems
Contests
National and Regional Contests
Turkey Contests
National Olympiad First Round
2000 National Olympiad First Round
2000 National Olympiad First Round
Part of
National Olympiad First Round
Subcontests
(36)
21
1
Hide problems
Turkey NMO 2000 1st Round - P21 (Geometry)
Let
A
B
C
D
ABCD
A
BC
D
be a cyclic quadrilateral with
∣
A
B
∣
=
26
|AB|=26
∣
A
B
∣
=
26
,
∣
B
C
∣
=
10
|BC|=10
∣
BC
∣
=
10
,
m
(
A
B
D
^
)
=
4
5
∘
m(\widehat{ABD})=45^\circ
m
(
A
B
D
)
=
4
5
∘
,
m
(
A
C
B
^
)
=
9
0
∘
m(\widehat{ACB})=90^\circ
m
(
A
CB
)
=
9
0
∘
. What is the area of
△
D
A
C
\triangle DAC
△
D
A
C
?
<
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>
(
A
)
<
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120
<
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>
(
B
)
<
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108
<
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c
l
a
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s
=
′
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a
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x
−
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o
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>
(
C
)
<
/
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>
90
<
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c
l
a
s
s
=
′
l
a
t
e
x
−
b
o
l
d
′
>
(
D
)
<
/
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>
84
<
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a
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s
=
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−
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>
(
E
)
<
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80
<span class='latex-bold'>(A)</span>\ 120 \qquad<span class='latex-bold'>(B)</span>\ 108 \qquad<span class='latex-bold'>(C)</span>\ 90 \qquad<span class='latex-bold'>(D)</span>\ 84 \qquad<span class='latex-bold'>(E)</span>\ 80
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−
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>
(
A
)
<
/
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>
120
<
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c
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a
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=
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a
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x
−
b
o
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d
′
>
(
B
)
<
/
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>
108
<
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p
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c
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a
ss
=
′
l
a
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x
−
b
o
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d
′
>
(
C
)
<
/
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p
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>
90
<
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p
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c
l
a
ss
=
′
l
a
t
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x
−
b
o
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d
′
>
(
D
)
<
/
s
p
an
>
84
<
s
p
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c
l
a
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=
′
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a
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x
−
b
o
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d
′
>
(
E
)
<
/
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an
>
80
20
1
Hide problems
Turkey NMO 2000 1st Round - P20 (Algebra)
For every real
x
x
x
, the polynomial
p
(
x
)
p(x)
p
(
x
)
whose roots are all real satisfies
p
(
x
2
−
1
)
=
p
(
x
)
p
(
−
x
)
p(x^2-1)=p(x)p(-x)
p
(
x
2
−
1
)
=
p
(
x
)
p
(
−
x
)
. What can the degree of
p
(
x
)
p(x)
p
(
x
)
be at most?
<
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(
A
)
<
/
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>
0
<
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c
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s
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>
(
B
)
<
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>
2
<
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c
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a
s
s
=
′
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a
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e
x
−
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o
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d
′
>
(
C
)
<
/
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>
4
<
s
p
a
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c
l
a
s
s
=
′
l
a
t
e
x
−
b
o
l
d
′
>
(
D
)
<
/
s
p
a
n
>
There is no upper bound for the degree of
p
(
x
)
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c
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(
E
)
<
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None
<span class='latex-bold'>(A)</span>\ 0 \qquad<span class='latex-bold'>(B)</span>\ 2 \qquad<span class='latex-bold'>(C)</span>\ 4 \qquad<span class='latex-bold'>(D)</span>\ \text{There is no upper bound for the degree of } p(x) \qquad<span class='latex-bold'>(E)</span>\ \text{None}
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(
A
)
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0
<
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p
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c
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a
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=
′
l
a
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e
x
−
b
o
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d
′
>
(
B
)
<
/
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>
2
<
s
p
an
c
l
a
ss
=
′
l
a
t
e
x
−
b
o
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d
′
>
(
C
)
<
/
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p
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>
4
<
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p
an
c
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a
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=
′
l
a
t
e
x
−
b
o
l
d
′
>
(
D
)
<
/
s
p
an
>
There is no upper bound for the degree of
p
(
x
)
<
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a
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=
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x
−
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>
(
E
)
<
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>
None
19
1
Hide problems
Turkey NMO 2000 1st Round - P19 (Combinatorics)
Let
P
P
P
be an arbitrary point inside
△
A
B
C
\triangle ABC
△
A
BC
with sides
3
,
7
,
8
3,7,8
3
,
7
,
8
. What is the probability that the distance of
P
P
P
to at least one vertices of the triangle is less than
1
1
1
?
<
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>
(
A
)
<
/
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>
π
36
2
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(
B
)
<
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>
π
36
3
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−
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>
(
C
)
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>
π
36
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s
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−
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>
(
D
)
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>
1
2
<
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(
E
)
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>
3
4
<span class='latex-bold'>(A)</span>\ \frac{\pi}{36}\sqrt 2 \qquad<span class='latex-bold'>(B)</span>\ \frac{\pi}{36}\sqrt 3 \qquad<span class='latex-bold'>(C)</span>\ \frac{\pi}{36} \qquad<span class='latex-bold'>(D)</span>\ \frac12 \qquad<span class='latex-bold'>(E)</span>\ \frac 34
<
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x
−
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>
(
A
)
<
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>
36
π
2
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x
−
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d
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>
(
B
)
<
/
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>
36
π
3
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(
C
)
<
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36
π
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x
−
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>
(
D
)
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2
1
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x
−
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d
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>
(
E
)
<
/
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>
4
3
18
1
Hide problems
Turkey NMO 2000 1st Round - P18 (Number Theory)
What is the least integer
n
≥
100
n\geq 100
n
≥
100
such that
77
77
77
divides
1
+
2
+
2
2
+
2
3
+
⋯
+
2
n
1+2+2^2+2^3+\dots + 2^n
1
+
2
+
2
2
+
2
3
+
⋯
+
2
n
?
<
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a
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s
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a
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−
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>
(
A
)
<
/
s
p
a
n
>
101
<
s
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c
l
a
s
s
=
′
l
a
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x
−
b
o
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d
′
>
(
B
)
<
/
s
p
a
n
>
105
<
s
p
a
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c
l
a
s
s
=
′
l
a
t
e
x
−
b
o
l
d
′
>
(
C
)
<
/
s
p
a
n
>
111
<
s
p
a
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c
l
a
s
s
=
′
l
a
t
e
x
−
b
o
l
d
′
>
(
D
)
<
/
s
p
a
n
>
119
<
s
p
a
n
c
l
a
s
s
=
′
l
a
t
e
x
−
b
o
l
d
′
>
(
E
)
<
/
s
p
a
n
>
None
<span class='latex-bold'>(A)</span>\ 101 \qquad<span class='latex-bold'>(B)</span>\ 105 \qquad<span class='latex-bold'>(C)</span>\ 111 \qquad<span class='latex-bold'>(D)</span>\ 119 \qquad<span class='latex-bold'>(E)</span>\ \text{None}
<
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c
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a
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=
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a
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x
−
b
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d
′
>
(
A
)
<
/
s
p
an
>
101
<
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p
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c
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a
ss
=
′
l
a
t
e
x
−
b
o
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d
′
>
(
B
)
<
/
s
p
an
>
105
<
s
p
an
c
l
a
ss
=
′
l
a
t
e
x
−
b
o
l
d
′
>
(
C
)
<
/
s
p
an
>
111
<
s
p
an
c
l
a
ss
=
′
l
a
t
e
x
−
b
o
l
d
′
>
(
D
)
<
/
s
p
an
>
119
<
s
p
an
c
l
a
ss
=
′
l
a
t
e
x
−
b
o
l
d
′
>
(
E
)
<
/
s
p
an
>
None
17
1
Hide problems
Turkey NMO 2000 1st Round - P17 (Geometry)
What is the largest possible area of a quadrilateral with sides
1
,
4
,
7
,
8
1,4,7,8
1
,
4
,
7
,
8
?
<
s
p
a
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c
l
a
s
s
=
′
l
a
t
e
x
−
b
o
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d
′
>
(
A
)
<
/
s
p
a
n
>
7
2
<
s
p
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c
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a
s
s
=
′
l
a
t
e
x
−
b
o
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d
′
>
(
B
)
<
/
s
p
a
n
>
10
3
<
s
p
a
n
c
l
a
s
s
=
′
l
a
t
e
x
−
b
o
l
d
′
>
(
C
)
<
/
s
p
a
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>
18
<
s
p
a
n
c
l
a
s
s
=
′
l
a
t
e
x
−
b
o
l
d
′
>
(
D
)
<
/
s
p
a
n
>
12
3
<
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c
l
a
s
s
=
′
l
a
t
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x
−
b
o
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d
′
>
(
E
)
<
/
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p
a
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>
9
5
<span class='latex-bold'>(A)</span>\ 7\sqrt 2 \qquad<span class='latex-bold'>(B)</span>\ 10\sqrt 3 \qquad<span class='latex-bold'>(C)</span>\ 18 \qquad<span class='latex-bold'>(D)</span>\ 12\sqrt 3 \qquad<span class='latex-bold'>(E)</span>\ 9\sqrt 5
<
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−
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>
(
A
)
<
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>
7
2
<
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=
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a
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x
−
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o
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d
′
>
(
B
)
<
/
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p
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>
10
3
<
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p
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c
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=
′
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a
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x
−
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>
(
C
)
<
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>
18
<
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a
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=
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a
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x
−
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′
>
(
D
)
<
/
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>
12
3
<
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c
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a
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=
′
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a
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x
−
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o
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d
′
>
(
E
)
<
/
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p
an
>
9
5
16
1
Hide problems
Turkey NMO 2000 1st Round - P16 (Algebra)
What is the sum of real roots of
(
2
+
(
2
+
(
2
+
x
)
2
)
2
)
2
=
2000
(2+(2+(2+x)^2)^2)^2=2000
(
2
+
(
2
+
(
2
+
x
)
2
)
2
)
2
=
2000
?
<
s
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c
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−
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(
A
)
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/
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>
−
4
<
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c
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s
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a
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x
−
b
o
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>
(
B
)
<
/
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a
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>
−
2
<
s
p
a
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c
l
a
s
s
=
′
l
a
t
e
x
−
b
o
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d
′
>
(
C
)
<
/
s
p
a
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>
0
<
s
p
a
n
c
l
a
s
s
=
′
l
a
t
e
x
−
b
o
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d
′
>
(
D
)
<
/
s
p
a
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>
2
<
s
p
a
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c
l
a
s
s
=
′
l
a
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e
x
−
b
o
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d
′
>
(
E
)
<
/
s
p
a
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>
4
<span class='latex-bold'>(A)</span>\ -4 \qquad<span class='latex-bold'>(B)</span>\ -2 \qquad<span class='latex-bold'>(C)</span>\ 0 \qquad<span class='latex-bold'>(D)</span>\ 2 \qquad<span class='latex-bold'>(E)</span>\ 4
<
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−
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>
(
A
)
<
/
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>
−
4
<
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
>
0
<
s
p
an
c
l
a
ss
=
′
l
a
t
e
x
−
b
o
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d
′
>
(
D
)
<
/
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>
2
<
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p
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c
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a
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=
′
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a
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x
−
b
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d
′
>
(
E
)
<
/
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>
4
15
1
Hide problems
Turkey NMO 2000 1st Round - P15 (Combinatorics)
A
,
B
,
C
A,B,C
A
,
B
,
C
are playing backgammon tournament. At first,
A
A
A
plays with
B
B
B
. Then the winner plays with
C
C
C
. As the tournament goes on, the last winner plays with the player who did not play in the previous game. When a player wins two successive games, he will win the tournament. If each player has equal chance to win a game, what is the probability that
C
C
C
wins the tournament?
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14
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<span class='latex-bold'>(A)</span>\ \frac27 \qquad<span class='latex-bold'>(B)</span>\ \frac13 \qquad<span class='latex-bold'>(C)</span>\ \frac3{14} \qquad<span class='latex-bold'>(D)</span>\ \frac 17 \qquad<span class='latex-bold'>(E)</span>\ \text{None}
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None
14
1
Hide problems
Turkey NMO 2000 1st Round - P14 (Number Theory)
What is the last two digits of the decimal representation of
9
8
7
⋅
⋅
⋅
2
9^{8^{7^{\cdot^{\cdot^{\cdot^{2}}}}}}
9
8
7
⋅
⋅
⋅
2
?
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81
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01
<span class='latex-bold'>(A)</span>\ 81 \qquad<span class='latex-bold'>(B)</span>\ 61 \qquad<span class='latex-bold'>(C)</span>\ 41 \qquad<span class='latex-bold'>(D)</span>\ 21 \qquad<span class='latex-bold'>(E)</span>\ 01
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13
1
Hide problems
Turkey NMO 2000 1st Round - P13 (Geometry)
Let
d
d
d
be one of the common tangent lines of externally tangent circles
k
1
k_1
k
1
and
k
2
k_2
k
2
.
d
d
d
touches
k
1
k_1
k
1
at
A
A
A
. Let
[
A
B
]
[AB]
[
A
B
]
be a diameter of
k
1
k_1
k
1
. The tangent from
B
B
B
to
k
2
k_2
k
2
touches
k
2
k_2
k
2
at
C
C
C
. If
∣
A
B
∣
=
8
|AB|=8
∣
A
B
∣
=
8
and the diameter of
k
2
k_2
k
2
is
7
7
7
, then what is
∣
B
C
∣
|BC|
∣
BC
∣
?
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3
<span class='latex-bold'>(A)</span>\ 7 \qquad<span class='latex-bold'>(B)</span>\ 6\sqrt 2 \qquad<span class='latex-bold'>(C)</span>\ 10 \qquad<span class='latex-bold'>(D)</span>\ 8 \qquad<span class='latex-bold'>(E)</span>\ 5\sqrt 3
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3
23
1
Hide problems
Turkey NMO 2000 1st Round - P23 (Combinatorics)
A committee with
20
20
20
members votes for the candidates
A
,
B
,
C
A,B,C
A
,
B
,
C
by a different election system. Each member writes his ordered prefer list to the ballot (e.g. if he writes
B
A
C
BAC
B
A
C
, he prefers
B
B
B
to
A
A
A
and
C
C
C
, and prefers
A
A
A
to
C
C
C
). After the ballots are counted, it is recognized that each of the six different permutations of three candidates appears in at least one ballot, and
11
11
11
members prefer
A
A
A
to
B
B
B
,
12
12
12
members prefer
C
C
C
to
A
A
A
,
14
14
14
members prefer
B
B
B
to
C
C
C
. How many members are there such that
B
B
B
is the first choice of them?
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More information is needed
<span class='latex-bold'>(A)</span>\ 5 \qquad<span class='latex-bold'>(B)</span>\ 7 \qquad<span class='latex-bold'>(C)</span>\ 8 \qquad<span class='latex-bold'>(D)</span>\ 10 \qquad<span class='latex-bold'>(E)</span>\ \text{More information is needed}
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More information is needed
24
1
Hide problems
Turkey NMO 2000 1st Round - P24 (Algebra)
Let
a
,
b
,
c
,
d
,
e
a,b,c,d,e
a
,
b
,
c
,
d
,
e
be non-negative real numbers such that
a
+
b
+
c
+
d
+
e
>
0
a+b+c+d+e>0
a
+
b
+
c
+
d
+
e
>
0
. What is the least real number
t
t
t
such that
a
+
c
=
t
b
a+c=tb
a
+
c
=
t
b
,
b
+
d
=
t
c
b+d=tc
b
+
d
=
t
c
,
c
+
e
=
t
d
c+e=td
c
+
e
=
t
d
?
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<span class='latex-bold'>(A)</span>\ \frac{\sqrt 2}2 \qquad<span class='latex-bold'>(B)</span>\ 1 \qquad<span class='latex-bold'>(C)</span>\ \sqrt 2 \qquad<span class='latex-bold'>(D)</span>\ \frac32 \qquad<span class='latex-bold'>(E)</span>\ 2
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22
1
Hide problems
Turkey NMO 2000 1st Round - P22 (Number Theory)
How many ordered positive integer triples
(
x
,
y
,
z
)
(x,y,z)
(
x
,
y
,
z
)
are there such that
3
x
2
−
2
y
2
−
4
z
2
+
54
=
0
5
x
2
−
3
y
2
−
7
z
2
+
74
=
0
\begin{array}{rcl} 3x^2-2y^2-4z^2+54 = 0 \\ 5x^2-3y^2-7z^2+74 = 0 \end{array}
3
x
2
−
2
y
2
−
4
z
2
+
54
=
0
5
x
2
−
3
y
2
−
7
z
2
+
74
=
0
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Infinitely Many
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<span class='latex-bold'>(A)</span>\ 0 \qquad<span class='latex-bold'>(B)</span>\ 2 \qquad<span class='latex-bold'>(C)</span>\ 3 \qquad<span class='latex-bold'>(D)</span>\ \text{Infinitely Many} \qquad<span class='latex-bold'>(E)</span>\ \text{None of the options}
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)
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Infinitely Many
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None of the options
25
1
Hide problems
Turkey NMO 2000 1st Round - P25 (Geometry)
The area of a convex quadrilateral
A
B
C
D
ABCD
A
BC
D
is
18
18
18
. If
∣
A
B
∣
+
∣
B
D
∣
+
∣
D
C
∣
=
12
|AB|+|BD|+|DC|=12
∣
A
B
∣
+
∣
B
D
∣
+
∣
D
C
∣
=
12
, then what is
∣
A
C
∣
|AC|
∣
A
C
∣
?
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9
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3
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2
<span class='latex-bold'>(A)</span>\ 9 \qquad<span class='latex-bold'>(B)</span>\ 6\sqrt 3 \qquad<span class='latex-bold'>(C)</span>\ 8 \qquad<span class='latex-bold'>(D)</span>\ 6 \qquad<span class='latex-bold'>(E)</span>\ 6\sqrt 2
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6
2
27
1
Hide problems
Turkey NMO 2000 1st Round - P27 (Combinatorics)
How many different permutations
(
α
1
α
2
α
3
α
4
α
5
)
(\alpha_1 \alpha_2\alpha_3\alpha_4\alpha_5)
(
α
1
α
2
α
3
α
4
α
5
)
of the set
{
1
,
2
,
3
,
4
,
5
}
\{1,2,3,4,5\}
{
1
,
2
,
3
,
4
,
5
}
are there such that
(
α
1
…
α
k
)
(\alpha_1\dots \alpha_k)
(
α
1
…
α
k
)
is not a permutation of the set
{
1
,
…
,
k
}
\{1,\dots ,k\}
{
1
,
…
,
k
}
, for every
1
≤
k
≤
4
1\leq k \leq 4
1
≤
k
≤
4
?
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13
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65
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71
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461
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None
<span class='latex-bold'>(A)</span>\ 13 \qquad<span class='latex-bold'>(B)</span>\ 65 \qquad<span class='latex-bold'>(C)</span>\ 71 \qquad<span class='latex-bold'>(D)</span>\ 461 \qquad<span class='latex-bold'>(E)</span>\ \text{None}
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461
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31
1
Hide problems
Turkey NMO 2000 1st Round - P31 (Combinatorics)
How many ten digit positive integers with distinct digits are multiples of
11111
11111
11111
?
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1264
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2842
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3456
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11111
<span class='latex-bold'>(A)</span>\ 0 \qquad<span class='latex-bold'>(B)</span>\ 1264 \qquad<span class='latex-bold'>(C)</span>\ 2842 \qquad<span class='latex-bold'>(D)</span>\ 3456 \qquad<span class='latex-bold'>(E)</span>\ 11111
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1264
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3456
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11111
29
1
Hide problems
Turkey NMO 2000 1st Round - P29 (Geometry)
One of the external common tangent lines of the two externally tangent circles with center
O
1
O_1
O
1
and
O
2
O_2
O
2
touches the circles at
B
B
B
and
C
C
C
, respectively. Let
A
A
A
be the common point of the circles. The line
B
A
BA
B
A
meets the circle with center
O
2
O_2
O
2
at
A
A
A
and
D
D
D
. If
∣
B
A
∣
=
5
|BA|=5
∣
B
A
∣
=
5
and
∣
A
D
∣
=
4
|AD|=4
∣
A
D
∣
=
4
, then what is
∣
C
D
∣
|CD|
∣
C
D
∣
?
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20
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)
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27
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2
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5
<span class='latex-bold'>(A)</span>\ \sqrt{20} \qquad<span class='latex-bold'>(B)</span>\ \sqrt{27} \qquad<span class='latex-bold'>(C)</span>\ 6 \qquad<span class='latex-bold'>(D)</span>\ \frac{15}2 \qquad<span class='latex-bold'>(E)</span>\ 4\sqrt5
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15
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5
33
1
Hide problems
Turkey NMO 2000 1st Round - P33 (Geometry)
Let
K
K
K
be a point on the side
[
A
B
]
[AB]
[
A
B
]
, and
L
L
L
be a point on the side
[
B
C
]
[BC]
[
BC
]
of the square
A
B
C
D
ABCD
A
BC
D
. If
∣
A
K
∣
=
3
|AK|=3
∣
A
K
∣
=
3
,
∣
K
B
∣
=
2
|KB|=2
∣
K
B
∣
=
2
, and the distance of
K
K
K
to the line
D
L
DL
D
L
is
3
3
3
, what is
∣
B
L
∣
:
∣
L
C
∣
|BL|:|LC|
∣
B
L
∣
:
∣
L
C
∣
?
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8
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2
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7
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8
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2
<span class='latex-bold'>(A)</span>\ \frac78 \qquad<span class='latex-bold'>(B)</span>\ \frac{\sqrt 3}2 \qquad<span class='latex-bold'>(C)</span>\ \frac 87 \qquad<span class='latex-bold'>(D)</span>\ \frac 38 \qquad<span class='latex-bold'>(E)</span>\ \frac{\sqrt 2}2
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7
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3
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2
2
36
1
Hide problems
Turkey NMO 2000 1st Round - P36 (Algebra)
x
n
+
1
=
(
1
+
2
n
)
x
n
+
4
n
x_{n+1}= \left ( 1+\frac2n \right )x_n+\frac4n
x
n
+
1
=
(
1
+
n
2
)
x
n
+
n
4
, for every positive integer
n
n
n
. If
x
1
=
−
1
x_1=-1
x
1
=
−
1
, what is
x
2000
x_{2000}
x
2000
?
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1999998
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2000998
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2000008
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<span class='latex-bold'>(A)</span>\ 1999998 \qquad<span class='latex-bold'>(B)</span>\ 2000998 \qquad<span class='latex-bold'>(C)</span>\ 2009998 \qquad<span class='latex-bold'>(D)</span>\ 2000008 \qquad<span class='latex-bold'>(E)</span>\ 1999999
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2000998
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1999999
28
1
Hide problems
Turkey NMO 2000 1st Round - P28 (Algebra)
f
1
(
x
)
=
x
2
+
x
f
2
(
x
)
=
2
x
2
−
x
f
3
(
x
)
=
x
2
+
x
g
1
(
x
)
=
x
−
2
g
2
(
x
)
=
2
x
g
3
(
x
)
=
x
+
2
\begin{array}{ rlrlrl} f_1(x)=&x^2+x & f_2(x)=&2x^2-x & f_3(x)=&x^2 +x \\ g_1(x)=&x-2 & g_2(x)=&2x \ \ & g_3(x)=&x+2 \\ \end{array}
f
1
(
x
)
=
g
1
(
x
)
=
x
2
+
x
x
−
2
f
2
(
x
)
=
g
2
(
x
)
=
2
x
2
−
x
2
x
f
3
(
x
)
=
g
3
(
x
)
=
x
2
+
x
x
+
2
If
h
(
x
)
=
x
h(x)=x
h
(
x
)
=
x
can be get from
f
i
f_i
f
i
and
g
i
g_i
g
i
by using only addition, substraction, multiplication defined on those functions where
i
∈
{
1
,
2
,
3
}
i\in\{1,2,3\}
i
∈
{
1
,
2
,
3
}
, then
F
i
=
1
F_i=1
F
i
=
1
. Otherwise,
F
i
=
0
F_i=0
F
i
=
0
. What is
(
F
1
,
F
2
,
F
3
)
(F_1,F_2,F_3)
(
F
1
,
F
2
,
F
3
)
?
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None
<span class='latex-bold'>(A)</span>\ (0,0,0) \qquad<span class='latex-bold'>(B)</span>\ (0,0,1) \qquad<span class='latex-bold'>(C)</span>\ (0,1,0) \qquad<span class='latex-bold'>(D)</span>\ (0,1,1) \qquad<span class='latex-bold'>(E)</span>\ \text{None}
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None
26
1
Hide problems
Turkey NMO 2000 1st Round - P26 (Number Theory)
Let
f
(
x
)
=
x
3
+
7
x
2
+
9
x
+
10
f(x)=x^3+7x^2+9x+10
f
(
x
)
=
x
3
+
7
x
2
+
9
x
+
10
. Which value of
p
p
p
satisfies the statement
f
(
a
)
≡
f
(
b
)
(
mod
p
)
⇒
a
≡
b
(
mod
p
)
f(a) \equiv f(b) \ (\text{mod } p) \Rightarrow a \equiv b \ (\text{mod } p)
f
(
a
)
≡
f
(
b
)
(
mod
p
)
⇒
a
≡
b
(
mod
p
)
for every integer
a
,
b
a,b
a
,
b
?
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5
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7
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11
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13
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17
<span class='latex-bold'>(A)</span>\ 5 \qquad<span class='latex-bold'>(B)</span>\ 7 \qquad<span class='latex-bold'>(C)</span>\ 11 \qquad<span class='latex-bold'>(D)</span>\ 13 \qquad<span class='latex-bold'>(E)</span>\ 17
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7
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11
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13
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17
11
1
Hide problems
Turkey NMO 2000 1st Round - P11 (Combinatorics)
In how many ways can
7
7
7
red,
7
7
7
white balls be distributed into
7
7
7
boxes such that every box contains exactly
2
2
2
balls?
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163
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s
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)
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393
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s
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858
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1716
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None
<span class='latex-bold'>(A)</span>\ 163 \qquad<span class='latex-bold'>(B)</span>\ 393 \qquad<span class='latex-bold'>(C)</span>\ 858 \qquad<span class='latex-bold'>(D)</span>\ 1716 \qquad<span class='latex-bold'>(E)</span>\ \text{None}
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858
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1716
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None
10
1
Hide problems
Turkey NMO 2000 1st Round - P10 (Number Theory)
N
N
N
is a
50
−
50-
50
−
digit number (in the decimal scale). All digits except the
2
6
th
26^{\text{th}}
2
6
th
digit (from the left) are
1
1
1
. If
N
N
N
is divisible by
13
13
13
, what is the
2
6
th
26^{\text{th}}
2
6
th
digit?
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)
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3
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8
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More information is needed
<span class='latex-bold'>(A)</span>\ 1 \qquad<span class='latex-bold'>(B)</span>\ 3 \qquad<span class='latex-bold'>(C)</span>\ 6 \qquad<span class='latex-bold'>(D)</span>\ 8 \qquad<span class='latex-bold'>(E)</span>\ \text{More information is needed}
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More information is needed
9
1
Hide problems
Turkey NMO 2000 1st Round - P09 (Geometry)
A
B
C
D
E
ABCDE
A
BC
D
E
is convex pentagon.
m
(
B
^
)
=
m
(
D
^
)
=
9
0
∘
m(\widehat{B})=m(\widehat{D})=90^\circ
m
(
B
)
=
m
(
D
)
=
9
0
∘
,
m
(
C
^
)
=
12
0
∘
m(\widehat{C})=120^\circ
m
(
C
)
=
12
0
∘
,
∣
A
B
∣
=
2
|AB|=2
∣
A
B
∣
=
2
,
∣
B
C
∣
=
∣
C
D
∣
=
3
|BC|=|CD|=\sqrt3
∣
BC
∣
=
∣
C
D
∣
=
3
, and
∣
E
D
∣
=
1
|ED|=1
∣
E
D
∣
=
1
.
∣
A
E
∣
=
?
|AE|=?
∣
A
E
∣
=
?
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2
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3
3
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1
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<span class='latex-bold'>(A)</span>\ \frac{3\sqrt3}{2} \qquad<span class='latex-bold'>(B)</span>\ \frac{2\sqrt3}{3} \qquad<span class='latex-bold'>(C)</span>\ \frac{3}{2} \qquad<span class='latex-bold'>(D)</span>\ \sqrt3 - 1 \qquad<span class='latex-bold'>(E)</span>\ \sqrt3
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8
1
Hide problems
Turkey NMO 2000 1st Round - P08 (Algebra)
(
x
+
y
)
5
=
z
(
y
+
z
)
5
=
x
(
z
+
x
)
5
=
y
\begin{array}{rcl} (x+y)^5 &=& z \\ (y+z)^5 &=& x \\ (z+x)^5 &=& y \end{array}
(
x
+
y
)
5
(
y
+
z
)
5
(
z
+
x
)
5
=
=
=
z
x
y
How many real triples
(
x
,
y
,
z
)
(x,y,z)
(
x
,
y
,
z
)
are there satisfying above equation system?
<
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1
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d
′
>
(
B
)
<
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n
>
2
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c
l
a
s
s
=
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′
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(
C
)
<
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>
3
<
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n
c
l
a
s
s
=
′
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a
t
e
x
−
b
o
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>
(
D
)
<
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a
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>
Infinitely many
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None
<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>\ \text{Infinitely many} \qquad<span class='latex-bold'>(E)</span>\ \text{None}
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2
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)
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3
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b
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d
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>
(
D
)
<
/
s
p
an
>
Infinitely many
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(
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)
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None
7
1
Hide problems
Turkey NMO 2000 1st Round - P07 (Combinatorics)
Some of
A
,
B
,
C
,
D
,
A,B,C,D,
A
,
B
,
C
,
D
,
and
E
E
E
are truth tellers, and the others are liars. Truth tellers always tell the truth. Liars always lie. We know
A
A
A
is a truth teller. According to below conversation,
B
:
B:
B
:
I'm a truth teller.
C
:
C:
C
:
D
D
D
is a truth teller.
D
:
D:
D
:
B
B
B
and
E
E
E
are not both truth tellers.
E
:
E:
E
:
A
A
A
and
B
B
B
are truth tellers.How many truth tellers are there?
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)
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)
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2
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)
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3
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c
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a
s
s
=
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x
−
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′
>
(
D
)
<
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>
4
<
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=
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>
(
E
)
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More information is needed
<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{More information is needed}
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1
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)
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(
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)
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3
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−
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>
(
D
)
<
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>
4
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>
(
E
)
<
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>
More information is needed
6
1
Hide problems
Turkey NMO 2000 1st Round - P06 (Number Theory)
What is the largest prime
p
p
p
that makes
17
p
+
625
\sqrt{17p+625}
17
p
+
625
an integer?
<
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>
(
A
)
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>
3
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>
(
B
)
<
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>
67
<
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(
C
)
<
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>
101
<
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c
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a
s
s
=
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−
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>
(
D
)
<
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>
151
<
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>
(
E
)
<
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211
<span class='latex-bold'>(A)</span>\ 3 \qquad<span class='latex-bold'>(B)</span>\ 67 \qquad<span class='latex-bold'>(C)</span>\ 101 \qquad<span class='latex-bold'>(D)</span>\ 151 \qquad<span class='latex-bold'>(E)</span>\ 211
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(
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)
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3
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−
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(
B
)
<
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>
67
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(
C
)
<
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101
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(
D
)
<
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>
151
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(
E
)
<
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>
211
5
1
Hide problems
Turkey NMO 2000 1st Round - P05 (Geometry)
[
B
D
]
[BD]
[
B
D
]
is a median of
△
A
B
C
\triangle ABC
△
A
BC
.
m
(
A
B
D
^
)
=
9
0
∘
m(\widehat{ABD})=90^\circ
m
(
A
B
D
)
=
9
0
∘
,
∣
A
B
∣
=
2
|AB|=2
∣
A
B
∣
=
2
, and
∣
A
C
∣
=
6
|AC|=6
∣
A
C
∣
=
6
.
∣
B
C
∣
=
?
|BC|=?
∣
BC
∣
=
?
<
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A
)
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3
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(
B
)
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3
2
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5
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)
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4
2
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2
6
<span class='latex-bold'>(A)</span>\ 3 \qquad<span class='latex-bold'>(B)</span>\ 3\sqrt2 \qquad<span class='latex-bold'>(C)</span>\ 5 \qquad<span class='latex-bold'>(D)</span>\ 4\sqrt2 \qquad<span class='latex-bold'>(E)</span>\ 2\sqrt6
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4
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2
6
4
1
Hide problems
Turkey NMO 2000 1st Round - P04 (Algebra)
What is the sum of real roots of
(
x
x
)
x
=
x
x
x
(x\sqrt{x})^x = x^{x\sqrt{x}}
(
x
x
)
x
=
x
x
x
?
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A
)
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18
7
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(
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)
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71
4
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9
4
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24
19
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13
4
<span class='latex-bold'>(A)</span>\ \frac{18}{7} \qquad<span class='latex-bold'>(B)</span>\ \frac{71}{4} \qquad<span class='latex-bold'>(C)</span>\ \frac{9}{4} \qquad<span class='latex-bold'>(D)</span>\ \frac{24}{19} \qquad<span class='latex-bold'>(E)</span>\ \frac{13}{4}
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18
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71
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9
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19
24
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4
13
3
1
Hide problems
Turkey NMO 2000 1st Round - P03 (Combinatorics)
In how many ways can the numbers
0
,
1
,
2
,
…
,
9
0,1,2,\dots , 9
0
,
1
,
2
,
…
,
9
be arranged in such a way that the odd numbers form an increasing sequence, also the even numbers form an increasing sequence?
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126
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(
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)
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189
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252
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)
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315
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None
<span class='latex-bold'>(A)</span>\ 126 \qquad<span class='latex-bold'>(B)</span>\ 189 \qquad<span class='latex-bold'>(C)</span>\ 252 \qquad<span class='latex-bold'>(D)</span>\ 315 \qquad<span class='latex-bold'>(E)</span>\ \text{None}
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126
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189
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252
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315
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None
2
1
Hide problems
Turkey NMO 2000 1st Round - P02 (Number Theory)
Discriminant of a second degree polynomial with integer coefficients cannot be
<
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23
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24
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)
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25
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)
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>
28
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E
)
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33
<span class='latex-bold'>(A)</span>\ 23 \qquad<span class='latex-bold'>(B)</span>\ 24 \qquad<span class='latex-bold'>(C)</span>\ 25 \qquad<span class='latex-bold'>(D)</span>\ 28 \qquad<span class='latex-bold'>(E)</span>\ 33
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)
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an
>
23
<
s
p
an
c
l
a
ss
=
′
l
a
t
e
x
−
b
o
l
d
′
>
(
B
)
<
/
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>
24
<
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p
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c
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a
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a
t
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−
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(
C
)
<
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25
<
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a
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x
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(
D
)
<
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28
<
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(
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)
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33
1
1
Hide problems
Turkey NMO 2000 1st Round - P01 (Geometry)
If the incircle of a right triangle with area
a
a
a
is the circumcircle of a right triangle with area
b
b
b
, what is the minimum value of
a
b
\frac{a}{b}
b
a
?
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(
A
)
<
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3
+
2
2
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−
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(
B
)
<
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>
1
+
2
<
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l
a
s
s
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2
2
<
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=
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−
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>
(
D
)
<
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>
2
+
3
<
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−
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(
E
)
<
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>
2
3
<span class='latex-bold'>(A)</span>\ 3 + 2\sqrt2 \qquad<span class='latex-bold'>(B)</span>\ 1+\sqrt2 \qquad<span class='latex-bold'>(C)</span>\ 2\sqrt2 \qquad<span class='latex-bold'>(D)</span>\ 2+\sqrt3 \qquad<span class='latex-bold'>(E)</span>\ 2\sqrt3
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(
A
)
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3
+
2
2
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(
B
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1
+
2
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−
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C
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2
2
<
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D
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<
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2
+
3
<
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(
E
)
<
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2
3
35
1
Hide problems
Turkish NMO First Round - 2000 P-35 (Combinatorics)
If every
k
−
k-
k
−
element subset of
S
=
{
1
,
2
,
…
,
32
}
S=\{1,2,\dots , 32\}
S
=
{
1
,
2
,
…
,
32
}
contains three different elements
a
,
b
,
c
a,b,c
a
,
b
,
c
such that
a
a
a
divides
b
b
b
, and
b
b
b
divides
c
c
c
,
k
k
k
must be at least ?
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(
A
)
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17
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(
B
)
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24
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a
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−
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(
C
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>
25
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−
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(
D
)
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>
29
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)
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None
<span class='latex-bold'>(A)</span>\ 17 \qquad<span class='latex-bold'>(B)</span>\ 24 \qquad<span class='latex-bold'>(C)</span>\ 25 \qquad<span class='latex-bold'>(D)</span>\ 29 \qquad<span class='latex-bold'>(E)</span>\ \text{None}
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17
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B
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24
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C
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25
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D
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29
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E
)
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None
34
1
Hide problems
Turkish NMO First Round - 2000 P-34 (Number Theory)
Which statement is not true for at least one prime
p
p
p
?
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(
A
)
<
/
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a
n
>
If
x
2
+
x
+
3
≡
0
(
m
o
d
p
)
has a solution, then
x
2
+
x
+
25
≡
0
(
m
o
d
p
)
has a solution.
<
s
p
a
n
c
l
a
s
s
=
′
l
a
t
e
x
−
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o
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>
(
B
)
<
/
s
p
a
n
>
If
x
2
+
x
+
3
≡
0
(
m
o
d
p
)
does not have a solution, then
x
2
+
x
+
25
≡
0
(
m
o
d
p
)
has no solution
<
s
p
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n
c
l
a
s
s
=
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a
t
e
x
−
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o
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>
(
C
)
<
/
s
p
a
n
>
If
x
2
+
x
+
25
≡
0
(
m
o
d
p
)
has a solution, then
x
2
+
x
+
3
≡
0
(
m
o
d
p
)
has a solution
.
<
s
p
a
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c
l
a
s
s
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l
a
t
e
x
−
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o
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>
(
D
)
<
/
s
p
a
n
>
If
x
2
+
x
+
25
≡
0
(
m
o
d
p
)
does not have a solution, then
x
2
+
x
+
3
≡
0
(
m
o
d
p
)
has no solution.
<
s
p
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n
c
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a
s
s
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>
(
E
)
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None
<span class='latex-bold'>(A)</span>\ \text{If } x^2+x+3 \equiv 0 \pmod p \text{ has a solution, then } \\ \qquad x^2+x+25 \equiv 0 \pmod p \text{ has a solution.} \\ \\ <span class='latex-bold'>(B)</span>\ \text{If } x^2+x+3 \equiv 0 \pmod p \text{ does not have a solution, then} \\ \qquad x^2+x+25 \equiv 0 \pmod p \text{ has no solution} \\ \\ \qquad<span class='latex-bold'>(C)</span>\ \text{If } x^2+x+25 \equiv 0 \pmod p \text{ has a solution, then} \\ \qquad x^2+x+3 \equiv 0 \pmod p \text{ has a solution}. \\ \\ \qquad<span class='latex-bold'>(D)</span>\ \text{If } x^2+x+25 \equiv 0 \pmod p \text{ does not have a solution, then} \\ \qquad x^2+x+3 \equiv 0 \pmod p \text{ has no solution. } \\ \\ \qquad<span class='latex-bold'>(E)</span>\ \text{None}
<
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(
A
)
<
/
s
p
an
>
If
x
2
+
x
+
3
≡
0
(
mod
p
)
has a solution, then
x
2
+
x
+
25
≡
0
(
mod
p
)
has a solution.
<
s
p
an
c
l
a
ss
=
′
l
a
t
e
x
−
b
o
l
d
′
>
(
B
)
<
/
s
p
an
>
If
x
2
+
x
+
3
≡
0
(
mod
p
)
does not have a solution, then
x
2
+
x
+
25
≡
0
(
mod
p
)
has no solution
<
s
p
an
c
l
a
ss
=
′
l
a
t
e
x
−
b
o
l
d
′
>
(
C
)
<
/
s
p
an
>
If
x
2
+
x
+
25
≡
0
(
mod
p
)
has a solution, then
x
2
+
x
+
3
≡
0
(
mod
p
)
has a solution
.
<
s
p
an
c
l
a
ss
=
′
l
a
t
e
x
−
b
o
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d
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>
(
D
)
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/
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an
>
If
x
2
+
x
+
25
≡
0
(
mod
p
)
does not have a solution, then
x
2
+
x
+
3
≡
0
(
mod
p
)
has no solution.
<
s
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−
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E
)
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None
32
1
Hide problems
Turkish NMO First Round - 2000 P-32 (Algebra)
Find the sum of all possible values of
f
(
2
)
f(2)
f
(
2
)
such that
f
(
x
)
f
(
y
)
−
f
(
x
y
)
=
y
x
+
x
y
f(x)f(y)-f(xy) = \frac{y}{x}+\frac{x}{y}
f
(
x
)
f
(
y
)
−
f
(
x
y
)
=
x
y
+
y
x
, for every positive real numbers
x
,
y
x,y
x
,
y
<
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(
A
)
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>
5
2
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(
B
)
<
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−
5
4
<
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4
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(
D
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>
3
2
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(
E
)
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None
<span class='latex-bold'>(A)</span>\ \frac{5}{2} \qquad<span class='latex-bold'>(B)</span>\ -\frac{5}{4} \qquad<span class='latex-bold'>(C)</span>\ \frac{5}{4} \qquad<span class='latex-bold'>(D)</span>\ \frac{3}{2} \qquad<span class='latex-bold'>(E)</span>\ \text{None}
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A
)
<
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2
5
<
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c
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a
t
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x
−
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>
(
B
)
<
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>
−
4
5
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a
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=
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−
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C
)
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4
5
<
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p
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a
ss
=
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l
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x
−
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>
(
D
)
<
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>
2
3
<
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−
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>
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E
)
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None
30
1
Hide problems
Turkish NMO First Round - 2000 P-30 (Number Theory)
How many ordered integer pairs
(
x
,
y
)
(x,y)
(
x
,
y
)
(
0
≤
x
,
y
<
31
0\leq x,y < 31
0
≤
x
,
y
<
31
) are there satisfying
(
x
2
−
18
)
2
≡
y
2
(
m
o
d
31
)
(x^2-18)^2\equiv y^2 (\mod 31)
(
x
2
−
18
)
2
≡
y
2
(
mod
31
)
?
<
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)
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59
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(
B
)
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60
<
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C
)
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61
<
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(
D
)
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62
<
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None
<span class='latex-bold'>(A)</span>\ 59 \qquad<span class='latex-bold'>(B)</span>\ 60 \qquad<span class='latex-bold'>(C)</span>\ 61 \qquad<span class='latex-bold'>(D)</span>\ 62 \qquad<span class='latex-bold'>(E)</span>\ \text{None}
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A
)
<
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59
<
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(
B
)
<
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60
<
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−
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o
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(
C
)
<
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61
<
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c
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a
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=
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a
t
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x
−
b
o
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d
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>
(
D
)
<
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>
62
<
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c
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a
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=
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a
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x
−
b
o
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d
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>
(
E
)
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>
None
12
1
Hide problems
Turkish NMO First Round - 2000 P-12 (Algebra)
(
a
n
)
(a_n)
(
a
n
)
is a sequence with
a
1
=
1
a_1=1
a
1
=
1
and
∣
a
n
∣
=
∣
a
n
−
1
+
2
∣
|a_n| = |a_{n-1}+2|
∣
a
n
∣
=
∣
a
n
−
1
+
2∣
for every positive integer
n
≥
2
n\geq 2
n
≥
2
. What is the minimum possible value of
∑
i
=
1
2000
a
i
\sum_{i = 1}^{2000}a_{i}
∑
i
=
1
2000
a
i
?
<
s
p
a
n
c
l
a
s
s
=
′
l
a
t
e
x
−
b
o
l
d
′
>
(
A
)
<
/
s
p
a
n
>
−
4000
<
s
p
a
n
c
l
a
s
s
=
′
l
a
t
e
x
−
b
o
l
d
′
>
(
B
)
<
/
s
p
a
n
>
−
3000
<
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
>
−
1000
<
s
p
a
n
c
l
a
s
s
=
′
l
a
t
e
x
−
b
o
l
d
′
>
(
E
)
<
/
s
p
a
n
>
None
<span class='latex-bold'>(A)</span>\ -4000 \qquad<span class='latex-bold'>(B)</span>\ -3000 \qquad<span class='latex-bold'>(C)</span>\ -2000 \qquad<span class='latex-bold'>(D)</span>\ -1000 \qquad<span class='latex-bold'>(E)</span>\ \text{None}
<
s
p
an
c
l
a
ss
=
′
l
a
t
e
x
−
b
o
l
d
′
>
(
A
)
<
/
s
p
an
>
−
4000
<
s
p
an
c
l
a
ss
=
′
l
a
t
e
x
−
b
o
l
d
′
>
(
B
)
<
/
s
p
an
>
−
3000
<
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
>
−
1000
<
s
p
an
c
l
a
ss
=
′
l
a
t
e
x
−
b
o
l
d
′
>
(
E
)
<
/
s
p
an
>
None