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2015 Balkan MO Shortlist
N5
N5
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2015 Balkan MO Shortlist
Problems
(1)
v_2 (\prod_{n=1}^{2^m}\binom{2n}{n} )=m2^{m-1}+1
Source: Balkan BMO Shortlist 2015 N5
8/5/2019
For a positive integer
s
s
s
, denote with
v
2
(
s
)
v_2(s)
v
2
(
s
)
the maximum power of
2
2
2
that divides
s
s
s
. Prove that for any positive integer
m
m
m
that:
v
2
(
∏
n
=
1
2
m
(
2
n
n
)
)
=
m
2
m
−
1
+
1.
v_2\left(\prod_{n=1}^{2^m}\binom{2n}{n}\right)=m2^{m-1}+1.
v
2
(
n
=
1
∏
2
m
(
n
2
n
)
)
=
m
2
m
−
1
+
1.
(FYROM)
number theory
maximum
power of 2
divides
Product
IMO Shortlist