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National and Regional Contests
Poland Contests
Poland - Second Round
1991 Poland - Second Round
3
3
Part of
1991 Poland - Second Round
Problems
(1)
a+b = c+d = e+f = 101
Source: Polish MO Recond Round 1991 p3
9/9/2024
There are positive integers
a
a
a
,
b
b
b
,
c
c
c
,
d
d
d
,
e
e
e
,
f
f
f
such that
a
+
b
=
c
+
d
=
e
+
f
=
101
a+b = c+d = e+f = 101
a
+
b
=
c
+
d
=
e
+
f
=
101
. Prove that the number
a
c
e
b
d
f
\frac{ace}{bdf}
b
df
a
ce
cannot be written as a fraction
m
n
\frac{m}{n}
n
m
where
m
m
m
,
n
n
n
are positive integers with a sum less than
101
101
101
.
number theory