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Problems
Contests
National and Regional Contests
Singapore Contests
Singapore Junior Math Olympiad
2017 Singapore Junior Math Olympiad
2017 Singapore Junior Math Olympiad
Part of
Singapore Junior Math Olympiad
Subcontests
(5)
5
1
Hide problems
1/a+1/b+1/c=0 , gcd=1, no of triples with 50 \ge |a| \ge |b| \ge |c| 1
Let
a
,
b
,
c
a, b, c
a
,
b
,
c
be nonzero integers, with
1
1
1
as their only positive common divisor, such that
1
a
+
1
b
+
1
c
=
0
\frac{1}{a}+\frac{1}{b}+\frac{1}{c}= 0
a
1
+
b
1
+
c
1
=
0
. Find the number of such triples
(
a
,
b
,
c
)
(a, b, c)
(
a
,
b
,
c
)
with
50
≥
∣
a
∣
≥
∣
b
∣
≥
∣
c
∣
1
50 \ge |a| \ge |b| \ge |c| 1
50
≥
∣
a
∣
≥
∣
b
∣
≥
∣
c
∣1
.
4
1
Hide problems
m blue and n red vertices on a m+n polygon, numbering sides
Consider a polygon with
m
+
n
m + n
m
+
n
sides where
m
,
n
m, n
m
,
n
are positive integers. Colour
m
m
m
of its vertices red and the remaining
n
n
n
vertices blue. A side is given the number
2
2
2
if both its end vertices are red, the number
1
/
2.
1/2.
1/2.
if both its end vertices are blue and the number
1
1
1
otherwise. Let the product of these numbers be
P
P
P
. Find the largest possible value of
P
P
P
.
2
1
Hide problems
|x-a_1| |x-a_2| ... |x-a_{2n}| =(n!)^2, diophantine
Let
n
n
n
be a positive integer and
a
1
,
a
2
,
.
.
.
,
a
2
n
a_1,a_2,...,a_{2n}
a
1
,
a
2
,
...
,
a
2
n
be
2
n
2n
2
n
distinct integers. Given that the equation
∣
x
−
a
1
∣
∣
x
−
a
2
∣
.
.
.
∣
x
−
a
2
n
∣
=
(
n
!
)
2
|x-a_1| |x-a_2| ... |x-a_{2n}| =(n!)^2
∣
x
−
a
1
∣∣
x
−
a
2
∣...∣
x
−
a
2
n
∣
=
(
n
!
)
2
has an integer solution
x
=
m
x = m
x
=
m
, find
m
m
m
in terms of
a
1
,
a
2
,
.
.
.
,
a
2
n
a_1,a_2,...,a_{2n}
a
1
,
a
2
,
...
,
a
2
n
1
1
Hide problems
square is cut into rectangles with // sides, sum of ratio of sides >=1
A square is cut into several rectangles, none of which is a square, so that the sides of each rectangle are parallel to the sides of the square. For each rectangle with sides
a
,
b
,
a
<
b
a, b,a<b
a
,
b
,
a
<
b
, compute the ratio
a
/
b
a/b
a
/
b
. Prove that sum of these ratios is at least
1
1
1
.
3
1
Hide problems
BD=2CD when <BED=< BAC=2 <CED, AB=AC
Let
A
B
C
ABC
A
BC
be a triangle with
A
B
=
A
C
AB=AC
A
B
=
A
C
. Let
D
D
D
be a point on
B
C
BC
BC
, and
E
E
E
a point on
A
D
AD
A
D
such that
∠
B
E
D
=
∠
B
A
C
=
2
∠
C
E
D
\angle BED=\angle BAC=2\angle CED
∠
BE
D
=
∠
B
A
C
=
2∠
CE
D
. Prove that
B
D
=
2
C
D
BD=2CD
B
D
=
2
C
D
.