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Modular Arithmetic
hard
0:00.0
Find
x
x
x
in
x
+
1
≡
0
(
m
o
d
5
)
x + 1 \equiv 0 \pmod 5
x
+
1
≡
0
(
mod
5
)
.
A
4
4
4
B
1
1
1
C
0
0
0
D
−
1
-1
−
1
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