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Problems
Contests
National and Regional Contests
Netherlands Contests
Dutch Mathematical Olympiad
2010 Dutch Mathematical Olympiad
4
4
Part of
2010 Dutch Mathematical Olympiad
Problems
(1)
119 pairs of (x,y) such that (x+my) and (mx+y) are integers , 0<x,y<1,
Source: Dutch NMO 2010 p4
9/6/2019
(a) Determine all pairs
(
x
,
y
)
(x, y)
(
x
,
y
)
of (real) numbers with
0
<
x
<
1
0 < x < 1
0
<
x
<
1
and
0
<
y
<
1
0 <y < 1
0
<
y
<
1
for which
x
+
3
y
x + 3y
x
+
3
y
and
3
x
+
y
3x + y
3
x
+
y
are both integer. An example is
(
x
,
y
)
=
(
8
3
,
7
8
)
(x,y) =( \frac{8}{3}, \frac{7}{8})
(
x
,
y
)
=
(
3
8
,
8
7
)
, because
x
+
3
y
=
3
8
+
21
8
=
24
8
=
3
x+3y =\frac38 +\frac{21}{8} =\frac{24}{8} = 3
x
+
3
y
=
8
3
+
8
21
=
8
24
=
3
and
3
x
+
y
=
9
8
+
7
8
=
16
8
=
2
3x+y = \frac98 + \frac78 =\frac{16}{8} = 2
3
x
+
y
=
8
9
+
8
7
=
8
16
=
2
.(b) Determine the integer
m
>
2
m > 2
m
>
2
for which there are exactly
119
119
119
pairs
(
x
,
y
)
(x,y)
(
x
,
y
)
with
0
<
x
<
1
0 < x < 1
0
<
x
<
1
and
0
<
y
<
1
0 < y < 1
0
<
y
<
1
such that
x
+
m
y
x + my
x
+
m
y
and
m
x
+
y
mx + y
m
x
+
y
are integers.Remark: if
u
≠
v
,
u \ne v,
u
=
v
,
the pairs
(
u
,
v
)
(u, v)
(
u
,
v
)
and
(
v
,
u
)
(v, u)
(
v
,
u
)
are different.
number theory
Integer