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2015 PAMO
Problem 3
Problem 3
Part of
2015 PAMO
Problems
(1)
Some sum is divisible by 2015 - 2015 PAMO Problem 3
Source: 2015 Pan-African Mathematics Olympiad Problem 3
8/26/2015
Let
a
1
,
a
2
,
.
.
.
,
a
11
a_1,a_2,...,a_{11}
a
1
,
a
2
,
...
,
a
11
be integers. Prove that there are numbers
b
1
,
b
2
,
.
.
.
,
b
11
b_1,b_2,...,b_{11}
b
1
,
b
2
,
...
,
b
11
, each
b
i
b_i
b
i
equal
−
1
,
0
-1,0
−
1
,
0
or
1
1
1
, but not all being
0
0
0
, such that the number
N
=
a
1
b
1
+
a
2
b
2
+
.
.
.
+
a
11
b
11
N=a_1b_1+a_2b_2+...+a_{11}b_{11}
N
=
a
1
b
1
+
a
2
b
2
+
...
+
a
11
b
11
is divisible by
2015
2015
2015
.
Combinatorial Number Theory
number theory