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National and Regional Contests
China Contests
China National Olympiad
1995 China National Olympiad
1995 China National Olympiad
Part of
China National Olympiad
Subcontests
(3)
3
2
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China mathematical olympiad 1995 problem3
Find the minimun value of
∑
i
=
1
10
∑
j
=
1
10
∑
k
=
1
10
∣
k
(
x
+
y
−
10
i
)
(
3
x
−
6
y
−
36
j
)
(
19
x
+
95
y
−
95
k
)
∣
\sum_{i=1}^{10} \sum_{j=1}^{10} \sum_{k=1}^{10}|k(x+y-10i)(3x-6y-36j)(19x+95y-95k)|
∑
i
=
1
10
∑
j
=
1
10
∑
k
=
1
10
∣
k
(
x
+
y
−
10
i
)
(
3
x
−
6
y
−
36
j
)
(
19
x
+
95
y
−
95
k
)
∣
, where
x
,
y
x,y
x
,
y
are integers.
China mathematical olympiad 1995 problem6
Let
n
(
n
>
1
)
n(n>1)
n
(
n
>
1
)
be an odd. We define
x
k
=
(
x
1
(
k
)
,
x
2
(
k
)
,
⋯
,
x
n
(
k
)
)
x_k=(x^{(k)}_1,x^{(k)}_2,\cdots ,x^{(k)}_n)
x
k
=
(
x
1
(
k
)
,
x
2
(
k
)
,
⋯
,
x
n
(
k
)
)
as follow:
x
0
=
(
x
1
(
0
)
,
x
2
(
0
)
,
⋯
,
x
n
(
0
)
)
=
(
1
,
0
,
⋯
,
0
,
1
)
x_0=(x^{(0)}_1,x^{(0)}_2,\cdots ,x^{(0)}_n)=(1,0,\cdots ,0,1)
x
0
=
(
x
1
(
0
)
,
x
2
(
0
)
,
⋯
,
x
n
(
0
)
)
=
(
1
,
0
,
⋯
,
0
,
1
)
;
x
i
(
k
)
=
{
0
,
e
m
s
p
;
x
i
(
k
−
1
)
=
x
i
+
1
(
k
−
1
)
,
1
,
e
m
s
p
;
x
i
(
k
−
1
)
≠
x
i
+
1
(
k
−
1
)
,
x^{(k)}_i =\begin{cases}0,   x^{(k-1)}_i=x^{(k-1)}_{i+1},\\ 1,   x^{(k-1)}_i\not= x^{(k-1)}_{i+1},\end{cases}
x
i
(
k
)
=
{
0
,
1
,
e
m
s
p
;
x
i
(
k
−
1
)
=
x
i
+
1
(
k
−
1
)
,
e
m
s
p
;
x
i
(
k
−
1
)
=
x
i
+
1
(
k
−
1
)
,
i
=
1
,
2
,
⋯
,
n
i=1,2,\cdots ,n
i
=
1
,
2
,
⋯
,
n
, where
x
n
+
1
(
k
−
1
)
=
x
1
(
k
−
1
)
x^{(k-1)}_{n+1}= x^{(k-1)}_1
x
n
+
1
(
k
−
1
)
=
x
1
(
k
−
1
)
. Let
m
m
m
be a positive integer satisfying
x
0
=
x
m
x_0=x_m
x
0
=
x
m
. Prove that
m
m
m
is divisible by
n
n
n
.
2
2
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China mathematical olympiad 1995 problem2
Let
f
:
N
→
N
f: \mathbb{N} \rightarrow \mathbb{N}
f
:
N
→
N
be a function satisfying the following conditions: (1)
f
(
1
)
=
1
f(1)=1
f
(
1
)
=
1
; (2)
∀
n
∈
N
\forall n\in \mathbb{N}
∀
n
∈
N
,
3
f
(
n
)
f
(
2
n
+
1
)
=
f
(
2
n
)
(
1
+
3
f
(
n
)
)
3f(n) f(2n+1) =f(2n) ( 1+3f(n) )
3
f
(
n
)
f
(
2
n
+
1
)
=
f
(
2
n
)
(
1
+
3
f
(
n
))
; (3)
∀
n
∈
N
\forall n\in \mathbb{N}
∀
n
∈
N
,
f
(
2
n
)
<
6
f
(
n
)
f(2n) < 6 f(n)
f
(
2
n
)
<
6
f
(
n
)
. Find all solutions of equation
f
(
k
)
+
f
(
l
)
=
293
f(k) +f(l)=293
f
(
k
)
+
f
(
l
)
=
293
, where
k
<
l
k<l
k
<
l
. (
N
\mathbb{N}
N
denotes the set of all natural numbers).
China mathematical olympiad 1995 problem5
Let
a
1
,
a
2
,
⋯
,
a
10
a_1,a_2,\cdots ,a_{10}
a
1
,
a
2
,
⋯
,
a
10
be pairwise distinct natural numbers with their sum equal to 1995. Find the minimal value of
a
1
a
2
+
a
2
a
3
+
⋯
+
a
9
a
10
+
a
10
a
1
a_1a_2+a_2a_3+\cdots +a_9a_{10}+a_{10}a_1
a
1
a
2
+
a
2
a
3
+
⋯
+
a
9
a
10
+
a
10
a
1
.
1
2
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China mathematical olympiad 1995 problem4
Given four spheres with their radii equal to
2
,
2
,
3
,
3
2,2,3,3
2
,
2
,
3
,
3
respectively, each sphere externally touches the other spheres. Suppose that there is another sphere that is externally tangent to all those four spheres, determine the radius of this sphere.
1995 China mathematical olympiad problem 1
Let
a
1
,
a
2
,
⋯
,
a
n
;
b
1
,
b
2
,
⋯
,
b
n
(
n
≥
3
)
a_1,a_2,\cdots ,a_n; b_1,b_2,\cdots ,b_n (n\ge 3)
a
1
,
a
2
,
⋯
,
a
n
;
b
1
,
b
2
,
⋯
,
b
n
(
n
≥
3
)
be real numbers satisfying the following conditions: (1)
a
1
+
a
2
+
⋯
+
a
n
=
b
1
+
b
2
+
⋯
+
b
n
a_1+a_2+\cdots +a_n= b_1+b_2+\cdots +b_n
a
1
+
a
2
+
⋯
+
a
n
=
b
1
+
b
2
+
⋯
+
b
n
; (2)
0
<
a
1
=
a
2
,
a
i
+
a
i
+
1
=
a
i
+
2
0<a_1=a_2, a_i+a_{i+1}=a_{i+2}
0
<
a
1
=
a
2
,
a
i
+
a
i
+
1
=
a
i
+
2
(
i
=
1
,
2
,
⋯
,
n
−
2
i=1,2,\cdots ,n-2
i
=
1
,
2
,
⋯
,
n
−
2
); (3)
0
<
b
1
≤
b
2
,
b
i
+
b
i
+
1
≤
b
i
+
2
0<b_1\le b_2, b_i+b_{i+1}\le b_{i+2}
0
<
b
1
≤
b
2
,
b
i
+
b
i
+
1
≤
b
i
+
2
(
i
=
1
,
2
,
⋯
,
n
−
2
i=1,2,\cdots ,n-2
i
=
1
,
2
,
⋯
,
n
−
2
). Prove that
a
n
−
1
+
a
n
≤
b
n
−
1
+
b
n
a_{n-1}+a_n\le b_{n-1}+b_n
a
n
−
1
+
a
n
≤
b
n
−
1
+
b
n
.