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Romania Contests
Romania Team Selection Test
1993 Romania Team Selection Test
1993 Romania Team Selection Test
Part of
Romania Team Selection Test
Subcontests
(4)
3
4
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Integral solutions
Prove that the equation (x\plus{}y)^n\equal{}x^m\plus{}y^m has a unique solution in integers with
x
>
y
>
0
x>y>0
x
>
y
>
0
and
m
,
n
>
1
m,n>1
m
,
n
>
1
.
f(A) = f(B) = max A\triangle B
Let
Y
Y
Y
be the family of all subsets of
X
=
{
1
,
2
,
.
.
.
,
n
}
X = \{1,2,...,n\}
X
=
{
1
,
2
,
...
,
n
}
(
n
>
1
n > 1
n
>
1
) and let
f
:
Y
→
X
f : Y \to X
f
:
Y
→
X
be an arbitrary mapping. Prove that there exist distinct subsets
A
,
B
A,B
A
,
B
of
X
X
X
such that
f
(
A
)
=
f
(
B
)
=
m
a
x
A
△
B
f(A) = f(B) = max A\triangle B
f
(
A
)
=
f
(
B
)
=
ma
x
A
△
B
, where
A
△
B
=
(
A
−
B
)
∪
(
B
−
A
)
A\triangle B = (A-B)\cup(B-A)
A
△
B
=
(
A
−
B
)
∪
(
B
−
A
)
.
max of \frac{n!}{n_1!n_2!n_3!n_4!}2^{ {n_1 \choose 2}+...}
For each integer
n
>
3
n > 3
n
>
3
find all quadruples
(
n
1
,
n
2
,
n
3
,
n
4
)
(n_1,n_2,n_3,n_4)
(
n
1
,
n
2
,
n
3
,
n
4
)
of positive integers with
n
1
+
n
2
+
n
3
+
n
4
=
n
n_1 +n_2 +n_3 +n_4 = n
n
1
+
n
2
+
n
3
+
n
4
=
n
which maximize the expression
n
!
n
1
!
n
2
!
n
3
!
n
4
!
2
(
n
1
2
)
+
(
n
2
2
)
+
(
n
3
2
)
+
(
n
4
2
)
+
n
1
n
2
+
n
2
n
3
+
n
3
n
4
\frac{n!}{n_1!n_2!n_3!n_4!}2^{ {n_1 \choose 2}+{n_2 \choose 2}+{n_3 \choose 2}+{n_4 \choose 2}+n_1n_2+n_2n_3+n_3n_4}
n
1
!
n
2
!
n
3
!
n
4
!
n
!
2
(
2
n
1
)
+
(
2
n
2
)
+
(
2
n
3
)
+
(
2
n
4
)
+
n
1
n
2
+
n
2
n
3
+
n
3
n
4
2
4
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