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Modular Arithmetic
easy
0:00.0
If
x
≡
3
(
m
o
d
7
)
x \equiv 3 \pmod 7
x
≡
3
(
mod
7
)
, then
2
x
(
m
o
d
7
)
2x \pmod 7
2
x
(
mod
7
)
is:
A
3
B
6
C
1
D
0
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