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Portugal MO
1999 Portugal MO
5
5
Part of
1999 Portugal MO
Problems
(1)
4 |n if a_1a_2 + a_2a_3 + ··· + a_{n-1}a_n + a_na_1 = 0, for a_i \in {1,-1}
Source: Portugal OPM 1999 p5
5/18/2024
Each of the numbers
a
1
,
.
.
.
,
a
n
a_1,...,a_n
a
1
,
...
,
a
n
is equal to
1
1
1
or
−
1
-1
−
1
. If
a
1
a
2
+
a
2
a
3
+
⋅
⋅
⋅
+
a
n
−
1
a
n
+
a
n
a
1
=
0
a_1a_2 + a_2a_3 + ··· + a_{n-1}a_n + a_na_1 = 0
a
1
a
2
+
a
2
a
3
+
⋅⋅⋅
+
a
n
−
1
a
n
+
a
n
a
1
=
0
, proves that
n
n
n
is divisible by
4
4
4
.
number theory
Sum
algebra