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Putnam
1994 Putnam
6
Putnam 1994 B6
Putnam 1994 B6
Source:
July 13, 2014
Putnam
modular arithmetic
college contests
Problem Statement
For
a
∈
Z
a\in \mathbb{Z}
a
∈
Z
define
n
a
=
101
a
−
100
⋅
2
a
n_a=101a-100\cdot 2^a
n
a
=
101
a
−
100
⋅
2
a
Show that, for
0
≤
a
,
b
,
c
,
d
≤
99
0\le a,b,c,d\le 99
0
≤
a
,
b
,
c
,
d
≤
99
n
a
+
n
b
≡
n
c
+
n
d
(
m
o
d
10100
)
⟹
{
a
,
b
}
=
{
c
,
d
}
n_a+n_b\equiv n_c+n_d\pmod{10100}\implies \{a,b\}=\{c,d\}
n
a
+
n
b
≡
n
c
+
n
d
(
mod
10100
)
⟹
{
a
,
b
}
=
{
c
,
d
}
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