MathDB
Problems
Contests
National and Regional Contests
USA Contests
USA - Other Middle and High School Contests
SDML/SDMO Contests
SDMO (Middle School)
2008 SDMO (Middle School)
2008 SDMO (Middle School)
Part of
SDMO (Middle School)
Subcontests
(5)
5
1
Hide problems
Sequence partial sums
For a positive integer
n
n
n
, let
f
(
n
)
f\left(n\right)
f
(
n
)
be the sum of the first
n
n
n
terms of the sequence
0
,
1
,
1
,
2
,
2
,
3
,
3
,
4
,
4
,
…
,
r
,
r
,
r
+
1
,
r
+
1
,
…
.
0,1,1,2,2,3,3,4,4,\ldots,r,r,r+1,r+1,\ldots.
0
,
1
,
1
,
2
,
2
,
3
,
3
,
4
,
4
,
…
,
r
,
r
,
r
+
1
,
r
+
1
,
…
.
For example,
f
(
5
)
=
0
+
1
+
1
+
2
+
2
=
6
f\left(5\right)=0+1+1+2+2=6
f
(
5
)
=
0
+
1
+
1
+
2
+
2
=
6
.(a) Find a formula for
f
(
n
)
f\left(n\right)
f
(
n
)
. (b) Prove that
f
(
s
+
t
)
−
f
(
s
−
t
)
=
s
t
f\left(s+t\right)-f\left(s-t\right)=st
f
(
s
+
t
)
−
f
(
s
−
t
)
=
s
t
for all positive integers
s
s
s
and
t
t
t
, where
s
>
t
s>t
s
>
t
.
4
1
Hide problems
Triples multiplying to square of 2008
Find the number of ordered triples of positive integers
(
a
,
b
,
c
)
\left(a,b,c\right)
(
a
,
b
,
c
)
such that
a
×
b
×
c
=
200
8
2
a\times b\times c=2008^2
a
×
b
×
c
=
200
8
2
.
3
1
Hide problems
Triangle areas
In the diagram,
A
D
:
D
B
=
1
:
1
AD:DB=1:1
A
D
:
D
B
=
1
:
1
,
B
E
:
E
C
=
1
:
2
BE:EC=1:2
BE
:
EC
=
1
:
2
, and
C
F
:
F
A
=
1
:
3
CF:FA=1:3
CF
:
F
A
=
1
:
3
. If the area of triangle
A
B
C
ABC
A
BC
is
120
120
120
, then find the area of triangle
D
E
F
DEF
D
EF
.(will insert image here later)
2
1
Hide problems
Binomial Sum
Find the sum of the first
55
55
55
terms of the sequence \binom{0}{0}, \binom{1}{0}, \binom{1}{1}, \binom{2}{0}, \binom{2}{1}, \binom{2}{2}, \binom{3}{0}, \ldots. Note: For nonnegative integers
n
n
n
and
k
k
k
where
0
≤
k
≤
n
0\leq k\leq n
0
≤
k
≤
n
,
(
n
k
)
=
n
!
k
!
(
n
−
k
)
!
.
\binom{n}{k}=\frac{n!}{k!\left(n-k\right)!}.
(
k
n
)
=
k
!
(
n
−
k
)
!
n
!
.
1
1
Hide problems
Reciprocal Diophantine
Find all ordered pairs of integers
(
m
,
n
)
\left(m,n\right)
(
m
,
n
)
such that
1
m
+
1
n
=
1
7
.
\frac{1}{m}+\frac{1}{n}=\frac{1}{7}.
m
1
+
n
1
=
7
1
.