MathDB
Problems
Contests
National and Regional Contests
Vietnam Contests
Hanoi Open Mathematics Competition
2010 Hanoi Open Mathematics Competitions
2010 Hanoi Open Mathematics Competitions
Part of
Hanoi Open Mathematics Competition
Subcontests
(10)
1
1
Hide problems
compare 888...888 x333..333, 444...444x666...667 (HOMC 2010 J Q1)
Compare the numbers: P = 888...888 \times 333 .. 333 (
2010
2010
2010
digits of
8
8
8
and
2010
2010
2010
digits of
3
3
3
) and
Q
=
444...444
×
666...6667
Q = 444...444\times 666...6667
Q
=
444...444
×
666...6667
(
2010
2010
2010
digits of
4
4
4
and
2009
2009
2009
digits of
6
6
6
)(A):
P
=
Q
P = Q
P
=
Q
, (B):
P
>
Q
P > Q
P
>
Q
, (C):
P
<
Q
P < Q
P
<
Q
.
4
1
Hide problems
a \in (1,9) such (a - 1/a) is an integer (HOMC 2010 J Q4)
How many real numbers
a
∈
(
1
,
9
)
a \in (1,9)
a
∈
(
1
,
9
)
such that the corresponding number
a
−
1
a
a- \frac1a
a
−
a
1
is an integer?(A):
0
0
0
, (B):
1
1
1
, (C):
8
8
8
, (D):
9
9
9
, (E) None of the above.
3
1
Hide problems
5 last digits of 5^{2010} (HOMC 2010 J Q3)
Find
5
5
5
last digits of the number
M
=
5
2010
M = 5^{2010}
M
=
5
2010
.(A):
65625
65625
65625
, (B):
45625
45625
45625
, (C):
25625
25625
25625
, (D):
15625
15625
15625
, (E) None of the above.
2
1
Hide problems
2^{2n} + 2^n + 5 divisible by 5 (HOMC 2010 J Q2)
Find the number of integer
n
n
n
from the set
{
2000
,
2001
,
.
.
.
,
2010
}
\{2000,2001,...,2010\}
{
2000
,
2001
,
...
,
2010
}
such that
2
2
n
+
2
n
+
5
2^{2n} + 2^n + 5
2
2
n
+
2
n
+
5
is divisible by
7
7
7
(A):
0
0
0
, (B):
1
1
1
, (C):
2
2
2
, (D):
3
3
3
, (E) None of the above.
6
2
Hide problems
greatest integer <= (2 +\sqrt3)^5 (HOMC 2010 J Q6)
Find the greatest integer less than
(
2
+
3
)
5
(2 +\sqrt3)^5
(
2
+
3
)
5
.(A):
721
721
721
(B):
722
722
722
(C):
723
723
723
(D):
724
724
724
(E) None of the above.
2(abc+bcd+cda+dab)/(p^2+q^2+r^2+s^2) \in Z (HOMC 2010 Q6)
Let
a
,
b
a,b
a
,
b
be the roots of the equation
x
2
−
p
x
+
q
=
0
x^2-px+q = 0
x
2
−
p
x
+
q
=
0
and let
c
,
d
c, d
c
,
d
be the roots of the equation
x
2
−
r
x
+
s
=
0
x^2 - rx + s = 0
x
2
−
r
x
+
s
=
0
, where
p
,
q
,
r
,
s
p, q, r,s
p
,
q
,
r
,
s
are some positive real numbers. Suppose that
M
=
2
(
a
b
c
+
b
c
d
+
c
d
a
+
d
a
b
)
p
2
+
q
2
+
r
2
+
s
2
M =\frac{2(abc + bcd + cda + dab)}{p^2 + q^2 + r^2 + s^2}
M
=
p
2
+
q
2
+
r
2
+
s
2
2
(
ab
c
+
b
c
d
+
c
d
a
+
d
ab
)
is an integer. Determine
a
,
b
,
c
,
d
a, b, c, d
a
,
b
,
c
,
d
.
5
1
Hide problems
different colorings of bw in a 2x2 table (HOMC 2010 J Q5)
Each box in a
2
x
2
2x2
2
x
2
table can be colored black or white. How many different colorings of the table are there?(A):
4
4
4
, (B):
8
8
8
, (C):
16
16
16
, (D):
32
32
32
, (E) None of the above.
7
1
Hide problems
integer a such 2x^2-30x+a=0 has prime roots (HOMC 2010 J Q7)
Determine all positive integer
a
a
a
such that the equation
2
x
2
−
30
x
+
a
=
0
2x^2 - 30x + a = 0
2
x
2
−
30
x
+
a
=
0
has two prime roots, i.e. both roots are prime numbers.
8
1
Hide problems
n and (n^3+2n^2+2n+4) both perfect squares (HOMC 2010 J Q8)
If
n
n
n
and
n
3
+
2
n
2
+
2
n
+
4
n^3+2n^2+2n+4
n
3
+
2
n
2
+
2
n
+
4
are both perfect squares, find
n
n
n
.
10
1
Hide problems
max of x/(2x + y) +y/(2y + z)+z/(2z + x) in R^+ (HOMC 2010 J Q10)
Find the maximum value of
M
=
x
2
x
+
y
+
y
2
y
+
z
+
z
2
z
+
x
M =\frac{x}{2x + y} +\frac{y}{2y + z}+\frac{z}{2z + x}
M
=
2
x
+
y
x
+
2
y
+
z
y
+
2
z
+
x
z
,
x
,
y
,
z
>
0
x,y, z > 0
x
,
y
,
z
>
0
9
2
Hide problems
area computational (2010 HOMC Junior Q9)
Let be given a triangle
A
B
C
ABC
A
BC
and points
D
,
M
,
N
D,M,N
D
,
M
,
N
belong to
B
C
,
A
B
,
A
C
BC,AB,AC
BC
,
A
B
,
A
C
, respectively. Suppose that
M
D
MD
M
D
is parallel to
A
C
AC
A
C
and
N
D
ND
N
D
is parallel to
A
B
AB
A
B
. If
S
△
B
M
D
=
9
S_{\vartriangle BMD} = 9
S
△
BM
D
=
9
cm
2
,
S
△
D
N
C
=
25
^2, S_{\vartriangle DNC} = 25
2
,
S
△
D
NC
=
25
cm
2
^2
2
, compute
S
△
A
M
N
S_{\vartriangle AMN}
S
△
A
MN
?
x,y\in N*, 3x^2+x=4y^2+y => x-y perfect integer (HOMC 2010 Q9)
Let
x
,
y
x,y
x
,
y
be the positive integers such that
3
x
2
+
x
=
4
y
2
+
y
3x^2 +x = 4y^2 + y
3
x
2
+
x
=
4
y
2
+
y
. Prove that
x
−
y
x - y
x
−
y
is a perfect (square).