MathDB
Problems
Contests
National and Regional Contests
Turkey Contests
National Olympiad First Round
1999 National Olympiad First Round
1999 National Olympiad First Round
Part of
National Olympiad First Round
Subcontests
(36)
33
1
Hide problems
Turkish NMO First Round - 1999 P-33 (Geometry)
\left|AC\right| \equal{} 8 \sqrt {2},
B
B
B
is the midpoint of
[
A
C
]
\left[AC\right]
[
A
C
]
,
E
E
E
is the midpoint of arc
A
B
AB
A
B
of a circle having chord
[
A
B
]
\left[AB\right]
[
A
B
]
, and
D
D
D
is the point of tangency drawing from
C
C
C
.(
D
D
D
lies on the opposite side of line
A
B
AB
A
B
to
E
E
E
). If \left[DE\right]\bigcap \left[AB\right] \equal{} \left\{F\right\}, \left|CF\right| \equal{} ?
<
s
p
a
n
c
l
a
s
s
=
′
l
a
t
e
x
−
b
o
l
d
′
>
(
A
)
<
/
s
p
a
n
>
5
2
<
s
p
a
n
c
l
a
s
s
=
′
l
a
t
e
x
−
b
o
l
d
′
>
(
B
)
<
/
s
p
a
n
>
4
2
<
s
p
a
n
c
l
a
s
s
=
′
l
a
t
e
x
−
b
o
l
d
′
>
(
C
)
<
/
s
p
a
n
>
8
<
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
>
4
3
<span class='latex-bold'>(A)</span>\ 5\sqrt {2} \qquad<span class='latex-bold'>(B)</span>\ 4\sqrt {2} \qquad<span class='latex-bold'>(C)</span>\ 8 \qquad<span class='latex-bold'>(D)</span>\ 6 \qquad<span class='latex-bold'>(E)</span>\ 4\sqrt {3}
<
s
p
an
c
l
a
ss
=
′
l
a
t
e
x
−
b
o
l
d
′
>
(
A
)
<
/
s
p
an
>
5
2
<
s
p
an
c
l
a
ss
=
′
l
a
t
e
x
−
b
o
l
d
′
>
(
B
)
<
/
s
p
an
>
4
2
<
s
p
an
c
l
a
ss
=
′
l
a
t
e
x
−
b
o
l
d
′
>
(
C
)
<
/
s
p
an
>
8
<
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
>
4
3
31
1
Hide problems
Turkish NMO First Round - 1999 P-31 (Combinatorics)
30
30
30
same balls are put into four boxes
A
A
A
,
B
B
B
,
C
C
C
,
D
D
D
in such a way that sum of number of balls in
A
A
A
and
B
B
B
is greater than sum of in
C
C
C
and
D
D
D
. How many possible ways 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
>
2472
<
s
p
a
n
c
l
a
s
s
=
′
l
a
t
e
x
−
b
o
l
d
′
>
(
B
)
<
/
s
p
a
n
>
2600
<
s
p
a
n
c
l
a
s
s
=
′
l
a
t
e
x
−
b
o
l
d
′
>
(
C
)
<
/
s
p
a
n
>
2728
<
s
p
a
n
c
l
a
s
s
=
′
l
a
t
e
x
−
b
o
l
d
′
>
(
D
)
<
/
s
p
a
n
>
2856
<
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>\ 2472 \qquad<span class='latex-bold'>(B)</span>\ 2600 \qquad<span class='latex-bold'>(C)</span>\ 2728 \qquad<span class='latex-bold'>(D)</span>\ 2856 \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
>
2472
<
s
p
an
c
l
a
ss
=
′
l
a
t
e
x
−
b
o
l
d
′
>
(
B
)
<
/
s
p
an
>
2600
<
s
p
an
c
l
a
ss
=
′
l
a
t
e
x
−
b
o
l
d
′
>
(
C
)
<
/
s
p
an
>
2728
<
s
p
an
c
l
a
ss
=
′
l
a
t
e
x
−
b
o
l
d
′
>
(
D
)
<
/
s
p
an
>
2856
<
s
p
an
c
l
a
ss
=
′
l
a
t
e
x
−
b
o
l
d
′
>
(
E
)
<
/
s
p
an
>
None
29
1
Hide problems
Turkish NMO First Round - 1999 P-29 (Geometry)
The length of the altitude of equilateral triangle
A
B
C
ABC
A
BC
is
3
3
3
. A circle with radius
2
2
2
, which is tangent to
[
B
C
]
\left[BC\right]
[
BC
]
at its midpoint, meets other two sides. If the circle meets
A
B
AB
A
B
and
A
C
AC
A
C
at
D
D
D
and
E
E
E
, at the outer of
△
A
B
C
\triangle ABC
△
A
BC
, find the ratio
A
r
e
a
(
A
B
C
)
A
r
e
a
(
A
D
E
)
\frac {Area\, \left(ABC\right)}{Area\, \left(ADE\right)}
A
re
a
(
A
D
E
)
A
re
a
(
A
BC
)
.
(A)
\ 2\left(5 \plus{} \sqrt {3} \right) \qquad
(B)
\ 7\sqrt {2} \qquad
(C)
\ 5\sqrt {3} \\ \qquad
(D)
\ 2\left(3 \plus{} \sqrt {5} \right) \qquad
(E)
\ 2\left(\sqrt {3} \plus{} \sqrt {5} \right)
26
1
Hide problems
Turkish NMO First Round - 1999 P-26 (Number Theory)
Let
x
x
x
,
y
y
y
,
z
z
z
be integers such that \begin{array}{l} {x \minus{} 3y \plus{} 2z \equal{} 1} \\ {2x \plus{} y \minus{} 5z \equal{} 7} \end{array} Then
z
z
z
can be
<
s
p
a
n
c
l
a
s
s
=
′
l
a
t
e
x
−
b
o
l
d
′
>
(
A
)
<
/
s
p
a
n
>
3
111
<
s
p
a
n
c
l
a
s
s
=
′
l
a
t
e
x
−
b
o
l
d
′
>
(
B
)
<
/
s
p
a
n
>
4
111
<
s
p
a
n
c
l
a
s
s
=
′
l
a
t
e
x
−
b
o
l
d
′
>
(
C
)
<
/
s
p
a
n
>
5
111
<
s
p
a
n
c
l
a
s
s
=
′
l
a
t
e
x
−
b
o
l
d
′
>
(
D
)
<
/
s
p
a
n
>
6
111
<
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>\ 3^{111} \qquad<span class='latex-bold'>(B)</span>\ 4^{111} \qquad<span class='latex-bold'>(C)</span>\ 5^{111} \qquad<span class='latex-bold'>(D)</span>\ 6^{111} \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
>
3
111
<
s
p
an
c
l
a
ss
=
′
l
a
t
e
x
−
b
o
l
d
′
>
(
B
)
<
/
s
p
an
>
4
111
<
s
p
an
c
l
a
ss
=
′
l
a
t
e
x
−
b
o
l
d
′
>
(
C
)
<
/
s
p
an
>
5
111
<
s
p
an
c
l
a
ss
=
′
l
a
t
e
x
−
b
o
l
d
′
>
(
D
)
<
/
s
p
an
>
6
111
<
s
p
an
c
l
a
ss
=
′
l
a
t
e
x
−
b
o
l
d
′
>
(
E
)
<
/
s
p
an
>
None
25
1
Hide problems
Turkish NMO First Round - 1999 P-25 (Geometry)
\angle BAC \equal{} 80{}^\circ, \left|AB\right| \equal{} \left|AC\right|,
K
∈
[
A
B
]
K\in \left[AB\right]
K
∈
[
A
B
]
,
L
∈
[
A
B
L\in \left[AB\right.
L
∈
[
A
B
, \left|AB\right|^{2} \equal{} \left|AK\right|\cdot \left|AL\right|, \left|BL\right| \equal{} \left|BC\right|, \angle KCB \equal{} ?
<
s
p
a
n
c
l
a
s
s
=
′
l
a
t
e
x
−
b
o
l
d
′
>
(
A
)
<
/
s
p
a
n
>
2
0
∘
<
s
p
a
n
c
l
a
s
s
=
′
l
a
t
e
x
−
b
o
l
d
′
>
(
B
)
<
/
s
p
a
n
>
2
5
∘
<
s
p
a
n
c
l
a
s
s
=
′
l
a
t
e
x
−
b
o
l
d
′
>
(
C
)
<
/
s
p
a
n
>
3
0
∘
<
s
p
a
n
c
l
a
s
s
=
′
l
a
t
e
x
−
b
o
l
d
′
>
(
D
)
<
/
s
p
a
n
>
3
5
∘
<
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>\ 20^\circ \qquad<span class='latex-bold'>(B)</span>\ 25^\circ \qquad<span class='latex-bold'>(C)</span>\ 30^\circ \qquad<span class='latex-bold'>(D)</span>\ 35^\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
0
∘
<
s
p
an
c
l
a
ss
=
′
l
a
t
e
x
−
b
o
l
d
′
>
(
B
)
<
/
s
p
an
>
2
5
∘
<
s
p
an
c
l
a
ss
=
′
l
a
t
e
x
−
b
o
l
d
′
>
(
C
)
<
/
s
p
an
>
3
0
∘
<
s
p
an
c
l
a
ss
=
′
l
a
t
e
x
−
b
o
l
d
′
>
(
D
)
<
/
s
p
an
>
3
5
∘
<
s
p
an
c
l
a
ss
=
′
l
a
t
e
x
−
b
o
l
d
′
>
(
E
)
<
/
s
p
an
>
4
0
∘
24
1
Hide problems
Turkish NMO First Round - 1999 P-24 (Algebra)
Polynomial
f
(
x
)
f\left(x\right)
f
(
x
)
satisfies \left(x \minus{} 1\right)f\left(x \plus{} 1\right) \minus{} \left(x \plus{} 2\right)f\left(x\right) \equal{} 0 for every
x
∈
ℜ
x\in \Re
x
∈
ℜ
. If f\left(2\right) \equal{} 6, f\left({\tfrac{3}{2}} \right) \equal{} ?
<
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
>
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
2
<
s
p
a
n
c
l
a
s
s
=
′
l
a
t
e
x
−
b
o
l
d
′
>
(
D
)
<
/
s
p
a
n
>
15
8
<
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>\ -6 \qquad<span class='latex-bold'>(B)</span>\ 0 \qquad<span class='latex-bold'>(C)</span>\ \frac {3}{2} \qquad<span class='latex-bold'>(D)</span>\ \frac {15}{8} \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
>
−
6
<
s
p
an
c
l
a
ss
=
′
l
a
t
e
x
−
b
o
l
d
′
>
(
B
)
<
/
s
p
an
>
0
<
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
>
8
15
<
s
p
an
c
l
a
ss
=
′
l
a
t
e
x
−
b
o
l
d
′
>
(
E
)
<
/
s
p
an
>
None
23
1
Hide problems
Turkish NMO First Round - 1999 P-23 (Combinatorics)
Hour part of a defective digital watch displays only the numbers from
1
1
1
to
12
12
12
. After one minute from
n
:
59
n: 59
n
:
59
, although it must display \left(n \plus{} 1\right): 00, it displays
2
n
:
00
2n: 00
2
n
:
00
(Think in
m
o
d
12
mod\, 12
m
o
d
12
). For example, after
7
:
59
7: 59
7
:
59
, it displays
2
:
00
2: 00
2
:
00
instead of
8
:
00
8: 00
8
:
00
. If we set the watch to an arbitrary time, what is the probability that hour part displays
4
4
4
after exactly one day?
<
s
p
a
n
c
l
a
s
s
=
′
l
a
t
e
x
−
b
o
l
d
′
>
(
A
)
<
/
s
p
a
n
>
1
12
<
s
p
a
n
c
l
a
s
s
=
′
l
a
t
e
x
−
b
o
l
d
′
>
(
B
)
<
/
s
p
a
n
>
1
4
<
s
p
a
n
c
l
a
s
s
=
′
l
a
t
e
x
−
b
o
l
d
′
>
(
C
)
<
/
s
p
a
n
>
1
3
<
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
>
None
<span class='latex-bold'>(A)</span>\ \frac {1}{12} \qquad<span class='latex-bold'>(B)</span>\ \frac {1}{4} \qquad<span class='latex-bold'>(C)</span>\ \frac {1}{3} \qquad<span class='latex-bold'>(D)</span>\ \frac {1}{2} \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
>
12
1
<
s
p
an
c
l
a
ss
=
′
l
a
t
e
x
−
b
o
l
d
′
>
(
B
)
<
/
s
p
an
>
4
1
<
s
p
an
c
l
a
ss
=
′
l
a
t
e
x
−
b
o
l
d
′
>
(
C
)
<
/
s
p
an
>
3
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
>
None
22
1
Hide problems
Turkish NMO First Round - 1999 P-22 (Number Theory)
If
m
,
n
∈
Z
m,n\in Z
m
,
n
∈
Z
, then m^{2} \plus{} 3mn \minus{} 4n^{2} cannot be
<
s
p
a
n
c
l
a
s
s
=
′
l
a
t
e
x
−
b
o
l
d
′
>
(
A
)
<
/
s
p
a
n
>
69
<
s
p
a
n
c
l
a
s
s
=
′
l
a
t
e
x
−
b
o
l
d
′
>
(
B
)
<
/
s
p
a
n
>
76
<
s
p
a
n
c
l
a
s
s
=
′
l
a
t
e
x
−
b
o
l
d
′
>
(
C
)
<
/
s
p
a
n
>
91
<
s
p
a
n
c
l
a
s
s
=
′
l
a
t
e
x
−
b
o
l
d
′
>
(
D
)
<
/
s
p
a
n
>
94
<
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>\ 69 \qquad<span class='latex-bold'>(B)</span>\ 76 \qquad<span class='latex-bold'>(C)</span>\ 91 \qquad<span class='latex-bold'>(D)</span>\ 94 \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
>
69
<
s
p
an
c
l
a
ss
=
′
l
a
t
e
x
−
b
o
l
d
′
>
(
B
)
<
/
s
p
an
>
76
<
s
p
an
c
l
a
ss
=
′
l
a
t
e
x
−
b
o
l
d
′
>
(
C
)
<
/
s
p
an
>
91
<
s
p
an
c
l
a
ss
=
′
l
a
t
e
x
−
b
o
l
d
′
>
(
D
)
<
/
s
p
an
>
94
<
s
p
an
c
l
a
ss
=
′
l
a
t
e
x
−
b
o
l
d
′
>
(
E
)
<
/
s
p
an
>
None
21
1
Hide problems
Turkish NMO First Round - 1999 P-21 (Geometry)
A
B
C
ABC
A
BC
is a triangle with \angle BAC \equal{} 10{}^\circ, \angle ABC \equal{} 150{}^\circ. Let
X
X
X
be a point on
[
A
C
]
\left[AC\right]
[
A
C
]
such that \left|AX\right| \equal{} \left|BC\right|. Find
∠
B
X
C
\angle BXC
∠
BXC
.
<
s
p
a
n
c
l
a
s
s
=
′
l
a
t
e
x
−
b
o
l
d
′
>
(
A
)
<
/
s
p
a
n
>
1
5
∘
<
s
p
a
n
c
l
a
s
s
=
′
l
a
t
e
x
−
b
o
l
d
′
>
(
B
)
<
/
s
p
a
n
>
2
0
∘
<
s
p
a
n
c
l
a
s
s
=
′
l
a
t
e
x
−
b
o
l
d
′
>
(
C
)
<
/
s
p
a
n
>
2
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
0
∘
<
s
p
a
n
c
l
a
s
s
=
′
l
a
t
e
x
−
b
o
l
d
′
>
(
E
)
<
/
s
p
a
n
>
3
5
∘
<span class='latex-bold'>(A)</span>\ 15^\circ \qquad<span class='latex-bold'>(B)</span>\ 20^\circ \qquad<span class='latex-bold'>(C)</span>\ 25^\circ \qquad<span class='latex-bold'>(D)</span>\ 30^\circ \qquad<span class='latex-bold'>(E)</span>\ 35^\circ
<
s
p
an
c
l
a
ss
=
′
l
a
t
e
x
−
b
o
l
d
′
>
(
A
)
<
/
s
p
an
>
1
5
∘
<
s
p
an
c
l
a
ss
=
′
l
a
t
e
x
−
b
o
l
d
′
>
(
B
)
<
/
s
p
an
>
2
0
∘
<
s
p
an
c
l
a
ss
=
′
l
a
t
e
x
−
b
o
l
d
′
>
(
C
)
<
/
s
p
an
>
2
5
∘
<
s
p
an
c
l
a
ss
=
′
l
a
t
e
x
−
b
o
l
d
′
>
(
D
)
<
/
s
p
an
>
3
0
∘
<
s
p
an
c
l
a
ss
=
′
l
a
t
e
x
−
b
o
l
d
′
>
(
E
)
<
/
s
p
an
>
3
5
∘
17
1
Hide problems
Turkish NMO First Round - 1999 P-17 (Geometry)
In a regular pyramid with top point
T
T
T
and equilateral base
A
B
C
ABC
A
BC
, let
P
P
P
,
Q
Q
Q
,
R
R
R
,
S
S
S
be the midpoints of
[
A
B
]
\left[AB\right]
[
A
B
]
,
[
B
C
]
\left[BC\right]
[
BC
]
,
[
C
T
]
\left[CT\right]
[
CT
]
and
[
T
A
]
\left[TA\right]
[
T
A
]
, respectively. If \left|AB\right| \equal{} 6 and the altitude of pyramid is equal to
2
15
2\sqrt {15}
2
15
, then area of
P
Q
R
S
PQRS
PQRS
will be
<
s
p
a
n
c
l
a
s
s
=
′
l
a
t
e
x
−
b
o
l
d
′
>
(
A
)
<
/
s
p
a
n
>
4
15
<
s
p
a
n
c
l
a
s
s
=
′
l
a
t
e
x
−
b
o
l
d
′
>
(
B
)
<
/
s
p
a
n
>
8
2
<
s
p
a
n
c
l
a
s
s
=
′
l
a
t
e
x
−
b
o
l
d
′
>
(
C
)
<
/
s
p
a
n
>
8
3
<
s
p
a
n
c
l
a
s
s
=
′
l
a
t
e
x
−
b
o
l
d
′
>
(
D
)
<
/
s
p
a
n
>
6
5
<
s
p
a
n
c
l
a
s
s
=
′
l
a
t
e
x
−
b
o
l
d
′
>
(
E
)
<
/
s
p
a
n
>
9
2
<span class='latex-bold'>(A)</span>\ 4\sqrt {15} \qquad<span class='latex-bold'>(B)</span>\ 8\sqrt {2} \qquad<span class='latex-bold'>(C)</span>\ 8\sqrt {3} \qquad<span class='latex-bold'>(D)</span>\ 6\sqrt {5} \qquad<span class='latex-bold'>(E)</span>\ 9\sqrt {2}
<
s
p
an
c
l
a
ss
=
′
l
a
t
e
x
−
b
o
l
d
′
>
(
A
)
<
/
s
p
an
>
4
15
<
s
p
an
c
l
a
ss
=
′
l
a
t
e
x
−
b
o
l
d
′
>
(
B
)
<
/
s
p
an
>
8
2
<
s
p
an
c
l
a
ss
=
′
l
a
t
e
x
−
b
o
l
d
′
>
(
C
)
<
/
s
p
an
>
8
3
<
s
p
an
c
l
a
ss
=
′
l
a
t
e
x
−
b
o
l
d
′
>
(
D
)
<
/
s
p
an
>
6
5
<
s
p
an
c
l
a
ss
=
′
l
a
t
e
x
−
b
o
l
d
′
>
(
E
)
<
/
s
p
an
>
9
2
13
1
Hide problems
Turkish NMO First Round - 1999 P-13 (Geometry)
Square
B
D
E
C
BDEC
B
D
EC
with center
F
F
F
is constructed to the out of triangle
A
B
C
ABC
A
BC
such that \angle A \equal{} 90{}^\circ, \left|AB\right| \equal{} \sqrt {12}, \left|AC\right| \equal{} 2. If \left[AF\right]\bigcap \left[BC\right] \equal{} \left\{G\right\} , then
∣
B
G
∣
\left|BG\right|
∣
BG
∣
will be
(A)
\ 6 \minus{} 2\sqrt {3} \qquad
(B)
\ 2\sqrt {3} \minus{} 1 \qquad
(C)
\ 2 \plus{} \sqrt {3} \\ \qquad
(D)
\ 4 \minus{} \sqrt {3} \qquad
(E)
\ 5 \minus{} 2\sqrt {2}
10
1
Hide problems
Turkish NMO First Round - 1999 P-10 (Number Theory)
For every integers
a
,
b
,
c
a,b,c
a
,
b
,
c
whose greatest common divisor is
n
n
n
, if \begin{array}{l} {x \plus{} 2y \plus{} 3z \equal{} a} \\ {2x \plus{} y \minus{} 2z \equal{} b} \\ {3x \plus{} y \plus{} 5z \equal{} c} \end{array} has a solution in integers, what is the smallest possible value of positive number
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
>
7
<
s
p
a
n
c
l
a
s
s
=
′
l
a
t
e
x
−
b
o
l
d
′
>
(
B
)
<
/
s
p
a
n
>
14
<
s
p
a
n
c
l
a
s
s
=
′
l
a
t
e
x
−
b
o
l
d
′
>
(
C
)
<
/
s
p
a
n
>
28
<
s
p
a
n
c
l
a
s
s
=
′
l
a
t
e
x
−
b
o
l
d
′
>
(
D
)
<
/
s
p
a
n
>
56
<
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>\ 7 \qquad<span class='latex-bold'>(B)</span>\ 14 \qquad<span class='latex-bold'>(C)</span>\ 28 \qquad<span class='latex-bold'>(D)</span>\ 56 \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
>
7
<
s
p
an
c
l
a
ss
=
′
l
a
t
e
x
−
b
o
l
d
′
>
(
B
)
<
/
s
p
an
>
14
<
s
p
an
c
l
a
ss
=
′
l
a
t
e
x
−
b
o
l
d
′
>
(
C
)
<
/
s
p
an
>
28
<
s
p
an
c
l
a
ss
=
′
l
a
t
e
x
−
b
o
l
d
′
>
(
D
)
<
/
s
p
an
>
56
<
s
p
an
c
l
a
ss
=
′
l
a
t
e
x
−
b
o
l
d
′
>
(
E
)
<
/
s
p
an
>
None
9
1
Hide problems
Turkish NMO First Round - 1999 P-09 (Geometry)
Find the area of inscribed convex octagon, if the length of four sides is
2
2
2
, and length of other four sides is
6
2
6\sqrt {2}
6
2
.
(A)
\ 120 \qquad
(B)
\ 24 \plus{} 68\sqrt {2} \qquad
(C)
\ 88\sqrt {2} \qquad
(D)
\ 124 \qquad
(E)
\ 72\sqrt {3}
5
1
Hide problems
Turkish NMO First Round - 1999 P-05 (Geometry)
Let
A
B
C
ABC
A
BC
be an isosceles triangle with \left|AB\right| \equal{} \left|AC\right| \equal{} 10 and \left|BC\right| \equal{} 12.
P
P
P
and
R
R
R
are points on
[
B
C
]
\left[BC\right]
[
BC
]
such that \left|BP\right| \equal{} \left|RC\right| \equal{} 3.
S
S
S
and
T
T
T
are midpoints of
[
A
B
]
\left[AB\right]
[
A
B
]
and
[
A
C
]
\left[AC\right]
[
A
C
]
, respectively. If
M
M
M
and
N
N
N
are the foot of perpendiculars from
S
S
S
and
R
R
R
to
P
T
PT
PT
, then find
∣
M
N
∣
\left|MN\right|
∣
MN
∣
.
(A)
\ \frac {9\sqrt {13} }{26} \qquad
(B)
\ \frac {12 \minus{} 2\sqrt {13} }{13} \qquad
(C)
\ \frac {5\sqrt {13} \plus{} 20}{13} \qquad
(D)
\ 15\sqrt {3} \qquad
(E)
\ \frac {10\sqrt {13} }{13}
3
1
Hide problems
Turkish NMO First Round - 1999 P-03 (Combinatorics)
Four boxes with ball capacity
3
,
5
,
7
,
3, 5, 7,
3
,
5
,
7
,
and
8
8
8
are given. In how many ways can
19
19
19
same balls be put into these boxes?
<
s
p
a
n
c
l
a
s
s
=
′
l
a
t
e
x
−
b
o
l
d
′
>
(
A
)
<
/
s
p
a
n
>
34
<
s
p
a
n
c
l
a
s
s
=
′
l
a
t
e
x
−
b
o
l
d
′
>
(
B
)
<
/
s
p
a
n
>
35
<
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
>
40
<
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>\ 34 \qquad<span class='latex-bold'>(B)</span>\ 35 \qquad<span class='latex-bold'>(C)</span>\ 36 \qquad<span class='latex-bold'>(D)</span>\ 40 \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
>
34
<
s
p
an
c
l
a
ss
=
′
l
a
t
e
x
−
b
o
l
d
′
>
(
B
)
<
/
s
p
an
>
35
<
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
>
40
<
s
p
an
c
l
a
ss
=
′
l
a
t
e
x
−
b
o
l
d
′
>
(
E
)
<
/
s
p
an
>
None
2
1
Hide problems
Turkish NMO First Round - 1999 P-02 (Number Theory)
How many ordered integer pairs
(
x
,
y
)
\left(x,y\right)
(
x
,
y
)
are there such that xy \equal{} 4\left(y^{2} \plus{} x\right)?
<
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
>
3
<
s
p
a
n
c
l
a
s
s
=
′
l
a
t
e
x
−
b
o
l
d
′
>
(
C
)
<
/
s
p
a
n
>
7
<
s
p
a
n
c
l
a
s
s
=
′
l
a
t
e
x
−
b
o
l
d
′
>
(
D
)
<
/
s
p
a
n
>
14
<
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>\ 0 \qquad<span class='latex-bold'>(B)</span>\ 3 \qquad<span class='latex-bold'>(C)</span>\ 7 \qquad<span class='latex-bold'>(D)</span>\ 14 \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
>
0
<
s
p
an
c
l
a
ss
=
′
l
a
t
e
x
−
b
o
l
d
′
>
(
B
)
<
/
s
p
an
>
3
<
s
p
an
c
l
a
ss
=
′
l
a
t
e
x
−
b
o
l
d
′
>
(
C
)
<
/
s
p
an
>
7
<
s
p
an
c
l
a
ss
=
′
l
a
t
e
x
−
b
o
l
d
′
>
(
D
)
<
/
s
p
an
>
14
<
s
p
an
c
l
a
ss
=
′
l
a
t
e
x
−
b
o
l
d
′
>
(
E
)
<
/
s
p
an
>
None
1
1
Hide problems
Turkish NMO First Round - 1999 P-01 (Geometry)
Let
A
B
C
ABC
A
BC
be a triangle with \left|AB\right| \equal{} 14, \left|BC\right| \equal{} 12, \left|AC\right| \equal{} 10. Let
D
D
D
be a point on
[
A
C
]
\left[AC\right]
[
A
C
]
and
E
E
E
be a point on
[
B
C
]
\left[BC\right]
[
BC
]
such that \left|AD\right| \equal{} 4 and Area\left(ABC\right) \equal{} 2Area\left(CDE\right). Find
A
r
e
a
(
A
B
E
)
Area\left(ABE\right)
A
re
a
(
A
BE
)
.
<
s
p
a
n
c
l
a
s
s
=
′
l
a
t
e
x
−
b
o
l
d
′
>
(
A
)
<
/
s
p
a
n
>
4
6
<
s
p
a
n
c
l
a
s
s
=
′
l
a
t
e
x
−
b
o
l
d
′
>
(
B
)
<
/
s
p
a
n
>
6
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
6
<
s
p
a
n
c
l
a
s
s
=
′
l
a
t
e
x
−
b
o
l
d
′
>
(
D
)
<
/
s
p
a
n
>
4
2
<
s
p
a
n
c
l
a
s
s
=
′
l
a
t
e
x
−
b
o
l
d
′
>
(
E
)
<
/
s
p
a
n
>
4
5
<span class='latex-bold'>(A)</span>\ 4\sqrt {6} \qquad<span class='latex-bold'>(B)</span>\ 6\sqrt {2} \qquad<span class='latex-bold'>(C)</span>\ 3\sqrt {6} \qquad<span class='latex-bold'>(D)</span>\ 4\sqrt {2} \qquad<span class='latex-bold'>(E)</span>\ 4\sqrt {5}
<
s
p
an
c
l
a
ss
=
′
l
a
t
e
x
−
b
o
l
d
′
>
(
A
)
<
/
s
p
an
>
4
6
<
s
p
an
c
l
a
ss
=
′
l
a
t
e
x
−
b
o
l
d
′
>
(
B
)
<
/
s
p
an
>
6
2
<
s
p
an
c
l
a
ss
=
′
l
a
t
e
x
−
b
o
l
d
′
>
(
C
)
<
/
s
p
an
>
3
6
<
s
p
an
c
l
a
ss
=
′
l
a
t
e
x
−
b
o
l
d
′
>
(
D
)
<
/
s
p
an
>
4
2
<
s
p
an
c
l
a
ss
=
′
l
a
t
e
x
−
b
o
l
d
′
>
(
E
)
<
/
s
p
an
>
4
5
36
1
Hide problems
Trigonometric Inequality
Let
x
1
,
x
2
,
…
,
x
9
x_{1} ,x_{2} ,\ldots ,x_{9}
x
1
,
x
2
,
…
,
x
9
be real numbers on \left[ \minus{} 1,1\right]. If \sum _{i \equal{} 1}^{9}x_{i}^{3} \equal{} 0, then what is the largest possible value of \sum _{i \equal{} 1}^{9}x_{i}?
<
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
>
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
<
s
p
a
n
c
l
a
s
s
=
′
l
a
t
e
x
−
b
o
l
d
′
>
(
D
)
<
/
s
p
a
n
>
9
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
<span class='latex-bold'>(A)</span>\ 1 \qquad<span class='latex-bold'>(B)</span>\ \frac {3}{2} \qquad<span class='latex-bold'>(C)</span>\ 3 \qquad<span class='latex-bold'>(D)</span>\ \frac {9}{2} \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
>
1
<
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
>
3
<
s
p
an
c
l
a
ss
=
′
l
a
t
e
x
−
b
o
l
d
′
>
(
D
)
<
/
s
p
an
>
2
9
<
s
p
an
c
l
a
ss
=
′
l
a
t
e
x
−
b
o
l
d
′
>
(
E
)
<
/
s
p
an
>
None
35
1
Hide problems
Combinatorics
Flights are arranged between 13 countries. For
k
≥
2
k\ge 2
k
≥
2
, the sequence
A
1
,
A
2
,
…
A
k
A_{1} ,A_{2} ,\ldots A_{k}
A
1
,
A
2
,
…
A
k
is said to a cycle if there exist a flight from
A
1
A_{1}
A
1
to
A
2
A_{2}
A
2
, from
A
2
A_{2}
A
2
to
A
3
A_{3}
A
3
,
…
\ldots
…
, from A_{k \minus{} 1} to
A
k
A_{k}
A
k
, and from
A
k
A_{k}
A
k
to
A
1
A_{1}
A
1
. What is the smallest possible number of flights such that how the flights are arranged, there exist a cycle?
<
s
p
a
n
c
l
a
s
s
=
′
l
a
t
e
x
−
b
o
l
d
′
>
(
A
)
<
/
s
p
a
n
>
14
<
s
p
a
n
c
l
a
s
s
=
′
l
a
t
e
x
−
b
o
l
d
′
>
(
B
)
<
/
s
p
a
n
>
53
<
s
p
a
n
c
l
a
s
s
=
′
l
a
t
e
x
−
b
o
l
d
′
>
(
C
)
<
/
s
p
a
n
>
66
<
s
p
a
n
c
l
a
s
s
=
′
l
a
t
e
x
−
b
o
l
d
′
>
(
D
)
<
/
s
p
a
n
>
79
<
s
p
a
n
c
l
a
s
s
=
′
l
a
t
e
x
−
b
o
l
d
′
>
(
E
)
<
/
s
p
a
n
>
156
<span class='latex-bold'>(A)</span>\ 14 \qquad<span class='latex-bold'>(B)</span>\ 53 \qquad<span class='latex-bold'>(C)</span>\ 66 \qquad<span class='latex-bold'>(D)</span>\ 79 \qquad<span class='latex-bold'>(E)</span>\ 156
<
s
p
an
c
l
a
ss
=
′
l
a
t
e
x
−
b
o
l
d
′
>
(
A
)
<
/
s
p
an
>
14
<
s
p
an
c
l
a
ss
=
′
l
a
t
e
x
−
b
o
l
d
′
>
(
B
)
<
/
s
p
an
>
53
<
s
p
an
c
l
a
ss
=
′
l
a
t
e
x
−
b
o
l
d
′
>
(
C
)
<
/
s
p
an
>
66
<
s
p
an
c
l
a
ss
=
′
l
a
t
e
x
−
b
o
l
d
′
>
(
D
)
<
/
s
p
an
>
79
<
s
p
an
c
l
a
ss
=
′
l
a
t
e
x
−
b
o
l
d
′
>
(
E
)
<
/
s
p
an
>
156
34
1
Hide problems
Mod p
For how many primes
p
p
p
, there exits unique integers
r
r
r
and
s
s
s
such that for every integer
x
x
x
x^{3} \minus{} x \plus{} 2\equiv \left(x \minus{} r\right)^{2} \left(x \minus{} s\right)\pmod p?
<
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
>
None
<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>\ \text{None}
<
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
>
None
32
1
Hide problems
Sequence
Let \left(a_{n} \right)_{n \equal{} 1}^{\infty } be a sequence on real numbers such that a{}_{n \plus{} 1} \equal{} a_{n} a_{n \plus{} 2} for every
n
≥
1
n\ge 1
n
≥
1
. The number of elements in the set
{
a
n
:
n
≥
1
}
\left\{a_{n} : n\ge 1\right\}
{
a
n
:
n
≥
1
}
cannot be
<
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
>
3
<
s
p
a
n
c
l
a
s
s
=
′
l
a
t
e
x
−
b
o
l
d
′
>
(
C
)
<
/
s
p
a
n
>
4
<
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
>
None
<span class='latex-bold'>(A)</span>\ 2 \qquad<span class='latex-bold'>(B)</span>\ 3 \qquad<span class='latex-bold'>(C)</span>\ 4 \qquad<span class='latex-bold'>(D)</span>\ 5 \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
>
2
<
s
p
an
c
l
a
ss
=
′
l
a
t
e
x
−
b
o
l
d
′
>
(
B
)
<
/
s
p
an
>
3
<
s
p
an
c
l
a
ss
=
′
l
a
t
e
x
−
b
o
l
d
′
>
(
C
)
<
/
s
p
an
>
4
<
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
>
None
30
1
Hide problems
Modular Arithmetics
a
i
∈
{
0
,
1
,
2
,
3
,
4
}
a_{i} \in \left\{0,1,2,3,4\right\}
a
i
∈
{
0
,
1
,
2
,
3
,
4
}
for every
0
≤
i
≤
9
0\le i\le 9
0
≤
i
≤
9
. If 6\sum _{i \equal{} 0}^{9}a_{i} 5^{i} \equiv 1\, \, \left(mod\, 5^{10} \right), a_{9} \equal{} ?
<
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
28
1
Hide problems
Function
Find the number of functions defined on positive real numbers such that f\left(1\right) \equal{} 1 and for every
x
,
y
∈
ℜ
x,y\in \Re
x
,
y
∈
ℜ
, f\left(x^{2} y^{2} \right) \equal{} f\left(x^{4} \plus{} y^{4} \right).
<
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
>
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>\ 0 \qquad<span class='latex-bold'>(B)</span>\ 1 \qquad<span class='latex-bold'>(C)</span>\ 2 \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
>
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
>
4
<
s
p
an
c
l
a
ss
=
′
l
a
t
e
x
−
b
o
l
d
′
>
(
E
)
<
/
s
p
an
>
Infinitely many
27
1
Hide problems
Coloring
Points on a square with side length
c
c
c
are either painted blue or red. Find the smallest possible value of
c
c
c
such that how the points are painted, there exist two points with same color having a distance not less than
5
\sqrt {5}
5
.
<
s
p
a
n
c
l
a
s
s
=
′
l
a
t
e
x
−
b
o
l
d
′
>
(
A
)
<
/
s
p
a
n
>
10
2
<
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
>
5
<
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
>
None
<span class='latex-bold'>(A)</span>\ \frac {\sqrt {10} }{2} \qquad<span class='latex-bold'>(B)</span>\ 2 \qquad<span class='latex-bold'>(C)</span>\ \sqrt {5} \qquad<span class='latex-bold'>(D)</span>\ 2\sqrt {2} \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
>
2
10
<
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
>
5
<
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
>
None
20
1
Hide problems
Floor
How many pairs of real numbers
(
x
,
y
)
\left(x,y\right)
(
x
,
y
)
are there such that x^{4} \minus{} 2^{ \minus{} y^{2} } x^{2} \minus{} \left\| x^{2} \right\| \plus{} 1 \equal{} 0, where
∥
a
∥
\left\| a\right\|
∥
a
∥
denotes the greatest integer not exceeding
a
a
a
.
<
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
>
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>\ 0 \qquad<span class='latex-bold'>(B)</span>\ 1 \qquad<span class='latex-bold'>(C)</span>\ 2 \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
>
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
>
4
<
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
Number Theory
Let
t
k
(
n
)
t_{k} \left(n\right)
t
k
(
n
)
show the sum of
k
t
h
k^{th}
k
t
h
power of digits of positive number
n
n
n
. For which
k
k
k
, the condition that
t
k
(
n
)
t_{k} \left(n\right)
t
k
(
n
)
is a multiple of 3 does not imply the condition that
n
n
n
is a multiple of 3?
<
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
>
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
<
s
p
a
n
c
l
a
s
s
=
′
l
a
t
e
x
−
b
o
l
d
′
>
(
D
)
<
/
s
p
a
n
>
15
<
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>\ 3 \qquad<span class='latex-bold'>(B)</span>\ 6 \qquad<span class='latex-bold'>(C)</span>\ 9 \qquad<span class='latex-bold'>(D)</span>\ 15 \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
>
3
<
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
>
9
<
s
p
an
c
l
a
ss
=
′
l
a
t
e
x
−
b
o
l
d
′
>
(
D
)
<
/
s
p
an
>
15
<
s
p
an
c
l
a
ss
=
′
l
a
t
e
x
−
b
o
l
d
′
>
(
E
)
<
/
s
p
an
>
None
19
1
Hide problems
A game
k
k
k
black pieces are placed on
k
k
k
consecutive squares of top row starting from upper left of a
2
×
5
2\times 5
2
×
5
board. We are placing white pieces on empty squares one by one in arbitrary order. Two squares is said to adjacent if they have common vertex. When a white piece is placed on a square, the pieces on adjacent squares change their color. For which
k
k
k
, when all the squares are filled, it is possible that color of every piece is white?
<
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
>
None
<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>\ \text{None}
<
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
>
None
16
1
Hide problems
Rational roots
How many pairs of rational number
(
x
,
y
)
\left(x,y\right)
(
x
,
y
)
are there satisfying the equation y \equal{} \sqrt {x^{2} \plus{} \frac {1}{1999} }?
<
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
>
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>\ 0 \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
>
0
<
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
15
1
Hide problems
Combination
2 squares are painted in blue and 2 squares are painted in red on a
3
×
3
3\times 3
3
×
3
board in such a way that two square with same color is neither at same row nor at same column. In how many different ways can these four squares be painted?
<
s
p
a
n
c
l
a
s
s
=
′
l
a
t
e
x
−
b
o
l
d
′
>
(
A
)
<
/
s
p
a
n
>
198
<
s
p
a
n
c
l
a
s
s
=
′
l
a
t
e
x
−
b
o
l
d
′
>
(
B
)
<
/
s
p
a
n
>
288
<
s
p
a
n
c
l
a
s
s
=
′
l
a
t
e
x
−
b
o
l
d
′
>
(
C
)
<
/
s
p
a
n
>
396
<
s
p
a
n
c
l
a
s
s
=
′
l
a
t
e
x
−
b
o
l
d
′
>
(
D
)
<
/
s
p
a
n
>
576
<
s
p
a
n
c
l
a
s
s
=
′
l
a
t
e
x
−
b
o
l
d
′
>
(
E
)
<
/
s
p
a
n
>
792
<span class='latex-bold'>(A)</span>\ 198 \qquad<span class='latex-bold'>(B)</span>\ 288 \qquad<span class='latex-bold'>(C)</span>\ 396 \qquad<span class='latex-bold'>(D)</span>\ 576 \qquad<span class='latex-bold'>(E)</span>\ 792
<
s
p
an
c
l
a
ss
=
′
l
a
t
e
x
−
b
o
l
d
′
>
(
A
)
<
/
s
p
an
>
198
<
s
p
an
c
l
a
ss
=
′
l
a
t
e
x
−
b
o
l
d
′
>
(
B
)
<
/
s
p
an
>
288
<
s
p
an
c
l
a
ss
=
′
l
a
t
e
x
−
b
o
l
d
′
>
(
C
)
<
/
s
p
an
>
396
<
s
p
an
c
l
a
ss
=
′
l
a
t
e
x
−
b
o
l
d
′
>
(
D
)
<
/
s
p
an
>
576
<
s
p
an
c
l
a
ss
=
′
l
a
t
e
x
−
b
o
l
d
′
>
(
E
)
<
/
s
p
an
>
792
14
1
Hide problems
Division
Find the sum of squares of the digits of the least positive integer having
72
72
72
positive divisors.
<
s
p
a
n
c
l
a
s
s
=
′
l
a
t
e
x
−
b
o
l
d
′
>
(
A
)
<
/
s
p
a
n
>
41
<
s
p
a
n
c
l
a
s
s
=
′
l
a
t
e
x
−
b
o
l
d
′
>
(
B
)
<
/
s
p
a
n
>
65
<
s
p
a
n
c
l
a
s
s
=
′
l
a
t
e
x
−
b
o
l
d
′
>
(
C
)
<
/
s
p
a
n
>
110
<
s
p
a
n
c
l
a
s
s
=
′
l
a
t
e
x
−
b
o
l
d
′
>
(
D
)
<
/
s
p
a
n
>
123
<
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>\ 41 \qquad<span class='latex-bold'>(B)</span>\ 65 \qquad<span class='latex-bold'>(C)</span>\ 110 \qquad<span class='latex-bold'>(D)</span>\ 123 \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
>
41
<
s
p
an
c
l
a
ss
=
′
l
a
t
e
x
−
b
o
l
d
′
>
(
B
)
<
/
s
p
an
>
65
<
s
p
an
c
l
a
ss
=
′
l
a
t
e
x
−
b
o
l
d
′
>
(
C
)
<
/
s
p
an
>
110
<
s
p
an
c
l
a
ss
=
′
l
a
t
e
x
−
b
o
l
d
′
>
(
D
)
<
/
s
p
an
>
123
<
s
p
an
c
l
a
ss
=
′
l
a
t
e
x
−
b
o
l
d
′
>
(
E
)
<
/
s
p
an
>
None
12
1
Hide problems
Equatiıon System
\begin{array}{c} {x^{2} \plus{} y^{2} \plus{} z^{2} \equal{} 21} \\ {x \plus{} y \plus{} z \plus{} xyz \equal{} \minus{} 3} \\ {x^{2} yz \plus{} y^{2} xz \plus{} z^{2} xy \equal{} \minus{} 40} \end{array} The number of real triples
(
x
,
y
,
z
)
\left(x,y,z\right)
(
x
,
y
,
z
)
satisfying above system is
<
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
>
3
<
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
>
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
<span class='latex-bold'>(A)</span>\ 0 \qquad<span class='latex-bold'>(B)</span>\ 3 \qquad<span class='latex-bold'>(C)</span>\ 6 \qquad<span class='latex-bold'>(D)</span>\ 12 \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
>
0
<
s
p
an
c
l
a
ss
=
′
l
a
t
e
x
−
b
o
l
d
′
>
(
B
)
<
/
s
p
an
>
3
<
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
>
12
<
s
p
an
c
l
a
ss
=
′
l
a
t
e
x
−
b
o
l
d
′
>
(
E
)
<
/
s
p
an
>
None
11
1
Hide problems
Put the numbers into boxes
Place all numbers from 1 to 10 to the boxes such that every number except the uppermost is equal to the difference between the two numbers on its top. [asy] unitsize(-4); draw((0,0)--(5,0)--(5,5)--(0,5)--cycle); draw((10,0)--(15,0)--(15,5)--(10,5)--cycle); draw((20,0)--(25,0)--(25,5)--(20,5)--cycle); draw((30,0)--(35,0)--(35,5)--(30,5)--cycle);draw((5,10)--(10,10)--(10,15)--(5,15)--cycle); draw((15,10)--(20,10)--(20,15)--(15,15)--cycle); draw((25,10)--(30,10)--(30,15)--(25,15)--cycle);draw((10,20)--(15,20)--(15,25)--(10,25)--cycle); draw((20,20)--(25,20)--(25,25)--(20,25)--cycle);draw((15,30)--(20,30)--(20,35)--(15,35)--cycle);draw((2.5,5)--(7.5, 10)); draw((12.5,5)--(17.5, 10)); draw((22.5,5)--(27.5, 10));draw((32.5,5)--(27.5, 10)); draw((22.5,5)--(17.5, 10)); draw((12.5,5)--(7.5, 10));draw((7.5,15)--(12.5, 20)); draw((17.5,15)--(22.5, 20));draw((27.5,15)--(22.5, 20)); draw((17.5,15)--(12.5, 20));draw((12.5,25)--(17.5, 30)); draw((22.5,25)--(17.5, 30)); [/asy]The number in the lower box is at most
<
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
8
1
Hide problems
Polynomial
If the polynomial
P
(
x
)
P\left(x\right)
P
(
x
)
satisfies 2P\left(x\right) \equal{} P\left(x \plus{} 3\right) \plus{} P\left(x \minus{} 3\right) for every real number
x
x
x
, degree of
P
(
x
)
P\left(x\right)
P
(
x
)
will be at most
<
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
>
None
<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>\ \text{None}
<
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
>
None
7
1
Hide problems
Probability
Six cards with numbers 1, 1, 3, 4, 4, 5 are given. We are drawing 3 cards from 6 given cards one by one and are forming a three-digit number with the numbers over the cards drawn according to the drawing order. Find the probability that this three-digit number is a multiple of 3. (The card drawn is not put back)
<
s
p
a
n
c
l
a
s
s
=
′
l
a
t
e
x
−
b
o
l
d
′
>
(
A
)
<
/
s
p
a
n
>
1
5
<
s
p
a
n
c
l
a
s
s
=
′
l
a
t
e
x
−
b
o
l
d
′
>
(
B
)
<
/
s
p
a
n
>
2
5
<
s
p
a
n
c
l
a
s
s
=
′
l
a
t
e
x
−
b
o
l
d
′
>
(
C
)
<
/
s
p
a
n
>
3
7
<
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
>
None
<span class='latex-bold'>(A)</span>\ \frac {1}{5} \qquad<span class='latex-bold'>(B)</span>\ \frac {2}{5} \qquad<span class='latex-bold'>(C)</span>\ \frac {3}{7} \qquad<span class='latex-bold'>(D)</span>\ \frac {1}{2} \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
>
5
1
<
s
p
an
c
l
a
ss
=
′
l
a
t
e
x
−
b
o
l
d
′
>
(
B
)
<
/
s
p
an
>
5
2
<
s
p
an
c
l
a
ss
=
′
l
a
t
e
x
−
b
o
l
d
′
>
(
C
)
<
/
s
p
an
>
7
3
<
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
>
None
4
1
Hide problems
Inequality
If inequality \frac {\sin ^{3} x}{\cos x} \plus{} \frac {\cos ^{3} x}{\sin x} \ge k is hold for every
x
∈
(
0
,
π
2
)
x\in \left(0,\frac {\pi }{2} \right)
x
∈
(
0
,
2
π
)
, what is the largest possible value of
k
k
k
?
<
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
4
<
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
>
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
<span class='latex-bold'>(A)</span>\ \frac {1}{2} \qquad<span class='latex-bold'>(B)</span>\ \frac {3}{4} \qquad<span class='latex-bold'>(C)</span>\ 1 \qquad<span class='latex-bold'>(D)</span>\ \frac {3}{2} \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
>
2
1
<
s
p
an
c
l
a
ss
=
′
l
a
t
e
x
−
b
o
l
d
′
>
(
B
)
<
/
s
p
an
>
4
3
<
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
>
2
3
<
s
p
an
c
l
a
ss
=
′
l
a
t
e
x
−
b
o
l
d
′
>
(
E
)
<
/
s
p
an
>
None
6
1
Hide problems
Chinese Remainder
If
a
,
b
,
c
∈
Z
a,b,c\in {\rm Z}
a
,
b
,
c
∈
Z
and
x
≡
a
(
m
o
d
14
)
x
≡
b
(
m
o
d
15
)
x
≡
c
(
m
o
d
16
)
\begin{array}{l} {x\equiv a\, \, \, \pmod{14}} \\ {x\equiv b\, \, \, \pmod {15}} \\ {x\equiv c\, \, \, \pmod {16}} \end{array}
x
≡
a
(
mod
14
)
x
≡
b
(
mod
15
)
x
≡
c
(
mod
16
)
, the number of integral solutions of the congruence system on the interval
0
≤
x
<
2000
0\le x < 2000
0
≤
x
<
2000
cannot be
<
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
>
None
<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>\ \text{None}
<
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
>
None