MathDB
Problems
Contests
National and Regional Contests
Turkey Contests
Turkey Team Selection Test
1990 Turkey Team Selection Test
1990 Turkey Team Selection Test
Part of
Turkey Team Selection Test
Subcontests
(6)
6
1
Hide problems
n_i | (2^n_(i-1) - 1)
Let
k
≥
2
k\geq 2
k
≥
2
and
n
1
,
…
,
n
k
∈
Z
+
n_1, \dots, n_k \in \mathbf{Z}^+
n
1
,
…
,
n
k
∈
Z
+
. If
n
2
∣
(
2
n
1
−
1
)
n_2 | (2^{n_1} -1)
n
2
∣
(
2
n
1
−
1
)
,
n
3
∣
(
2
n
2
−
1
)
n_3 | (2^{n_2} -1)
n
3
∣
(
2
n
2
−
1
)
,
…
\dots
…
,
n
k
∣
(
2
n
k
−
1
−
1
)
n_k | (2^{n_{k-1}} -1)
n
k
∣
(
2
n
k
−
1
−
1
)
,
n
1
∣
(
2
n
k
−
1
)
n_1 | (2^{n_k} -1)
n
1
∣
(
2
n
k
−
1
)
, show that
n
1
=
⋯
=
n
k
=
1
n_1 = \dots = n_k =1
n
1
=
⋯
=
n
k
=
1
.
5
1
Hide problems
m-b_m = 1990
Let
b
m
b_m
b
m
be numbers of factors
2
2
2
of the number
m
!
m!
m
!
(that is,
2
b
m
∣
m
!
2^{b_m}|m!
2
b
m
∣
m
!
and
2
b
m
+
1
∤
m
!
2^{b_m+1}\nmid m!
2
b
m
+
1
∤
m
!
). Find the least
m
m
m
such that
m
−
b
m
=
1990
m-b_m = 1990
m
−
b
m
=
1990
.
3
1
Hide problems
Two n-digit numbers sharing some digits in base 2
Let
n
n
n
be an odd integer greater than
11
11
11
;
k
∈
N
k\in \mathbb{N}
k
∈
N
,
k
≥
6
k \geq 6
k
≥
6
,
n
=
2
k
−
1
n=2k-1
n
=
2
k
−
1
. We define
d
(
x
,
y
)
=
∣
{
i
∈
{
1
,
2
,
…
,
n
}
∣
x
i
≠
y
i
}
∣
d(x,y) = \left | \{ i\in \{1,2,\dots, n \} \bigm | x_i \neq y_i \} \right |
d
(
x
,
y
)
=
{
i
∈
{
1
,
2
,
…
,
n
}
x
i
=
y
i
}
for
T
=
{
(
x
1
,
x
2
,
…
,
x
n
)
∣
x
i
∈
{
0
,
1
}
,
i
=
1
,
2
,
…
,
n
}
T=\{ (x_1, x_2, \dots, x_n) \bigm | x_i \in \{0,1\}, i=1,2,\dots, n \}
T
=
{(
x
1
,
x
2
,
…
,
x
n
)
x
i
∈
{
0
,
1
}
,
i
=
1
,
2
,
…
,
n
}
and
x
=
(
x
1
,
x
2
,
…
,
x
n
)
,
y
=
(
y
1
,
y
2
,
…
,
y
n
)
∈
T
x=(x_1,x_2,\dots, x_n), y=(y_1, y_2, \dots, y_n) \in T
x
=
(
x
1
,
x
2
,
…
,
x
n
)
,
y
=
(
y
1
,
y
2
,
…
,
y
n
)
∈
T
. Show that
n
=
23
n=23
n
=
23
if
T
T
T
has a subset
S
S
S
satisfying[*]
∣
S
∣
=
2
k
|S|=2^k
∣
S
∣
=
2
k
[*]For each
x
∈
T
x \in T
x
∈
T
, there exists exacly one
y
∈
S
y\in S
y
∈
S
such that
d
(
x
,
y
)
≤
3
d(x,y)\leq 3
d
(
x
,
y
)
≤
3
2
1
Hide problems
x_1 + x_2 + ... + x_n = 0
For real numbers
x
i
x_i
x
i
, the statement
x
1
+
x
2
+
x
3
=
0
⇒
x
1
x
2
+
x
2
x
3
+
x
3
x
1
≤
0
x_1 + x_2 + x_3 = 0 \Rightarrow x_1x_2 + x_2x_3 + x_3x_1 \leq 0
x
1
+
x
2
+
x
3
=
0
⇒
x
1
x
2
+
x
2
x
3
+
x
3
x
1
≤
0
is always true. (Prove!) For which
n
≥
4
n\geq 4
n
≥
4
integers, the statement
x
1
+
x
2
+
⋯
+
x
n
=
0
⇒
x
1
x
2
+
x
2
x
3
+
⋯
+
x
n
−
1
x
n
+
x
n
x
1
≤
0
x_1 + x_2 + \dots + x_n = 0 \Rightarrow x_1x_2 + x_2x_3 + \dots + x_{n-1}x_n + x_nx_1 \leq 0
x
1
+
x
2
+
⋯
+
x
n
=
0
⇒
x
1
x
2
+
x
2
x
3
+
⋯
+
x
n
−
1
x
n
+
x
n
x
1
≤
0
is always true. Justify your answer.
4
1
Hide problems
That area problem
Let
A
B
C
D
ABCD
A
BC
D
be a convex quadrilateral such that
E
,
F
∈
[
A
B
]
,
A
E
=
E
F
=
F
B
G
,
H
∈
[
B
C
]
,
B
G
=
G
H
=
H
C
K
,
L
∈
[
C
D
]
,
C
K
=
K
L
=
L
D
M
,
N
∈
[
D
A
]
,
D
M
=
M
N
=
N
A
\begin{array}{rl} E,F \in [AB],& AE = EF = FB \\ G,H \in [BC],& BG = GH = HC \\ K,L \in [CD],& CK = KL = LD \\ M,N \in [DA],& DM = MN = NA \end{array}
E
,
F
∈
[
A
B
]
,
G
,
H
∈
[
BC
]
,
K
,
L
∈
[
C
D
]
,
M
,
N
∈
[
D
A
]
,
A
E
=
EF
=
FB
BG
=
G
H
=
H
C
C
K
=
K
L
=
L
D
D
M
=
MN
=
N
A
Let
[
N
G
]
∩
[
L
E
]
=
{
P
}
,
[
N
G
]
∩
[
K
F
]
=
{
Q
}
,
[NG] \cap [LE] = \{P\}, [NG]\cap [KF] = \{Q\},
[
NG
]
∩
[
L
E
]
=
{
P
}
,
[
NG
]
∩
[
K
F
]
=
{
Q
}
,
[
M
H
]
∩
[
K
F
]
=
{
R
}
,
[
M
H
]
∩
[
L
E
]
=
{
S
}
{[}MH] \cap [KF] = \{R\}, [MH]\cap [LE]=\{S\}
[
M
H
]
∩
[
K
F
]
=
{
R
}
,
[
M
H
]
∩
[
L
E
]
=
{
S
}
Prove that [*]
A
r
e
a
(
A
B
C
D
)
=
9
⋅
A
r
e
a
(
P
Q
R
S
)
Area(ABCD) = 9 \cdot Area(PQRS)
A
re
a
(
A
BC
D
)
=
9
⋅
A
re
a
(
PQRS
)
[*]
N
P
=
P
Q
=
Q
G
NP=PQ=QG
NP
=
PQ
=
QG
1
1
Hide problems
Three tangent circles
The circles
k
1
,
k
2
,
k
3
k_1, k_2, k_3
k
1
,
k
2
,
k
3
with radii (
a
>
c
>
b
a>c>b
a
>
c
>
b
)
a
,
b
,
c
a,b,c
a
,
b
,
c
are tangent to line
d
d
d
at
A
,
B
,
C
A,B,C
A
,
B
,
C
, respectively.
k
1
k_1
k
1
is tangent to
k
2
k_2
k
2
, and
k
2
k_2
k
2
is tangent to
k
3
k_3
k
3
. The tangent line to
k
3
k_3
k
3
at
E
E
E
is parallel to
d
d
d
, and it meets
k
1
k_1
k
1
at
D
D
D
. The line perpendicular to
d
d
d
at
A
A
A
meets line
E
B
EB
EB
at
F
F
F
. Prove that
A
D
=
A
F
AD=AF
A
D
=
A
F
.