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Singapore MO Open
2012 Singapore MO Open
4
congruent to mod p
congruent to mod p
Source: SMO 2012 Q4
June 30, 2012
modular arithmetic
floor function
number theory
relatively prime
number theory proposed
Problem Statement
Let
p
p
p
be an odd prime. Prove that
1
p
−
2
+
2
p
−
2
+
⋯
+
(
p
−
1
2
)
p
−
2
≡
2
−
2
p
p
(
m
o
d
p
)
.
1^{p-2}+2^{p-2}+\cdots+\left(\frac{p-1}{2}\right)^{p-2}\equiv\frac{2-2^p}{p}\pmod p.
1
p
−
2
+
2
p
−
2
+
⋯
+
(
2
p
−
1
)
p
−
2
≡
p
2
−
2
p
(
mod
p
)
.
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