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Problems
Contests
National and Regional Contests
Saudi Arabia Contests
Saudi Arabia Pre-TST + Training Tests
2013 Saudi Arabia Pre-TST
2013 Saudi Arabia Pre-TST
Part of
Saudi Arabia Pre-TST + Training Tests
Subcontests
(16)
4.3
1
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no of permutations of 1,2,..., n with s_i > s_j$ for all i >= j + 3 when n =5, 7
How many permutations
(
s
1
,
s
2
,
.
.
.
,
s
n
)
(s_1, s_2,...,s_n)
(
s
1
,
s
2
,
...
,
s
n
)
of
(
1
,
2
,
.
.
.
,
n
)
(1,2 ,...,n)
(
1
,
2
,
...
,
n
)
are there satisfying the condition
s
i
>
s
j
s_i > s_j
s
i
>
s
j
for all
i
≥
j
+
3
i \ge j + 3
i
≥
j
+
3
when
n
=
5
n = 5
n
=
5
and when
n
=
7
n = 7
n
=
7
?
4.2
1
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if 2013 divides $x^{1433} + y^{1433} then divides also x^7 + y^7
Let
x
,
y
x, y
x
,
y
be two integers. Prove that if
2013
2013
2013
divides
x
1433
+
y
1433
x^{1433} + y^{1433}
x
1433
+
y
1433
then
2013
2013
2013
divides
x
7
+
y
7
x^7 + y^7
x
7
+
y
7
.
4.1
1
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a_{n+1} =\sqrt{a_n^2 + 1}, a_{2n_0} = 3a_{n_0}, a_{46} wanted
Let
a
1
,
a
2
,
a
3
,
.
.
.
a_1,a_2, a_3,...
a
1
,
a
2
,
a
3
,
...
be a sequence of real numbers which satisfy the relation
a
n
+
1
=
a
n
2
+
1
a_{n+1} =\sqrt{a_n^2 + 1}
a
n
+
1
=
a
n
2
+
1
Suppose that there exists a positive integer
n
0
n_0
n
0
such that
a
2
n
0
=
3
a
n
0
a_{2n_0} = 3a_{n_0}
a
2
n
0
=
3
a
n
0
. Find the value of
a
46
a_{46}
a
46
.
3.3
1
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points of the plane have been colored by 2013 different colors
The points of the plane have been colored by
2013
2013
2013
different colors. We say that a triangle
△
A
B
C
\vartriangle ABC
△
A
BC
has the color
X
X
X
if its three vertices
A
,
B
,
C
A,B,C
A
,
B
,
C
has the color
X
X
X
. Prove that there are in nitely many triangles with the same color and the same area.
3.2
1
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if 19 divides a_1^9+a_2^9+...+a_9^9 then also divides a_1a_2...a_9
Let
a
1
,
a
2
,
.
.
.
,
a
9
a_1, a_2,..., a_9
a
1
,
a
2
,
...
,
a
9
be integers. Prove that if
19
19
19
divides
a
1
9
+
a
2
9
+
.
.
.
+
a
9
9
a_1^9+a_2^9+...+a_9^9
a
1
9
+
a
2
9
+
...
+
a
9
9
then
19
19
19
divides the product
a
1
a
2
.
.
.
a
9
a_1a_2...a_9
a
1
a
2
...
a
9
.
3.1
1
Hide problems
f(x)=x has unique solution when f(f(x)) = 4x + 1
Let
f
:
R
→
R
f : R \to R
f
:
R
→
R
be a function satisfying
f
(
f
(
x
)
)
=
4
x
+
1
f(f(x)) = 4x + 1
f
(
f
(
x
))
=
4
x
+
1
for all real number
x
x
x
. Prove that the equation
f
(
x
)
=
x
f(x) = x
f
(
x
)
=
x
has a unique solution.
2.3
1
Hide problems
exists a power of a whose last n digits are 0...01
The positive integer
a
a
a
is relatively prime with
10
10
10
. Prove that for any positive integer
n
n
n
, there exists a power of
a
a
a
whose last
n
n
n
digits are
0...0
⏟
n-1
1
\underbrace{0...0}_\text{n-1}1
n-1
0...0
1
.
2.2
1
Hide problems
max of (a - b)(2a - b)}/a(a - b + c) when ax^2 + bx + c = 0 has roots in [0, 1]
The quadratic equation
a
x
2
+
b
x
+
c
=
0
ax^2 + bx + c = 0
a
x
2
+
b
x
+
c
=
0
has its roots in the interval
[
0
,
1
]
[0, 1]
[
0
,
1
]
. Find the maximum of
(
a
−
b
)
(
2
a
−
b
)
a
(
a
−
b
+
c
)
\frac{(a - b)(2a - b)}{a(a - b + c)}
a
(
a
−
b
+
c
)
(
a
−
b
)
(
2
a
−
b
)
.
2.1
1
Hide problems
(a^4 - 1)(a^4 + 15a^2 + 1) = 0 mod 35
Prove that if
a
a
a
is an integer relatively prime with
35
35
35
then
(
a
4
−
1
)
(
a
4
+
15
a
2
+
1
)
≡
0
(a^4 - 1)(a^4 + 15a^2 + 1) \equiv 0
(
a
4
−
1
)
(
a
4
+
15
a
2
+
1
)
≡
0
mod
35
35
35
.
1.3
1
Hide problems
10 students take a test consisting of 4 different papers
Ten students take a test consisting of
4
4
4
different papers in Algebra, Geometry, Number Theory and Combinatorics. First, the proctor distributes randomly the Algebra paper to each student. Then the remaining papers are distributed one at a time in the following order: Geometry, Number Theory, Combinatorics in such a way that no student receives a paper before he fi nishes the previous one. In how many ways can the proctor distribute the test papers given that a student may for example nish the Number Theory paper before another student receives the Geometry paper, and that he receives the Combinatorics paper after that the same other student receives the Combinatorics papers.
1.2
1
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47 divides 3^x - 2^y iff 23 divides 4x + y.
Let
x
,
y
x, y
x
,
y
be two non-negative integers. Prove that
47
47
47
divides
3
x
−
2
y
3^x - 2^y
3
x
−
2
y
if and only if
23
23
23
divides
4
x
+
y
4x + y
4
x
+
y
.
1.1
1
Hide problems
\sqrt{(1- x^2)(1 - y^2) } <=2(1 - x)(1 - y) + 1
Let
−
1
≤
x
,
y
≤
1
-1 \le x, y \le 1
−
1
≤
x
,
y
≤
1
. Prove the inequality
2
(
1
−
x
2
)
(
1
−
y
2
)
≤
2
(
1
−
x
)
(
1
−
y
)
+
1
2\sqrt{(1- x^2)(1 - y^2) } \le 2(1 - x)(1 - y) + 1
2
(
1
−
x
2
)
(
1
−
y
2
)
≤
2
(
1
−
x
)
(
1
−
y
)
+
1
4.4
1
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lines AE,BD are perpendicular iff AB = AC
△
A
B
C
\vartriangle ABC
△
A
BC
is a triangle,
M
M
M
the midpoint of
B
C
,
D
BC, D
BC
,
D
the projection of
M
M
M
on
A
C
AC
A
C
and
E
E
E
the midppoint of
M
D
MD
M
D
. Prove that the lines
A
E
,
B
D
AE,BD
A
E
,
B
D
are orthogonal if and only if
A
B
=
A
C
AB = AC
A
B
=
A
C
.
3.4
1
Hide problems
concurrent wanted, incircle, arc midpoint of circumcircle, perpendicular related
△
A
B
C
\vartriangle ABC
△
A
BC
is a triangle with
A
B
<
B
C
,
Γ
AB < BC, \Gamma
A
B
<
BC
,
Γ
its circumcircle,
K
K
K
the midpoint of the minor arc
C
A
CA
C
A
of the circle
C
C
C
and
T
T
T
a point on
Γ
\Gamma
Γ
such that
K
T
KT
K
T
is perpendicular to
B
C
BC
BC
. If
A
′
,
B
′
A',B'
A
′
,
B
′
are the intouch points of the incircle of
△
A
B
C
\vartriangle ABC
△
A
BC
with the sides
B
C
,
A
C
BC,AC
BC
,
A
C
, prove that the lines
A
T
,
B
K
,
A
′
B
′
AT,BK,A'B'
A
T
,
B
K
,
A
′
B
′
are concurrent.
2.4
1
Hide problems
triangle area criterion for right triangle with excenters, 1/2 AI_b AI_c.
△
A
B
C
\vartriangle ABC
△
A
BC
is a triangle and
I
b
.
I
c
I_b. I_c
I
b
.
I
c
its excenters opposite to
B
,
C
B,C
B
,
C
. Prove that
△
A
B
C
\vartriangle ABC
△
A
BC
is right at
A
A
A
if and only if its area is equal to
1
2
A
I
b
⋅
A
I
c
\frac12 AI_b \cdot AI_c
2
1
A
I
b
⋅
A
I
c
.
1.4
1
Hide problems
AB x CC' = AC xBB' if AC'GB' is cyclic
A
B
C
ABC
A
BC
is a triangle,
G
G
G
its centroid and
A
′
,
B
′
,
C
′
A',B',C'
A
′
,
B
′
,
C
′
the midpoints of its sides
B
C
,
C
A
,
A
B
BC,CA,AB
BC
,
C
A
,
A
B
, respectively. Prove that if the quadrilateral
A
C
′
G
B
′
AC'GB'
A
C
′
G
B
′
is cyclic then
A
B
⋅
C
C
′
=
A
C
⋅
B
B
′
AB \cdot CC' = AC \cdot BB'
A
B
⋅
C
C
′
=
A
C
⋅
B
B
′
: