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Poland - Second Round
1995 Poland - Second Round
3
3
Part of
1995 Poland - Second Round
Problems
(1)
a+b = 1. Show that c+d = 1 iff [na]+[nb] = [nc]+[nd] for all n
Source: Polish second round 1995 p3
1/19/2020
Let
a
,
b
,
c
,
d
a,b,c,d
a
,
b
,
c
,
d
be positive irrational numbers with
a
+
b
=
1
a+b = 1
a
+
b
=
1
. Show that
c
+
d
=
1
c+d = 1
c
+
d
=
1
if and only if
[
n
a
]
+
[
n
b
]
=
[
n
c
]
+
[
n
d
]
[na]+[nb] = [nc]+[nd]
[
na
]
+
[
nb
]
=
[
n
c
]
+
[
n
d
]
for all positive integers
n
n
n
.
irrational
floor function
algebra